Aaa Fundamentals Of Probability Solutions Manual
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Instructor's Solutions Manual
Third Edition
Fundamentals of
ProbabilitY
With
Stochastic Processes
SAEED GHAHRAMANI
Western New England College
Upper Saddle River, New Jersey 07458
C ontents
1.2
1.4
1.7
2.2
2.3
2.4
2.5
1
Axioms of Probability
Sample Space and Events
1
Basic Theorems
2
Random Selection of Points from Intervals
Review Problems
9
2
7
Combinatorial Methods
13
Counting Principle
13
Permutations
16
Combinations
18
Stirling’ Formula
31
Review Problems
31
3
Conditional Probability and Independence
3.1 Conditional Probability
35
3.2 Law of Multiplication
39
3.3 Law of Total Probability
41
3.4 Bayes’ Formula
46
3.5 Independence
48
3.6 Applications of Probability to Genetics
Review Problems
59
1
4
35
56
Distribution Functions and
Discrete Random Variables
4.2 Distribution Functions
63
4.3 Discrete Random Variables
66
4.4 Expectations of Discrete Random Variables
71
4.5 Variances and Moments of Discrete Random Variables
4.6 Standardized Random Variables
83
Review Problems
83
63
77
iv
5.1
5.2
5.3
6.1
6.2
6.3
7.1
7.2
7.3
7.4
7.5
7.6
8.1
8.2
8.3
8.4
9.1
9.2
9.3
Contents
5
Special Discrete Distributions
87
Bernoulli and Binomial Random Variables
Poisson Random Variable
94
Other Discrete Random Variables
99
Review Problems
106
6
87
Continuous Random Variables
111
Probability Density Functions
111
Density Function of a Function of a Random Variable
Expectations and Variances
116
Review Problems
123
7
Special Continuous Distributions
Uniform Random Variable
126
Normal Random Variable
131
Exponential Random Variables
139
Gamma Distribution
144
Beta Distribution
147
Survival Analysis and Hazard Function
Review Problems
153
8
126
152
Bivariate Distributions
Joint Distribution of Two Random Variables
Independent Random Variables
166
Conditional Distributions
174
Transformations of Two Random Variables
Review Problems
191
9
113
157
157
183
Multivariate Distributions
Joint Distribution of n > 2 Random Variables
Order Statistics
210
Multinomial Distributions
215
Review Problems
218
200
200
Contents
10.1
10.2
10.3
10.4
10.5
11.1
11.2
11.3
11.4
11.5
12.2
12.3
12.4
12.5
10
More Expectations and Variances
Expected Values of Sums of Random Variables
Covariance
227
Correlation
237
Conditioning on Random Variables
239
Bivariate Normal Distribution
251
Review Problems
254
11
Sums of Independent Random
Variables and Limit Theorems
v
222
222
261
Moment-Generating Functions
261
Sums of Independent Random Variables
269
Markov and Chebyshev Inequalities
274
Laws of Large Numbers
278
Central Limit Theorem
282
Review Problems
287
12
291
Stochastic Processes
More on Poisson Processes
291
Markov Chains
296
Continuous-Time Markov Chains
Brownian Motion
326
Review Problems
331
315
Chapter 1
A xioms
1.2
of
Probability
SAMPLE SPACE AND EVENTS
1. For 1 ≤ i, j ≤ 3, by (i, j ) we mean that Vann’s card number is i, and
Paul’s card number is
j . Clearly, A = (1, 2), (1, 3), (2, 3) and B = (2, 1), (3, 1), (3, 2) .
(a) Since A ∩ B = ∅, the events A and B are mutually exclusive.
(b) None of (1, 1), (2, 2), (3, 3) belongs to A ∪ B. Hence A ∪ B not being the sample space
shows that A and B are not complements of one another.
2. S = {RRR, RRB, RBR, RBB, BRR, BRB, BBR, BBB}.
3. {x : 0 < x < 20}; {1, 2, 3, . . . , 19}.
4. Denote the dictionaries by d1 , d2 ; the third book by a. The answers are
{d1 d2 a, d1 ad2 , d2 d1 a, d2 ad1 , ad1 d2 , ad2 d1 } and {d1 d2 a, ad1 d2 }.
5. EF : One 1 and one even.
E c F : One 1 and one odd.
E c F c : Both even or both belong to {3, 5}.
6. S = {QQ, QN, QP , QD, DN, DP , N P , N N, P P }. (a) {QP }; (b) {DN, DP , NN}; (c) ∅.
7. S = x : 7 ≤ x ≤ 9 16 ; x : 7 ≤ x ≤ 7 41 ∪ x : 7 43 ≤ x ≤ 8 41 ∪ x : 8 43 ≤ x ≤ 9 16 .
8. E ∪ F ∪ G = G: If E or F occurs, then G occurs.
EF G = G: If G occurs, then E and F occur.
9. For 1 ≤ i ≤ 3, 1 ≤ j ≤ 3, by ai bj we mean passenger a gets off at hotel i and passenger b
gets off at hotel j . The answers are {ai bj : 1 ≤ i ≤ 3, 1 ≤ j ≤ 3} and {a1 b1 , a2 b2 , a3 b3 },
respectively.
10. (a) (E ∪ F )(F ∪ G) = (F ∪ E)(F ∪ G) = F ∪ EG.
2
Chapter 1
(b)
Axioms of Probability
Using part (a), we have
(E ∪ F )(E c ∪ F )(E ∪ F c ) = (F ∪ EE c )(E ∪ F c ) = F (E ∪ F c ) = F E ∪ F F c = F E.
(b) A ∪ B ∪ C;
11. (a) AB c C c ;
(e) AB c C c ∪ Ac B c C ∪ Ac BC c ;
(c) Ac B c C c ;
(d) ABC c ∪ AB c C ∪ Ac BC;
(f) (A − B) ∪ (B − A) = (A ∪ B) − AB.
12. If B = ∅, the relation is obvious. If the relation is true for every event A, then it is true for S,
the sample space, as well. Thus
S = (B ∩ S c ) ∪ (B c ∩ S) = ∅ ∪ B c = B c ,
showing that B = ∅.
13. Parts (a) and (d) are obviously true; part (c) is true by DeMorgan’s law; part (b) is false: throw
a four-sided die; let F = {1, 2, 3}, G = {2, 3, 4}, E = {1, 4}.
14. (a)
∞
n=1
An ; (b)
37
n=1
An .
15. Straightforward.
16. Straightforward.
17. Straightforward.
18. Let a1 , a2 , and a3 be the first, the second, and the third volumes of the dictionary. Let a4 , a5 ,
a6 , and a7 be the remaining books. Let A = {a1 , a2 , . . . , a7 }; the answers are
S = x1 x2 x3 x4 x5 x6 x7 : xi ∈ A, 1 ≤ i ≤ 7, and xi = xj if i = j
and
x1 x2 x3 x4 x5 x6 x7 ∈ S : xi xi+1 xi+2 = a1 a2 a3 for some i, 1 ≤ i ≤ 5 ,
respectively.
19.
∞ ∞
m=1
n=m
An .
20. Let B1 = A1 , B2 = A2 − A1 , B3 = A3 − (A1 ∪ A2 ), . . . , Bn = An −
1.4
BASIC THEOREMS
1. No; P (sum 11) = 2/36 while P (sum 12) = 1/36.
2. 0.33 + 0.07 = 0.40.
n−1
i=1
Ai , . . . .
Section 1.4
Basic Theorems
3
3. Let E be the event that an earthquake will damage the structure next year. Let H be the
event that a hurricane will damage the structure next year. We are given that P (E) = 0.015,
P (H ) = 0.025, and P (EH ) = 0.0073. Since
P (E ∪ H ) = P (E) + P (H ) − P (EH ) = 0.015 + 0.025 − 0.0073 = 0.0327,
the probability that next year the structure will be damaged by an earthquake and/or a hurricane
is 0.0327. The probability that it is not damaged by any of the two natural disasters is 0.9673.
4. Let A be the event of a randomly selected driver having an accident during the next 12 months.
Let B be the event that the person is male. By Theorem 1.7, the desired probability is
P (A) = P (AB) + P (AB c ) = 0.12 + 0.06 = 0.18.
5. Let A be the event that a randomly selected investor invests in traditional annuities. Let B be
the event that he or she invests in the stock market. Then P (A) = 0.75, P (B) = 0.45, and
P (A ∪ B) = 0.85. Since,
P (AB) = P (A) + P (B) − P (A ∪ B) = 0.75 + 0.45 − 0.85 = 0.35,
35% invest in both stock market and traditional annuities.
6. The probability that the first horse wins is 2/7. The probability that the second horse wins
is 3/10. Since the events that the first horse wins and the second horse wins are mutually
exclusive, the probability that either the first horse or the second horse will win is
2
3
41
+
=
.
7 10
70
7. In point of fact Rockford was right the first time. The reporter is assuming that both autopsies
are performed by a given doctor. The probability that both autopsies are performed by the same
doctor–whichever doctor it may be–is 1/2. Let AB represent the case in which Dr. A performs
the first autopsy and Dr. B performs the second autopsy, with similar representations for other
cases. Then the sample space is S = {AA, AB, BA, BB}. The event that both autopsies are
performed by the same doctor is {AA, BB}. Clearly, the probability of this event is 2/4=1/2.
8. Let m be the probability that Marty will be hired. Then m + (m + 0.2) + m = 1 which gives
m = 8/30; so the answer is 8/30 + 2/10 = 7/15.
9. Let s be the probability that the patient selected at random suffers from schizophrenia. Then
s + s/3 + s/2 + s/10 = 1 which gives s = 15/29.
10. P (A ∪ B) ≤ 1 implies that P (A) + P (B) − P (AB) ≤ 1.
11. (a) 2/52 + 2/52 = 1/13;
(b) 12/52 + 26/52 − 6/53 = 8/13;
(c) 1 − (16/52) = 9/13.
4
Chapter 1
Axioms of Probability
12. (a) False; toss a die and let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
False; toss a die and let A = {1, 2, 3, 4}, B = {1, 2, 3, 4, 5}, C = {1, 2, 3, 4, 5, 6}.
(b)
13. A simple Venn diagram shows that the answers are 65% and 10%, respectively.
14. Applying Theorem 1.6 twice, we have
P (A ∪ B ∪ C) = P (A ∪ B) + P (C) − P (A ∪ B)C
= P (A) + P (B) − P (AB) + P (C) − P (AC ∪ BC)
= P (A) + P (B) − P (AB) + P (C) − P (AC) − P (BC) + P (ABC)
= P (A) + P (B) + P (C) − P (AB) − P (AC) − P (BC) + P (ABC).
15. Using Theorem 1.5, we have that the desired probability is
P (AB − ABC) + P (AC − ABC) + P (BC − ABC)
= P (AB) − P (ABC) + P (AC) − P (ABC) + P (BC) − P (ABC)
= P (AB) + P (AC) + P (BC) − 3P (ABC).
16. 7/11.
17.
n
i=1
pij .
18. Let M and F denote the events that the randomly selected student earned an A on the midterm
exam and an A on the final exam, respectively. Then
P (MF ) = P (M) + P (F ) − P (M ∪ F ),
where P (M) = 17/33, P (F ) = 14/33, and by DeMorgan’s law,
P (M ∪ F ) = 1 − P (M c F c ) = 1 −
Therefore,
P (MF ) =
22
11
=
.
33
33
17 14 22
3
+
−
= .
33 33 33
11
19. A Venn diagram shows that the answers are 1/8, 5/24, and 5/24, respectively.
20. The equation has real roots if and only if b2 ≥ 4c. From the 36 possible outcomes for (b, c),
in the following 19 cases we have that b2 ≥ 4c: (2, 1), (3, 1), (3, 2), (4, 1), . . . , (4, 4), (5, 1),
. . . , (5, 6), (6, 1), . . . , (6, 6). Therefore, the answer is 19/36.
21. The only prime divisors of 63 are 3 and 7. Thus the number selected is relatively prime to 63
if and only if it is neither divisible by 3 nor by 7. Let A and B be the events that the outcome
Section 1.4
Basic Theorems
5
is divisible by 3 and 7, respectively. The desired quantity is
P (Ac B c ) = 1 − P (A ∪ B) = 1 − P (A) − P (B) + P (AB)
=1−
21
9
3
4
−
+
= .
63 63 63
7
22. Let T and F be the events that the number selected is divisible by 3 and 5, respectively.
(a)
The desired quantity is the probability of the event T F c :
P (T F c ) = P (T ) − P (T F ) =
333
66
267
−
=
.
1000 1000
1000
(b) The desired quantity is the probability of the event T c F c :
P (T c F c ) = 1 − P (T ∪ F ) = 1 − P (T ) − P (F ) + P (T F )
=1−
200
66
533
333
−
+
=
.
1000 1000 1000
1000
23. (Draw a Venn diagram.) From the data we have that 55% passed all three, 5% passed calculus
and physics but not chemistry, and 20% passed calculus and chemistry but not physics. So at
least (55 + 5 + 20)% = 80% must have passed calculus. This number is greater than the given
78% for all of the students who passed calculus. Therefore, the data is incorrect.
24. By symmetry the answer is 1/4.
25. Let A, B, and C be the events that the number selected is divisible by 4, 5, and 7, respectively.
We are interested in P (AB c C c ). Now AB c C c = A − A(B ∪ C) and A(B ∪ C) ⊆ A. So by
Theorem 1.5,
P (AB c C c ) = P (A) − P A(B ∪ C) = P (A) − P (AB ∪ AC)
= P (A) − P (AB) − P (AC) + P (ABC)
=
50
35
7
172
250
−
−
+
=
.
1000 1000 1000 1000
1000
26. A Venn diagram shows that the answer is 0.36.
27. Let A be the event that the first number selected is greater than the second; let B be the
event that the second number selected is greater than the first; and let C be the event that
the two numbers selected are equal. Then P (A) + P (B) + P (C) = 1, P (A) = P (B), and
P (C) = 1/100. These give P (A) = 99/200.
n−1
28. Let B1 = A1 , and for n ≥ 2,Bn = An
−
∞
∞
mutually exclusive events and
i=1 Ai =
Ai . Then {B1 , B2 , . . . } is a sequence of
B
.
i=1 i Hence
i=1
6
Chapter 1
Axioms of Probability
∞
An = P
P
n=1
∞
∞
Bn =
n=1
∞
P (Bn ) ≤
n=1
P (An ),
n=1
since Bn ⊆ An , n ≥ 1.
29. By Boole’s inequality (Exercise 28),
∞
An = 1 − P
P
n=1
∞
∞
Acn ≥ 1 −
n=1
P (Acn ).
n=1
30. She is wrong! Consider the next 50 flights. For 1≤ i ≤ 50, let Ai be the event that the ith
50
mission will be completed without mishap. Then
Ai is the
event that all of the next 50
50
missions will be completed successfully. We will show that P
i=1 Ai > 0. This proves
that Mia is wrong. Note that the probability of the simultaneous occurrence of any number of
Aci ’s is nonzero. Furthermore, consider any set E consisting of n (n ≤ 50) of the Aci ’s. It is
reasonable to assume that the probability of the simultaneous occurrence of the events of E is
strictly less than the probability of the simultaneous occurrence of the events of any subset of
E. Using these facts, it is straightforward to conclude from the inclusion–exclusion principle
that,
50
50
50
1
P
Aci <
P (Aci ) =
= 1.
50
i=1
i=1
i=1
i=1
Thus, by DeMorgan’s law,
50
Ai = 1 − P
P
i=1
50
Aci > 1 − 1 = 0.
i=1
31. Q satisfies Axioms 1 and 2, but not
necessarily
Axiom 3. So it is not, in general,
a probability
on S. Let S = {1, 2, 3, }. Let P {1} = P {2} = P {3} = 1/3. Then Q {1} = Q {2} =
2
1/9, whereas Q {1, 2} = P {1, 2} = 4/9. Therefore,
Q {1, 2, } = Q {1} + Q {2} .
R is not a probability on S because it does not satisfy Axiom 2; that is, R(S) = 1.
32. Let BRB mean that a blue hat is placed on the first player’s head, a red hat on the second
player’s head, and a blue hat on the third player’s head, with similar representations for other
cases. The sample space is
S = {BBB, BRB, BBR, BRR, RRR, RRB, RBR, RBB}.
This shows that the probability that two of the players will have hats of the same color and
the third player’s hat will be of the opposite color is 6/8 = 3/4. The following improvement,
Section 1.7
Random Selection of Points from Intervals
7
based on this observation, explained by Sara Robinson in Tuesday, April 10, 2001 issue of
the New York Times, is due to Professor Elwyn Berlekamp of the University of California at
Berkeley.
Three-fourths of the time, two of the players will have hats of the same color and
the third player’s hat will be the opposite color. The group can win every time this
happens by using the following strategy: Once the game starts, each player looks
at the other two players’ hats. If the two hats are different colors, he [or she] passes.
If they are the same color, the player guesses his [or her] own hat is the opposite
color. This way, every time the hat colors are distributed two and one, one player
will guess correctly and the others will pass, and the group will win the game. When
all the hats are the same color, however, all three players will guess incorrectly and
the group will lose.
1.7
RANDOM SELECTION OF POINTS FROM INTERVALS
1.
30 − 10
2
= .
30 − 0
3
2.
0.0635 − 0.04
= 0.294.
0.12 − 0.04
3. (a) False; in the experiment of choosing a point at random from the interval (0, 1), let
A = (0, 1) − {1/2}. A is not the sample
space
but P (A) = 1.
(b) False; in the same experiment P {1/2} = 0 while { 21 } = ∅.
4. P (A ∪ B) ≥ P (A) = 1, so P (A ∪ B) = 1. This gives
P (AB) = P (A) + P (B) − P (A ∪ B) = 1 + 1 − 1 = 1.
5. The answer is
1999 1999
P {1, 2, . . . , 1999} =
P {i} =
0 = 0.
i=1
i=1
6. For i = 0, 1, 2, . . . , 9, the probability that i appears as the first digit of the decimal represeni i+1
tation of the selected point is the probability that the point falls into the interval
,
.
10 10
Therefore, it equals
i+1
i
−
1
10
10
.
=
10
1−0
This shows that all numerals are equally likely to appear as the first digit of the decimal
representation of the selected point.
8
Chapter 1
Axioms of Probability
7. No, it is not. Let S = {w1 , w2 ,
. . . }. Suppose that for some p > 0, P {wi } = p, i = 1, 2,
. . . . Then, by Axioms 2 and 3,
∞
i=1
p = 1. This is impossible.
8. Use induction. For n = 1, the theorem is trivial. Exercise 4 proves the theorem for n = 2.
Suppose that the theorem is true for n. We show it for n + 1,
P (A1 A2 · · · An An+1 ) = P (A1 A2 · · · An ) + P (An+1 ) − P (A1 A2 · · · An ∪ An+1 )
= 1 + 1 − 1 = 1,
where P (A1 A2 · · · An ) = 1 is true by the induction hypothesis, and
P (A1 A2 · · · An ∪ An+1 ) ≥ P (An+1 ) = 1,
implies that P (A1 A2 · · · An ∪ An+1 ) = 1.
1
1
1
1 1
1
1 1
1
9. (a) Clearly, ∈
−
, +
. If x ∈
−
, +
, then, for all n ≥ 1,
2 n=1 2 2n 2 2n
2 2n 2 2n
n=1
∞
∞
1
1
1
1
−
c, and a = c, it can be checked that there are 73, 73, and 27 cases
in which b2 < 4ac, respectively. Therefore, the desired probability is
173
73 + 73 + 27
=
.
216
216
Chapter 2
C ombinatorial Methods
2.2
COUNTING PRINCIPLES
1. The total number of six-digit numbers is 9×10×10×10×10×10 = 9×105 since the first digit
cannot be 0. The number of six-digit numbers without the digit five is 8 × 9 × 9 × 9 × 9 × 9 =
8 × 95 . Hence there are 9 × 105 − 8 × 95 = 427, 608 six-digit numbers that contain the digit
five.
2. (a)
55 = 3125.
(b) 53 = 125.
3. There are 26 × 26 × 26 = 17, 576 distinct sets of initials. Hence in any town with more than
17,576 inhabitants, there are at least two persons with the same initials. The answer to the
question is therefore yes.
4. 415 = 1, 073, 741, 824.
5.
2
1
= 22 ≈ 0.00000024.
23
2
2
6. (a) 525 = 380, 204, 032. (b) 52 × 51 × 50 × 49 × 48 = 311, 875, 200.
7. 6/36 = 1/6.
8. (a)
9.
4×3×2×2
1
= .
12 × 8 × 8 × 4
64
(b)
1−
8×5×6×2
27
=
.
12 × 8 × 8 × 4
32
1
≈ 0.00000000093.
415
10. 26 × 25 × 24 × 10 × 9 × 8 = 11, 232, 000.
11. There are 263 × 102 = 1, 757, 600 such codes; so the answer is positive.
12. 2nm .
13. (2 + 1)(3 + 1)(2 + 1) = 36. (See the solution to Exercise 24.)
14
Chapter 2
Combinatorial Methods
14. There are (26 − 1)23 = 504 possible sandwiches. So the claim is true.
15. (a) 54 = 625. (b) 54 − 5 × 4 × 3 × 2 = 505.
16. 212 = 4096.
17. 1 −
48 × 48 × 48 × 48
= 0.274.
52 × 52 × 52 × 52
18. 10 × 9 × 8 × 7 = 5040.
19. 1 −
(a) 9 × 9 × 8 × 7 = 4536; (b) 5040 − 1 × 1 × 8 × 7 = 4984.
(N − 1)n
.
Nn
20. By Example 2.6, the probability is 0.507 that among Jenny and the next 22 people she meets
randomly there are two with the same birthday. However, it is quite possible that one of these
two persons is not Jenny. Let n be the minimum number of people Jenny must meet so that
the chances are better than even that someone shares her birthday. To find n, let A denote the
event that among the next n people Jenny meets randomly someone’s birthday is the same as
Jenny’s. We have
364n
P (A) = 1 − P (Ac ) = 1 −
.
365n
To have P (A) > 1/2, we must find the smallest n for which
1−
or
1
364n
> ,
365n
2
364n
1
< .
n
365
2
This gives
1
2
= 252.652.
n>
364
log
365
Therefore, for the desired probability to be greater than 0.5, n must be 253. To some this might
seem counterintuitive.
log
21. Draw a tree diagram for the situation in which the salesperson goes from I to B first. In
this situation, you will find that in 7 out of 23 cases, she will end up staying at island I . By
symmetry, if she goes from I to H , D, or F first, in each of these situations in 7 out of 23
cases she will end up staying at island I . So there are 4 × 23 = 92 cases altogether and in
4 × 7 = 28 of them the salesperson will end up staying at island I . Since 28/92 = 0.3043, the
answer is 30.43%. Note that the probability that the salesperson will end up staying at island
I is not 0.3043 because not all of the cases are equiprobable.
Section 2.2
Counting Principle
15
22. He is at 0 first, next he goes to 1 or −1. If at 1, then he goes to 0 or 2. If at −1, then he goes
to 0 or −2, and so on. Draw a tree diagram. You will find that after walking 4 blocks, he is at
one of the points 4, 2, 0, −2, or −4. There are 16 possible cases altogether. Of these 6 end up
at 0, none at 1, and none at −1. Therefore, the answer to (a) is 6/16 and the answer to (b) is 0.
23. We can think of a number less than 1,000,000 as a six-digit number by allowing it to start with
0 or 0’s. With this convention, it should be clear that there are 96 such numbers without the
digit five. Hence the desired probability is 1 − (96 /106 ) = 0.469.
24. Divisors of N are of the form p1e1 p2e2 · · · pkek , where ei = 0, 1, 2, . . . , ni , 1 ≤ i ≤ k. Therefore,
the answer is (n1 + 1)(n2 + 1) · · · (nk + 1).
25. There are 64 possibilities altogether. In 54 of these possibilities there is no 3. In 53 of these
possibilities only the first die lands 3. In 53 of these possibilities only the second die lands 3,
and so on. Therefore, the answer is
54 + 4 × 5 3
= 0.868.
64
26. Any subset of the set {salami, turkey, bologna, corned beef, ham, Swiss cheese, American
cheese} except the empty set can form a reasonable sandwich. There are 27 − 1 possibilities.
To every sandwich a subset of the set {lettuce, tomato, mayonnaise} can also be added. Since
there are 3 possibilities for bread, the final answer is (27 − 1) × 23 × 3 = 3048 and the
advertisement is true.
27.
11 × 10 × 9 × 8 × 7 × 6 × 5 × 4
= 0.031.
118
28. For i = 1, 2, 3, let Ai be the event that no one departs at stop i. The desired quantity is
P (Ac1 Ac2 Ac3 ) = 1 − P (A1 ∪ A2 ∪ A3 ). Now
P (A1 ∪ A2 ∪ A3 ) = P (A1 ) + P (A2 ) + P (A3 )
− P (A1 A2 ) − P (A1 A3 ) − P (A2 A3 ) + P (A1 A2 A3 )
=
1
1
1
7
26 26 26
+ 6 + 6 − 6 − 6 − 6 +0= .
6
3
3
3
3
3
3
27
Therefore, the desired probability is 1 − (7/27) = 20/27.
29. For 0 ≤ i ≤ 9, the sum of the first two digits is i in (i + 1) ways. Therefore, there are (i + 1)2
numbers in the given set with the sum of the first two digits equal to the sum of the last two
digits and equal to i. For i = 10, there are 92 numbers in the given set with the sum of the first
two digits equal to the sum of the last two digits and equal to 10. For i = 11, the corresponding
numbers are 82 and so on. Therefore, there are altogether
12 + 22 + · · · + 102 + 92 + 82 + · · · + 12 = 670
16
Chapter 2
Combinatorial Methods
numbers with the desired probability and hence the answer is 670/104 = 0.067.
30. Let A be the event that the number selected contains at least one 0. Let B be the event that it
contains at least one 1 and C be the event that it contains at least one 2. The desired quantity
is P (ABC) = 1 − P (Ac ∪ B c ∪ C c ), where
P (Ac ∪ B c ∪ C c ) = P (Ac ) + P (B c ) + P (C c )
− P (Ac B c ) − P (Ac C c ) − P (B c C c ) + P (Ac B c C c )
=
8 × 9r−1
8 × 9r−1
8r
8r
9r
+
+
−
−
9 × 10r−1 9 × 10r−1 9 × 10r−1 9 × 10r−1 9 × 10r−1
−
2.3
7 × 8r−1
7r
+
.
9 × 10r−1 9 × 10r−1
PERMUTATIONS
1. The answer is
1
1
=
≈ 0.0417.
4!
24
2. 3! = 6.
3.
8!
= 56.
3! 5!
4. The probability that John will arrive right after Jim is 7!/8! (consider Jim and John as one
arrival). Therefore, the answer is 1 − (7!/8!) = 0.875.
Another Solution: If Jim is the last person, John will not arrive after Jim. Therefore, the
remaining seven can arrive in 7! ways. If Jim is not the last person, the total number of
possibilities in which John will not arrive right after Jim is 7 × 6 × 6!. So the answer is
7! + 7 × 6 × 6!
= 0.875.
8!
5. (a) 312 = 531, 441. (b)
12!
= 924.
6! 6!
(c)
12!
= 27, 720.
3! 4! 5!
6. 6 P2 = 30.
7.
20!
= 3, 491, 888, 400.
4! 3! 5! 8!
8.
(5 × 4 × 7) × (4 × 3 × 6) × (3 × 2 × 5)
= 50, 400.
3!
Section 2.3
Permutations
17
9. There are 8! schedule possibilities. By symmetry, in 8!/2 of them Dr. Richman’s lecture
precedes Dr. Chollet’s and in 8!/2 ways Dr. Richman’s lecture precedes Dr. Chollet’s. So the
answer is 8!/2 = 20, 160.
10.
11!
= 92, 400.
3! 2! 3! 3!
11. 1 − (6!/66 ) = 0.985.
12. (a)
11!
= 34, 650.
4! 4! 2!
(b) Treating all P ’s as one entity, the answer is
(c) Treating all I ’s as one entity, the answer is
10!
= 6300.
4! 4!
8!
= 840.
4! 2!
(d) Treating all P ’s as one entity, and all I ’s as another entity, the answer is
7!
= 210.
4!
(e) By (a) and (c), The answer is 840/34650 = 0.024.
8!
68 = 0.000333.
2! 3! 3!
9!
14.
529 = 6.043 × 10−13 .
3! 3! 3!
13.
15.
m!
.
(n + m)!
16. Each girl and each boy has the same chance of occupying the 13th chair. So the answer is
12/20 = 0.6. This can also be seen from
17.
12 × 19!
12
=
= 0.6.
20!
20
12!
= 0.000054.
1212
18. Look at the five math books as one entity. The answer is
19. 1 −
20.
9 P7
97
= 0.962.
2 × 5! × 5!
= 0.0079.
10!
21. n!/nn .
5! × 18!
= 0.00068.
22!
18
Chapter 2
Combinatorial Methods
22. 1 − (6!/66 ) = 0.985.
23. Suppose that A and B are not on speaking terms.
134 P4 committees can be formed in which
neither A serves nor B; 4 ×134 P3 committees can be formed in which A serves and B does not.
The same numbers of committees can be formed in which B serves and A does not. Therefore,
the answer is 134 P4 + 2(4 ×134 P3 ) = 326, 998, 056.
24. (a) mn .
25. 3 ·
26. (a)
(b)
(b)
m Pn .
(c) n!.
8!
68 = 0.003.
2! 3! 2! 1!
20!
= 7.61 × 10−6 .
39 × 37 × 35 × · · · × 5 × 3 × 1
1
= 3.13 × 10−24 .
39 × 37 × 35 × · · · × 5 × 3 × 1
27. Thirty people can sit in 30! ways at a round table. But for each way, if they rotate 30 times
(everybody move one chair to the left at a time) no new situations will be created. Thus in
30!/30 = 29! ways 15 married couples can sit at a round table. Think of each married couple
as one entity and note that in 15!/15 = 14! ways 15 such entities can sit at a round table. We
have that the 15 couples can sit at a round table in (2!)15 · 14! different ways because if the
couples of each entity change positions between themselves, a new situation will be created.
So the desired probability is
14!(2!)15
= 3.23 × 10−16 .
29!
The answer to the second part is
24!(2!)5
= 2.25 × 10−6 .
29!
28. In 13! ways the balls can be drawn one after another. The number of those in which the first
white appears in the second or in the fourth or in the sixth or in the eighth draw is calculated
as follows. (These are Jack’s turns.)
8 × 5 × 11! + 8 × 7 × 6 × 5 × 9! + 8 × 7 × 6 × 5 × 4 × 5 × 7!
+ 8 × 7 × 6 × 5 × 4 × 3 × 2 × 5 × 5! = 2, 399, 846, 400.
Therefore, the answer is 2, 399, 846, 400/13! = 0.385.
Section 2.4
2.4
Combinations
19
COMBINATIONS
1.
20
= 38, 760.
6
100
2.
i=51
100
= 583, 379, 627, 841, 332, 604, 080, 945, 354, 060 ≈ 5.8 × 1029 .
i
3.
4.
5.
6.
7.
8.
9.
20 25
= 6, 864, 396, 000.
6
6
12 40
3
2
= 0.066.
52
5
N −1 N
n
= .
n−1
n
N
5 2
= 10.
3 2
8 5 3
= 560.
3 2 3
18
18
+
= 21, 624.
6
4
10
12
= 0.318.
5
7
12
10. The coefficient of 2 x in the expansion of (2 + x) is
. Therefore, the coefficient of x 9
9
12
= 1760.
is 23
9
7
3
4
7
11. The coefficient of (2x) (−4y) in the expansion of (2x − 4y) is
. Thus the coefficient
4
7
of x 3 y 2 in this expansion is 23 (−4)4
= 71, 680.
4
9
6
6
12.
+2
= 4620.
3
4
3
3 9
12
20
Chapter 2
13. (a)
Combinatorial Methods
10 10
2 = 0.246;
5
10
(b)
i=5
10 10
2 = 0.623.
i
14. If their
is larger than 5, they are all from the set {6, 7, 8, . . . , 20}. Hence the answer
minimum
15
20
= 0.194.
5
5
6 28
2
4
= 0.228;
15. (a)
34
6
50 150
5
45
16.
= 0.00206.
200
50
is
n
n
=
2
i
n
i
17.
i=0
n
i=0
i n
=
x
i
i=0
n
i=0
(b)
6
6
10
12
+
+
+
6
6
6
6
= 0.00084.
34
6
n i n−i
2 1 = (2 + 1)n = 3n .
i
n i n−i
x 1 = (x + 1)n .
i
6 4
5
66 = 0.201.
2
24
12
19. 2
= 0.00151.
12
18.
20. Royal Flush:
Straight flush:
Four of a kind:
4
= 0.0000015.
52
5
36
= 0.000014.
52
5
4
13 × 12
1
= 0.00024.
52
5
Section 2.4
Combinations
21
4
4
13
· 12
3
2
= 0.0014.
52
5
Full house:
Flush:
13
4
− 40
5
= 0.002.
52
5
Straight:
10(4)5 − 40
= 0.0039.
52
5
4
12 2
13
·
4
3
2
= 0.021.
52
5
Three of a kind:
Two pairs:
4 4
4
· 11
2 2
1
= 0.048.
52
5
13
2
4
12 3
·
4
2
3
= 0.42.
52
5
13
One pair:
None of the above:
1− the sum of all of the above cases = 0.5034445.
21. The desired probability is
12 12
6
6
= 0.3157.
24
12
x
22. The answer is the solution of the equation
= 20. This equation is equivalent to
3
x(x − 1)(x − 2) = 120 and its solution is x = 6.
22
Chapter 2
Combinatorial Methods
23. There are 9×103 = 9000 four-digit numbers. From every 4-combination of the set {0, 1, . . . , 9},
exactly one four-digit number can be constructed in which its ones place is less than its tens
place, its tens place is less than its hundreds place, and its hundreds
place
is less than its
10
thousands place. Therefore, the number of such four-digit numbers is
= 210. Hence
4
the desired probability is 0.023333.
24.
(x + y + z)2 =
n1 +n2 +n3
=
n!
x n1 y n2 zn3
n
!
n
!
n
!
1
2
3
=2
2!
2!
2!
x 2 y 0 z0 +
x 0 y 2 z0 +
x 0 y 0 z2
2! 0! 0!
0! 2! 0!
0! 0! 2!
+
2!
2!
2!
x 1 y 1 z0 +
x 1 y 0 z1 +
x 0 y 1 z1
1! 1! 0!
1! 0! 1!
0! 1! 1!
= x 2 + y 2 + z2 + 2xy + 2xz + 2yz.
25. The coefficient of (2x)2 (−y)3 (3z)2 in the expansion of (2x − y + 3z)7 is
coefficient of x 2 y 3 z2 in this expansion is 22 (−1)3 (3)2
7!
= −7560.
2! 3! 2!
7!
. Thus the
2! 3! 2!
13!
. Therefore,
3! 7! 3!
= −7, 413, 120.
26. The coefficient of (2x)3 (−y)7 (3)3 in the expansion of (2x − y + 3)13 is
the coefficient of x 3 y 7 in this expansion is 23 (−1)7 (3)3
13!
3! 7! 3!
52!
52!
ways 52 cards can be dealt among four people. Hence the sample
=
13! 13! 13! 13!
(13!)4
space contains 52!/(13!)4 points. Now in 4! ways the four different suits can be distributed
among the players; thus the desired probability is 4!/[52!/(13!)4 ] ≈ 4.47 × 10−28 .
27. In
28. The theorem is valid for k = 2; it is the binomial expansion. Suppose that it is true for all
integers ≤ k − 1. We show it for k. By the binomial expansion,
n n1
(x1 + x2 + · · · + xk ) =
x1 (x2 + · · · + xk )n−n1
n
1
n1 =0
n
(n − n1 )!
n n1
x1
=
x2n2 x3n3 · · · xknk
n
n
!
n
!
·
·
·
n
!
1
2
3
k
n2 +n3 +···+nk =n−n1
n1 =0
n
(n − n1 )!
=
x1n1 x2n2 · · · xknk
n
n
!
n
!
·
·
·
n
!
1
2 3
k
n +n +···+n =n
n
n
1
2
k
Section 2.4
=
n1 +n2 +···+nk
Combinations
23
n!
x1n1 x2n2 · · · xknk .
n
!
n
!
·
·
·
n
!
k
=n 1 2
29. We must have 8 steps. Since the distance from M to L is ten 5-centimeter intervals and the
first step is made
at M, there are 9 spots left at which the remaining 7 steps can be made. So
9
the answer is
= 36.
7
2 98
98
+
100
1 49
48
50
= 0.753; (b) 2
= 1.16 × 10−14 .
30. (a)
100
50
50
31. (a) It must be clear that
n1
n2
n3
n4
..
.
n
=
2
n1
+ nn1
=
2
n2
+ n2 (n + n1 )
=
2
n3
+ n3 (n + n1 + n2 )
=
2
nk−1
nk =
+ nk−1 (n + n1 + · · · + nk−1 ).
2
(b)
For n = 25, 000, successive calculations of nk ’s yield,
n1 = 312, 487, 500,
n2 = 48, 832, 030, 859, 381, 250,
n3 = 1, 192, 283, 634, 186, 401, 370, 231, 933, 886, 715, 625,
n4 = 710, 770, 132, 174, 366, 339, 321, 713, 883, 042, 336, 781, 236,
550, 151, 462, 446, 793, 456, 831, 056, 250.
For n = 25, 000, the total number of all possible hybrids in the first four generations,
n1 + n2 + n3 + n4 , is 710,770,132,174,366,339,321,713,883,042,337,973,520,184,337,
863,865,857,421,889,665,625. This number is approximately 710 × 1063 .
32. For n = 1, we have the trivial identity
1 0 1−0
1 1 1−1
x+y =
x y
+
x y .
0
1
24
Chapter 2
Combinatorial Methods
Assume that
n−1
(x + y)
n−1
=
i=0
n − 1 i n−1−i
.
xy
i
This gives
n−1
(x + y) = (x + y)
n − 1 i n−1−i
xy
i
n
i=0
n−1
n − 1 i+1 n−1−i
n − 1 i n−i
+
x y
xy
i
i
i=0
n−1
=
i=0
n−1
n − 1 i n−i
n − 1 i n−i
+
xy
xy
i−1
i
i=0
n
=
i=1
n−1
=x +
n
i=1
n−1
= xn +
i=1
n−1
n − 1 i n−i
+ yn
+
xy
i−1
i
n i n−i
+ yn =
xy
i
n
i=0
n i n−i
xy .
i
33. The desired probability is computed as follows.
12
6
30
2
28
2
26
2
24
2
22
2
20 18 15 12 9 6 3
1230 ≈ 0.000346.
2
3
3
3
3 3 3
10 6
10 9 4
2
2
6
1
4
34. (a) = 0.347; (b)
= 0.520;
20
20
6
6
10
10 8 2
2
3
2
2
= 0.130; (d) = 0.0031.
(c)
20
20
6
6
26 26
13 13
35.
= 0.218.
52
26
Section 2.4
Combinations
25
36. Let a 6-element combination of a set of integers be denoted by {a1 , a2 , . . . , a6 }, where a1 <
a2 < · · · < a6 . It can be easily verified that the function h : B → A defined by
h {a1 , a2 , . . . , a6 } = {a1 , a2 + 1, . . . , a6 + 5}
is one-to-one and onto. Therefore, there is a one-to-one
correspondence between B and
44
A . This shows that the number of elements in A is
. Thus the probability that no
6
44
49
consecutive integers are selected among the winning numbers is
≈ 0.505. This
6
6
implies that the probability of at least two consecutive integers among the winning numbers
is approximately 1 − 0.505 = 0.495. Given that there are 47 integers between 1 and 49, this
high probability might be counter-intuitive. Even without knowledge of expected value, a
keen student might observe that, on the average, there should be (49 − 1)/7 = 6.86 numbers
between each ai and ai+1 , 1 ≤ i ≤ 5. Thus he or she might erroneously think that it is unlikely
to obtain consecutive integers frequently.
37. (a) Let Ei be the event that car i remains unoccupied. The desired probability is
P (E1c E2c · · · Enc ) = 1 − P (E1 ∪ E2 ∪ · · · ∪ En ).
Clearly,
P (Ei ) =
(n − 1)m
,
nm
1 ≤ i ≤ n;
P (Ei Ej ) =
(n − 2)m
,
nm
1 ≤ i, j ≤ n, i = j ;
P (Ei Ej Ek ) =
(n − 3)m
,
nm
1 ≤ i, j, k ≤ n, i = j = k;
and so on. Therefore, by the inclusion-exclusion principle,
n
n (n − i)m
P (E1 ∪ E2 ∪ · · · ∪ En ) =
(−1)i−1
.
m
i
n
i=1
So
n
P (E1c E2c
· · · Enc )
=1−
i−1
(−1)
i=1
=
1
nm
n
(−1)i
i=0
n (n − i)m
=
i
nm
n (n − i)m
(−1)
i
nm
i=0
n
i
n
(n − i)m .
i
Let F be the event that cars 1, 2, . .. ,n − r are all occupied and the remaining cars are
n
unoccupied. The desired probability is
P (F ). Now by part (a), the number of ways m
r
(b)
26
Chapter 2
Combinatorial Methods
passengers can be distributed among n − r cars, no car remaining unoccupied is
n−r
(n − r − i)m .
(−1)
i
i=0
n−r
i
So
1
P (F ) = m
n
n−r
(n − r − i)m
(−1)
i
i=0
n−r
i
and hence the desired probability is
n−r
1 n
i n−r
(n − r − i)m .
(−1)
i
nm r i=0
38. Let the n indistinguishable balls be represented by n identical oranges and the n distinguishable
cells be represented by n persons. We should count the number of different ways that the n
oranges can be divided among the n persons, and the number of different ways in which exactly
one person does not get an orange. The answer to the latter part is n(n − 1) since in this case
one person does not get an orange, one person gets exactly two oranges, and the remaining
persons each get exactly one orange. There are n choices for the person who does not get
an orange and n − 1 choices for the person who gets exactly two oranges; n(n − 1) choices
altogether. To count the number of different ways that the n oranges can be divided among the
n persons, add n − 1 identical apples to the oranges and note that by Theorem 2.4, the total
(2n − 1)!
number of permutations of these n − 1 apples and n oranges is
. (We can arrange
n! (n − 1)!
n − 1 identical apples and n identical
oranges
in a row in (2n − 1)!/ n! (n − 1)! ways.) Now
(2n − 1)!
2n − 1
each one of these
=
permutations corresponds to a way of dividing the
n! (n − 1)!
n
n oranges among the n persons and vice versa. Give all of the oranges preceding the first apple
to the first person, the oranges between the first and the second apples to the second person,
the oranges between the second and the third apples to the third person and so on. Therefore,
if, for example, an apple appears in the beginning of the permutation, the first person does not
get an orange, and if two apples are at the end of the permutations,
the (n − 1)st and the nth
2n − 1
persons get no oranges. Thus the answer is n(n − 1)
.
n
39. The left side of the identity is the binomial expansion of (1 − 1)n = 0.
Section 2.4
Combinations
27
40. Using the hint, we have
n
n+1
n+2
n+r
+
+
+ ··· +
0
1
2
r
n
n+2
n+1
n+3
n+2
=
+
−
+
−
0
1
0
2
1
n+4
n+3
n+r +1
n+r
−
+ ··· +
−
+
3
2
r
r −1
n
n+1
n+r +1
n+r +1
=
−
+
=
.
0
0
r
r
41. The identity expresses that to choose r balls from n red and m blue balls, we must choose
either r red balls, 0 blue balls or r − 1 red balls, one blue ball or r − 2 red balls, two blue balls
or · · · 0 red balls, r blue balls.
1
1
n
n+1
42. Note that
=
. Hence
i+1 i
n+1 i+1
1
1
n+1
n+1
n+1
The given sum =
+
+ ··· +
=
(2n+1 − 1).
n+1
1
2
n+1
n+1
5 3
43.
3
45 = 0.264.
2
t
N −t
m n−m
.
44. (a) PN =
N
n
(b) From part (a), we have
PN
(N − t)(N − n)
=
.
PN −1
N(N − t − n + m)
This implies PN > PN−1 if and only if (N − t)(N − n) > N (N − t − n + m) or, equivalently,
if and only if N ≤ nt/m. So PN is increasing if and only if N ≤ nt/m. This shows that the
maximum of PN is at [nt/m], where by [nt/m] we mean the greatest integer ≤ nt/m.
45. The sample space consists of (n + 1)4 elements. Let the elements of the sample be denoted by
x1 , x2 , x3 , and x4 . To count the number of samples (x1 , x2 , x3 , x4 ) for which x1 + x2 = x3 + x4 ,
let y3 = n − x3 and y4 = n − x4 . Then y3 and y4 are also random elements from the set
{0, 1, 2, . . . , n}. The number of cases in which x1 + x2 = x3 + x4 is identical to the number of
cases in which x1 + x2 + y3 + y4 = 2n. By Example 2.23, the number of nonnegative integer
28
Chapter 2
Combinatorial Methods
2n + 3
solutions to this equation is
. However, this also counts the solutions in which one
3
of x1 , x2 , y3 , and y4 is greater than n. Because of the restrictions 0 ≤ x1 , x2 , y3 , y4 ≤ n,
we must subtract, from this number, the total number of the solutions in which one of x1 , x2 ,
y3 , and y4 is greater than n. Such solutions are obtained by finding all nonnegative integer
solutions of the equation x1 + x2 + y3 + y4 = n − 1, and then adding n + 1 to exactly one
of x1 , x2 , y3 , and y4 . Their count is
4 timesthe number of nonnegative integer solutions of
n+2
x1 + x2 + y3 + y4 = n − 1; that is, 4
. Therefore, the desired probability is
3
2n + 3
n+2
−4
2n2 + 4n + 3
3
3
=
.
(n + 1)4
3(n + 1)3
46. (a) The n − m unqualified applicants are “ringers.” The experiment is not affected by their
inclusion, so that the probability of any one of the qualified applicants being selected is the
same as it would be if there were only qualified applicants. That is, 1/m. This is because in
a random arrangement of m qualified applicants, the probability that a given applicant is the
first one is 1/m.
(b) Let A be the event that a given qualified applicant is hired. We will show that P (A) =
1/m. Let Ei be the event that the given qualified applicant is the ith applicant interviewed,
and he or she is the first qualified applicant to be interviewed. Clearly,
n−m+1
P (A) =
P (Ei ),
i=1
where
P (Ei ) =
n−m Pi−1
· 1 · (n − i)!
.
n!
Therefore,
n−m+1
P (A) =
· (n − i)!
n!
n−m Pi−1
i=1
n−m+1
=
(n − m)!
(n − i)!
(n − m − i + 1)!
n!
i=1
n−m+1
1
(n − i)!
(m − 1)!
·
(n − m − i + 1)! (m − 1)!
n!
i=1
m! (n − m)!
n−m+1
1
1
n−i
=
·
n
m
m−1
i=1
m
=
1
·
m!
Section 2.4
1
1
=
·
n
m
m
Combinations
n−m+1
i=1
n−i
.
m−1
29
(4)
n−m+1
n−i
n−i
, note that
is the coefficient of x m−1 in the expansion
To calculate
m
−
1
m
−
1
i=1
n−m+1
n−i
n−i
of (1 + x) . Therefore,
is the coefficient of x m−1 in the expansion of
m
−
1
i=1
n−m+1
(1 + x)n−i =
i=1
(1 + x)n − (1 + x)m−1
.
x
n−m+1
n−i
is the coefficient of x m in the expansion of
This shows that
m
−
1
i=1
n
n
m−1
(1 + x) − (1 + x) , which is
. So (4) implies that
m
n
1
1
1
= .
· ·
P (A) =
n
m
m
m
m
6
47. Clearly, N = 610 , N(Ai ) = 510 , N (Ai Aj ) = 410 , i = j , and so on. So S1 has
equal
1
6
terms, S2 has
equal terms, and so on. Therefore, the solution is
2
6 10
6 10
6 10
6 10
6 10
6 10
10
5 +
4 −
3 +
2 −
1 +
0 = 16, 435, 440.
6 −
1
2
3
4
5
6
1 n n−3
1 n 3 n−3
1 n 3 n−3
48. |A0 | =
, |A1 | =
, |A2 | =
.
2 3
3
2 3 1
2
2 3 2
1
The answer is
|A0 |
(n − 4)(n − 5)
=
.
|A0 | + |A1 | + |A2 |
n2 + 2
2n
n
2n
49. The coefficient of x in (1 + x) is
. Its coefficient in (1 + x)n (1 + x)n is
n
n n
n
n
n
n
n n
+
+
+ ··· +
0 n
1 n−1
2 n−2
n 0
2
2 2 2
n
n
n
n
+
+
+ ··· +
,
=
1
2
n
0
30
Chapter 2
Combinatorial Methods
n
n
since
=
, 0 ≤ i ≤ n.
i
n−1
50. Consider a particular set of k letters. Let M be the number of possibilitiesinwhich only
n
these k letters are addressed correctly. The desired probability is the quantity
M n!. All
k
we got to do is to find M. To do so, note that the remaining n − k letters are all addressed
incorrectly. For these n − k letters, there are n − k addresses. But the addresses are written
on the envelopes at random. The probability that none is addressed correctly on one hand is
M/(n − k)!, and on the other hand, by Example 2.24, is
n−k
1−
i=1
n
(−1)i−1
=
i!
So M satisfies
M
=
(n − k)!
n
i=2
and hence
i=2
(−1)i−1
.
i!
(−1)i−1
,
i!
n
M = (n − k)!
i=2
(−1)i−1
.
i!
The final answer is
n
M
k
=
n!
n
(n − k)!
k
n
i=2
n!
(−1)i−1
i!
=
1
k!
n
i=2
(−1)i−1
.
i!
51. The set of all sequences of H’s and T’s of length i with no successive H’s are obtained either
by adding a T to the tails of all such sequences of length i − 1, or a TH to the tails of all such
sequences of length i − 2. Therefore,
xi = xi−1 + xi−2 , i ≥ 2.
Clearly, x1 = 2 and x3 = 3. For consistency, we define x0 = 1. From the theory of recurrence
relations we know that the solution of xi = xi−1 + xi−2 is of the√form xi = Ar1i +√
Br2i , where
1+ 5
1− 5
and r2 =
and so
r1 and r2 are the solutions of r 2 = r + 1. Therefore, r1 =
2
2
1 + √5 i
1 − √5 i
xi = A
+B
.
2
2
√
√
5+3 5
5−3 5
Using the initial conditions x0 = 1 and x2 = 2, we obtain A =
and B =
.
10
10
Section 2.5
Stirling’s Formula
31
Hence the answer is
√
√
√
√
xn
1 5 + 3 5 1 + 5 n 5 − 3 5 1 − 5 n
= n
+
2n
2
10
2
10
2
√
√ n
√
√ n
1
5
+
3
=
5
1
+
5
+
5
−
3
5
1
−
5 .
10 × 22n
52. For this exercise, a solution is given by Abramson and Moser in the October 1970 issue of the
American Mathematical Monthly.
2.5
STIRLING’s FORMULA
√
4π n (2n)2n e−2n
2n 1
(2n)! 1
1
=
∼
∼√ .
2n
2n
2n
−2n
2n
n! n! 2
(2π n) n e
2
n 2
πn
√
3
3
√
4π n (2n)2n e−2n
(2n)!
2
∼√
= n.
2
4n
−4n
2n
−2n
(4n)! (n!)
4
8π n (4n) e
(2π n) n e
1. (a)
(b)
REVIEW PROBLEMS FOR CHAPTER 2
1. The desired quantity is equal to the number of subsets of all seven varieties of fruit minus 1
(the empty set); so it is 27 − 1 = 127.
2. The number of choices Virginia has is equal to the number of subsets of {1, 2, 5, 10, 20} minus
1 (for empty set). So the answer is 25 − 1 = 31.
3. (6 × 5 × 4 × 3)/64 = 0.278.
4. 10
5.
10
2
= 0.222.
9!
= 7560.
3! 2! 2! 2!
6. 5!/5 = 4! = 24.
7. 3! · 4! · 4! · 4! = 82, 944.
23
6
8. 1 − = 0.83.
30
6
32
Chapter 2
Combinatorial Methods
9. Since the refrigerators are identical, the answer is 1.
10. 6! = 720.
11. (Draw a tree diagram.) In 18 out of 52 possible cases the tournament ends because John wins
4 games without winning 3 in a row. So the answer is 34.62%.
12. Yes, it is because the probability of what happened is 1/72 = 0.02.
13. 9 8 = 43, 046, 721.
14. (a) 26 × 25 × 24 × 23 × 22 × 21 = 165, 765, 600;
(b) 26 × 25 × 24 × 23 × 22 × 5 = 39, 468, 000;
5
3
2
1
(c)
26
25
24
23 = 21, 528, 000.
2
1
1
1
6
6
6
6 2 2
+
+
+
3
1
1
1 1 1
= 0.467.
15.
10
3
6
6 4
+
3
1 2
Another Solution:
= 0.467.
10
3
16.
8 × 4 ×6 P4
= 0.571.
8 P6
17. 1 −
18.
278
= 0.252.
288
(3!/3)(5!)3
= 0.000396.
15!/15
19. 312 = 531, 441.
20.
4 48 3 36 2 24 1 12
1 12 1 12 1 12 1 12
52!
13! 13! 13! 13!
= 0.1055.
Chapter 2
Review Problems
33
21. Let A1 , A2 , A3 , and A4 be the events that there is no professor, no associate professor, no
assistant professor, and no instructor in the committee, respectively. The desired probability
is
P (Ac1 Ac2 Ac3 Ac4 ) = 1 − P (A1 ∪ A2 ∪ A3 ∪ A4 ),
where P (A1 ∪ A2 ∪ A3 ∪ A4 ) is calculated using the inclusion-exclusion principle:
P (A1 ∪ A2 ∪ A3 ∪ A4 ) = P (A1 ) + P (A2 ) + P (A3 ) + P (A4 )
− P (A1 A2 ) − P (A1 A3 ) − P (A1 A4 ) − P (A2 A3 ) − P (A2 A4 ) − P (A3 A4 )
+ P (A1 A2 A3 ) + P (A1 A3 A4 ) + P (A1 A2 A4 ) + P (A2 A3 A4 ) − P (A1 A2 A3 A4 )
34
28
28
24
22
22
18
16
18
= 1
+
+
+
−
−
−
−
6
6
6
6
6
6
6
6
6
16
12
12
6
10
6
−
−
+
+
+
+
− 0 = 0.621.
6
6
6
6
6
6
Therefore, the desired probability equals 1 − 0.621 = 0.379.
(15!)2
= 0.0002112.
30!/(2!)15
N
23. (N − n + 1)
.
n
4 48
40
2 24
1
= 0.390; (b) = 6.299 × 10−11 ;
24. (a)
52
52
26
13
13 39 8 31
5
8
8
5
= 0.00000261.
(c)
52 39
13 13
22.
25. 12!/(3!)4 = 369, 600.
26. There is a one-to-one correspondence between all cases in which the eighth outcome obtained
is not a repetition and all cases in which the first outcome obtained will not be repeated. The
answer is
6 × 5 × 5 × 5 × 5 × 5 × 5 × 5 5 7
= 0.279.
=
6
6×6×6×6×6×6×6×6
27. There are 9 × 103 = 9, 000 four-digit numbers. To count the number of desired four-digit
numbers, note that if 0 is to be one of the digits, then the thousands place of the number must be
34
Chapter 2
Combinatorial Methods
0, but this cannot be the case since the first digit of an n-digit number is nonzero. Keeping this
in mind, it must be clear that from every 4-combination of the set {1, 2, . . . , 9}, exactly one
four-digit number can be constructed in which its ones place is greater than its tens place, its
tens place is greater than it hundreds place, and its hundreds place
is greater than its thousands
9
place. Therefore, the number of such four-digit numbers is
= 126. Hence the desired
4
probability is = 0.014.
28. Since the sum of the digits of 100,000 is 1, we ignore 100,000 and assume that all of the numbers
have five digits by placing 0’s in front of those with less than five digits. The following
process
establishes a one-to-one correspondence between such numbers, d1 d2 d3 d4 d5 , 5i=1 di = 8,
and placement of 8 identical objects into 5 distinguishable cells: Put d1 of the objects into
the first cell, d2 of the
into the
cell, d3 into the third cell, and so on. Since
objects
second
8+5−1
12
this can be done in
=
= 495 ways, the number of integers from the set
5−1
8
{1, 2, 3, . . . , 100000} in which the sum of the digits is 8 is 495. Hence the desired probability
is 495/100, 000 = 0.00495.
Chapter 3
C onditional Probability
and I ndependence
3.1
CONDITIONAL PROBABILITY
1. P (W | U ) =
P (U W )
0.15
=
= 0.60.
P (U )
0.25
2. Let E be the event that in the blood of the randomly selected soldier A antigen is found. Let
F be the event that the blood type of the soldier is A. We have
P (F | E) =
3.
P (F E)
0.41
=
= 0.911.
P (E)
0.41 + 0.04
0.20
= 0.625.
0.32
4. The reduced sample space is (1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4) ; therefore, the
desired probability is 1/7.
5.
2
30 − 20
= .
30 − 15
3
6. Both of the inequalities are equivalent to P (AB) > P (A)P (B).
7.
2
1/3
= .
(1/3) + (1/2)
5
8. 4/30 = 0.133.
36
Chapter 3
Conditional Probability and Independence
40 65
2
6
105
8
9.
= 0.239.
40
65
2
8−i
i
1−
105
i=0
8
⎧
⎪
1/19 if i = 0
⎪
⎨
10. P (α = i | β = 0) = 2/19 if i = 1, 2, 3, . . . , 9
⎪
⎪
⎩
0
if i = 10, 11, 12, . . . , 18.
11. Let b∗ gb mean that the oldest child of the family is a boy, the second oldest is a girl, the youngest
is a boy, and the boy found in the family is the oldest child, with similar representations for
other cases. The reduced sample space is
S = ggb∗ , gb∗ g, b∗ gg, b∗ bg, bb∗ g, gb∗ b, gbb∗ , bgb∗ , b∗ gb, b∗ bb, bb∗ b, bbb∗ .
Note that the outcomes of the sample space are not equiprobable. We have that
P {ggb∗ } = P {gb∗ g} = P {b∗ gg} = 1/7
P {b∗ bg} = P {bb∗ g} = 1/14
P {gb∗ b} = P {gbb∗ } = 1/14
P {bgb∗ } = P {b∗ gb} = 1/14
P {b∗ bb} = P {bb∗ b} = P {bbb∗ } = 1/21.
The solutions to (a), (b), (c) are as follows.
(a) P {bb∗ g} = 1/14;
(b) P {bb∗ g, gbb∗ , bgb∗ , bb∗ b, bbb∗ } = 13/42;
(c) P {b∗ bg, bb∗ g, gb∗ b, gbb∗ , bgb∗ , b∗ gb} = 3/7.
12. P (A) = 1 implies that P (A ∪ B) = 1. Hence, by
P (A ∪ B) = P (A) + P (B) − P (AB),
we have that P (B) = P (AB). Therefore,
P (B | A) =
P (B)
P (AB)
=
= P (B).
P (A)
1
Section 3.1
Conditional Probability
37
P (AB)
, where
b
P (AB) = P (A) + P (B) − P (A ∪ B) ≥ P (A) + P (B) − 1 = a + b − 1.
13. P (A | B) =
14. (a) P (AB) ≥ 0, P (B) > 0. Therefore, P (A | B) =
P (AB)
≥ 0.
P (B)
P (SB)
P (B)
=
= 1.
P (B)
P (B)
∞
∞
P
∞
i=1 Ai B
P
i=1 Ai B
(c) P
Ai B =
=
P (B)
P (B)
i=1
(b)
P (S | B) =
∞
P (Ai B)
=
i=1
P (B)
∞
=
i=1
P (Ai B)
=
P (B)
∞
P (Ai | B).
i=1
∞
Note that P (∪∞
i=1 Ai B) =
i=1 P (Ai B), since mutual exclusiveness of Ai ’s imply that of
Ai B’s; i.e., Ai Aj = ∅, i = j , implies that (Ai B)(Aj B) = ∅, i = j .
15. The given inequalities imply that P (EF ) ≥ P (GF ) and P (EF c ) ≥ P (GF c ). Thus
P (E) = P (EF ) + P (EF c ) ≥ P (GF ) + P (GF c ) = P (G).
16. Reduce the sample space: Marlon chooses from six dramas and sevencomedies
two
at random.
What is the probability that they are both comedies? The answer is
7 13
= 0.269.
2
2
17. Reduce the sample space: There are 21 crayons of which three are red. Seven of these crayons
are selected at
random
and given to Marty. What is the probability that three of them are red?
18 21
The answer is
= 0.0263.
4
7
18. (a) The reduced sample space is S = {1, 3, 5, 7, 9, . . . , 9999}. There are 5000 elements in
S. Since the set {5, 7, 9, 11, 13, 15, . . . , 9999} includes exactly 4998/3 = 1666 odd numbers
that are divisible by three, the reduced sample space has 1667 odd numbers that are divisible
by 3. So the answer is 1667/5000 = 0.3334.
(b) Let O be the event that the number selected at random is odd. Let F be the event that it is
divisible by 5 and T be the event that it is divisible by 3. The desired probability is calculated
as follows.
P (F c T c | O) = 1 − P (F ∪ T | O) = 1 − P (F | O) − P (T | O) + P (F T | O)
=1−
1000 1667
333
−
+
= 0.5332.
5000 5000 5000
38
Chapter 3
Conditional Probability and Independence
19. Let A be the event that during this period he has hiked in Oregon Ridge Park at least once. Let
B be the event that during this period he has hiked in this park at least twice. We have
P (B | A) =
where
P (A) = 1 −
and
P (B) = 1 −
P (B)
,
P (A)
510
= 0.838
610
510 10 × 59
−
= 0.515.
610
610
So the answer is 0.515/0.838 = 0.615.
20. The numbers of 333 red and 583 blue chips are divisible by 3. Thus the reduced sample space
has 333 + 583 = 916 points. Of these numbers, [1000/15] = 66 belong to red balls and
are divisible by 5 and [1750/15] = 116 belong to blue balls and are divisible by 5. Thus the
desired probability is 182/916 = 0.199.
21. Reduce the sample space: There are two types of animals in a laboratory, 15 type I and 13
type II. Six animals are selected at random; what is the probability that at least two of them
are Type II? The answer is
15
13 15
+
6
1
5
1−
= 0.883.
28
6
22. Reduce the sample space: 30 students of which 12 are French and nine are Korean are divided
randomly into two classes of 15 each. What is the probability that one of them has exactly
four French and exactly three Korean students? The solution to this problem is
12 9 9
4
3 8
= 0.00241.
30 15
15 15
23. This sounds puzzling because apparently the only deduction from the name “Mary” is that one
of the children is a girl. But the crucial difference between this and Example 3.2 is reflected
in the implicit assumption that both girls cannot be Mary. That is, the same name cannot be
used for two children in the same family. In fact, any other identifying feature that cannot be
shared by both girls would do the trick.
Section 3.2
3.2
Law of Multiplication
39
LAW OF MULTIPLICATION
1. Let G be the event that Susan is guilty. Let L be the event that Robert will lie. The probability
that Robert will commit perjury is
P (GL) = P (G)P (L | G) = (0.65)(0.25) = 0.1625.
2. The answer is
11 10
9
8
7
6
×
×
×
×
× = 0.15.
14 13 12 11 10 9
3. By the law of multiplication, the answer is
52 50 48 46 44 42
×
×
×
×
×
= 0.72.
52 51 50 49 48 47
4. (a)
(b)
5. (a)
(b)
6.
7
6
8
5
×
×
×
= 0.0144;
20 19 18 17
8
7
12
8
12
7
12
8
7
8
7
6
×
×
+
×
×
+
×
×
+
×
×
= 0.344.
20 19 18 20 19 18 20 19 18 20 19 18
5
5 4 4 3 3 2 2 1 1
6
×
× × × × × × × × × = 0.00216.
11 10 9 8 7 6 5 4 3 2 1
5
4
3 2 1
×
× × × = 0.00216.
11 10 9 8 7
5
5
8
5
3
8
5
3
×
×
×
+ ×
×
×
= 0.0712.
8 10 13 15 8 11 13 16
7. Let Ai be the event that the ith person draws the “you lose” paper. Clearly,
P (A1 ) =
1
,
200
1
199 1
·
=
,
200 199
200
199 198 1
1
P (A3 ) = P (Ac1 Ac2 A3 ) = P (Ac1 )P (Ac2 | Ac1 )P (A3 | Ac1 Ac2 ) =
·
·
=
,
200 199 198
200
P (A2 ) = P (Ac1 A2 ) = P (Ac1 )P (A2 | Ac1 ) =
and so on. Therefore, P (Ai ) = 1/200 for 1 ≤ i ≤ 200. This means that it makes no difference
if you draw first, last or anywhere in the middle. Here is Marilyn Vos Savant’s intuitive solution
to this problem:
40
Chapter 3
Conditional Probability and Independence
It makes no difference if you draw first, last, or anywhere in the middle. Look at it
this way: Say the robbers make everyone draw at once. You’d agree that everyone
has the same change of losing (one in 200), right? Taking turns just makes that
same event happen in a slow and orderly fashion. Envision a raffle at a church with
200 people in attendance, each person buys a ticket. Some buy a ticket when they
arrive, some during the event, and some just before the winner is drawn. It doesn’t
matter. At the party the end result is this: all 200 guests draw a slip of paper, and,
regardless of when they look at the slips, the result will be identical: one will lose.
You can’t alter your chances by looking at your slip before anyone else does, or
waiting until everyone else has looked at theirs.
8. Let B be the event that a randomly selected person from the population at large has poor credit
report. Let I be the event that the person selected at random will improve his or her credit
rating within the next three years. We have
P (B | I ) =
P (BI )
P (I | B)P (B)
(0.30)(0.18)
=
=
= 0.072.
P (I )
P (I )
0.75
The desired probability is 1−0.072 = 0.928. Therefore, 92.8% of the people who will improve
their credit records within the next three years are the ones with good credit ratings.
9. For 1 ≤ n ≤ 39, let En be the event that none of the first n − 1 cards is a heart or the ace
of spades. Let Fn be the event that the nth
card drawn is the ace of spades. Then the event
of “no heart before the ace of spades” is 39
n=1 En Fn . Clearly, {En Fn , 1 ≤ n ≤ 39} forms a
sequence of mutually exclusive events. Hence
39
39
n=1
39
P (En Fn ) =
En Fn =
P
n=1
39
=
n=1
P (En )P (Fn | En )
n=1
38
1
1
n−1
×
= ,
52
53 − n
14
n−1
a result which is not unexpected.
13 39
10
3
6
= 0.059.
10. P (F )P (E | F ) = ×
52
43
9
11. By the law of multiplication,
P (An ) =
2 3 4
n+1
2
× × × ··· ×
=
.
3 4 5
n+2
n+2
Section 3.3
Law of Total Probability
41
Now since A1 ⊇ A2 ⊇ A3 ⊇ · · · ⊇ An ⊇ An+1 ⊇ · · · , by Theorem 1.8,
∞
Ai = lim P (An ) = 0.
P
i=1
3.3
1.
n→∞
LAW OF TOTAL PROBABILITY
1
1
× 0.05 + × 0.0025 = 0.02625.
2
2
2. (0.16)(0.60) + (0.20)(0.40) = 0.176.
3.
1
1
1
(0.75) + (0.68) + (0.47) = 0.633.
3
3
3
1
12 13 13 39
×
+
×
= .
51 52 51 52
4
13
13 39
39
12
13
1
11
2
1
1
2
× +
×
× = .
5.
+
52
52
52
50
50
50
4
2
2
2
4.
6. (0.20)(0.40) + (0.35)(0.60) = 0.290.
7. (0.37)(0.80) + (0.63)(0.65) = 0.7055.
8.
1
1
1
1
1
1
(0.6) + (0.5) + (0.7) + (0.9) + (0.7) + (0.8) = 0.7.
6
6
6
6
6
6
9. (0.50)(0.04) + (0.30)(0.02) + (0.20)(0.04) = 0.034.
10. Let B be the event that the randomly selected child from the countryside is a boy. Let E be
the event that the randomly selected child is the first child of the family and F be the event
that he or she is the second child of the family. Clearly, P (E) = 2/3 and P (F ) = 1/3. By
the law of total probability,
P (B) = P (B | E)P (E) + P (B | F )P (F ) =
1 2 1 1
1
× + × = .
2 3 2 3
2
Therefore, assuming that sex distributions are equally probable, in the Chinese countryside,
the distribution of sexes will remain equal. Here is Marilyn Vos Savant’s intuitive solution to
this problem:
42
Chapter 3
Conditional Probability and Independence
The distribution of sexes will remain roughly equal. That’s because–no matter how
many or how few children are born anywhere, anytime, with or without restriction–
half will be boys and half will be girls: Only the act of conception (not the government!) determines their sex.
One can demonstrate this mathematically. (In this example, we’ll assume that
women with firstborn girls will always have a second child.) Let’s say 100 women
give birth, half to boys and half to girls. The half with boys must end their families.
There are now 50 boys and 50 girls. The half with girls (50) give birth again, half
to boys and half to girls. This adds 25 boys and 25 girls, so there are now 75 boys
and 75 girls. Now all must end their families. So the result of the policy is that there
will be fewer children in number, but the boy/girl ratio will not be affected.
11. The probability that the first person gets a gold coin is 3/5. The probability that the second
person gets a gold coin is
2 3 3 2
3
× + × = .
4 5 4 5
5
The probability that the third person gets a gold coin is
3 2 1 3 2 2 2 3 2 2 1 3
3
× × + × × + × × + × × = ,
5 4 3 5 4 3 5 4 5 5 4 3
5
and so on. Therefore, they are all equal.
12. A Probabilistic Solution: Let n be the number of adults in the town. Let x be the number
of men in the town. Then n − x is the number of women in the town. Since the number of
married men and married women are equal, we have
x·
7
3
= (n − x) · .
9
5
This relation implies that x = (27/62)n. Therefore, the probability that a randomly selected
adult is male is (27/62)n n = 27/62. The probability that a randomly selected adult is female
is 1 − (27/62) = 35/62. Let A be the event that a randomly selected adult is married. Let M
be the event that the randomly selected adult is a man, and let W be the event that the randomly
selected adult is a woman. By the law of total probability,
P (A) = P (A | M)P (M) + P (A | W )P (W )
=
7 27 3 35
42
21
·
+ ·
=
=
≈ 0.677.
9 62 5 62
62
31
Therefore, 21/31st of the adults are married.
An Arithmetical Solution:
The common numerator of the two fractions is 21. Hence
21/27th of the men and 21/35th of the women are married. We find the common numerator
because the number of married men and the number of married women are equal. This shows
that of every 27 + 35 = 62 adults, 21 + 21 = 42 are married. Hence 42/62th = 21/31st of the
adults in the town are married.
Section 3.3
Law of Total Probability
43
13. The answer is clearly 0.40. This can also be computed from
(0.40)(0.75) + (0.40)(0.25) = 0.40.
14. Let A be the event that a randomly selected child is the kth born of his or her family. Let Bj
be the event that he or she is from a family with j children. Then
c
P (A) =
P (A | Bj )P (Bj ),
j =k
where, clearly, P (A | Bj ) = 1/j . To find P (Bj ), note that
there are αi N families with j
children. Therefore, the total number of children in the world is ci=0 i(αi N) of which j (N αj )
are from families with j children. Hence
j αj
j (N αj )
= c
P (Bj ) = c
.
i=0 i(αi N )
i=0 iαi
This shows that the desired fraction is given by
c
P (A) =
c
P (A | Bj )P (Bj ) =
j =k
c
=
j =k
15. Q(E | F ) =
αj
c
i=0
c
iαi
j =k
j =k
αj
i=0
iαi
= c
j αj
1
· c
j
i=0 iαi
.
P (EF B)
P (B)
Q(EF )
P (EF | B)
P (EF B)
=
=
= P (E | F B).
=
Q(F )
P (F | B)
P (F B)
P (F B)
P (B)
16. Let M, C, and F denote the events that the random student is married, is married to a student
at the same campus, and is female, respectively. We have that
1
2
P (F | M) = P (F | MC)P (C | M)+P (F | MC c )P (C c | M) = (0.40) +(0.30) = 0.333.
3
3
17. Let p(k, n) be the probability that exactly k of the first n seeds planted in the farm germinated.
Using induction on n, we will show that p(k, n) = 1/(n − 1) for all k < n. For n = 2,
p(1, 2) = 1 = 1/(2 − 1) is true. If p(k, n − 1) = 1/(n − 2) for all k < n − 1, then, by the
law of total probability,
p(k, n) =
=
k−1
n−k−1
p(k − 1, n − 1) +
p(k, n − 1)
n−1
n−1
k−1
1
n−k−1
1
1
·
+
·
=
.
n−1 n−2
n−1
n−2
n−1
This proves the induction hypothesis.
44
Chapter 3
Conditional Probability and Independence
18. Reducing the sample space, we have that the answer is 7/10.
10
7
10 8
6
10 8
5
8
8
3
3
2
1
3
1
2
3
3
3
19. × + × + × + × = 0.0383.
18
18
18
18
18
18
18
18
3
3
3
3
3
3
3
3
20. We have that
P (A | G) = P (A | GO)P (O | G) + P (A | GM)P (M | G) + P (A | GY )P (Y | G)
1 1 1 3 1
5
=0× + × + × =
.
3 2 3 4 3
12
21. Let E be the event that the third number falls between the first two. Let A be the event that
the first number is smaller than the second number. We have that
P (E | A) =
P (EA)
1/6
1
=
= .
P (A)
1/2
3
Intuitively, the fact that P (A) = 1/2 and P (EA) = 1/6 should be clear (say, by symmetry).
However, we can prove these rigorously. We show that P (A) = 1/2; P (EA) = 1/6 can be
proved similarly. Let B be the event that the second number selected is smaller than the first
number. Clearly A = B c and we only need to show that P (B) = 1/2. To do this, let Bi be
the event that the first number drawn is i, 1 ≤ i ≤ n. Since {B1 , B2 , . . . , Bn } is a partition of
the sample space,
n
P (B | Bi )P (Bi ).
P (B) =
i=1
Now P (B | B1 ) = 0 because if the first number selected is 1, the second number selected
i−1
cannot be smaller. P (B | Bi ) =
, 1 ≤ i ≤ n since if the first number is i, the second
n−1
number must be one of 1, 2, 3, . . . , i − 1 if it is to be smaller. Thus
n
n
P (B) =
P (B | Bi )P (Bi ) =
i=1
=
i=2
i−1 1
1
· =
n−1 n
(n − 1)n
n
(i − 1)
i=2
1
1
(n − 1)n
1
1 + 2 + 3 + · · · + (n − 1) =
·
= .
(n − 1)n
(n − 1)n
2
2
22. Let Em be the event that Avril selects the best suitor given her strategy. Let Bi be the event
that the best suitor is the ith of Avril’s dates. By the law of total probability,
n
P (Em ) =
i=1
1
P (Em | Bi )P (Bi ) =
n
n
P (Em | Bi ).
i=1
Section 3.3
Law of Total Probability
45
Clearly, P (Em | Bi ) = 0 for 1 ≤ i ≤ m. For i > m, if the ith suitor is the best, then Avril
chooses him if and only if among the first i − 1 suitors Avril dates, the best is one of the first
m. So
m
P (Em | Bi ) =
.
i−1
Therefore,
n
n
m
1
m
1
=
.
P (Em ) =
n i=m+1 i − 1
n i=m+1 i − 1
Now
n
1
≈
i
−
1
i=m+1
n
m
n
1
dx = ln
.
x
m
m n
ln
.
n
m
To find the maximum of P (Em ), consider the differentiable function
x n
h(x) = ln
.
n
x
Thus
P (Em ) ≈
Since
1 n
1
ln
− =0
n
x
n
implies that x = n/e, the maximum of P (Em ) is at m = [n/e], where [n/e] is the greatest
integer less than or equal to n/e. Hence Avril should dump the first [n/e] suitors she dates
and marry the first suitor she dates afterward who is better than all those preceding him. The
probability that with such a strategy she selects the best suitor of all n is approximately
h (x) =
n
h
e
=
1
1
ln e = ≈ 0.368.
e
e
23. Let N be the set of nonnegative integers. The domain of f is
(g, r) ∈ N × N : 0 ≤ g ≤ N, 0 ≤ r ≤ N, 0 < g + r < 2N .
∂f
∂f
=
= 0 gives
∂g
∂r
g = r = N/2 and f (N/2, N/2) = 1/2. However, this is not the maximum value because on
the boundary of the domain of f along r = 0, we find that
Extending the domain of f to all points (g, r) ∈ R × R, we find that
f (g, 0) =
1
N −g
1+
2
2N − g
f (1, 0) =
1 3N − 2
1
≥ .
2 2N − 1
2
is maximum at g = 1 and
46
Chapter 3
Conditional Probability and Independence
We also find that on the boundary along r = N,
1 g
+1
f (g, N ) =
2 g+N
is maximum at g = N − 1 and
f (N − 1, N ) =
1
1 3N − 2
≥ .
2 2N − 1
2
1 3N − 2
. Therefore,
2 2N − 1
there are exactly two maximums and they occur at (1, 0) and (N −1, N). That is, the maximum
of f occurs if one urn contains one green and 0 red balls and the other one contains N −1 green
1 3N − 2
3
and N red balls. For large N, the probability that the prisoner is freed is
≈ .
2 2N − 1
4
The maximums of f along other sides of the boundary are all less than
3.4
BAYES’ FORMULA
1.
(3/4)(0.40)
3
= .
(3/4)(0.40) + (1/3)(0.60)
5
2.
1(2/3)
8
= .
1(2/3) + (1/4)(1/3)
9
3. Let G and I be the events that the suspect is guilty and innocent, respectively. Let A be the
event that the suspect is left-handed. Since {G, I } is a partition of the sample space, we can
use Bayes’ formula to calculate P (G | A), the probability that the suspect has committed the
crime in view of the new evidence.
P (G | A) =
P (A | G)P (G)
(0.85)(0.65)
=
≈ 0.87.
P (A | G)P (G) + P (A | I )P (I )
(0.85)(0.65) + (0.23)(0.35)
4. Let G be the event that Susan is guilty. Let C be the event that Robert and Julie give conflicting
testimony. By Bayes’ formula,
P (G | C) =
(0.25)(0.65)
P (C | G)P (G)
=
= 0.607.
c
c
P (C | G)P (G) + P (C | G )P (G )
(0.25)(0.65) + (0.30)(0.35)
(0.02)(0.30)
= 0.1463.
(0.02)(0.30) + (0.05)(0.70)
6
11
1
4
2
3
3
6.
= .
6
11
1
1
37
+1
3
3
2
2
5.
Section 3.4
7.
Bayes’ Formula
47
(0.92)(1/5000)
= 0.084.
(0.92)(1/5000) + (1/500)(4999/5000)
8. Let A be the event that two of the three coins are dimes. Let B be the event that the coin
selected from urn I is a dime. Then
5 3
· +
P (A | B)P (B)
7 4
P (B | A) =
=
P (A | B)P (B) + P (A | B c )P (B c )
5 3 2 1
· + ·
7 4 7 4
9.
2 1 4
·
68
7 4 7
= .
83
4
5 1 3
+
·
7
7 4 7
(0.15)(0.25)
= 0.056.
(0.15)(0.25) + (0.85)(0.75)
10. Let R be the event that the upper side of the card selected is red. Let BB be the event that the
card with both sides black is selected. Define RR and RB similarly. By Bayes’ Formula,
P (RB | R) =
=
P (R | RB)P (RB)
P (R | RB)P (RB) + P (R | RR)P (RR) + P (R | BB)P (BB)
(1/2)(1/3)
1
= .
(1/2)(1/3) + 1(1/3) + 0(1/3)
3
1
1
11.
5
i=0
6
1000 − i
1000 1
100
100
= 0.21.
6
12. Let A be the event that the wallet originally contained a $2 bill. Let B be the event that the
bill removed is a $2 bill. The desired probability is given by
P B | A P (A)
P (A | B) =
P B | A P (A) + P B | Ac P (Ac )
1
1×
2
2
=
= .
3
1 1 1
1× + ×
2 2 2
13. By Bayes’ formula, the probability that the horse that comes out is from stable I equals
(20/33)(1/2)
4
= .
(20/33)(1/2) + (25/33)(1/2)
9
The probability that it is from stable II is 5/9; hence the desired probability is
20 4 25 5
205
· +
· =
= 0.69.
33 9 33 9
297
48
Chapter 3
Conditional Probability and Independence
5 3
2
2 2
·
8
4
4
14.
= 0.571.
5 3
5 3
5 3
5
1
2
3
3 1
2 2
1 3
4
0· + · + · + ·
8
8
8
8
4
4
4
4
4
4
4
15. Let I be the event that the person is ill with the disease, N be the event that the result of the
test on the person is negative, and R denote the event that the person has the rash. We are
interested in P (I | R):
P (I | R) = P (I N | R) + P (I N c | R) = 0 + P (I N c | R).
Since {I N, I N c , I c N, I c N c } is a partition of the sample space, by Bayes’ Formula,
P (I | R) = P (I N c | R)
3.5
=
P (R | I N c )P (I N c )
P (R | I N)P (I N) + P (R | I N c )P (I N c ) + P (R | I c N )P (I c N ) + P (R | I c N c )P (I c N c )
=
(0.2)(0.30 × 0.90)
= 0.61.
0(0.30 × 0.10) + (0.2)(0.30 × 0.90) + 0(0.70 × 0.75) + (0.2)(0.70 × 0.25)
INDEPENDENCE
1. No, because by independence, regardless of the number of heads that have previously occurred,
the probability of tails remains to be 1/2 on each flip.
2. A and B are mutually exclusive; therefore, they are dependent. If A occurs, then the probability
that B occurs is 0 and vice versa.
3. Neither. Since the probability that a fighter plane returns from a mission without mishap is
49/50 independent of other missions, the probability that a pilot who flew 49 consecutive
missions without mishap making another successful flight is still 49/50=0.98; neither higher
nor lower than the probability of success in any other mission.
4. P (AB) = 1/12 = (1/2)(1/6); so A and B are independent.
5. (3/8)3 (5/8)5 = 0.00503.
6. (3/4)2 = 0.5625.
Section 3.5
Independence
49
7. (a) (0.725)2 = 0.526; (b) (1 − 0.725)2 = 0.076.
8. Suppose that for an event A, P (A) = 3/4. Then the probability that A occurs in two consecutive independent experiments is 9/16. So the correct odds are 9 to 7, not 9 to 1. In later
computations, Cardano, himself, had realized that the correct answer is 9 to 7 and not 9 to 1.
9. We have that
4
P (A beats B) = P (A rolls 4) = ,
6
P (B beats A) = 1 − P (A beats B) = 1 −
2
4
= ,
6
6
4
P (B beats C) = P (C rolls 2) = ,
6
P (C beats B) = 1 − P (B beats C) = 1 −
4
2
= ,
6
6
P (C beats D) = P (C rolls 6) + P (C rolls 2 and D rolls 1) =
P (D beats C) = 1 − P (C beats D) = 1 −
4
2 4 3
+ × = ,
6 6 6
6
2
4
= ,
6
6
P (D beats A) = P (D rolls 5) + P (D rolls 1 and A rolls 0) =
3 3 2
4
+ × = .
6 6 6
6
10. For 1 ≤ i ≤ 4, let Ai be the event of obtaining 6 on the ith toss. Chevalier de Méré had
implicitly thought that Ai ’s are mutually exclusive and so
1 1 1 1
1
P A1 ∪ A2 ∪ A3 ∪ A4 = + + + = 4 × .
6 6 6 6
6
Clearly Ai ’s are not mutually exclusive. The correct answers are 1 − (5/6)4 = 0.5177 and
1 − (35/36)24 = 0.4914.
11. (1 − 0.0001)64 = 0.9936.
12. In the experiment of tossing a coin, let A be the event of obtaining heads and B be the event
of obtaining tails.
13. (a) P (A ∪ B) ≥ P (A) = 1, so P (A ∪ B) = 1. Now
1 = P (A ∪ B) = P (A) + P (B) − P (AB) = 1 + P (B) − P (AB)
gives P (B) = P (AB).
(b) If P (A) = 0, then P (AB) = 0; so P (AB) = P (A)P (B) is valid. If P (A) = 1, by
part (a), P (AB) = P (B) = P (A)P (B).
2
14. P (AA) = P (A)P (A) implies that P (A) = P (A) . This gives P (A) = 0 or P (A) = 1.
50
Chapter 3
Conditional Probability and Independence
15. P (AB) = P (A)P (B) implies that P (A) = P (A)P (B). This gives P (A) 1 − P (B) = 0;
so P (A) = 0 or P (B) = 1.
16. 1 − (0.45)6 = 0.9917.
17. 1 − (0.3)(0.2)(0.1) = 0.994.
18. There are
(100 × 10 9 ) × (300 × 10 9 ) − 1 = 30 × 10 21 − 1
other stars in the universe. Provided that Aczel’s estimate is correct, the probability of no life
in orbit around any one given star in the known universe is
0.99999999999995
independently of other stars. Therefore, the probability of no life in orbit around any other
star is
(0.99999999999995)30,000,000,000,000,000,000,000 −1 .
Using Aczel’s words, “this number is indistinguishable from 0 at any level of decimal accuracy
reported by the computer.” Hence the probability that there is life in orbit around at least one
other star is 1 for all practical purposes. If there were only a billion galaxies each having 10
billion stars, still the probability of life would have been indistinguishable from 1.0 at any level
of accuracy reported by the computer. In fact, if we divide the stars into mutually exclusive
groups with each group containing billions of stars, then the argument above and Exercise 8
of Section 1.7 imply that the probability of life in orbit around many other stars is a number
practically indistinguishable from 1.
19. 1 − (0.94)15 − 15(0.94)14 (0.06) = 0.226.
20. A and B are independent if and only if P (AB) = P (A)P (B), or, equivalently, if and only if
m
M
m+w
=
·
.
M +W
M +W M +W
This implies that m/M = w/W. Therefore, A and B are independent if and only if the fraction
of the men who smoke is equal to the fraction of the women who smoke.
21. (a) By Theorem 1.6,
P A(B ∪ C) = P (AB ∪ AC) = P (AB) + P (AC) − P (ABC)
= P (A)P (B) + P (A)P (C) − P (A)P (B)P (C)
= P (A) P (B) + P (C) − P (B)P (C) = P (A)P (B ∪ C).
(b) P (A − B)C = P (AB c C) = P (A)P (B c )P (C) = P (AB c )P (C) = P (A − B)P (C).
22. 1 − (5/6)6 = 0.6651.
Section 3.5
n
23. (a) 1 − (n − 1)/n .
24.
Independence
51
(b) As n → ∞, this approaches 1 − (1/e) = 0.6321.
1 − (0.85)10 − 10(0.85)9 (0.15)
= 0.567.
1 − (0.85)10
25. No. In the experiment of choosing a random number from (0, 1), let A, B, and C denote the
events that the point lies in (0, 1/2), (1/4, 3/4), and (1/2, 1), respectively.
26. Denote a family with two girls and one boy by ggb, with similar representations for other
cases. The sample space is S = {ggg, bbb, ggb, gbb}. we have
P {ggg} = P {bbb} = 1/8,
P {ggb} = P {gbb} = 3/8.
Clearly, P (A) = 6/8 = 3/4, P (B) = 4/8 = 1/2, and P (AB) = 3/8. Since P (AB) =
P (A)P (B), the events A and B are independent. Using the same method, we can show that
for families with two children and for families with four children, A and B are not independent.
27. If p is the probability of its occurrence in one trial, 1 − (1 − p)4 = 0.59. This implies that
p = 0.2.
28. (a) 1 − (1 − p1 )(1 − p2 ) · · · (1 − pn ).
(1 − p1 )(1 − p2 ) · · · (1 − pn ).
(b)
29. Let Ei be the event that the switch located at i is closed. The desired probability is
P (E1 E2 E4 E6 ∪E1 E3 E5 E6 ) = P (E1 E2 E4 E6 )+P (E1 E3 E5 E6 )−P (E1 E2 E3 E4 E5 E6 ) = 2p4 −p6 .
5 2
30.
3 3
3 1 2
3
= 0.329.
31. For n = 3, the probabilities of the given events, respectively, are
3 1
2 2
and
3 1 1
1 2 2
2 1
2
2
+
1
2
3 1
+
2 2
3
=
1
,
2
2 1
3
= .
2
4
The probability of their joint occurrence is
3 1
2 2
2 1
2
=
3
1 3
= · .
8
2 4
So the given events are independent. For n = 4, similar calculations show that the given
events are not independent.
52
Chapter 3
Conditional Probability and Independence
32. (a) 1 − (1/2) .
n
n 1
(b)
k 2
n
.
(c) Let An be the event of getting n heads in the first n flips. We have
A1 ⊇ A2 ⊇ A3 ⊇ · · · ⊇ An ⊇ An+1 ⊇ · · · .
The event of getting heads in all of the flips indefinitely is ∞
n=1 An . By the continuity property
of probability function (Theorem 1.8), its probability is
∞
An = lim P (An ) = lim
P
n=1
n→∞
n→∞
1
2
n
= 0.
33. Let Ai be the event that the sixth sum obtained is i, i = 2, 3, . . . , 12. Let B be the event that
the sixth sum obtained is not a repetition. By the law of total probability,
12
P (B) =
P (B | Ai )P (Ai ).
i=2
Note that in this sum, the terms for i = 2 and i = 12 are equal. This is true also for the terms
for i = 3 and 11, for the terms for i = 4 and 10, for the terms for i = 5 and 9, and for the
terms for i = 6 and 8. So
6
P (B) = 2
i=2
=2
35
P (B | Ai )P (Ai ) + P (B | A7 )P (A7 )
34 5 2
33 5 3
32
1
+
+
+
36
36
36
36
36
36
36
31 5 5 30 5 6
+
+
= 0.5614.
36
36
36
36
5
5
4
36
34. (a) Let E be the event that Dr. May’s suitcase does not reach his destination with him. We
have
P (E) = (0.04) + (0.96)(0.05) + (0.96)(0.95)(0.05) + (0.96)(0.95)(0.95)(0.04) = 0.168,
or simply, P (E) = 1 − (0.96)(0.95)(0.96) = 0.168.
(b) Let D be the event that the suitcase is lost in Da Vinci airport in Rome. Then, by Bayes’
formula,
P (D)
(0.96)(0.05)
P (D | E) =
=
= 0.286.
P (E)
0.168
35. Let E be the event of obtaining heads on the coin before an ace from the cards. Let H , T , A,
and N denote the events of heads, tails, ace, and not ace in the first experiment, respectively.
We use two different techniques to solve this problem.
Section 3.5
Independence
53
Technique 1: By the law of total probability,
P (E) = P (E | H )P (H ) + P (E | T )P (T ) = 1 ·
1
1
+ P (E | T ) · ,
2
2
where
P (E | T ) = P (E | T A)P (A | T ) + P (E | T N )P (N | T ) = 0 ·
Thus
P (E) =
1
12
+ P (E) · .
13
13
12 1
1
+ P (E)
,
2
13 2
which gives P (E) = 13/14.
Technique 2: We have that
P (E) = P (E | H A)P (H A)+P (E | T A)P (T A)+P (E | H N )P (H N )+P (E | T N )P (T N).
Thus
1
1
1
1
1 12
1 12
×
+0× ×
+1× ×
+ P (E) × × .
2 13
2 13
2 13
2 13
This gives P (E) = 13/14.
P (E) = 1 ×
36. Let P (A) = p and P (B) = q. Let An be the event that none of A and B occurs in the first
n − 1 trials and the outcome of the nth experiment is A. The desired probability is
∞
∞
An =
P
n=1
∞
P (An ) =
n=1
(1 − p − q)n−1 p =
n=1
p
p
=
.
1 − (1 − p − q)
p+q
37. The probability of sum 5 is 1/9 and the probability of sum 7 is 1/6. Therefore, by the result of
Exercise 36, the desired probability is
1/9
= 2/5.
1/6 + 1/9
38. Let A be the event that one of them is red and the other one is blue. Let RB represent the
event that the ball drawn from urn I is red and the ball drawn form urn II is blue, with similar
representations for RR, BB, and BR. We have that
P (A) = P (A | RB)P (RB) + P (A | RR)P (RR) + P (A | BB)P (BB) + P (A | BR)P (BR)
8 6
10 4
9 5
9 5
9 5
9 1
1 5
1 1
1
1
1 1 1 1
1 1
·
+
·
+
·
+
·
=
14
14
14
14
10 6
10 6
10 6
10 6
2
2
2
2
= 0.495.
54
Chapter 3
Conditional Probability and Independence
39. For convenience, let p0 = 0; the desired probability is
1−
n
i=1
n
(1 − pi ) −
(1 − p1 )(1 − p2 ) · · · (1 − pi−1 )pi (1 − pi+1 ) · · · (1 − pn ).
i=1
40. Let p be the probability that a randomly selected person was born on one of the first 365 days;
then 365p + (p/4) = 1 implies that p = 4/1461. Let E be the event that exactly four people
of this group have the same birthday and that all the others have different birthdays. E is the
union of the following three mutually exclusive events:
F : Exactly four people of this group have the same birthday, all the others have different
birthdays, and none of the birthdays is on the 366th day.
G: Exactly four people of this group have the same birthday, all the others have different
birthdays, and exactly one has his/her birthday on the 366th day.
H : Exactly four people of this group have their birthday on the 366th day and all the others
have different birthdays.
We have that
P (E) = P (F ) + P (G) + P (H )
365 30 4 4 364
4 26
=
·
26!
1
4
26
1461
1461
30
1
4
365 29
4 4 364
+
·
25!
·
1
25
1461
1
4
1461
1461
30
1 4 365
4 26
+
·
26!
= 0.00020997237.
4
26
1461
1461
25
If we were allowed to ignore the effect of the leap year, the solution would have been as
follows.
365 30 1 4 364
1 26
·
26!
= 0.00021029.
1
1
26
365
365
41. Let Ei be the event that the switch located at i is closed. We want to calculate the probability of
E2 E4 ∪ E1 E5 ∪ E2 E3 E5 ∪ E1 E3 E4 . Using the rule to calculate the probability of the union of
several events (the inclusion-exclusion principle) we get that the answer is 2p 2 +2p3 −5p4 +p5 .
42. Let E be the event that A will answer correctly to his or her first question. Let F and G be
the corresponding events for B and C, respectively. Clearly,
P (ABC) = P (ABC | EF G)P (EF G) + P (ABC | E c F G)P (E c F G)
+ P (ABC | E c F c )P (E c F c ).
(5)
Now
P (ABC | EF G) = P (ABC),
(6)
Section 3.5
Independence
55
and
P (ABC | E c F c ) = 1.
(7)
To calculate P (ABC | E c F G), note that since A has already lost, the game continues between
B and C. Let BC be the event that B loses and C wins. Then
P (ABC | E c F G) = P (BC).
(8)
Let F2 be the event that B answers the second question correctly; then
P (BC) = P (BC | F2 )P (F2 ) + P (BC | F2C )P (F2C ).
(9)
To find P (BC | F2 ), note that this quantity is the probability that B loses to C given that B
did not lose the first play. So, by independence, this is the probability that B loses to C given
that C plays first. Now by symmetry, this quantity is the same as C losing to B if B plays first.
Thus it is equal to P (CB), and hence (9) gives
P (BC) = P (CB) · p + 1 · (1 − p);
noting that P (CB) = 1 − P (BC), this gives
P (BC) =
1
.
1+p
Therefore, by (8),
P (ABC | E c F G) =
1
.
1+p
substituting this, (8), and (7) in (5), yields
P (ABC) = P (ABC) · p 3 +
1
(1 − p)p 2 + (1 − p)2 .
1+p
Solving this for P (ABC), we obtain
P (ABC) =
1
.
(1 + p)(1 + p + p 2 )
Now we find P (BCA) and P (CAB).
P (BCA) = P (BCA | E)P (E) + P (BCA | E c )P (E c )
p
= P (ABC) · p + 0 · (1 − p) =
,
(1 + p)(1 + p + p 2 )
P (CAB) = P (CAB | E)P (E) + P (CAB | E c )P (E c )
= P (BCA) · p + 0 · (1 − p) =
p2
.
(1 + p)(1 + p + p 2 )
56
Chapter 3
Conditional Probability and Independence
43. We have that
1 1
3
1
· +0· = .
2 4
4
8
P (H1 ) = P (H1 | H )P (H ) + P (H1 | H c )P (H c ) =
Similarly, P (H2 ) = 1/8. To calculate P (H1c H2c ), the probability that none of her sons is
hemophiliac, we condition on H again.
P (H1c H2c ) = P (H1c H2c | H )P (H ) + P (H1c H2c | H c )P (H c ).
Clearly, P (H1c H2c | H c ) = 1. To find P (H1c H2c | H ), we use the fact that H1 and H2 are
conditionally independent given H .
P (H1c H2c | H ) = P (H1c | H )P (H2c | H ) =
Thus
P (H1c H2c ) =
1
1 1
· = .
2 2
4
3
13
1 1
· +1· = .
4 4
4
16
44. The only quantity not calculated in the hint is P (Ui | Rm ). By Bayes’ Formula,
i
m
1
P (Rm | Ui )P (Ui )
n
n+1
P (Ui | Rm ) = n
= n
k m 1
P (Rm | Uk )P (Uk )
n
n+1
k=0
k=0
3.6
i
=
n
k=0
m
n
k
.
m
n
APPLICATIONS OF PROBABILITY TO GENETICS
1. Clearly, Kim and Dan both have genotype OO. With a genotype other than AO for John, it is
impossible for Dan to have blood type O. Therefore, the probability is 1 that John’s genotype
is AO.
k
k(k + 1)
2. The answer is
+k =
.
2
2
3. The genotype of the parent with wrinkled shape is necessarily rr. The genotype of the other
parent is either Rr or RR. But, RR will never produce wrinkled offspring. So it must be Rr.
Therefore, the parents are rr and Rr.
4. Let A represent the dominant allele for free earlobes and a represent the recessive allele for
attached earlobes. Let B represent the dominant allele for freckles and b represent the recessive
allele for no freckles. Since Dan has attached earlobes and no freckles, Kim and John both
must be AaBb. This implies that Kim and John’s next child is AA with probability 1/4, Aa
Section 3.6
Applications of Probability to Genetics
57
with probability 1/2, and aa with probability 1/4. Therefore, the next child has free earlobes
with probability 3/4. Similarly, the next child is BB with probability 1/4, Bb with probability
1/2, and bb with probability 1/4. Hence he or she will have no freckles with probability 1/4.
By independence, the desired probability is (3/4)(1/4) = 3/16.
5. If the genes are not linked, 25% of the offspring are expected to be BbV v, 25% are expected
to be bbvv, 25% are expected to be Bbvv, and 25% are expected to be bbV v. The observed
data shows that the genes are linked.
6. Clearly, John’s genotype is either Dd or dd. Let E be the event that it is dd. Then E c is the
event that John’s genotype is Dd. Let F be the event that Dan is deaf. That is, his genotype
is dd. We use Bayes’ theorem to calculate the desired probability.
P (E | F ) =
P (F | E)P (E)
P (F | E)P (E) + P (F | E c )P (E c )
=
1 · (0.01)
= 0.0198.
1 · (0.01) + (1/2)(0.99)
Therefore, the probability is 0.0198 that John is also deaf.
7. A person who has cystic fibrosis carries two mutant alleles. Applying the Hardy-Weinberg
law, we have that q 2 = 0.0529, or q = 0.23. Therefore, p = 0.77. Since q 2 + 2pq =
1 − p2 = 0.4071, the percentage of the people who carry at least one mutant allele of the
disease is 40.71%.
8. Dan inherits all of his sex-linked genes from his mother. Therefore, John being normal has no
effect on whether or not Dan has hemophilia or not. Let E be the event that Kim is H h. Then
E c is the event that Kim is H H . Let F be the event that Dan has hemophilia. By the law of
total probability,
P (F ) = P (F | E)P (E) + P (F | E c )P (E c )
= (1/2) 2(0.98)(0.02) + 0 · (0.98)(0.98) = 0.0196.
9. Dan has inherited all of his sex-linked genes from his mother. Let E1 be the event that Kim is
CC, E2 be the event that she is Cc, and E3 be the event that she is cc. Let F be the event that
Dan is color-blind. By Bayes’ formula, the desired probability is
P (E3 | F ) =
=
P (F | E3 )P (E3 )
P (F | E1 )P (E1 ) + P (F | E2 )P (E2 ) + P (F | E3 )P (E3 )
1 · (0.17)(0.17)
= 0.17.
0 · (0.83)(0.83) + (1/2) 2(0.83)(0.17) + 1 · (0.17)(0.17)
10. Since Ann is hh and John is hemophiliac, Kim is either H h or hh. Let E be the event that she
is H h. Then E c is the event that she is hh. Let F be the event that Ann has hemophilia. By
58
Chapter 3
Conditional Probability and Independence
Bayes’ formula, the desired probability is
P (F | E)P (E)
P (F | E)P (E) + P (F | E c )P (E c )
(1/2) 2(0.98)(0.02)
=
= 0.98.
(1/2) 2(0.98)(0.02) + 1 · (0.02)(0.02)
P (E | F ) =
11. Clearly, both parents of Mr. J must be Cc. Since Mr. J has survived to adulthood, he is not cc.
Therefore, he is either CC or Cc. We have
P (he is CC | he is CC or Cc) =
P (he is CC)
1/4
1
=
= .
P (he is CC or Cc)
3/4
3
P (he is Cc | he is CC or Cc) =
2
.
3
Mr. J’s wife is either CC with probability 1 − p or Cc with probability p. Let E be the event
that Mr. J is Cc, F be the event that his wife is Cc, and H be the event that their next child is
cc. The desired probability is
P (H ) = P (H EF ) = P (H | EF )P (EF )
= P (H | EF )P (E)P (F ) =
1 2
p
· ·p = .
4 3
6
12. Let E1 be the event that both parents are of genotype AA, let E2 be the event that one parent
is of genotype Aa and the other of genotype AA, and let E3 be the event that both parents are
of genotype Aa. Let F be the event that the man is of genotype AA. By Bayes’ formula,
P (E1 | F ) =
P (F | E1 )P (E1 )
P (F | E1 )P (E1 ) + P (F | E2 )P (E2 ) + P (F | E3 )P (E3 )
=
p2
1 · p4
=
= p2 .
1 · p 4 + (1/2) · 4p 3 q + (1/4) · 4p 2 q 2
(p + q)2
Similarly, P (E2 | F ) = 2pq and P (E3 | F ) = q 2 . Let B be the event that the brother is AA.
We have
P (B | F ) = P (B | F E1 )P (E1 | F ) + P (B | F E2 )P (E2 | F ) + P (B | F E3 )P (E3 | F )
= P (B | E1 )P (E1 | F ) + P (B | E2 )P (E2 | F ) + P (B | E3 )P (E3 | F )
= 1 · p2 +
1
(1 + p)2
1
(2p + q)2
· 2pq + · q 2 =
=
.
2
4
4
4
Chapter 3
Review Problems
59
REVIEW PROBLEMS FOR CHAPTER 3
1.
12 13 13 12
26
·
+
·
=
= 0.347.
30 30 30 30
75
2. 1 − (0.97)6 = 0.167.
3. (0.48)(0.30) + (0.67)(0.53) + (0.89)(0.17) = 0.65.
4. (0.5)(0.05) + (0.7)(0.02) + (0.8)(0.035) = 0.067.
5. (a) (0.95)(0.97)(0.85) = 0.783; (b) 1 − (0.05)(0.03)(0.05) = 0.999775;
(c)
1 − (0.95)(0.97)(0.85) = 0.217; (d)
(0.05)(0.03)(0.15) = 0.000225.
6. 103/132 = 0.780.
(0.08)(0.20)
= 0.0796.
(0.2)(0.3) + (0.25)(0.5) + (0.08)(0.20)
26
39
= 0.929.
8. 1 −
6
6
7.
9. 1/6.
1−
10.
5
6
10
− 10
1−
5
5
9 1
6
6
10
= 0.615.
6
2 4
·
8
7 7
= 0.35.
11.
=
23
2 4 5 3
· + ·
7 7 7 7
12. Let A be the event of “head on the coin.” Let B be the event of “tail on the coin and 1 or 2 on
the die.” Then A and B are mutually exclusive, and by the result of Exercise 36 of Section 3.5,
1/2
3
the answer is
= .
(1/2) + (1/6)
4
13. The probability that the number of 1’s minus the number of 2’s will be 3 is
P (four 1’s and one 2) + P (three 1’s and no 2’s)
6 1 4 2 1 4
6 1 3 4 3
=
= 0.03.
+
4 6
1 6 6
3 6
6
60
Chapter 3
Conditional Probability and Independence
14. The probability that the first urn was selected in the first place is
20
·
45
20 1
· +
45 2
1
10
2
.
=
19
10 1
·
25 2
The desired probability is
20 10 10 9
·
+
·
≈ 0.42.
45 19 25 19
15. Let B be the event that the ball removed from the third urn is blue. Let BR be the event that
the ball drawn from the first urn is blue and the ball drawn from the second urn is red. Define
BB, RB, and RR similarly. We have that
P (B) = P (B | BB)P (BB) + P (B | RB)P (RB) + P (B | RR)P (RR) + P (B | BR)P (BR)
4 1 5
5 9 5
6 9 1
5 1 1
38
=
·
+
·
+
·
+
·
=
= 0.36.
14 10 6 14 10 6 14 10 6 14 10 6
105
16. Let E be the event that Lorna guesses correctly. Let R be the event that a red hat is placed
on Lorna’s head, and B be the event that a blue hat is placed on her head. By the law of total
probability,
P (E) = P (E | R)P (R) + P (E | B)P (B)
1
1
1
= α · + (1 − α) · =
2
2
2
This shows that Lorna’s chances are 50% to guess correctly no matter what the value of α is.
This should be intuitively clear.
17. Let F be the event that the child is found; E be the event that he is lost in the east wing, and
W be the event that he is lost in the west wing. We have
P (F ) = P (F | E)P (E) + P (F | W )P (W )
= 1 − (0.6)3 (0.75) + 1 − (0.6)2 (0.25) = 0.748.
18. The answer is that it is the same either way. Let W be the event that they win one of the nights
to themselves. Let F be the event that they win Friday night to themselves. Then
P (W ) = P (W | F )P (F ) + P (W | F c )P (F c ) = 1 ·
2
1 1 2
+ · = .
3 2 3
3
19. Let A be the event that Kevin is prepared. We have that
P (R | B c S c ) =
=
P (RB c S c | A)P (A) + P (RB c S c | Ac )P (Ac )
P (RB c S c )
=
P (B c S c )
P (B c S c | A)P (A) + P (B c S c | Ac )P (Ac )
(0.85)(0.15)2 (0.85) + (0.20)(0.80)2 (0.15)
= 0.308.
(0.15)2 (0.85) + (0.80)2 (0.15)
Chapter 3
Review Problems
61
Note that
P (R) = P (R | A)P (A) + P (R | Ac )P (Ac ) = (0.85)(0.85) + (0.20)(0.15) = 0.7525.
Since P (R | B c S c ) = P (R), the events R, B, and S are not independent. However, it must be
clear that R, B, and S are conditionally independent given that Kevin is prepared and they are
conditionally independent given that Kevin is unprepared. To explain this, suppose that we are
given that, for example, Smith and Brown both failed a student. This information will increase
the probability that the student was unprepared. Therefore, it increases the probability that
Rose will also fails the student. However, if we know that the student was unprepared, the
knowledge that Smith and Brown failed the student does not affect the probability that Rose
will also fail the student.
20. (a) Let A be the event that Adam has at least one king; B be the event that he has at least
two kings. We have
P (B | A) =
P (Adam has at least two kings)
P (AB)
=
P (A)
P (Adam has at least one king)
48
48 4
13
12
1
1− −
52
52
13
13
=
= 0.3696.
48
13
1−
52
13
(b) Let A be the event that Adam has the king of diamonds. Let B be the event that he has
the king of diamonds and at least one other king. Then
48 3
48 3
48 3
+
+
11 1
10 2
9
3
52
13
P (BA)
P (B | A) =
=
= 0.5612.
P (A)
51
12
52
13
Knowing that Adam has the king of diamonds reduces the sample space to a size considerably
smaller than the case in which we are given that he has a king. This is why the answer to
62
Chapter 3
Conditional Probability and Independence
part (b) is larger than the answer to part (a). If one is not convinced of this, he or she should
solve the problem in a simpler case. For example, a case in which there are four cards, say,
king of diamonds, king of hearts, jack of clubs, and eight of spade. If two cards are drawn,
the reduced sample space in the case Adam announces that he has a king is
{Kd Kh , Kd Jc , Kd 8s , Kh Jc , Kh 8s },
while the reduced sample space in the case Adam announces that he has the king of diamonds
is
{Kd Kh , Kd Jc , Kd 8s }.
In the first case, the probability of more kings is 1/5; in the second case the probability of
more kings is 1/3.
Chapter 4
D istribution F unctions and
Discrete R andom Variables
4.2
DISTRIBUTION FUNCTIONS
1. The set of possible values of X is {0, 1, 2, 3, 4, 5}. The probabilities associated with these
values are
x
P (X = x)
0
6/36
1
10/36
2
8/36
3
6/36
4
4/36
5
2/36
2. The set of possible values of X is {−6, −2, −1, 2, 3, 4}. The probabilities associated with
these values are
5
2
P (X = −6) = P (X = 2) = P (X = 4) = = 0.095,
15
2
5 5
1 1
P (X = −2) = P (X = −1) = P (X = 3) = = 0.238.
15
2
3. The set of possible values of X is {0, 1, 2 . . . , N}. Assuming that people have the disease
independent of each other,
P (X = i) =
(1 − p)i−1 p
1≤i≤N
(1 − p)N
i = 0.
4. Let X be the length of the side of a randomly chosen plastic die manufactured by the factory,
then
P (X 3 > 1.424) = P (X > 1.125) =
1.25 − 1.125
1
= .
1.25 − 1
2
64
Chapter 4
Distribution Functions and Discrete Random Variables
5. P (X < 1) = F (1−) = 1/2.
P (X = 1) = F (1) − F (1−) = 1/6.
P (1 ≤ X < 2) = F (2−) − F (1−) = 1/4.
P (X > 1/2) = 1 − F (1/2) = 1 − 1/2 = 1/2.
P (X = 3/2) = 0.
P (1 < X ≤ 6) = F (6) − F (1) = 1 − 2/3 = 1/3.
6. Let F be the distribution function of X. Then
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪
⎪
1/8
⎪
⎨
F (t) = 1/2
⎪
⎪
⎪
⎪
7/8
⎪
⎪
⎪
⎪
⎩
1
t <0
0≤t <1
1≤t <2
2≤t <3
t ≥ 3.
7. Note that X is neither continuous nor discrete. The answers are
(a) F (6−) = 1 implies that k(−36 + 72 − 3) = 1; so k = 1/33.
(b) F (4) − F (2) = 29/33 − 4/33 = 25/33.
(c) 1 − F (3) = 1 − (24/33) = 9/33.
(d)
9
29
−
F (4) − F (3−)
5
33 33
=
P (X ≤ 4 | X ≥ 3) =
= .
1 − F (3−)
6
9
1−
33
8. F (Q0.5 ) = 1/2 implies that 1 + e−x = 2. The only solution of this question is x = 0. So
x = 0 is the median of F . Similarly, F (Q0.25 ) = 1/4 implies that 1 + e−x = 4, the solution
of which is x = − ln 3. F (Q0.75 ) = 3/4 implies that 1 + e−x = 4/3, the solution of which is
x = ln 3. So − ln 3 and ln 3 are the first and the third quartiles of F , respectively. Therefore,
50% of the years the rate at which the price of oil per gallon changes is negative or zero, 25%
of the years the rate is − ln 3 ≈ −1.0986 or less, and 75% of the years the rate is ln 3 ≈ 1.0986
or less.
9. (a)
P (|X| ≤ t) = P (−t ≤ X ≤ t) = P (X ≤ t) − P (X < −t)
= F (t) − 1 − P (X ≥ −t) = F (t) − 1 − P (x ≤ t) = 2F (t) − 1.
(b) Using part (a), we have
P (|X| > t) = 1 − P (|X| ≤ t) = 1 − 2F (t) − 1 = 2 1 − F (t) .
Section 4.2
Distribution Functions
65
(c)
P (X = t) = 1 + P (X = t) − 1 = P (X ≤ t) + P (X > t) + P (X = t) − 1
= P (X ≤ t) + P (X ≥ t) − 1 = P (X ≤ t) + P (X ≤ −t) − 1
= F (t) + F (−t) − 1.
10. F is a distribution function because F (−∞) = 0, F (∞) = 1, F is right continuous, and
F (t) =
1 −t
e > 0 implies that F is nondecreasing.
π
11. F is a distribution function because F (−∞) = 0, F (∞) = 1, F is right continuous, and
F (t) =
1
> 0 implies that it is nondecreasing.
(1 + t)2
12. Clearly, F is right continuous. On t < 0 and on t ≥ 0, it is increasing, limt→∞ F (t) = 1,
and limt→−∞ F (t) = 0. It looks like F satisfies all of the conditions necessary to make
it a distribution function. However, F (0−) = 1/2 > F (0+) = 1/4 shows that F is not
nondecreasing. Therefore, F is not a probability distribution function.
13. Let the departure time of the last flight before the passenger arrives be 0. Then Y , the arrival
time of the passenger is a random number from (0, 45). The waiting time is X = 45 − Y . We
have that for 0 ≤ t ≤ 45,
P (X ≤ t) = P (45 − Y ≤ t) = P (Y ≥ 45 − t) =
t
45 − (45 − t)
= .
45
45
So F , the distribution function of X is
⎧
⎪
0
t <0
⎪
⎨
F (t) = t/45 0 ≤ t < 45
⎪
⎪
⎩
1
t ≥ 45.
14. Let X be the first two-digit number selected from the set {00, 01, 02, . . . , 99} which is between
4 and 18. Since for i = 4, 5, . . . , 18,
P (X = i | 4 ≤ X ≤ 18) =
1/100
1
P (X = i)
=
= ,
P (4 ≤ X ≤ 18)
15/100
15
we have that X is chosen randomly from the set {4, 5, . . . , 18}.
15. Let X be the minimum of the three numbers,
36
3
P (X < 5) = 1 − P (X ≥ 5) = 1 − = 0.277.
40
3
66
Chapter 4
Distribution Functions and Discrete Random Variables
16.
2−0
2
P (X2 −5X +6 > 0) = P (X −2)(X −3) > 0 = P (X < 2)+P (X > 3) =
+0 = .
3−0
3
17.
F (t) =
⎧
⎪
⎪
⎪0
⎪
⎨
t
⎪
⎪
1−t
⎪
⎪
⎩1
t <0
0 ≤ t < 1/2
t ≥ 1/2.
18. The distribution function of X is F (t) = 0 if t < 1; F (t) = 1 − (89/90)n if n ≤ t < n + 1,
n ≥ 1. Since
F (26−) = 1 −
89
25
= 0.244 < 0.25 < 1 −
89
90
90
26 is the first quartile. Since
89 62
89
F (63−) = 1 −
= 0.4998 < 0.5 < 1 −
90
90
63 is the median of X. Similarly,
89 124
89
F (125−) = 1 −
= 0.7498 < 0.75 < 1 −
90
90
implies that 125 is the third quartile of X.
19.
G(t) =
4.3
⎧
⎨F (t)
t <5
⎩1
t ≥ 5.
DISCRETE RANDOM VARIABLES
1. F , the distribution functions of X is given by
F (x) =
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪
⎪
1/15
⎪
⎪
⎪
⎪
⎨3/15
if x < 1
if 1 ≤ x < 2
if 2 ≤ x < 3
⎪
6/15
if 3 ≤ x < 4
⎪
⎪
⎪
⎪
⎪
⎪
10/15 if 4 ≤ x < 5
⎪
⎪
⎪
⎪
⎩
1
if x ≥ 5.
26
63
= 0.252 = F (26),
= 0.505 = F (63),
125
= 0.753 = F (125),
Section 4.3
Discrete Random Variables
67
2. p, the probability mass function of X, is given by
x
p(x)
1
11/36
2
9/36
3
7/36
4
5/36
5
3/36
6
1/36
F , the probability distribution function of X, is given by
⎧
⎪
0
if x < 1
⎪
⎪
⎪
⎪
⎪
⎪
11/36 if 1 ≤ x < 2
⎪
⎪
⎪
⎪
⎪
⎪
20/36 if 2 ≤ x < 3
⎪
⎨
F (x) = 27/36 if 3 ≤ x < 4
⎪
⎪
⎪
⎪
32/36 if 4 ≤ x < 5
⎪
⎪
⎪
⎪
⎪
⎪
35/36 if 5 ≤ x < 6
⎪
⎪
⎪
⎪
⎩1
if x ≥ 6.
3. The possible values of X are 2, 3, . . . , 12. The sample space of this experiment consists of 36
equally likely outcomes. Hence the probability of any of them is 1/36. Thus
(1, 1) = 1/36,
p(3) = P (X = 3) = P (1, 2), (2, 1) = 2/36,
p(4) = P (X = 4) = P (1, 3), (2, 2), (3, 1) = 3/36.
p(2) = P (X = 2) = P
Similarly,
i
p(i)
5
4/36
6
5/36
7
6/36
8
5/36
9
4/36
10
3/36
11
2/36
12
1/36
4. Let p be the probability mass function of X. We have
x
p(x)
−2
1/2
2
1/10
4
13/45
6
1/9
5. Let p be the probability mass function of X and q be the probability mass function of Y . We
have
9
10
i−1
1
, i = 1, 2, . . . .
10
9 (j −3)/2 1
j −1
=
,
q(j ) = P (Y = j ) = P X =
2
10
10
p(i) =
6. Mode of p = 1; mode of q = 1.
j = 3, 5, 7, . . . .
68
Chapter 4
7. (a)
(b)
Distribution Functions and Discrete Random Variables
5
k=1
k(−1)2 + k + 4k + 9k = 1 ⇒ k = 1/15.
1
1/9
= 8.
=
1
x
9
1 − (1/9)
x=1 (1/9)
x=1
1
2
=
k(1 + 2 + · · · + n) = 1 ⇒ k =
.
n(n + 1)
n(n + 1) /2
∞
(c)
(d)
(e)
kx = 1 ⇒ k = 1/15.
k
1
x
= 1 ⇒ k = ∞
k(12 + 22 + · · · + n2 ) = 1 ⇒ k =
6
.
n(n + 1)(2n + 1)
8. Let p be the probability mass function of X; then
18
28
i
12 − i
p(i) = P (X = i) =
i = 0, 1, 2, . . . , 12.
46
12
9. For x < 0, F (x) = 0. If x ≥ 0, for some nonnegative integer n, n ≤ x < n + 1, and we have
that
n
F (x) =
i=0
=
31
4 4
i
=
1
1
3
1+
+
4
4
4
1
3 1 − (1/4)n+1
·
=1−
4
1 − (1/4)
4
2
+ ··· +
1
n
4
n+1
.
Thus
F (x) =
0
if x < 0
1 − (1/4)n+1
if n ≤ x < n + 1, n = 0, 1, 2, . . . .
10. Let p be the probability mass function of X and F be its distribution function. We have
p(i) =
5
i−1 1
6
6
,
i = 1, 2, 3, . . . .
F (x) = 0 for x < 1. If x ≥ 1, for some positive integer n, n ≤ x < n + 1, and we have that
n
F (x) =
i=1
=
5
i−1 1
6
6
=
5
5
1
1+
+
6
6
6
5
1 1 − (5/6)n
·
=1−
6 1 − (5/6)
6
n
.
2
+ ··· +
5
6
n−1
Section 4.3
Hence
F (x) =
⎧
⎪
⎨0
Discrete Random Variables
69
if x < 1
⎪
⎩1 −
5
n
if n ≤ x < n + 1,
6
n = 1, 2, 3, . . . .
11. The set of possible values of X is {2, 3, 4, . . . }. For n ≥ 2, X = n if and only if either all of
the first n − 1 bits generated are 0 and the nth bit generated is 1, or all of the first n − 1 bits
generated are 1 and the nth bit generated is 0. Therefore, by independence,
P (X = n) =
1
2
n−1
·
1 1
+
2
2
n−1
·
1 1
=
2
2
n−1
n ≥ 2.
,
12. The event Z > i occurs if and only if Liz has not played with Bob since i Sundays ago, and
the earliest she will play with him is next Sunday. Now the probability is i/k that Liz will
play with Bob if last time they played was i Sundays ago; hence
i
P (Z > i) = 1 − ,
k
i = 1, 2, . . . , k − 1.
Let p be the probability mass function of Z. Then, using this fact for 1 ≤ i ≤ k, we obtain
i
1
i−1
− 1−
= .
p(i) = P (Z = i) = P (Z > i − 1) − P (Z > i) = 1 −
k
k
k
13. The possible values of X are 0, 1, 2, 3, 4, and 5. For i, 0 ≤ i ≤ 5,
5
6 Pi · 9 P5−i · 10!
i
.
P (X = i) =
15!
The numerical values of these probabilities are as follows.
i
P (X = i)
0
42/1001
1
252/1001
2
420/1001
3
240/1001
4
45/1001
14. For i = 0, 1, 2, and 3, we have
10 10 − i 6−2i
2
i
6 − 2i
.
P (X = i) =
20
6
The numerical values of these probabilities are as follows.
i
p(i)
0
112/323
1
168/323
2
42/323
3
1/323
5
2/1001
70
Chapter 4
Distribution Functions and Discrete Random Variables
15. Clearly,
P (X > n) = P
6
Ei ·
i=1
To calculate P E1 ∪ E2 ∪ · · · ∪ E6 , we use the inclusion-exclusion principle. To do so, we
must calculate the probabilities of all possible intersections of the events from E1 , . . . , E6 ,
add the probabilities that are obtained by intersecting an odd number of events, and subtract
all the
probabilities that are obtained by
Clearly, there
intersecting an even number of events.
6
6
6
are
terms of the form P (Ei ),
terms of the form P (Ei Ej ),
terms of the form
1
2
3
P (Ei Ej Ek ), and so on. Now for all i, P (Ei ) = (5/6)n ; for all i and j , P (Ei Ej ) = (4/6)n ;
for all i, j , and k, P (Ei Ej Ek ) = (3/6)n ; and so on. Thus
P (X > n) = P (E1 ∪ E2 ∪ · · · ∪ E6 )
6 5 n
6 4 n
6 3 n
6 2
=
−
+
−
1 6
2 6
3 6
4 6
5 n
4 n
3 n
2 n
1
=6
− 15
+ 20
− 15
+6
6
6
6
6
6
n
6 1
+
5 6
n
n
.
Let p be the probability mass function of X. The set of all possible values of X is {6, 7, 8, . . . },
and
p(n) = P (X = n) = P (X > n − 1) − P (X > n)
4 n−1
3 n−1
2
5 n−1
−5
+ 10
− 10
=
6
6
6
6
n−1
1
+5
6
n−1
,
n ≥ 6.
16. Put the students in some random order. Suppose that the first two students form the first team,
the third and fourth students form the second team, the fifth and sixth students form the third
team, and so on. Let F stand for “female” and M stand for “male.” Since our only concern
is gender of the students, the total number of ways we can form 13 teams, each consisting of
two students, is equal to the number of distinguishable permutations
of a sequence of 23 M’s
26!
26
and three F ’s. By Theorem 2.4, this number is
=
. The set of possible values of
23! 3!
3
the random variable X is {2, 4, . . . , 26}. To calculate the probabilities associated with these
values, note that for k = 1, 2, . . . , 13, X = 2k if and only if one of the following events
occurs:
A:
One of the first k −1 teams is a female-female team, the kth team is either a male-female
or a female-male team, and the remaining teams are all male-male teams.
B:
The first k − 1 teams are all male-male teams, and the kth team is either a male-female
team or a female-male team.
Section 4.4
Expectations of Discrete Random Variables
71
To find P (A), note that for A to occur, there are k −1 possibilities for one of the first k −1 teams
to be a female-female team, two possibilities for the kth team (male-female and female-male),
and one possibility for the remaining teams to be all male-male teams. Therefore,
2(k − 1)
P (A) = .
26
3
To find P (B), note that for B to occur, there is one possibility for the first k − 1 teams to
be all male-male, and two possibilities for the kth team: male-female and female-male. The
number of possibilities for the remaining 13−k teams is equal to the number of distinguishable
26 − 2k)!
permutations of two F ’s and (26−2k)−2 M’s, which, by Theorem 2.4, is
=
2! (26 − 2k − 2)!
26 − 2k
. Therefore,
2
26 − 2k
2
2
.
P (B) =
26
3
Hence, for 1 ≤ k ≤ 13,
26 − 2k
2(k − 1) + 2
1
1 2
1
2
=
k − k+ .
P (X = 2k) = P (A) + P (B) =
26
650
26
4
3
4.4
EXPECTATIONS OF DISCRETE RANDOM VARIABLES
1. Yes, of course there is a fallacy in Dickens’ argument. If, in England, at that time there were
exactly two train accidents each month, then Dickens would have been right. Usually, for all
n > 0 and for any two given days, the probability of n train accidents in day 1 is equal to the
probability of n accidents in day 2. Therefore, in all likelihood the risk of train accidents on
the final day in March and the risk of such accidents on the first day in April would have been
about the same. The fact that train accidents occurred at random days, two per month on the
average, imply that in some months more than two and in other months two or less accidents
were occurring.
2. Let X be the fine that the citizen pays on a random day. Then
E(X) = 25(0.60) + 0(0.40) = 15.
Therefore, it is much better to park legally.
72
Chapter 4
Distribution Functions and Discrete Random Variables
3. The expected value of the winning amount is
30
500
1
4000
+ 800
+ 1, 200, 000
= 0.86.
2, 000, 000
2, 000, 000
2, 000, 000
Considering the cost of the ticket, the expected value of the player’s gain in one game is
−1 + 0.86 = −0.14.
4. Let X be the amount that the player gains in one game, then
4 6
3 1
P (X = 4) = = 0.114,
10
4
1
= 0.005,
10
4
P (X = 9) =
and P (X = −1) = 1 − 0.114 − 0.005 = 0.881. Thus
E(X) = −1(0.881) + 4(0.114) + 9(0.005) = −0.38.
Therefore, on the average, the player loses 38 cents per game.
5. Let X be the net gain in one play of the game. The set of possible values of X is {−8, −4, 0, 6, 10}.
The probabilities associated with these values are
1
1
,
p(−8) = p(0) = =
5
10
2
2
2
1
. Hence
and p(6) = p(10) = =
5
10
2
E(X) = −8 ·
2 2
4
1 1
p(−4) = =
,
5
10
2
4
1
2
1
2
4
−4·
+0·
+6·
+ 10 ·
= .
10
10
10
10
10
5
Since E(X) > 0, the game is not fair.
6. The expected number of defective items is
3
i=0
5
15
i 5−i
i·
= 0.75.
20
3
Section 4.4
Expectations of Discrete Random Variables
73
7. For i = 4, 5, 6, 7, let Xi be the profit if i magazines are ordered. Then
E(X4 ) =
4a
,
3
E(X5 ) =
5a 12
4a
2a 6
·
+
·
=
,
3 18
3 18
3
E(X6 ) = 0 ·
E(X7 ) = −
5
6a 7
19a
6
+a·
+
·
=
,
18
18
3 18
18
2a 6
a 5
4a 4
7a 3
10a
·
+ ·
+
·
+
·
=
.
3 18 3 18
3 18
3 18
18
Since 4a/3 > 19a/18 and 4a/3 > 10a/18, either 4, or 5 magazines should be ordered to
maximize the profit in the long run.
∞
8. (a)
x=1
6
6
= 2
π 2x2
π
∞
(b)
E(X) =
x
x=1
∞
x=1
1
6 π2
= 1.
=
·
x2
π2 6
6
6
= 2
2
2
π x
π
∞
x=1
1
= ∞.
x
2
4
1
4
9
9
+
+
+
+
= 1.
27
27
27
27
27
i=−2
(b) E(X) = 2x=−2 xp(x) = 0, E(|X|) = 2x=−2 |x|p(x) = 44/27,
E(X2 ) = 2x=−2 x 2 p(x) = 80/27. Hence
9. (a)
p(x) =
E(2X2 − 5X + 7) = 2(80/27) − 5(0) + 7 = 349/27.
10
10. Let R be the radius of the randomly selected disk; then E(2π R) = 2π
i
i=1
11. p(x) the probability mass function of X is given by
x
p(x)
−3
3/8
0
1/8
3
1/4
4
1/4
Hence
1
1
1
5
3
+0· +3· +4· = ,
8
8
4
4
8
3
1
1
1
77
E(X2 ) = 9 · + 0 · + 9 · + 16 · = ,
8
8
4
4
8
E(X) = −3 ·
1
= 11π.
10
74
Chapter 4
Distribution Functions and Discrete Random Variables
3
1
1
1
23
+0· +3· +4· = ,
8
8
4
4
8
23
31
77
E(X2 − 2|X|) =
−2
= ,
8
8
8
3
1
1
1
23
E(X|X|) = −9 · + 0 · + 9 · + 16 · = .
8
8
4
4
8
E(|X|) = 3 ·
10
12. E(X) =
i·
i=1
11
1
=
and E(X2 ) =
10
2
10
i2 ·
i=1
1
77
= . So
10
2
11 77
−
= 22.
E X(11 − X) = E(11X − X2 ) = 11 ·
2
2
13. Let X be the number of different birthdays; we have
365 × 364 × 363 × 362
= 0.9836,
3654
4
365 × 364 × 363
2
P (X = 3) =
= 0.0163,
3654
4
4
365 × 364 +
365 × 364
2
3
P (X = 2) =
= 0.00007,
3654
P (X = 4) =
P (X = 1) =
365
= 0.000000021.
3654
Thus
E(X) = 4(0.9836) + 3(0.0163) + 2(0.00007) + 1(0.000, 000, 021) = 3.98.
14. Let X be the number of children they should continue to have until they have one of each sex.
For i ≥ 2, clearly, X = i if and only if either all of their first i − 1 children are boys and the ith
child is a girl, or all of their first i − 1 children are girls and the ith child is a boy. Therefore,
by independence,
1 i−1 1 1 i−1 1 1 i−1
· +
· =
, i ≥ 2.
P (X = i) =
2
2
2
2
2
So
∞
E(X) =
i
i=2
Note that for |r| < 1,
1
i−1
2
∞
i=1
ir
∞
= −1 +
i
i=1
i−1
1
2
= 1/[(1 − r)2 ].
i−1
= −1 +
1
= 3.
(1 − 1/2)2
Section 4.4
Expectations of Discrete Random Variables
75
15. Let Aj be the event that the person belongs to a family with j children. Then
c
c
1
αj .
j
P (K = k|Aj )P (Aj ) =
P (K = k) =
j =0
j =k
Therefore,
c
c
c
kP (K = k) =
E(K) =
k=1
k
αj
=
j
k=1
j =k
c
c
k=1 j =k
kαj
.
j
16. Let X be the number of cards to be turned face up until an ace appears. Let A be the event
that no ace appears among the first i − 1 cards that are turned face up. Let B be the event that
the ith card turned face up is an ace. We have
48
4
i−1
.
P (X = i) = P (AB) = P (B|A)P (A) =
·
52
52 − (i − 1)
i−1
Therefore,
49
E(X) =
i=1
48
4
i−1
= 10.6.
52
(53 − i)
i−1
i
To some, this answer might be counterintuitive.
17. Let X be the largest number selected. Clearly,
P (X = i) = P (X ≤ i) − P (X ≤ i − 1) =
i
N
n
−
i − 1
n
N
,
i = 1, 2, . . . , N.
Hence
N
E(X) =
i=1
i n+1 i(i − 1)n
1
−
= n
n
n
N
N
N
N
i n+1 − i(i − 1)n
i=1
N
=
1
Nn
N
i n+1 − (i − 1)n+1 − (i − 1)
n
N
=
i=1
For large N,
N
(i − 1) ≈
i=1
N
x n dx =
n
0
N n+1
.
n+1
n+1
−
(i − 1)n
i=1
Nn
.
76
Chapter 4
Distribution Functions and Discrete Random Variables
Therefore,
N n+1
n + 1 = nN .
n
N
n+1
N n+1 −
E(X) ≈
18. (a)
Note that
1
1
1
= −
.
n(n + 1)
n n+1
So
k
n=1
1
=
n(n + 1)
k
1
n=1
n
−
1
1
=1−
.
n+1
k+1
This implies that
∞
k
p(n) = lim
k→∞
n=1
n=1
1
1
= 1 − lim
= 1.
k→∞ k + 1
n(n + 1)
Therefore, p is a probability mass function.
∞
∞
np(n) =
(b) E(X) =
n=1
n=1
1
= ∞,
n+1
where the last equality follows since we know from calculus that the harmonic series,
1 + 1/2 + 1/3 + · · · , is divergent. Hence E(X) does not exist.
19. By the solution to Exercise 16, Section 4.3, it should be clear that for 1 ≤ k ≤ n,
2n − 2k
2(k − 1) + 2
2
.
P (X = 2k) =
2n
3
Hence
n
n
2kP (X = 2k) =
E(X) =
k=1
k=1
2n − 2k
4k(k − 1) + 4k
2
=
2n
3
=
4
2
k 3 − (4n − 2)
k 2 + (2n2 − n − 1)
k
2n
n=1
k=1
k=1
3
=
n(n + 1)
n(n + 1)(2n + 1)
n2 (n + 1)2
4
2·
− (4n − 2) ·
+ (2n2 − n − 1)
2n
4
6
2
3
n
=
(n + 1)2
.
2n − 1
n
n
Section 4.5
Variances and Moments of Discrete Random Variables
77
4.5 VARIANCES AND MOMENTS OF DISCRETE RANDOM VARIABLES
1. On average, in the long run, the two businesses have the same profit. The one that has a profit
with lower standard deviation should be chosen by Mr. Jones because he’s interested in steady
income. Therefore, he should choose the first business.
2. The one with lower standard deviation, namely, the second device.
3. E(X) =
3
x=−3
xp(x) = −1, E(X2 ) =
3
x=−3
x 2 p(x) = 4. Therefore, Var(X) = 4−1 = 3.
4. p, the probability mass function of X is given by
x
p(x)
−3
3/8
0
3/8
6
2/8
Thus
9 12
3
E(X) = − +
= ,
8
8
8
99
9
783
Var(X) =
−
=
= 12.234,
8
64
64
27 72
99
+
= ,
8
8
8
√
σX = 12.234 = 3.498.
E(X2 ) =
5. By straightforward calculations,
N
i·
E(X) =
i=1
1
1 N(N + 1)
N +1
=
·
=
,
N
N
2
2
N
E(X2 ) =
i2 ·
i=1
1 N(N + 1)(2N + 1)
(N + 1)(2N + 1)
1
=
·
=
,
N
N
6
6
(N + 1)(2N + 1) (N + 1)2
N2 − 1
−
=
,
6
4
12
!
N2 − 1
σX =
.
12
Var(X) =
6. Clearly,
5
E(X) =
i=0
5
E(X ) =
2
i=0
39
5−i
= 1.25,
i·
52
5
13
39
i
5−i
i2 ·
= 2.426.
52
5
13
i
78
Chapter 4
Distribution Functions and Discrete Random Variables
Therefore, Var(X) = 2.426 − (1.25)2 = 0.864, and hence σX =
√
0.864 = 0.9295.
7. By the Corollary of Theorem 4.2, E(X2 − 2X) = 3 implies that E(X2 ) − 2E(X) = 3.
Substituting E(X) = 1 in this relation gives E(X 2 ) = 5. Hence, by Theorem 4.3,
2
Var(X) = E(X2 ) − E(X) = 5 − 1 = 4.
By Theorem 4.5,
Var(−3X + 5) = 9Var(X) = 9 × 4 = 36.
8. Let X be Harry’s net gain. Then
⎧
⎪
−2
⎪
⎪
⎪
⎨0.25
X=
⎪
0.50
⎪
⎪
⎪
⎩0.75
with probability 1/8
with probability 3/8
with probability 3/8
with probability 1/8.
Thus
3
3
1
1
+ 0.25 · + 0.50 · + 0.75 · = 0.125
8
8
8
8
1
3
3
1
E(X2 ) = (−2)2 · + 0.252 · + 0.502 · + 0.752 · = 0.6875.
8
8
8
8
E(X) = −2 ·
These show that the expected value of Harry’s net gain is 12.5 cents. Its variance is
Var(X) = 0.6875 − 0.1252 = 0.671875.
9. Note that E(X) = E(Y ) = 0. Clearly,
0
P |X − 0| ≤ t =
1
if t < 1
0
P |Y − 0| ≤ t =
1
if t < 10
if t ≥ 10.
if t ≥ 1,
These relations, clearly, show that for all t > 0,
P |Y − 0| ≤ t ≤ P |X − 0| ≤ t .
Therefore, X is more concentrated about 0 than Y is.
10. (a) Let X be the number of trials required to open the door. Clearly,
1
P (X = x) = 1 −
n
x−1 1
n
,
x = 1, 2, 3, . . . .
Section 4.5
Variances and Moments of Discrete Random Variables
79
Thus
1
x 1−
n
x=1
∞
E(X) =
x−1 1
=
n
1
n
1
x 1−
n
x=1
∞
x−1
(10)
.
We know from calculus that ∀r, |r| < 1,
∞
1
.
(1 − r)2
xr x−1 =
x=1
(11)
Thus
1
x 1−
n
x=1
∞
x−1
=
1
= n2 .
1 2
1− 1−
n
(12)
Substituting (12) in (10), we obtain E(X) = n. To calculate Var(X), first we find E(X 2 ). We
have
∞
E(X ) =
2
x
x=1
2
1
1−
n
x−1 1
1
=
n
n
1
x2 1 −
n
x=1
∞
x−1
.
(13)
Now to calculate this sum, we multiply both sides of (11) by r and then differentiate it with
respect to r; we get
∞
x 2 r x−1 =
x=1
1+r
.
(1 − r)3
Using this relation in (13), we obtain
1
E(X 2 ) = ·
n
1
1+1−
n
= 2n2 − n.
1 3
1− 1−
n
Therefore,
Var(X) = (2n2 − n) − n2 = n(n − 1).
(b) Let Ai be the event that on the ith trial the door opens. Let X be the number of trials
required to open the door. Then
1
P (X = 1) = ,
n
80
Chapter 4
Distribution Functions and Discrete Random Variables
P (X = 2) = P (Ac1 A2 ) = P (A2 |Ac1 )P (Ac1 )
=
1
n−1
1
·
= ,
n−1
n
n
P (X = 3) = P (Ac1 Ac2 A3 ) = P (A3 |Ac2 Ac1 )P (Ac2 Ac1 )
= P (A3 |Ac2 Ac1 )P (Ac2 |Ac1 )P (Ac1 )
=
1
n−2 n−1
1
·
·
= .
n−2 n−1
n
n
Similarly, P (X = i) = 1/n for 1 ≤ i ≤ n. Therefore, X is a random number selected from
{1, 2, 3, . . . , n}. By Exercise 5, E(X) = (n + 1)/2 and Var(X) = (n2 − 1)/12.
11. For E(X3 ) to exist, we must have E |X3 | < ∞. Now
∞
xn3
n=1
whereas
6
p(xn ) = 2
π
n=1
∞
E |X | =
3
∞
|xn3 |p(xn )
n=1
√
(−1)n n n
6
= 2
2
n
π
6
= 2
π
∞
n=1
∞
n=1
√
6
n n
= 2
2
n
π
(−1)n
√ < ∞,
n
∞
n=1
1
√ = ∞.
n
12. For 0 < s < r, clearly,
|x|s ≤ max 1, |x|r ≤ 1 + |x|r ,
∀x ∈ R .
Let A be the set of possible values
of Xr and p be its probability mass function. Since the rth
absolute moment of X exists, x∈A |x| p(x) < ∞. Now
1 + |x|r p(x)
|x|s p(x) ≤
x∈A
x∈A
p(x) +
=
x∈A
|x|r p(x) = 1 +
x∈A
|x|r p(x) < ∞,
x∈A
implies that the absolute moment of order s of X also exists.
13. Var(X)=Var(Y ) implies that
2
2
E(X2 ) − E(X) = E(Y 2 ) − E(Y ) .
Since E(X) = E(Y ), this implies that E(X2 ) = E Y 2 . Let
P (X = a) = p1 ,
P (X = b) = p2 ,
P (X = c) = p3 ;
P (Y = a) = q1 ,
P (Y = b) = q2 ,
P (Y = c) = q3 .
Section 4.5
Variances and Moments of Discrete Random Variables
81
Clearly,
p1 + p2 + p3 = q1 + q2 + q3 = 1.
This implies
(p1 − q1 ) + (p2 − q2 ) + (p3 − q3 ) = 0.
(14)
The relations E(X) = E(Y ) and E(X2 ) = E(Y 2 ) imply that
ap1 + bp2 + cp3 = aq1 + bq2 + cq3
a p1 + b2 p2 + c2 p3 = a 2 q1 + b2 q2 + c2 q3 .
2
These and equation (14) give us the following system of 3 equations in the 3 unknowns p1 −q1 ,
p2 − q2 , and p3 − q3 .
⎧
⎪
⎨ (p1 − q1 ) + (p2 − q2 ) + (p3 − q3 ) = 0
a(p1 − q1 ) + b(p2 − q2 ) + c(p3 − q3 ) = 0
⎪
⎩ 2
a (p1 − q1 ) + b2 (p2 − q2 ) + c2 (p3 − q3 ) = 0.
In matrix form, this is equivalent to
⎛
⎞⎛
⎞ ⎛ ⎞
1 1 1
p1 − q1
0
⎝ a b c ⎠ ⎝p2 − q2 ⎠ = ⎝0⎠ .
a 2 b2 c2
0
p3 − q3
Now
⎛
⎞
1 1 1
det ⎝ a b c ⎠ = bc2 + ca 2 + ab2 − ba 2 − cb2 − ac2
a 2 b2 c2
= (c − a)(c − b)(b − a) = 0,
since a, b, and c are three different real numbers. This implies that the matrix
⎛
⎞
1 1 1
⎝a b c⎠
a 2 b2 c2
is invertible. Hence the solution to (15) is
p1 − q1 = p2 − q2 = p3 − q3 = 0.
Therefore, p1 = q1 , p2 = q2 , p3 = q3 implying that X and Y are identically distributed.
(15)
82
Chapter 4
Distribution Functions and Discrete Random Variables
14. Let
P (X = a1 ) = p1 ,
P (X = a2 ) = p2 ,
... ,
P (X = an ) = pn ;
P (Y = a1 ) = q1 ,
P (Y = a2 ) = q2 ,
... ,
P (Y = an ) = qn .
Clearly,
p1 + p2 + · · · + pn = q1 + q2 + · · · + qn = 1.
This implies that
(p1 − q1 ) + (p2 − q2 ) + · · · + (pn − qn ) = 0.
The relations E(Xr ) = E(Y r ), for r = 1, 2, . . . , n − 1 imply that
a1 p1 + a2 p2 + · · · + an pn = a1 q1 + a2 q2 + · · · + an qn ,
a12 p1 + a22 p2 + · · · + an2 pn = a12 q1 + a22 q2 + · · · + an2 qn ,
..
.
a1n−1 p1 + a2n−1 p2 + · · · + ann−1 pn = a1n−1 q1 + a2n−1 q2 + · · · + ann−1 qn .
These and the previous relation give us the following n equations in the n unknowns p1 − q1 ,
p2 − q2 , . . . , pn − qn .
⎧
(p1 − q1 ) + (p2 − q2 ) + · · · + (pn − qn ) = 0
⎪
⎪
⎪
⎪
⎪
⎪
⎨ a1 (p1 − q1 ) + a2 (p2 − q2 ) + · · · + an (pn − qn ) = 0
a12 (p1 − q1 ) + a22 (p2 − q2 ) + · · · + an2 (pn − qn ) = 0
⎪
⎪
⎪
⎪
......................................................
⎪
⎪
⎩ n−1
a1 (p1 − q1 ) + a2n−1 (p2 − q2 ) + · · · + ann−1 (pn − qn ) = 0
In matrix form, this is equivalent to
⎛
⎜
⎜
⎜
⎜
⎜
⎝
1
a1
a12
..
.
1
a2
a22
..
.
···
···
···
a1n−1 a2n−1 · · ·
Now
⎛
1
a1
a12
..
.
1
a2
a22
..
.
···
···
···
⎜
⎜
⎜
det ⎜
⎜
⎝
a1n−1 a2n−1 · · ·
1
an
an2
..
.
⎞⎛
⎞ ⎛ ⎞
p1 − q1
0
⎟ ⎜p2 − q2 ⎟ ⎜0⎟
⎟⎜
⎟ ⎜ ⎟
⎟ ⎜ p3 − q3 ⎟ ⎜0⎟
⎟⎜
⎟ = ⎜ ⎟.
⎟ ⎜ .. ⎟ ⎜ .. ⎟
⎠ ⎝ . ⎠ ⎝.⎠
ann−1
1
an
an2
..
.
ann−1
pn − qn
0
⎞
⎟
⎟
⎟
(aj − ai ) = 0,
⎟=
⎟
⎠ j =n,n−1,... ,2
i n) ≥ 0.50 or 1 − (0.98)n ≥ 0.50.
This gives (0.98)n ≤ 0.50 or n ≥ ln 0.50/ ln 0.98 = 34.31. Therefore, n = 35.
4. Let F be the distribution function of X, then
t
e−t/200 ,
F (t) = 1 − 1 +
200
t ≥ 0.
Using this, we obtain
P (200 ≤ X ≤ 300) = P (X ≤ 300) − P (X < 200) = F (300) − F (200−)
= F (300) − F (200) = 0.442 − 0.264 = 0.178.
5. Let X be the number of sections that will get a hard test. We want to calculate E(X). The
random variable X can only assume the values 0, 1, 2, 3, and 4; its probability mass function
is given by
8
22
i 4−i
p(i) = P (X = i) =
, i = 0, 1, 2, 3, 4,
30
4
where the numerical values of p(i)’s are as follows.
i
p(i)
0
0.2669
1
0.4496
2
0.2360
3
0.0450
4
0.0026
Thus
E(X) = 0(0.2669) + 1(0.4496) + 2(0.2360) + 3(0.0450) + 4(0.00026) = 1.067.
6. (a) 1 − F (6) = 5/36.
(b) F (9) = 76/81.
(c) F (7) − F (2) = 44/49.
7. We have that
E(X) = (15.85)(0.15) + (15.9)(0.21) + (16)(0.35) + (16.1)(0.15) + (16.2)(0.14) = 16,
Var(X) = (15.85 − 16)2 (0.15) + (15.9 − 16)2 (0.21) + (16 − 16)2 (0.35)
+ (16.1 − 16)2 (0.15) + (16.2 − 16)2 (0.14) = 0.013.
E(Y ) = (15.85)(0.14) + (15.9)(0.05) + (16)(0.64) + (16.1)(0.08) + (16.2)(0.09) = 16,
Var(Y ) = (15.85 − 16)2 (0.14) + (15.9 − 16)2 (0.05) + (16 − 16)2 (0.64)
+ (16.1 − 16)2 (0.08) + (16.2 − 16)2 (0.09) = 0.008.
Chapter 4
Review Problems
85
These show that, on the average, companies A and B fill their bottles with 16 fluid ounces of
soft drink. However, the amount of soda in bottles from company A vary more than in bottles
from company B.
8. Let F be the distribution function of X, Then
⎧
⎪
⎪0
⎪
⎪
⎪
⎪
⎪7/30
⎪
⎪
⎪
⎪
⎨13/30
F (t) =
⎪
18/30
⎪
⎪
⎪
⎪
⎪
⎪
23/30
⎪
⎪
⎪
⎪
⎩
1
∞
9. (a) To determine the value of k, note that
i=0
implies that ke = 1 or k = e
2t
−2t
t < 58
58 ≤ t < 62
62 ≤ t < 64
64 ≤ t < 76
76 ≤ t < 80
t ≥ 80.
(2t)i
k
= 1. Therefore, k
i!
. Thus p(i) = e
−2t
∞
i=0
(2t)i
= 1. This
i!
(2t)i
.
i!
(b)
3
P (X < 4) =
P (X = i) = e−2t 1 + 2t + 2t 2 + (4t 3 /3) ,
i=0
P (X > 1) = 1 − P (X = 0) − P (X = 1) = 1 − e−2t − 2te−2t .
10. Let p be the probability mass function, and F be the distribution function of X. We have
1
3
p(0) = p(3) = , p(1) = p(2) = , and
8
8
⎧
⎪
⎪0
⎪
⎪
⎪
⎪
⎪
1/8
⎪
⎪
⎨
F (t) = 4/8
⎪
⎪
⎪
⎪
⎪
7/8
⎪
⎪
⎪
⎪
⎩
1
t <0
0≤t <1
1≤t <2
2≤t <3
t ≥ 3.
11. (a) The sample space has 52! elements because when the cards are dealt face down, any
ordering of the cards is a possibility. To find
p(j
), the probability that the 4th king
4
will appear on the j th card, we claim that in
· (j − 1) P · 48! ways the 4th king
3
1
will appear on the j th card, and the remaining 3 kings earlier. To see this, note that
86
Chapter 4
(b)
(c)
Distribution Functions and Discrete Random Variables
4
we have
combinations for the king that appears on the j th card, and (j − 1) P
3
1
different permutations for the remaining 3 kings that appear earlier. The last term 48!,
is for the remaining 48 cards that can appear in any order in the remaining 48 positions.
Therefore,
4
j −1
j −1
· (j − 1) P · 48!
3
1
3
3
p(j ) =
=
= .
52!
52
52!
4! 48!
4
51
52
The probability that the player wins is p(52) =
= 1/13.
3
4
To find
52
1
j −1
E=
jp(j ) =
j
,
52
3
j =4
j =4
4
the expected length of the game, we use a technique introduced by Jenkyns and Muller
in Mathematics Magazine, 54, (1981), page 203. We have the following relation which
can be readily checked.
j
j −1
j −1
4
(j + 1)
−j
, j ≥ 5.
j
=
5
4
4
3
52
This gives
52
52
j −1
4
j
j −1
j
=
(j + 1)
−
j
5 j =5
3
4
4
j =5
j =5
4
52
4
=
53
−5
= 11, 478, 736,
5
4
4
52
where the next-to-the-last equality follows because terms cancel out in pairs. Thus
52
52
j −1
1
j −1
1
j
= 4+
j
52
52
3
3
j =4
j =5
4
4
1
= (4 + 11, 478, 736) = 42.4.
52
4
E=
As Jenkyns and Muller have noted, “This relatively high expectation value is what makes the
game interesting. However, the low probability of winning makes it frustrating!”
Chapter 5
Special Discrete
Distributions
5.1
1.
BERNOULLI AND BINOMIAL RANDOM VARIABLES
8 1
4 4
4 3 4
4
= 0.087.
1
= 32.
2
1
+ 1 = 4 (note that we should count the mother of the family as well).
2
3 5 3
= 0.054.
6
2. (a) 64 ×
(b) 6 ×
6 1
3.
3 6
6
1
4.
2 10
5 10
5.
2 30
2
9
10
4
2 20 3
30
= 0.098.
= 0.33.
6. Let X be the number of defective nails. If the manufacturer’s claim is true, we have
P (X ≥ 2) = 1 − P (X = 0) − P (X = 1)
24
24
0
24
=1−
(0.03) (0.97) −
(0.03)(0.97)23 = 0.162.
0
1
This shows that there is 16.2% chance that two or more defective nails is found. Therefore, it
is not fair to reject company’s claim.
7. Let p and q be the probability mass functions of X and Y , respectively. Then
4
p(x) =
(0.60)x (0.40)4−x ,
x
x = 0, 1, 2, 3, 4;
88
Chapter 5
Special Discrete Distributions
y−1
q(y) = P (Y = y) = P X =
2
4
= y−1 (0.60)(y−1)/2 (0.40)4−[(y−1)/2] ,
y = 1, 3, 5, 7, 9.
2
8
8.
i=0
15
(0.8)i (0.2)15−i = 0.142.
i
10 11 5 25 5
9.
= 0.108.
5
36
36
5 1 0 2 5
5 1
10. (a) 1 −
−
0 3
1 3
3
1 2 4
3
= 0.539.
5
1
(b)
2 10
2
9
10
3
= 0.073.
11. We know that p(x) is maximum at [(n + 1)p]. If (n + 1)p is an integer, p(x) is maximum at
[(n + 1)p] = np + p. But in such a case, some straightforward algebra shows that
n
n
np+p
n−np−p
(1 − p)
=
p np+p−1 (1 − p)n−np−p+1 ,
p
np + p − 1
np + p
implying that p(x) is also maximum at np + p − 1.
52
12. The probability of royal or straight flush is 40
. If Ernie plays n games, he will get, on
5
52
52
the average, n 40
royal or straight flushes. We want to have 40n
= 1; this
5
5
52
gives n =
40 = 64, 974.
5
6 1 3 2 3
13.
= 0.219.
3
3 3
14. 1 − (999/1000)100 = 0.095.
15. The maximum occurs at k = [11(0.45)] = 4. The maximum probability is
10
(0.45)4 (0.55)6 = 0.238.
4
16. Call the event of obtaining a full house success. X, the number of full houses is n independent
poker hands is a binomial random variable with parameters (n, p), where p is
the probability
52
that a random poker hand is a full house. To calculate p, note that there are
possible
5
52
4 4 13!
poker hands and
= 3744 full houses. Thus p = 3744
≈ 0.0014. Hence
3 2 11!
5
Section 5.1
Bernoulli and Binomial Random Variables
89
E(X) = np ≈ 0.0014n and Var(X) = np(1−p) ≈ 0.00144n. Note that if n is approximately
715, then E(X) = 1. Thus we should expect to find, on the average, one full house in every
715 random poker hands.
6 1 6 3 0
6 1 5 3
17. 1 −
≈ 0.995.
−
6 4
4
4
5 4
3000
3000
0
3000
18. 1 −
(0.0005) (0.9995)
−
(0.0005)(0.9995)2999 ≈ 0.442.
0
1
19. The expected value of the expenses if sent in one parcel is
45.20 × 0.07 + 5.20 × 0.93 = 8.
The expected value of the expenses if sent in two parcels is
2
2
(23.30 × 2)(0.07) + (23.30 + 3.30)
(0.07)(0.93) + (6.60)(0.93)2 = 9.4.
1
Therefore, it is preferable to send in a single parcel.
20. Let n be the minimum number of children they should plan to have. Since the probability of all
girls is (1/2)n and the probability of all boys is (1/2)n , we must have 1−(1/2)n −(1/2)n ≥ 0.95.
ln 0.05
This gives (1/2)n−1 ≤ 0.05 or n − 1 ≥
= 4.32 or n ≥ 5.32. Therefore, n = 6.
ln(0.5)
21. (a) For this to happen, exactly
one of the N stations has to attempt transmitting a message.
N
p(1 − p)N −1 = Np(1 − p)N −1 .
1
The probability of this is
(b) Let f (p) = Np(1 − p)N−1 . The value of p which maximizes the probability of a message
going through with no collision is the root of the equation f (p) = 0. Now
f (p) = N(1 − p)N −1 − Np(N − 1)(1 − p)N −2 = 0.
Noting that p = 1, this equation gives p = 1/N. This answer makes a lot of sense because at
every “suitable instance,” on average, Np = 1 station will transmit a message.
(c) By part (b), the maximum probability is
f
1
N
=N
1
N
1−
1
N
N −1
1
= 1−
N
N −1
.
As N → ∞, this probability approaches 1/e, showing that for large numbers of stations
(in reality 20 or more), the probability of a successful transmission is approximately 1/e
independently of the number of stations if p = 1/N .
90
Chapter 5
Special Discrete Distributions
22. The k students whose names have been called are not standing. Let A1 , A2 , . . . , An−k be the
students whose names have not been called. For i, 1 ≤ i ≤ n − k, call Ai a “success,” if he or
she is standing; failure, otherwise. Therefore, whether Ai is standing or sitting is a Bernoulli
trial, and hence the random variable X is the number of successes in n − k Bernoulli trials.
For X to be binomial, for i = j , the event that Ai is a success must be independent of the
event that Aj is a success. Furthermore, the probability that Ai is a success must be the same
for all i, 1 ≤ i ≤ n − k. The latter condition is satisfied since Ai is standing if and only if his
original seat was among the first k. This happens with probability p = k/n regardless of i .
However, the former condition is not valid. The relation
k−1
,
P Aj is standing | Ai is standing =
n
shows that given Ai is a success changes the probability that Aj is success. That is, Ai being a
success is not independent of Aj being a success. This shows that X is not a binomial random
variable.
23. Let X be the number of undecided voters who will vote for abortion. The desired probability
is
[
n + (b − a)
P b + (n − X) > a + X = P X <
=
2
=
1
n+(b−a)
]
2
i=0
n 1 i1
i 2
2
n−i
[ n+(b−a)
]
2
n
n
2
i=0
i
.
24. Let X be the net gain of the player per unit of stake. X is a discrete random variable with
possible values −1, 1, 2, and 3. We have
3 1 0 5 3 125
=
,
P (X = −1) =
6
216
0 6
75
3 1 5 2
=
,
P (X = 1) =
216
1 6 6
15
3 1 2 5
=
,
P (X = 2) =
6
216
2 6
1
3 1 3 5 0
=
P (X = 3) =
.
3 6
6
216
Hence
75
15
1
125
+1·
+2·
+3·
≈ −0.08.
216
216
216
216
Therefore, the player loses 0.08 per unit stake.
E(X) = −1 ·
Section 5.1
25.
n x
E(X ) =
x
p (1 − p)n−x =
x
x=1
n
2
2
Bernoulli and Binomial Random Variables
91
n x
(x − x + x)
p (1 − p)n−x
x
x=1
n
2
n
n x
n x
n−x
=
x(x − 1)
p (1 − p)
+
x
p (1 − p)n−x
x
x
x=1
x=1
n
n
=
x=2
n!
px (1 − p)n−x + E(X)
(x − 2)! (n − x)!
n
= n(n − 1)p2
x=2
n − 2 x−2
p (1 − p)n−x + np
x−2
2
n−2
= n(n − 1)p p + (1 − p)
+ np = n2 p2 − np2 + np.
26. (a) A four-engine plane is preferable to a two-engine plane if and only if
4 0
4
2 0
4
3
1−
p (1 − p) −
p(1 − p) > 1 −
p (1 − p)2 .
0
1
0
This inequality gives p > 2/3. Hence a four-engine plane is preferable if and only if p > 2/3.
If p = 2/3, it makes no difference.
(b) A five-engine plane is preferable to a three-engine plane if and only if
5 5
5 4
5 3
3 2
0
2
p (1 − p) +
p (1 − p) +
p (1 − p) >
p (1 − p) + p 3 .
5
4
3
2
Simplifying this inequality, we get 3(p − 1)2 (2p − 1) ≥ 0 which implies that a five-engine
plane is preferable if and only if 2p − 1 ≥ 0. That is, for p > 1/2, a five-engine plane is
preferable; for p < 1/2, a three-engine plane is preferable; for p = 1/2 it makes no difference.
27. Clearly, 8 bits are transmitted. A parity check will not detect an error in the 7–bit character
received erroneously if and only if the number of bits received incorrectly is even. Therefore,
the desired probability is
4
8
(1 − 0.999)2n (0.999)8−2n = 0.000028.
2n
n=1
28. The message is erroneously received but the errors are not detected by the parity-check if for
1 ≤ j ≤ 6, j of the characters are erroneously received but not detected by the parity–check,
and the remaining 6−j characters are all transmitted correctly. By the solution of the previous
exercise, the probability of this event is
6
(0.000028)j (0.999)8(6−j ) = 0.000161.
j =1
92
Chapter 5
Special Discrete Distributions
29. The probability of a straight flush is 40
52
5
≈ 0.000015391. Hence we must have
n
3
1−
(0.000015391)0 (1 − 0.000015391)n ≥ .
0
4
This gives
1
(1 − 0.000015391)n ≤ .
4
So
n≥
log(1/4)
≈ 90071.06.
log(1 − 0.000015391)
Therefore, n ≈ 90, 072.
30. Let p, q, and r be the probabilities that a randomly selected offspring is AA, Aa, and aa,
respectively. Note that both parents of the offspring are AA with probability (α/n)2 , they are
2
both Aa with probability 1 − (α/n) , and the probability is 2(α/n) 1 − (α/n) that one
parent is AA and the other is Aa. Therefore, by the law of total probability,
p =1·
q =0·
r =0·
α
n
α
n
α
n
α 2 1 α
1
α
1α 2 1α
1
· 1−
+ ·2
+
1−
=
+ ,
4
n
2
n
n
4 n
2 n
4
2
2
2
1
α
α
α
1 1 α
1
+ 1−
1−
= −
+ ·2
,
2
n
2
n
n
2 2 n
α
2
α 2
α
1
α 2
1
1−
=
1−
+ 1−
+0·2
.
4
n
n
n
4
n
2
+
The probability that at most two of the offspring are aa is
2
i=0
m i
r (1 − r)m−i .
i
The probability that exactly i of the offspring are AA and the remaining are all Aa is
m i m−i
pq .
i
31. The desired probability is the sum of three probabilities: probability of no customer served and
two new arrivals, probability of one customer served and three new arrivals, and probability
4
of
These quantities,
two
customers served
and four new
arrivals.
respectively, are (0.4) ·
4
4
4
4
(0.45)2 (0.55)2 ,
(0.6)(0.4)3 ·
(0.45)3 (0.55), and
(0.6)2 (0.4)2 · (0.45)4 . The
2
1
3
2
sum of these quantities, which is the answer, is 0.054.
Section 5.1
93
Bernoulli and Binomial Random Variables
32. (a) Let S be the event that the first trial is a success and E be the event that in n trials, the
number of successes is even. Then
P (E) = P (E|S)P (S) + P (E|S c )P (S c ).
Thus
rn = (1 − rn−1 )p + rn−1 (1 − p).
Using this relation, induction, and r0 = 1, we find that
rn =
1
1 + (1 − 2p)n .
2
(b) The left sum is the probability of 0, 2, 4, . . . , or [n/2] successes. Thus it is the probability
of an even number of successes in n Bernoulli trials and hence it is equal to rn .
33. For 0 ≤ i ≤ n, let Bi be the event that i of the balls are red. Let A be the event that in drawing
k balls from the urn, successively, and with replacement, no red balls appear. Then
P (B0 |A) =
P (A|B0 )P (B0 )
=
n
1×
n
P (A|Bi )P (Bi )
i=0
i=0
n − i
n
1
n
2
k n 1
i 2
n
=
n
i=0
1
n n−i
i
n
.
k
34. Let E be the event that Albert’s statement is the truth and F be the event that Donna tells the
truth. Since Rose agrees with Donna and Rose always tells the truth, Donna is telling the truth
as well. Therefore, the desired probability is P (E | F ) = P (EF )/P (F ). To calculate P (F ),
observe that for Rose to agree with Donna, none, two, or all four of Albert, Brenda, Charles,
and Donna should have lied. Since these four people lie independently, this will happen with
probability
1 4 4 2 2 1 2 2 4 41
+
+
= .
2 3
3
3
3
81
To calculate P (EF ), note that EF is the event that Albert tells the truth and Rose agrees with
Donna. This happens if all of them tell the truth, or Albert tells the truth but exactly two of
Brenda, Charles and Donna lie. Hence
1 4 1 3 2 2 1
13
P (EF ) =
+ ·
= .
3
3
2 3
3
81
Therefore,
P (E | F ) =
P (EF )
13/81
13
=
=
= 0.317.
P (F )
41/81
41
94
Chapter 5
5.2
Special Discrete Distributions
POISSON RANDOM VARIABLES
1. λ = (0.05)(60) = 3; the answer is 1 −
2. λ = 1.8; the answer is
3
i=0
e−3 30
= 1 − e−3 = 0.9502.
0!
e−1.8 (1.8)i
≈ 0.89.
i!
3. λ = 0.025 × 80 = 2; the answer is 1 −
e−2 20 e−2 21
−
= 1 − 3e−2 = 0.594.
0!
1!
4. λ = (500)(0.0014) = 0.7. The answer is 1 −
e−0.7 (0.7)0 e−0.7 (0.7)1
−
≈ 0.156.
0!
1!
5. We call a room “success” if it is vacant next Saturday; we call it “failure” if it is occupied.
Assuming that next Saturday is a random day, X, the number of vacant rooms on that day is
approximately Poisson with rate λ = 35. Thus the desired probability is
29
1−
i=0
e−35 (35)i
= 0.823.
i!
6. λ = (3/10)35 = 10.5. The probability of 10 misprints in a given chapter is
0.124. Therefore, the desired probability is (0.124)2 = 0.0154.
7. P (X = 1) = P (X = 3) implies that e−λ λ =
√ √
6
is
e−
5!
6
5
e−10.5 (10.5)10
=
10!
√
e−λ λ3
from which we get λ = 6. The answer
3!
= 0.063.
8. The probability that a bun contains no raisins is
4 −2n/k
e
(1 − e−n/k )2 .
2
e−n/k (n/k)0
= e−n/k . So the answer is
0!
9. Let X be the number of times the randomly selected kid has hit the target. We are given that
P (X = 0) = 0.04; this implies that
Now
e−λ 20
= 0.04 or e−λ = 0.04. So λ = − ln 0.04 = 3.22.
0!
P (X ≥ 2) = 1 − P (X = 0) − P (X = 1) = 1 − 0.04 −
= 1 − 0.04 − (0.04)(3.22) = 0.83.
Therefore, 83% of the kids have hit the target at least twice.
e−λ λ
1!
Section 5.2
Poisson Random Variables
95
10. First we calculate pi ’s from binomial probability mass function with n = 26 and p = 1/365.
Then we calculate them from Poisson probability mass function with parameter λ = np =
26/365. For different values of i, the results are as follows.
i
0
1
2
3
Binomial
0.93115
0.06651
0.00228
0.00005
Poisson
0.93125
0.06634
0.00236
0.00006.
Remark: In this example, since success is very rare, even for small n’s Poisson gives good
approximation for binomial. The following table demonstrates this fact for n = 5.
i
0
1
2
Binomial
0.9874
0.0136
0.00007
Poisson
0.9864
0.0136
0.00009.
11. Let N(t) bethe number of shooting stars observed up to time t. Let one minute be the unit of
time. Then N(t) : t ≥ 0 is a Poisson process with λ = 1/12. We have that
e−30/12 (30/12)3
= 0.21.
P N(30) = 3 =
3!
12. P N(2) = 0 = e−3(2) = e−6 = 0.00248.
13. Let N(t) be thenumber of wrong
calls up to t. If one day is taken as the time unit, it is reasonable
to assume that N(t) : t ≥ 0 is a Poisson process with λ = 1/7. By the independent increment
property and stationarity, the desired probability is
P N(1) = 0 = e−(1/7)·1 = 0.87.
14. Choose one month as the unit of time. Then
λ = 5 and the probability of no crimes during
any given month of a year is P N(1) = 0 = e−5 = 0.0067. Hence the desired probability is
12
(0.0067)2 (1 − 0.0067)10 = 0.0028.
2
15. Choose one day as the unit of time. Then λ = 3 and the probability of no accidents in one day
is
P N(1) = 0 = e−3 = 0.0498.
The number of days without any accidents in January is approximately another Poisson random
variable with approximate rate 31(0.05) = 1.55. Hence the desired probability is
e−1.55 (1.55)3
≈ 0.13.
3!
96
Chapter 5
Special Discrete Distributions
16. Choosing one hours as time unit, we have that λ = 6. Therefore, the desired probability is
P N(0.5) = 1 and N(2.5) = 10 = P N(0.5) = 1 and N(2.5) − N (0.5) = 9
= P N (0.5) = 1 P N(2.5) − N (0.5) = 9
= P N(0.5) = 1 P N (2) = 9
=
31 e−3 129 e−12
·
≈ 0.013.
1!
9!
17. The expected number of fractures per meter is λ = 1/60. Let N(t) be the number of fractures
in t meters of wire. Then
e−t/60 (t/60)n
, n = 0, 1, 2, . . . .
P N(t) = n =
n!
Ina ten minute
period, the machine turns out 70 meters of wire. The desired probability,
P N(70) > 1 is calculated as follows:
P N(70) > 1 = 1 − P N(70) = 0 − P N (70) = 1
70
= 1 − e−70/60 − e−70/60 ≈ 0.325.
60
18. Let the epoch at which the traffic light for the left–turn lane turns red be labeled t = 0. Let
N(t) be the number of cars that arrive at the junction at or prior to t trying to turn left. Since
cars arrive at the junction according to a Poisson process, clearly, N(t) : t ≥0 is a stationary
and orderly process which possesses independent increments.
Therefore,
N(t) : t ≥ 0 is
also a Poisson process. Its parameter is given by λ = E N(1) = 4(0.22) = 0.88. (For a
rigorous proof, see the solution to Exercise 9, Section 12.2.) Thus
n
e−(0.88)t (0.88)t
P N (t) = n =
,
n!
and the desired probability is
n
3
e−(0.88)3 (0.88)3
≈ 0.273.
P N(3) ≥ 4 = 1 −
n!
n=0
19. Let X be the number of earthquakes of magnitude 5.5 or higher on the Richter scale during the
next 60 years. Clearly, X is a Poisson random variable with parameter λ = 6(1.5) = 9. Let A
be the event that the earthquakes will not damage the bridge during
∞the next 60 years. Since
the events {X = i}, i = 0, 1, 2, . . . , are mutually exclusive and i=1 {X = i} is the sample
space, by the Law of Total Probability (Theorem 3.4),
∞
P (A) =
∞
P (A | X = i)P (X = i) =
i=0
∞
=
ie
(0.985)
i=0
−9
i!
9i
=e
−9
∞
i=0
(1 − 0.015)i
i=0
e−9 9i
i!
i
(0.985)(9)
i!
= e−9 e(0.985)(9) = 0.873716.
Section 5.2
Poisson Random Variables
97
20. Let N be the total number of letter carriers in America. Let n be the total number of dog bites
letter carriers sustain. Let X be the number of bites a randomly selected letter carrier, say Karl,
sustains on a given year. Call a bite “success,” if it is Karl that is bitten and failure if anyone
but Karl is bitten. Since the letter carriers are bitten randomly, it is reasonable to assume that
X is approximately a binomial random variable with parameters n and p = 1/N . Given that
n is large (it was more than 7000 in 1983 and at least 2,795 in 1997), 1/N is small, and n/N is
moderate, X can be approximated by a Poisson random variable with parameter λ = n/N. We
know that P (X = 0) = 0.94. This implies that (e−λ · λ0 )/0! = 0.94. Thus e−λ = 0.94, and
hence λ = − ln 0.94 = 0.061875. Therefore, X is a Poisson random variable with parameter
0.061875. Now
P (X > 1)
1 − P (X = 0) − P (X = 1)
=
P X>1|X≥1 =
P (X ≥ 1)
1 − P (X = 0)
=
1 − 0.94 − 0.0581625
= 0.030625,
1 − 0.94
where
e−λ · λ1
= λe−λ = (0.061875)(0.94) = 0.0581625.
1!
Therefore, approximately 3.06% of the letter carriers who sustained one bite, will be bitten
again.
P (X = 1) =
e−nM/N (nM/N )0
≥ α. This gives n ≥ −N ln(1 − α)/M. The
0!
answer is the least integer greater than or equal to −N ln(1 − α)/M.
21. We should find n so that 1 −
22. (a) For each k-combination n1 , n2 , . . . , nk of 1, 2, . . . , n, there are (n − 1)n−k distributions
with exactly k matches, where the matches occur at n1 , n2 , . . . , nk . This is because each of
the remaining n − k balls can be placed
into any of the cells except the cell that has the same
n
number as the ball. Since there are
k-combinations n1 , n2 , . . . , nk of 1, 2, . . . , n, the total
k
number of ways we can place the n balls into the ncells
so that there are exactly k matches is
n
(n − 1)n−k
n
k
n−k
(n − 1) . Hence the desired probability is
.
k
nn
(b) Let X be the number of matches. We will show that limn→∞ P (X = k) = e−1 /k!; that is,
X is Poisson with parameter 1. We have
n
(n − 1)n−k
n n−1 n
k
= lim
(n − 1)−k
lim P (X = k) = lim
n→∞
n→∞
n→∞ k
nn
n
1
1
1
n!
1 n
= lim
·
= e−1 ·
·
· 1−
k
n→∞ k! (n − k)!
n
(n − 1)
k!
98
Chapter 5
Special Discrete Distributions
Note that limn→∞
1
1−
n
n
= e−1 , and lim
n→∞
formula,
n!
= 1, since by Stirling’s
(n − k)! (n − 1)k
√
n!
2π n · nn · e−n
lim
=
lim
√
n→∞ (n − k)! (n − 1)k
n→∞
2π(n − k) · (n − k)n−k · e−(n−k) · (n − 1)k
(
(n − k)k 1
n
nn
= lim
·
·
·
n→∞
n − k (n − k)n (n − 1)k ek
1
= 1,
ek
(n − k)n
k n
nn
k
→
e
because
=
1
−
→ e−k .
where
(n − k)n
nn
n
= 1· · ek · 1 ·
23. (a) The probability of an even number of events in (t, t + α) is
∞
n=0
∞
e−λα (λα)2n
= e−λα
(2n)!
(λα)2n
= e−αλ
(2n)!
n=0
1
−αλ 1 λα
e + e−λα =
=e
2
2
1
2
∞
n=0
(λα)n 1
+
n!
2
∞
n=0
(−λα)n
n!
1
(1 + e−2λα ).
2
(b) The probability of an odd number of events in (t, t + α) is
∞
n=1
e−λα (λα)2n−1
= e−λα
(2n − 1)!
= e−λα
∞
∞
(λα)2n−1
(λα)n 1
1
= e−λα
−
(2n − 1)!
2 n=0 n!
2
n=1
1 λα 1 −λα
1
= 1 − e−2λα .
e − e
2
2
2
∞
n=0
(−λα)n
n!
24. We have that
P N1 (t) = n, N2 (t) = m
∞
=
P N1 (t) = n, N2 (t) = m | N (t) = i P N (t) = i
i=0
= P N1 (t) = n, N2 (t) = m | N (t) = n + m P N (t) = n + m
e−λt (λt)n+m
n+m n
.
=
p (1 − p)m ·
(n + m)!
n
Therefore,
∞
P N1 (t) = n =
m=0
P N1 (t) = n, N2 (t) = m
Section 5.3
Other Discrete Random Variables
99
∞
=
m=0
∞
n+m n
e−λt (λt)n+m
p (1 − p)m ·
n
(n + m)!
(n + m)! n
e−λtp e−λt (1−p) (λt)n (λt)m
p (1 − p)m
n! m!
(n + m)!
m=0
m
∞
e−λtp e−λt (1−p) (λtp)n λt (1 − p)
=
n! m!
m=0
m
∞
e−λt (1−p) λt (1 − p)
e−λtp (λtp)n
=
n!
m!
m=0
=
=
e−λtp (λtp)n
.
n!
It can easily
be argued that the
of Poisson process are also satisfied for the
other properties
process
N1 (t): t ≥ 0 . So N1 (t) : t ≥ 0 is a Poisson process with rate λp. By symmetry,
N2 (t) : t ≥ 0 is a Poisson process with rate λ(1 − p).
25. Let N(t) be the number of females entering the store between 0 and t. By Exercise 24,
N(t) : t ≥ 0 is a Poisson process with rate 1 · (2/3) = 2/3. Hence the desired probability is
15
e−15(2/3) 15(2/3)
P N(15) = 15 =
= 0.035.
15!
26. (a) Let A be the region whose points have a (positive) distance d or less from the given tree.
The desired probability is the probability of no trees in this region and is equal to
2
e−λπ d (λπ d 2 )0
= e−λπ d .
0!
2
(b) We want to find the probability that the region A has at most n − 1 trees. The desired
quantity is
n−1 −λπ d 2
e
(λπ d 2 )i
.
i!
i=0
27. p(i) = (λ/i)p(i − 1) implies that for i < λ, the function p is increasing and for i > λ it is
decreasing. Hence i = [λ] is the maximum.
5.3
OTHER DISCRETE RANDOM VARIABLES
a defective item drawn, and N denote a nondefective
item drawn. The answer
1. Let D denote
is S = N N N, DN N, N DN, N N D, N DD, DND, DDN .
100
Chapter 5
Special Discrete Distributions
2. S = ss, f ss, sf s, sff s, ff ss, f sf s, sfff s, f sff s, fff ss, ff sf s, . . . .
3. (a) 1/(1/12) = 12. (b)
11
2
12
1
≈ 0.07.
12
4. (a) (1 − pq)r−1 pq. (b) 1/pq.
7
(0.2)3 (0.8)5 ≈ 0.055.
5.
2
6. (a) (0.55)5 (0.45) ≈ 0.023. (b) (0.55)3 (0.45)(0.55)3 (0.45) ≈ 0.0056.
5 45
50
= 0.42.
7.
1
7
8
8. The probability that at least n light bulbs are required is equal to the probability that the first
n − 1 light bulbs are all defective. So the answer is p n−1 .
9. We have
n−1 x
p (1 − p)n−x
x
x−1
= .
n x
n
p (1 − p)n−x
x
P (N = n)
=
P (X = x)
10. Let X be the number of the words the student had to spell until spelling a word correctly. The
random variable X is geometric with parameter 0.70. The desired probability is given by
4
(0.30)i−1 (0.70) = 0.9919.
P (X ≤ 4) =
i=1
11. The average number of digits until the fifth 3 is 5/(1/10) = 50. So the average number of
digits before the fifth 3 is 49.
12. The probability that a random bridge hand has three aces is
4 48
3 10
p = = 0.0412.
52
13
Therefore, the average number of bridge hands until one has three aces is 1/p = 1/0.0412 =
24.27.
13. Either the (N + 1)st success must occur on the (N + M − m + 1)st trial, or the (M + 1)st
Section 5.3
Other Discrete Random Variables
failure must occur on the (N + M − m + 1)st trial. The answer is
N + M − m 1
N + M − m 1 N +M−m+1
+
2
2
M
N
N +M−m+1
101
.
14. We have that X + 10 is negative binomial with parameters (10, 0.15). Therefore, ∀i ≥ 0,
i+9
P (X = i) = P (X + 10 = i + 10) =
(0.15)10 (0.85)i .
9
15. Let X be the number of good diskettes in the sample. The desired probability is
90 10
10 90
10
0
1
9
+
≈ 0.74.
P (X ≥ 9) = P (X = 9) + P (X = 10) =
100
100
10
10
16. We have that 560(0.35) = 196 persons make contributions. So the answer is
364 196
364
14
1
15
−
= 0.987.
1−
560
560
15
15
17. The transmission of a message takes more than t minutes, if the first [t/2] + 1 times it is sent
it will be garbled, where [t/2] is the greatest integer less than or equal to t/2. The probability
of this is p[t/2]+1 .
18. The probability that the sixth coin is accepted on the nth try is
n−1
(0.10)6 (0.90)n−6 .
5
Therefore, the desired probability is
∞
n=50
49
n−1
n−1
6
n−6
(0.10) (0.90)
(0.10)6 (0.90)n−6 = 0.6346.
=1−
5
5
n=6
19. The probability that the station will successfully transmit or retransmit a message is (1−p)N −1 .
This is because for the station to successfully transmit or retransmit its message, none of the
other stations should transmit messages at the same instance. The number of transmissions
and retransmissions of a message until the success is geometric with parameter (1 − p)N −1 .
Therefore, on average, the number of transmissions and retransmissions is 1/(1 − p)N −1 .
102
Chapter 5
Special Discrete Distributions
20. If the fifth tail occurs after the 14th trial, ten or more heads have occurred. Therefore, the fifth
tail occurs before the tenth head if and only if the fifth tail occurs before or on the 14th flip.
Calling tails success, X, the number of flips required to get the fifth tail is negative binomial
with parameters 5 and 1/2. The desired probability is given by
14
14
P (X = n) =
n=5
n=5
n − 1 1
4
2
5 1 n−5
2
≈ 0.91.
21. The probability of a straight is
10 45 − 40
= 0.003924647.
52
5
Therefore, the expected number of poker hands required until the first straight is
1/0.003924647 = 254.80.
22. (a) Since
P (X = n − 1)
1
=
> 1,
P (X = n)
1−p
P (X = n) is a decreasing function of n; hence its maximum is at n = 1.
(b) The probability that X is even is given by
∞
∞
P (X = 2k) =
k=1
p(1 − p)2k−1 =
k=1
p(1 − p)
1−p
.
=
2
1 − (1 − p)
2−p
(c) We want to show the following:
Let X be a discrete random variable with the set of possible values 1, 2, 3 . . . .
If for all positive integers n and m,
P (X > n + m | X > m) = P (X > n),
(17)
then X is a geometric random variable. That is, there exists a number p,
0 < p < 1, such that
P (X = n) = p(1 − p)n−1 .
(18)
To prove this, note that (17) implies that for all positive integers n and m,
P (X > n + m)
= P (X > n).
P (X > m)
Therefore,
P (X > n + m) = P (X > n)P (X > m).
(19)
Section 5.3
Other Discrete Random Variables
103
Let p = P (X = 1); using induction, we prove that (18) is valid for all positive integers n. To
show (18) for n = 2, note that (19) implies that
P (X > 2) = P (X > 1)P (X > 1).
Since P (X > 1) = 1 − P (X = 1) = 1 − p, this relation gives
1 − P (X = 1) − P (X = 2) = (1 − p)2 ,
or
1 − p − P (X = 2) = (1 − p)2 ,
which yields
P (X = 2) = p(1 − p),
so (18) is also true for n = 2. Now assume that (18) is valid for all positive integers i, i ≤ n;
that is, assume that
P (X = i) = p(1 − p)i−1 ,
i ≤ n.
(20)
We will show that (18) is true for n + 1. The induction hypothesis [relation (20)] implies that
n
n
P (X = i) =
P (X ≤ n) =
i=1
p(1 − p)i−1 = p
i=1
1 − (1 − p)n
= 1 − (1 − p)n .
1 − (1 − p)
So P (X > n) = (1 − p)n and, similarly, P (X > n − 1) = (1 − p)n−1 . Now (19) yields
P (X > n + 1) = P (X > n)P (X > 1),
which implies that
1 − P (X ≤ n) − P (X = n + 1) = (1 − p)n (1 − p).
Substituting P (X ≤ n) = 1 − (1 − p)n in this relation, we obtain
P (X = n + 1) = p(1 − p)n ,
which establishes (18) for n + 1. Therefore, we have what we wanted to show.
23. Consider a coin for which the probability of tails is 1 − p and the probability of heads is p.
In successive and independent flips of the coin, let X1 be the number of flips until the first
head, X2 be the total number of flips until the second head, X3 be the total number of flips
until the third head, and so on. Then the length of the first character of the message and X1
are identically distributed. The total number of the bits forming the first two characters of
the message and X2 are identically distributed. The total number of the bits forming the first
three characters of the message and X3 are identically distributed, and so on. Therefore, the
total number of the bits forming the message has the same distribution as Xk . This is negative
binomial with parameters k and p.
104
Chapter 5
Special Discrete Distributions
24. Let X be the number of cartons to be opened before finding one without rotten eggs. X is not a
geometric random variable because the number of cartons is limited, and one carton not having
rotten eggs is not independent of another carton not having rotten
eggs.
However,
it should be
1000 1200
obvious that a geometric random variable with parameter p =
= 0.1109 is
12
12
a good approximation for X. Therefore, we should expect approximately 1/p = 1/0.1109 =
9.015 cartons to be opened before finding one without rotten eggs.
25. Either the Nth success should occur on the (2N − M)th trial or the Nth failure should occur
on the (2N − M)th trial. By symmetry, the answer is
2N − M − 1 1 N 1 N −M
2N − M − 1 1
2·
=
N −1
2
2
2
N −1
2N −M−1
.
26. The desired quantity is 2 times the probability of exactly N successes in (2N − 1) trials and
failures on the (2N)th and (2N + 1)st trials:
2N − 1 1 N
1
1 (2N −1)−N
2
· 1−
1−
N
2
2
2
2
2N − 1 1
=
N
2
2N
.
27. Let X be the number of rolls until Adam gets a six. Let Y be the number of rolls of the die
until Andrew rolls an odd number. Since the events (X = i), 1 ≤ i < ∞, form a partition of
the sample space, by Theorem 3.4,
P Y >X =
∞
P Y >X|X=i P X=i =
i=1
∞
=
i=1
∞
P Y >i P X=i
i=1
1
2
i
·
5
i−1
6
6 1
1
= ·
6
5 6
∞
i=1
5
12
i
1
= ·
5
5
12
1
= ,
7
5
1−
12
where P (Y > i) = (1/2)i since for Y to be greater than i, Andrew must obtain an even number
on each of the the first i rolls.
28. The probability of 4 tagged trout among the second 50 trout caught is
pn =
50
4
n − 50
46
.
n
50
It is logical to find the value of n for which pn is maximum. (In statistics this value is called
the maximum likelihood estimate for the number of trout in the lake.) To do this, note that
(n − 50)2
pn
=
.
pn−1
n(n − 96)
Section 5.3
Other Discrete Random Variables
105
Now pn ≥ pn−1 if and only if (n − 50)2 ≥ n(n − 96), or n ≤ 625. Therefore, n = 625 makes
pn maximum, and hence there are approximately 625 trout in the lake.
29. (a) Intuitively, it should be clear that the answer is D/N . To prove this, let Ej be the event of
obtaining exactly j defective items among the first (k − 1) draws. Let Ak be the event that the
kth item drawn is defective. We have
D
N −D
k−1
k−1
D−j
j
k−1−j
P (Ak ) =
.
P (Ak | Ej )P (Ej ) =
·
N
N
−
k
+
1
j =0
j =0
k−1
Now
D
D−1
(D − j )
=D
j
j
and
N
N −1
(N − k + 1)
=N
.
k−1
k−1
Therefore,
k−1
P (Ak ) =
j =0
D−1
N −D
D
D
j
k−1−j
=
N −1
N
N
k−1
where
k−1
j =0
D−1
N −D
D
j
k−1−j
= ,
N −1
N
k−1
k−1
j =0
D−1
N −D
j
k−1−j
=1
N −1
k−1
D−1
N −D
j
k−1−j
since
is the probability mass function of a hypergeometric random
N −1
k−1
variable with parameters N − 1, D − 1, and k − 1.
(b) Intuitively, it should be clear that the answer is (D − 1)/(N − 1). To prove this, let Ak be
as before and let Fj be the event of exactly j defective items among the first (k − 2) draws.
Let B be the event that the (k − 1)st and the kth items drawn are defective. We have
k−2
P (B) =
P (B | Fj )P (Fj )
j =0
106
Chapter 5
Special Discrete Distributions
k−2
=
j =0
k−2
=
j =0
D
N −D
(D − j )(D − j − 1)
j
k−2−j
·
N
(N − k + 2)(N − k + 1)
k−2
D−2
N −D
D(D − 1)
j
k−2−j
N −2
N (N − 1)
k−2
=
=
D(D − 1)
N(N − 1)
k−2
j =0
D(D − 1)
.
N(N − 1)
D−2
N −D
j
k−2−j
N −2
k−2
Using this, we have that the desired probability is
D(D − 1)
N (N − 1)
P (B)
P (Ak Ak−1 )
D−1
=
=
.
P (Ak | Ak−1 ) =
=
P (Ak−1 )
P (Ak−1 )
N −1
D
N
REVIEW PROBLEMS FOR CHAPTER 5
20
20
(0.25)i (0.75)20−i = 0.0009.
i
1.
i=12
2. N(t), the number of customers arriving at the post office at or prior to t is a Poisson process
with λ = 1/3. Thus
6
P N(30) ≤ 6 =
P N (30) = i =
i=0
3. 4 ·
2
4.
i=0
8
= 1.067.
30
12
(0.30)i (0.70)12−i = 0.253.
i
6
i=0
i
e−(1/3)30 (1/3)30
= 0.130141.
i!
Chapter 5
Review Problems
107
5
5.
(0.18)2 (0.82)3 = 0.179.
2
1999
6.
i=2
i−1 1
2 − 1 1000
2
999
1000
i−2
= 0.59386.
12
7.
i=7
160
200
i
12 − i
= 0.244.
360
12
8. Call a train that arrives between 10:15 A.M. and 10:28 A.M. a success. Then p, the probability
of success is
p=
28 − 15
13
=
.
60
60
Therefore, the expected value and the variance of the number of trains that arrive in the given
period are 10(13/60) = 2.167 and 10(13/60)(47/60) = 1.697, respectively.
9. The number of checks returned during the next two days is Poisson with λ = 6. The desired
probability is
4
P (X ≤ 4) =
i=0
e−6 6i
= 0.285.
i!
10. Suppose that 5% of the items are defective. Under this hypothesis, there are 500(0.05) = 25
defective items. The probability of two defective items among 30 items selected at random is
25 475
2
28
= 0.268.
500
30
Therefore, under the above hypothesis, having two defective items among 30 items selected
at random is quite probable. The shipment should not be rejected.
11. N is a geometric
random variable with p = 1/2. So E(N) = 1/p = 2, and Var(N ) =
(1 − p)/p2 = 1 − (1/2) /(1/4) = 2.
12.
5
5 1
6
6
= 0.067.
13. The number of times a message is transmitted or retransmitted is geometric with parameter
1 − p. Therefore, the expected value of the number of transmissions and retransmissions of a
108
Chapter 5
Special Discrete Distributions
message is 1/(1 − p). Hence the expected number of retransmissions of a message is
p
1
−1=
.
1−p
1−p
14. Call a customer a “success,” if he or she will make a purchase using a credit card. Let E
be the event that a customer entering the store will make a purchase. Let F be the event that
the customer will use a credit card. To find p, the probability of success, we use the law of
multiplication:
p = P (EF ) = P (E)P F | E = (0.30)(0.85) = 0.255.
The random variable X is binomial with parameters 6 and 0.255. Hence
6−i
i
6
,
0.255 1 − 0.255
P X=i =
i
i = 0, 1, . . . , 6.
Clearly, E(X) = np = 6(0.255) = 1.53 and
Var(X) = np(1 − p) = 6(0.255)(1 − 0.255) = 1.13985.
5
15.
18
i
i=3
10
5−i
28
5
= 0.772.
16. By the formula for the expected value of a hypergeometric random variable, the desired quantity
is (5 × 6)/16 = 1.875.
17. We want to find the probability that at most 4 of the seeds do not germinate:
4
i=0
2
18. 1 −
i=0
40
(0.06)i (0.94)40−i = 0.91.
i
20
(0.06)i (0.94)20−i = 0.115.
i
Let X be the number of requests for reservations at the end of the second day. It is reasonable
to assume that X is Poisson with parameter 3 × 3 × 2 = 18. Hence the desired probability is
23
23
P (X = i) = 1 −
P (X ≥ 24) = 1 −
i=0
i=0
e−18 (18)i
= 1 − 0.89889 = 0.10111.
i!
Chapter 5
Review Problems
109
19. Suppose that the company’s claim is correct. Then the probability of 12 or less drivers using
seat belts regularly is
12
20
(0.70)i (0.30)20−i ≈ 0.228.
i
i=0
Therefore, under the assumption that the company’s claim is true, it is quite likely that out of
20 randomly selected drivers, 12 use seat belts. This is not a reasonable evidence to conclude
that the insurance company’s claim is false.
2999
999
1
20. (a) (0.999) (0.001) = 0.000368. (b)
(0.001)3 (0.999)2997 = 0.000224.
2
21. Let X be the number of children having the disease. We have that the desired probability is
5
(0.23)3 (0.77)2
P (X = 3)
3
P (X = 3 | X ≥ 1) =
=
= 0.0989.
P (X ≥ 1)
1 − (0.77)5
22. (a)
w
w+b
n−1
b
.
w+b
w
w+b
(b)
n−1
.
23. Let n be the desired number of seeds to be planted. Let X be the number of seeds which
will germinate. We have that X is binomial with parameters n and 0.75. We want to find the
smallest n for which
P (X ≥ 5) ≥ 0.90.
or, equivalently,
P (X < 5) ≤ 0.10.
That is, we want to find the smallest n for which
4
i=0
n
(0.75)i (.25)n−i ≤ 0.10.
i
By trial and error, as the following table shows, we find that the smallest n satisfying
P (X < 5) ≤ 0.10 is 9. So at least nine seeds is to be planted.
n
5
6
7
8
9
4
n
i=0 i
(0.75)i (.25)n−i
0.7627
0.4661
0.2436
0.1139
0.0489
110
Chapter 5
Special Discrete Distributions
24. Intuitively, it must be clear that the answer is k/n. To prove this, let B be the event that the ith
baby born is blonde. Let A be the event that k of the n babies are blondes. We have
n−1
n − 1 k−1
n−k
p·
p (1 − p)
P (AB)
k
k−1
k−1
= = .
P (B | A) =
=
n k
n
P (A)
n
p (1 − p)n−k
k
k
25. The size of a seed is a tiny fraction of the size of the area. Let us divide the area up into many
small cells each about the size of a seed. Assume that, when the seeds are distributed, each
of them will land in a single cell. Accordingly, the number of seeds distributed will equal
the number of nonempty cells. Suppose that each cell has an equal chance of having a seed
independent of other cells (this is only approximately true). Since λ is the average number of
seeds per unit area, the expected number of seeds in the area, A, is λA. Let us call a cell in
A a “success” if it is occupied by a seed. Let n be the total number of cells in A and p be
the probability that a cell will contain a seed. Then X, the number of cells in A with seeds
is a binomial random variable with parameters n and p. Using the formula for the expected
number of successes in a binomial distribution (= np), we see that np = λA and p = λA/n.
As n goes to infinity, p approaches zero while np remains finite. Hence the number of seeds
that fall on the area A is a Poisson random variable with parameter λA and
P (X = i) =
e−λA (λA)i
.
i!
26. Let D/N → p, then by the Remark 5.2, for all n,
D N −D
n x
x
n−x
≈
p (1 − p)n−x .
N
x
n
Now since n → ∞ and nD/N → λ, n is large and np is appreciable, thus
e−λ λx
n x
.
p (1 − p)n−x ≈
x!
x
Chapter 6
C ontinuous R andom
Variables
6.1
PROBABILITY DENSITY FUNCTIONS
∞
ce−3x dx = 1 ⇒ c = 3.
1/2
3e−3x dx = 1 − e−3/2 ≈ 0.78.
(b) P (0 < X ≤ 1/2) =
1. (a)
0
0
⎧
⎨ 32
2. (a) f (x) = x 3
⎩
0
x≥4
x < 4.
(b) P (X ≤ 5) = 1 − (16/25) = 9/25,
P (X ≥ 6) = 16/36
= 4/9,
P (5 ≤ X ≤ 7) = 1 − (16/49) − 1 − (16/25) = 0.313,
P (1 ≤ X < 3.5) = 0 − 0 = 0.
2
2
x 3 3x 2
3. (a)
c(x − 1)(2 − x) dx = 1 ⇒ c −
+
− 2x = 1 ⇒ c = 6.
1
3
2
1
x
(b) F (x) =
6(x − 1)(2 − x) dx, 1 ≤ x < 2. Thus
1
⎧
⎪
⎨0
F (x) = −2x 3 + 9x 2 − 12x + 5
⎪
⎩
1
x<1
1≤x<2
x ≥ 2.
(c) P (X < 5/4) = F (5/4) = 5/32,
P (3/2 ≤ X ≤ 2) = F (2) − F (3/2) = 1 − (1/2) = 1/2.
4. (a) P (X < 1.5) =
1
1.5
2
2
dx = .
2
x
3
112
Chapter 6
Continuous Random Variables
1.25
2
dx
x2
1
(b) P (1 < X < 1.25 | X < 1.5) =
1.5
2
dx
x2
1
1
5. (a)
−1
=
2/5
3
= .
2/3
5
1
c
dx = 1 ⇒ c · arcsin x
= 1 ⇒ c = 1/π.
√
−1
1 − x2
(b) For −1 < x < 1,
F (x) =
x
−1
1
1
1
dx = arcsin x + .
)
π
2
π 1 − x2
⎧
⎪
0
⎪
⎪
⎪
⎨
1
1
F (x) =
arcsin x +
⎪
π
2
⎪
⎪
⎪
⎩
1
Thus
x < −1
−1 ≤ x < 1
x ≥ 1.
6. Since h(x) ≥ 0 and
α
∞
1
f (x)
dx =
1 − F (α)
1 − F (α)
∞
f (x) dx =
α
1
1 − F (α) = 1,
1 − F (α)
h is a probability density function.
7. (a) Let F be the distribution function of X. Then X is symmetric about α if and only if for all
x, 1 − F (α + x) = F (α − x), or upon differentiation f (α + x) = f (α − x).
(b) f (α + x) = f (α − x) if and only if (α − x − 3)2 = (α + x − 3)2 . This is true for all x, if
and only if α − x − 3 = −(α + x − 3) which gives α = 3. A similar argument shows that g
is symmetric about α = 1.
∞
8. (a) Since f is a probability density function,
f (x) dx = 1. But
−∞
∞
−∞
f (x) dx =
0
−1
k(2x − 3x 2 ) dx = k
0
−1
(2x − 3x 2 ) dx = k x 2 − x 3
0
−1
= −2k.
So −2k = 1 or k = −1/2.
(b) The loss is at most $500 if and only if X ≥ −1/2. Therefore, the desired probability is
1
P X≥−
=
2
0
1
1
3
− (2x − 3x 2 ) dx = − x 2 − x 3
= .
−1/2
2
2
16
−1/2
0
Section 6.2
9. P (X > 15) =
∞
15
Density Function of a Function of a Random Variable
113
1 −x/15
1
e
dx = . Thus the answer is
15
e
8
i=4
8 1 i
1
1−
e
i e
8−i
= 0.3327.
10. Since αf + βg ≥ 0 and
∞
−∞
αf (x) + βg(x) dx = α
∞
−∞
f (x) dx + β
∞
−∞
g(x) dx = α + β = 1,
αf + βg is also a probability density function.
11. Since F (−∞) = 0 and F (∞) = 1, We have that
α + β(−π/2) = 0
α + β(π/2) = 1.
Solving this system of two equations in two unknown, we obtain α = 1/2 and β = 1/π. Thus
f (x) = F (x) =
6.2
2
, −∞ < x < ∞.
π(4 + x 2 )
DENSITY FUNCTION OF A FUNCTION OF A RANDOM VARIABLE
1. Let G be the distribution function of Y ; for −8 < y < 8,
G(y) = P (Y ≤ y) = P (X ≤ y) = P (X ≤
3
Therefore,
This gives
⎧
⎪
0
⎪
⎪
⎪
⎨
1
1√
3
G(y) =
y+
⎪
2
4
⎪
⎪
⎪
⎩1
√
3
y)=
√
3y
−2
1
1√
1
dx = 3 y + .
4
4
2
y < −8
−8 ≤ y < 8
y ≥ 8.
⎧
⎪
⎨ 1 y −2/3
g(y) = G (y) = 12
⎪
⎩0
−8 < y < 8
otherwise.
114
Chapter 6
Continuous Random Variables
Let H be the distribution function of Z; for 0 ≤ z < 16,
√
√
H (z) = P (X ≤ z) = P (− 4 z ≤ x ≤ 4 z ) =
√
4z
4
Thus
This gives
⎧
⎪
0
⎪
⎪
⎪
⎨ √
14
H (z) =
z
⎪
2
⎪
⎪
⎪
⎩1
√
4
− z
1√
1
dx = 4 z.
4
2
z<0
0 ≤ z < 16
z ≥ 16.
⎧
⎪
⎨ 1 z−3/4
h(z) = H (z) = 8
⎪
⎩0
0 < z < 16
otherwise.
2. Let G be the probability distribution function of Y and g be its probability density function.
For t > 0,
G(t) = P eX ≤ t = P (X ≤ ln t) = F (ln t).
For t ≤ 0, G(t) = 0. Therefore,
⎧
⎪
⎨ 1 f (ln t) t > 0
g(t) = G (t) = t
⎪
⎩0
t ≤ 0.
√
3. The set of possible values of X is A = (0, ∞). Let h : (0, ∞) → R be defined by h(x) = x x.
The set of possible values of h is B = (0, ∞). The inverse of h is g, where g(y) = y 2/3 . Thus
√
g (y) = 2/(3 3 y ) and hence
2 −y 2/3
e
, y ∈ (0, ∞).
fY (y) = √
33y
To find the probability density function of e−X , let h : (0, ∞) → R be defined by h(x) = e−x ;
h is an invertible function with the set of possible values B = (0, 1). The inverse of h is
g(z) = − ln z. So g (z) = −1/z. Therefore,
fZ (z) = e
0, otherwise.
1
1
− = z · = 1, z ∈ (0, 1);
z
z
−(− ln z)
Section 6.2
Density Function of a Function of a Random Variable
115
4. The set of possible values of X is A = (0, ∞). Let h : (0, ∞) → R be defined by h(x) =
log2 x. The set of possible values of h is B = (−∞, ∞). h is invertible and its inverse is
g(y) = 2y , where g (y) = (ln 2)2y . Thus
y
−3
2 (ln 2)2y = (3 ln 2)2y e−3(2y ) , y ∈ (−∞, ∞).
fY (y) = 3e
5. Let G and g be the probability distribution and the probability density functions of Y , respectively. Then
√
√
3
G(y) = P (Y ≤ y) = P X2 ≤ y = P (X ≤ y y )
√
y y
√
λe−λx dx = 1 − e−λy y , y ∈ [0, ∞).
=
0
So
g(y) = G (y) =
3λ √ −λy √y
ye
, y ≥ 0;
2
0, otherwise.
6. Let G and g be the probability distribution and density functions of X2 , respectively. For
t ≥ 0,
√
√
√
√
G(t) = P (X2 ≤ t) = P (− t < X < t ) = F ( t ) − F (− t ).
Thus
√
√
√
√
1
1
1
g(t) = G (t) = √ f ( t ) + √ f (− t ) = √ f ( t ) + f (− t ) , t ≥ 0.
2 t
2 t
2 t
For t < 0, g(t) = 0.
7. Let G and g be the distribution and density functions of Z, respectively. For −π/2 < z < π/2,
G(z) = P (arctan X ≤ z) = P (X ≤ tan z) =
tan z
1
1
1
=
= z+ .
arctan x
−∞
π
π
2
Thus
⎧
⎪
⎨1
g(z) = π
⎪
⎩0
−
tan z
−∞
1
dx
π(1 + x 2 )
π
π
1)P (X > 1)
1
= P (X ≤ t | X ≤ 1)P (X ≤ 1) + P X ≥ X > 1 P (X > 1).
t
116
Chapter 6
Continuous Random Variables
For t ≥ 1, this gives
G(t) = 1 ·
1
e−x dx + 1 ·
0
∞
e−x dx = 1.
1
For 0 < t < 1, this gives
1
=
G(t) = P (X ≤ t) + P X ≥
t
Hence
t
e
−x
dx +
0
∞
e−x dx = 1 − e−t + e−1/t .
1/t
⎧
⎪
⎨0
G(t) = 1 − e−t + e−1/t
⎪
⎩
1
Therefore,
6.3
t ≤0
0 α.
Therefore,
E(Y ) =
α
−∞
k −kα/A A ky/A A2 ky/A α
A
k −k(α−y)/A
dy = e
− 2e
=α− .
ye
ye
A
A
k
k
k
−∞
7. Let H be the distribution function of C; then
t − 32
t − 32
P (F ≤ t) = P C ≤
=H
.
1.8
1.8
Hence the probability density function of F is
1 t − 32
5 t − 32
d
P (F ≤ t) =
h
= h
.
dt
1.8
1.8
9
1.8
The expected value of F is given by
E(F ) = 1.8E(C) + 32 = 1.8
∞
−∞
xh(x) dx + 32.
2
2 ln x
dx. To calculate this integral, let U = ln x, dV = 1/x 2 , and use
x2
1
integration by parts:
8. E(ln X) =
1
2
2
2 ln x 2
2
2 ln x
dx = −
−
− 2 dx = 1 − ln 2 = 0.3069.
2
x
x 1
x
1
9. The expected value of the length of the other side is given by
E
)
81 − X 2 =
4
)
81 − x 2 ·
2
x
dx.
6
Letting u = 81 − x 2 , we get du = −2x dx and
)
1
E 81 − X 2 =
12
77
65
√
u du ≈ 8.4.
118
Chapter 6
10. E(X) =
Continuous Random Variables
∞
−∞
1 −|x|
xe
dx = 0, because the integrand is an odd function. Now
2
∞
∞
1 2 −|x|
x e
dx =
x 2 e−x dx
E(X2 ) =
2
−∞
0
since the integrand is an even function; applying integration by parts to the last integral twice,
we obtain E(X2 ) = 2. Hence Var(X) = 2 − 02 = 2.
11. Note that
E |X|
α
=
∞
−∞
|x|α
2
dx =
2
π(1 + x )
π
∞
0
xα
dx
(1 + x 2 )
since the integrand is an even function. Now for 0 < α < 1,
1
∞
∞
xα
xα
xα
dx
=
dx
+
dx.
2
1 + x2
1 + x2
0
0 1+x
1
Clearly, the first integral in the right side is convergent. To show that the second one is also
convergent, note that.
xα
1
xα
≤
= 2−α .
2
2
1+x
x
x
Therefore,
∞
∞
∞
xα
1
1
1
dx
≤
dx
=
=
< ∞.
2
2−α
1−α
1
1+x
x
(α − 1)x
1−α
1
1
For α ≥ 1,
∞
So
0
0
∞
xα
≥
1 + x2
∞
1
xα
dx ≥
1 + x2
∞
1
∞
1
x
2
dx
=
)
= ∞.
ln(1
+
x
1
1 + x2
2
α
x
dx diverges.
1 + x2
12. By Remark 6.4,
E(X) =
0
∞
P (X > t) dt =
∞
β
α
+ .
λ µ
∞
c1
cn
dx = 1. For n > 1,
dx =
2
n+1
x
cn x
(αe−λt + βe−µt ) dt =
0
13. (a) c1 is an arbitrary positive number because ∀c1 ,
∞
c1
.
1 implies that cn = n−1/(n−1)⎧
∞
⎨∞
if n = 1
cn
(b) E(Xn ) =
dx
=
n
⎩n(n−2)/(n−1) /(n − 1) if n > 1.
cn x
et
cn
cn 1
1
t
, where
(c) P (Zn ≤ t) = P (ln Xn ≤ t) = P (Xn ≤ e ) =
dx
=
−
n+1
n cnn ent
cn x
Section 6.3
Expectations and Variances
119
cn = n−1/(n−1) . Let gn be the probability density function of Zn . Then gn (t) = cn e−nt ,
t ≥ ln cn .
(d)
E(Xnm+1 )
=
∞
cn
cn x m+1
dx. This integral exists if and only if m − n < −1.
x n+1
14. Using integration by parts twice, we obtain
E(X
n+1
1 π n+1
x
sin x dx = π
+ (n + 2)
x
cos x dx
π 0
0
1 π n
x sin x dx
= π n+1 + (n + 2) − (n + 1)
π 0
n+1
+ (n + 2) − (n + 1)E(X n−1 ) .
=π
1
)=
π
π
n+2
n+1
Hence
E(Xn+1 ) + (n + 1)(n + 2)E(X n−1 ) = π n+1 .
15. Since X is symmetric about α, for all x ∈ (−∞, ∞), f (α +x) = f (α −x). Letting y = x +α,
we have
E(X) =
−∞
=
∞
∞
−∞
yf (y) dy =
∞
−∞
(x + α)f (x + α) dx
xf (x + α) dx + α
∞
−∞
f (x + α) dx.
Now since f is symmetric about α, xf (x + α) is an odd function,
−xf (−x + α) = − xf (x + α) .
∞
∞
∞
xf (x + α) = 0. Since
f (x + α) dx =
f (y) dy = 1, we have
Therefore,
−∞
−∞
E(X) = 0 + α · 1 = α.
−∞
To show that the median of X is α, we will show that P (X ≤ α) = P (X ≥ α). This also
shows that the value of these two probabilities is 1/2. Letting u = α − x, we have
α
∞
P (X ≤ α) =
f (x) dx =
f (α − u) du.
−∞
0
Letting u = x − α, we have that
P (X ≥ α) =
α
∞
f (x) dx =
0
∞
f (u + α) du.
120
Chapter 6
Continuous Random Variables
Since for all u,
f (α − u) = f (α + u),
we have that
P (X ≤ α) = P (X ≥ α) = 1/2.
16. By Theorem 6.3,
E |X − y| =
∞
−∞
=y
|x − y|f (x)dx =
y
−∞
f (x) dx −
y
−∞
(x − y)f (x) dx
y
y
−∞
∞
(y − x)f (x) dx +
∞
xf (x) dx +
xf (x) dx − y
y
∞
f (x) dx.
y
Hence
y
∞
dE |X − y|
=
f (x) dx + yf (y) − yf (y) − yf (y) −
f (x) dx + yf (y)
dy
−∞
y
y
∞
=
f (x) dx −
f (x) dx.
−∞
y
dE |X − y|
= 0, we obtain that y is the solution of the following equation:
Setting
dy
∞
y
f (x) dx =
f (x) dx.
−∞
y
By the definition of the median
random variable, the solution to this equation
of a continuous
is y = median(X). Hence E |X − y| is minimum for y = median(X).
17. (a)
∞
I (t) dt =
0
X
I (t) dt +
0
∞
X
I (t) dt =
dt +
0
X
∞
0 dt = X.
X
∞
I (t) dt is a random variable.)
∞
∞
(b) E(X) = E
I (t) dt =
E I (t) dt =
(Note that
0
0
0
(c) By part (b),
E(X ) =
r
∞
0
P (X > t) dt =
0
P (X > t) dt =
r
0
=
∞
∞
∞
0
√
P X > r t dt
0
∞
1−F
√
r
t dt = r
0
where the last equality follows by the substitution y =
∞
y r−1 1 − F (y) dy,
√
r
t.
1 − F (t) dt.
Section 6.3
Expectations and Variances
121
18. On the interval [n, n + 1),
P |X| ≥ n + 1 ≤ P |X| > t ≤ P |X| ≥ n .
Therefore,
n+1
P |X| ≥ n + 1 dt ≤
n
n+1
P |X| > t dt ≤
n
or
P |X| ≥ n + 1 ≤
n+1
P |X| ≥ n dt,
n
n+1
P |X| > t dt ≤ P |X| ≥ n .
n
So
∞
∞
P |X| ≥ n + 1 ≤
n=0
n=0
and hence
∞
n+1
P |X| > t dt ≤
n
∞
P |X| > n ,
n=0
P |X| ≥ n ≤ E |X| ≤ 1 +
n=1
∞
P |X| ≥ n .
n=1
19. By Exercise 12,
E(X) =
α
β
+ .
λ µ
Using Exercise 16, we obtain
∞
E(X ) = 2
2
x(αe−λx + βe−µx ) dx =
0
2α 2β
+ 2.
λ2
µ
Hence
Var(X) =
2α
λ2
α
2β
β
−
+
2
µ
λ µ
+
2
=
2α − α 2 2β − β 2 2αβ
+
−
.
λ2
µ2
λµ
20. X ≥st Y implies that for all t,
P (X > t) ≥ P (Y > t).
Taking integrals of both sides of (21) yields,
∞
P (X > t) dt ≥
0
(21)
∞
P (Y > t) dt.
0
Relation (21) also implies that
1 − P (X ≤ t) ≥ 1 − P (Y ≤ t),
or, equivalently,
P (X ≤ t) ≤ P (Y ≤ t)·
(22)
122
Chapter 6
Continuous Random Variables
Since this is true for all t, we have
P (X ≤ −t) ≤ P (Y ≤ −t)·
Taking integrals of both sides of this inequality, we have
∞
∞
P (X ≤ −t) ≤
P (Y ≤ −t) dt,
0
0
or, equivalently,
∞
−
0
P (Y ≤ −t) dt.
(23)
0
Adding (22) and (23) yields
∞
∞
P (X > t) dt −
P (X ≤ −t) dt ≥
0
∞
P (X ≤ −t) ≥ −
0
∞
P (Y > t) dt −
0
∞
P (Y ≤ −t) dt·
0
By Theorem 6.2, this gives E(X) ≥ E(Y ). To show that the converse of this theorem is false,
let X and Y be discrete random variables both with set of possible values {1, 2, 3}. Let the
probability mass functions of X and Y be defined by
pX (1) = 0.3
pX (2) = 0.4
pX (3) = 0.3
pY (1) = 0.5
pY (2) = 0.1
pY (3) = 0.4
We have that E(X) = 2 > E(Y ) = 1.9. However, since
P (X > 2) = 0.3 < P (Y > 2) = 0.4,
we see that X is not stochastically larger than Y .
21. First, we show that limx→−∞ xP X ≤ x = 0. To do so, since x → −∞, we concentrate on
negative values of x. Letting u = −t, we have
∞
∞
x
f (t) dt = x
f (−u) du = −
−xf (−u) du.
xP X ≤ x = x
−∞
−x
So it suffices to show that as x → −∞,
∞
−x
∞
−∞
−x
−x
−xf (−u) du → 0. Now
−xf (−u) du ≤
Therefore, it remains to prove that
*∞
*∞
−x
∞
uf (−u) du.
−x
uf (−u) du → 0 as x → −∞. But this is true because
|u|f (−u) du =
∞
−∞
|x|f (x) dx < ∞.
Chapter 6
Review Problems
Next, we will show that limx→∞ xP X > x = 0. To do so, note that
lim xP X > x = lim x
x→∞
since
*∞
−∞
x→∞
∞
f (t) dt ≤ lim
x→∞
x
∞
tf (t) dt = 0
x
|tf (t)| dt < ∞.
REVIEW PROBLEMS FOR CHAPTER 6
1. Let F be the distribution function of Y . Clearly, F (y) = 0 if y ≤ 1. For y > 1,
1
1−
1
1
y
F (y) = P
≤y =P X≥
=
=1− .
X
y
1−0
y
1
So
f (y) = F (y) =
⎧
⎨1/y 2
y>1
⎩0
elsewhere.
∞
2
2
2 ∞
2. E(X) =
x · 3 dx =
dx = − = 2,
x
x2
x 1
1
1
∞
∞
2
x 2 · 3 dx = 2 ln x = ∞. So Var(X) does not exist.
E(X2 ) =
1
x
1
∞
1
6 1 1
(6x 2 − 6x 3 ) dx = 2x 3 − x 4 = ,
0
4
2
0
1
1
6
6
3
E(X2 ) =
,
(6x 3 − 6x 4 ) dx = x 4 − x 5 =
0
4
5
10
0
1 2
3
1
1
Var(X) =
=
−
, σX = √ .
10
2
20
2 5
3. E(X) =
Therefore,
1
1
2
2
P
− √ 0,
0
1
e−|x|
1
x
dx =
e dx +
e−x dx
P (−2 < X < 1) =
2 −2
−2 2
0
1
1
=1−
−
= 0.748.
2e 2e2
1
∞
c
dx = c ln(1 + x)
= ∞.
0
1+x
0
So, for no value of c, f (x) is a probability density function.
∞
6. The set of possible values of X is A = [1, 2]. Let h : [1, 2] → R be defined by h(x) = ex . The
set of possible values of eX is B = [e, e2 ]; the inverse of h is g(y) = ln y, where g (y) = 1/y.
Therefore,
4(ln y)3
4(ln y)3
fY (y) =
|g (y)| =
, y ∈ [e, e2 ].
15
15y
Applying the same procedure to Z and W , we obtain
√
4( z )3 1 2z
fZ (z) =
√ = , z ∈ [1, 4].
15
2 z
15
√
2(1 + w )3
fW (w) =
w ∈ [0, 1].
√
15 w
7. The set of possible values of X is A = (0, 1). Let h : (0, 1) → R be defined by h(x) = x 4 .
The set of possible values of X4 is B = (0, 1). The inverse of h(x) = x 4 is g(y) =
1
1
. We have that
g (y) = y −3/4 = √ √
4
4 y4y
1
√
√ 1
√
√
2
4
=
30
y(1
−
y
)
fY (y) = 30( 4 y )2 (1 − 4 y )2 )
√
√
4 y4y
4 4 y3
√
15(1 − 4 y )2
=
, y ∈ (0, 1).
√
24y
8. We have that
⎧
1
⎪
⎨ √
f (x) = F (x) = π 1 − x 2
⎪
⎩
0
Therefore,
E(X) =
since the integrand is an odd function.
1
−1
−1 < x < 1
otherwise.
x
dx = 0
√
π 1 − x2
√
4 y. So
Chapter 6
9. Clearly
n
i=1
∞
−∞
i=1
125
αi fi ≥ 0. Since
n
Review Problems
n
n
αi fi (x) dx =
i=1
αi
i=1
∞
n
−∞
fi (x) dx =
αi = 1,
i=1
αi fi is a probability density function.
10. Let U = x and dV = f (x)dx. Then dU = dx and V = F (x). Since F (α) = 1,
α
α
α
xf (x) dx = xF (x) −
F (x) dx
0
0
0 α
α
F (x) dx = α −
F (x) dx
= αF (α) −
0 α
α 0
α
1 − F (x) dx.
dx −
F (x) dx =
=
E(X) =
0
0
0
11. Let X be the lifetime of a random light bulb. The probability that it lasts over 1000 hours is
P (X > 1000) =
∞
1000
5 × 105
1 ∞
1
5
−
dx
=
5
×
10
= .
x3
2x 2 1000 4
Thus the probability that out of six such light bulbs two last over 1000 hours is
6 1 2 3 4
≈ 0.3
2 4
4
12. Since Y ≥ 0, P (Y ≤ t) = 0 for t < 0. For t ≥ 0,
P (Y ≤ t) = P |X| ≤ t = P (−t ≤ X ≤ t) = P (X ≤ t) − P (X < −t)
= P (X ≤ t) − P (X ≤ −t) = F (t) − F (−t).
Hence G, the probability distribution function of |X| is given by
G(t) =
F (t) − F (−t)
0
if t ≥ 0
if t < 0;
g, the probability density function of |X| is obtained by differentiating G:
g(t) = G (t) =
f (t) + f (−t)
0
if t ≥ 0
if t < 0.
Chapter 7
Special C ontinuous
Distributions
7.1
UNIFORM RANDOM VARIABLES
1. (23 − 20)/(27 − 20) = 3/7.
2. 15(1/4) = 3.75.
3. Let 2:00 P.M. be the origin, then a and b satisfy the following system of two equations in two
⎧
a+b
⎪
⎪
=0
⎨
2
2
⎪
⎪
⎩ (b − a) = 12.
12
Solving this system, we obtain a = −6 and b = 6. So the bus arrives at a random time
between 1:54 P.M. and 2:06 P.M.
unknown.
4. P (b2 − 4 ≥ 0) = P (b > 2 or b < −2) = 2/6 = 1/3.
5. The probability density function of R, the radius of the sphere is
⎧
1
1
⎪
⎨
=
2
f (r) = 4 − 2
⎪
⎩0
Thus
4
E(V ) =
P
4
3
2
4
3
πr3
2 0, H (x) = P (Z ≤ x) = 0, x < 0;
√
√
H (x) = P (Z ≤ x) = P (X ≤ n x ) = n x, 0 < x < 1;
H (x) = 1, if x ≥ 1. Therefore,
⎧
⎪
⎨ 1 x n1 −1
h(x) = H (x) = n
⎪
⎩0
0 t .
Then F (x0 ) = t and
Therefore,
We have shown that
F (x) ≤ t
if and only if
x ≤ x0 .
P F (X) ≤ t = P X ≤ x0 = F (x0 ) = t.
⎧
⎪0
⎨
P F (X) ≤ t = t
⎪
⎩
1
if t ≤ 0
if 0 ≤ t ≤ 1
if t ≥ 1,
meaning that F (X) is uniform over (0, 1).
15. We are given that Y is a uniform random variable. First we show that Y is uniform over the
interval (0, 1). To do this, it suffices to show that P (Y ≤ 1) =
1∞and P (Y < 0) = 0. These
are obvious implications of the fact that g is nonnegative and
g(x) dx = 1:
−∞
P (Y ≤ 1) = P
P (Y < 0) = P
X
−∞
g(t) dt ≤ 1 = 1.
X
−∞
g(t) dt < 0 = 0,
The following relation shows that the probability density function of X is g.
⎛ u
⎞
g(t) dt − 0
u
⎟
d
d ⎜
d
⎜ −∞
⎟ = g(u),
g(t) dt =
P (X ≤ u) =
P Y ≤
⎝
⎠
du
du
du
1−0
−∞
where the last equality follows from the fundamental theorem of calculus.
130
Chapter 7
Special Continuous Distributions
16. Let F be the distribution function of X, then F (t) = P (X ≤ t) is 0 for t < −1 and is 1 for
t ≥ 4. Let −1 ≤ t < 4; we have that
t +1
F (t) = P (X ≤ t) = P (5ω − 1 ≤ t) = P ω ≤
5
(t+1)/5
t +1
t +1
= P ω ∈ 0,
=
.
dx =
5
5
0
Therefore,
⎧
⎪
0
⎪
⎪
⎪
⎨
t +1
F (t) =
⎪
5
⎪
⎪
⎪
⎩1
t < −1
−1 ≤ t < 4
t ≥ 4.
This is the distribution function of a uniform random variable over (−1, 4).
√
17. We√have that X = n if and only if Y = 0.y1 ny3 y4 y5 · · · , or, equivalently,
if and only
if,
10 Y = y1 .ny3 y4 y5 · · · . Therefore, X = n if and only if for some k ∈ 0, 1, 2, . . . , 9 ,
k+
√
n+1
n
≤ 10 Y < k +
.
10
10
This is equivalent to
n
1
k+
100
10
2
≤Y <
n+1
1
k+
100
10
2
.
Therefore, the desired probability is
9
P
k=0
1
n
k+
100
10
9
=
k=0
9
=
k=0
2
≤Y <
1
n+1
k+
100
10
2
n+1
1
k+
100
10
−
n
1
k+
100
10
20k + 2n + 1
= 0.091 + 0.002n.
10, 000
We see that this quantity increases as n does.
2
2
Section 7.2
7.2
Normal Random Variables
131
NORMAL RANDOM VARIABLES
)
1. Since np = (0.90)(50) = 45 and np(1 − p) = 2.12,
44.5 − 45
= P (Z ≥ −0.24)
P (X ≥ 44.5) = P Z ≥
2.12
= 1 − (−0.24) = (0.24) = 0.5948.
)
2. np = 1095/365 = 3 and np(1 − p) =
(
364
3
= 1.73. Therefore,
365
5.5 − 3
=1−
P (X ≥ 5.5) = P Z ≥
1.73
(1.45) = 0.0735.
3. We have that
P (|Z|) ≤ x) = P (−x ≤ Z ≤ x) = (x) − (−x)
= (x) − 1 − (x) = 2 (x) − 1 =
4. Let
1
g(x) = P (x < Z < x + α) = √
2π
x+α
e−y
2 /2
(x).
dy.
x
The number x that maximizes P (x < Z < x + α) is the root of g (x) = 0; that is, it is the
solution of
1 −(x+α)2 /2
2
e
g (x) = √
− e−x /2 = 0,
2π
which is x = −α/2.
∞
X
1
2
are,
respectively,
5. E(X cos X), E(sin X), and E
(x cos x)e−x /2 dx,
√
1 + X2
2π −∞
∞
∞
1
x
1
2
2
(sin x)e−x /2 dx, and √
e−x /2 dx. Since these are integrals of
√
2π −∞
2π −∞ 1 + x 2
odd functions from −∞ to ∞, all three of them are 0.
6. (a) P (X > 35.5) = P
X − 35.5
4.8
>
35.5 − 35.5
=1−
4.8
(0) = 0.5.
(b) The desired probability is given by
P (30 < X < 40) = P
=
30 − 35.5
4.8
(0.94) +
40 − 35.5
= (0.94) − (−1.15)
4.8
(1.15) − 1 = 0.8264 + 0.8749 − 1 = 0.701.
95) = P Z >
kσ ) = P (X − µ > kσ ) + P (X − µ < −kσ ) = P (Z > k) + P (Z < −k)
= 1 − (k) + 1 − (k) = 2 1 − (k) .
This shows that P (|X − µ| > kσ ) does not depend on µ or σ .
15. Let X be the lifetime of a randomly selected light bulb.
900 − 1000
=1−
P (X ≥ 900) = P Z ≥
100
Hence the company’s claim is false.
(−1) =
(1) = 0.8413.
16. Let X be the lifetime of the light bulb manufactured by the first company. Let Y be the
lifetime of the light bulb manufacturedby the second company.
Assuming that X and Y are
independent, the desired probability, P max(X, Y ) ≥ 980 , is calculated as follows.
P max(X, Y ) ≥ 980 = 1 − P max(X, Y ) < 980 = 1 − P (X < 980, Y < 980)
= 1 − P (X < 980) P (Y < 980)
980 − 900
980 − 1000
P Z<
=1−P Z <
100
150
= 1 − P (Z < −0.2)P (Z < 0.53) = 1 − 1 − (0.2) (0.53)
= 1 − (1 − 0.5793)(0.7019) = 0.7047.
17. Let r be the rate of return of this stock; r is a normal random variable with mean µ = 0.12
and standard deviation σ = 0.06. Let n be the number of shares Mrs. Lovotti should purchase.
We want to find the smallest n for which the probability of profit in one year is at least $1000.
Let X be the current price of the total shares of the stock that Mrs. Lovotti buys this year,
and Y be the total price of the shares next year. We want to find the smallest n for which
P (Y − X ≥ 1000). We have
Y − X
1000
1000
P (Y − X ≥ 1000) = P
≥
=P r≥
X
X
X
⎛
⎞
1000
− 0.12 ⎟
⎜
1000
35n
⎟ ≥ 0.90.
=P r≥
=P⎜
Z
≥
⎝
⎠
35n
0.06
134
Chapter 7
Special Continuous Distributions
Therefore, we want to find the smallest n for which
⎛
⎞
1000
− 0.12 ⎟
⎜
35n
⎟ ≤ 0.10.
P⎜
Z
≤
⎝
⎠
0.06
By Table 1 of the Appendix, this is satisfied if
1000
− 0.12
35n
≤ −1.29.
0.06
This gives n ≥ 670.69. Therefore, Mrs. Lovotti should buy 671 shares of the stock.
18. We have that
1
(x − 1)2
1
(x − 1)2
f (x) = √ √ exp −
=
.
√ exp −
1/2
2(1/4)
1/2 π
(1/2) 2π
This shows that f is the probability density function of a normal random variable with mean
1 and standard deviation 1/2 (variance 1/4).
19. Let F be the distribution function of |X − µ|. F (t) = 0 if t < 0; for t ≥ 0,
F (t) = P |X − µ| ≤ t = P (−t ≤ X − µ ≤ t)
t
X−µ
t
= P (µ − t ≤ X ≤ µ + t) = P − ≤
≤
σ
σ
σ
t
t
t
t
t
−
−
=
− 1−
=2
=
σ
σ
σ
σ
σ
Therefore,
⎧ t
⎨2
σ
F (t) =
⎩
0
−1
otherwise.
This gives
F (t) =
Hence
2
σ
t
E |X − µ| =
∞
t
0
substituting u = t/σ , we obtain
t ≥ 0.
σ
t ≥0
2
σ
t
σ
dt.
∞
2σ
2
(u) du = √
ue−u /2 du
2π 0
0
(
∞
2
2σ
2σ
−u2 /2
=σ
=√
.
=√
−e
0
π
2π
2π
E(|X − µ|) = 2σ
∞
u
− 1.
Section 7.2
Normal Random Variables
135
20. The general form of the probability density function of a normal random variable is
f (x) =
1
1
µ
µ2
(x − µ)2
1 2
=
.
x
+
x
−
exp
−
√ exp −
√
2σ 2
2σ 2
σ2
2σ 2
σ 2π
σ 2π
Comparing this with the given probability density function, we see that
⎧√
1
⎪
k= √
⎪
⎪
⎪
σ 2π
⎪
⎪
⎪
⎪
⎪
1
⎪
⎪k 2 =
⎨
2σ 2
µ
⎪
⎪
2k = − 2
⎪
⎪
σ
⎪
⎪
⎪
2
⎪
⎪µ
⎪
⎪
⎩ 2 = 1.
2σ
√
Solving the first two equations for k and σ , we obtain k = π and σ = 1/(π 2). These and
the third equation give µ = −1/π which satisfy the fourth equation. So k = π and f is the
1 1
probability density function of N − , 2 .
π 2π
21. Let X be the viscosity of the given brand. We must find the smallest x for which P (X ≤ x) ≥
x − 37
0.90 or P Z ≤
≥ 0.90. This gives
10
x = 49.9.
x − 37
10
≥ 0.90 or (x − 37)/10 = 1.29; so
22. Let X be the length of the residence of a family selected at random from this town. Since
96 − 80
= 0.298,
P (X ≥ 96) = P Z ≥
30
using binomial distribution, the desired probability is
2
12
(0.298)i (1 − 0.298)12−i = 0.742.
i
1−
i=0
23. We have
∞
1
2
eαx · √ e−x /2 dx
2π
−∞
∞
1
1 2
1 2
2
= eα /2
√ e− 2 α +αx− 2 x dx
2π
−∞
∞
1
1
2
2
2
= eα /2
√ e− 2 (x−α) dx = eα /2 ,
2π
−∞
E(eαZ ) =
136
Chapter 7
Special Continuous Distributions
∞
1
1
1
1
2
2
√ e− 2 (x−α) dx = 1, since √ e− 2 (x−α) is the probability density function
2π
2π
−∞
of a normal random variable with mean α and variance 1.
where
24. For t ≥ 0,
√
√
P (Y ≤ t) = P − t ≤ X ≤ t = P −
√
√
t
t
≤Z≤
σ
σ
=2
√t
σ
− 1.
Let f be the probability density function of Y . Then
d
1
f (t) = P (Y ≤ t) = 2 √
dt
2σ t
√t
σ
⎧
1
t
⎪
⎨ √
exp −
2σ 2
f (t) = σ 2π t
⎪
⎩
0
So
,
t ≥ 0.
t ≥0
t ≤ 0.
25. For t ≥ 0,
ln t − µ
P (Y ≤ t) = P eX ≤ t = P (X ≤ ln t) = P Z ≤
=
σ
ln t − µ
.
σ
Let f be the probability density function of Y . We have
f (t) =
So
f (t) =
d
1
P (Y ≤ t) =
dt
σt
⎧
⎪
⎨
⎪
⎩
ln t − µ
,
σ
t ≥ 0.
1
(ln t − µ)2
√ exp −
2σ 2
σ t 2π
t ≥0
0
otherwise.
26. Let f be the probability density function of Y . Since for t ≥ 0,
P (Y ≤ t) = P
we have that
)
|X| ≤ t = P |X| ≤ t 2 = P − t 2 ≤ X ≤ t 2 = 2 (t 2 ) − 1,
⎧
1
4
⎪
⎨4t √ e−t /2
d
2π
f (t) = P (Y ≤ t) =
⎪
dt
⎩
0
t ≥0
otherwise.
27. Suppose that X is the number of books sold in a month. The random variable X is binomial
with parameters
n = (800)(30) = 24, 000 and p = 1/5001. Moreover, E(X) = np = 4.8
√
and σX = np(1 − p) = 2.19. Let k be the number of copies of the bestseller to be ordered
Section 7.2
Normal Random Variables
137
every month. We want to have P (X < k) > 0.98 or P (X ≤ k − 1) > 0.98. Using
De Moivre-Laplace theorem and making correction for continuity, this inequality is valid if
X − 4.8
k − 1 + 0.5 − 4.8
P
<
> 0.98.
2.19
2.19
From Table 1 of the appendix, we have (k − 1 + 0.5 − 4.8)/2.19 = 2.06, or k = 9.81.
Therefore, the store should order 10 copies a month.
28. Let X be the number of light bulbs of type I. We want to calculate P (18 ≤ X ≤ 22).
Since the number of light bulbs is large and half of the light bulbs are type I, we can assume
that X is approximately
binomial with parameters 40 and 1/2. Note that np = 20 and
√
√
np(1 − p) = 10. Using De Moivre-Laplace theorem and making correction for continuity,
we have
17.5 − 20
X − 20
22.5 − 20
P (17.5 ≤ X ≤ 22.5) = P
≤ √
≤ √
√
10
10
10
= (0.79) − (−0.79) = 2 (0.79) − 1 = 0.5704.
Remark: Using binomial distribution, the solution to this problem is
22
40 1 i 1 40−i
= 0.5704.
i
2
2
i=18
As we see, up to at least 4 decimal places, this solution gives the same answer as obtained
above. This indicates the importance of correction for continuity; if it is ignored, we obtain
0.4714, an answer which is almost 10% lower than the actual answer.
29. Let X be the number
√ of 1’s selected; X is binomial with parameters 100, 000 and 1/40. Thus
np = 2500 and
np(1 − p) = 49.37. So
3499.50 − 2500
=1−
P (X ≥ 3500) ≈ P Z ≥
49.37
(20.25) = 0.
Hence it is fair to say that the algorithm is not accurate.
30. Note that
2
ka −x = k exp − x 2 ln a = k exp −
x2
.
1/ ln a
Comparing this with the probability density function of a normal random
variable with pa√
2
rameters µ and σ , we see that µ = 0 and 2σ = 1/ ln a. Thus σ = 1/(2 ln a), and hence
(
ln a
1
=
.
k= √
π
σ 2π
So, for this value of k, the function f is the√probability density function a normal random
variable with mean 0 and standard deviation 1/(2 ln a).
138
Chapter 7
Special Continuous Distributions
31. (a) The derivation of these inequalities from the hint is straightforward.
(b) By part (a),
1−
1
1 − (x)
<
√
2 < 1.
2
x
1/(x 2π ) e−x /2
Thus
1 − (x)
√
2 ≤ 1,
1/(x 2π ) e−x /2
1 ≤ lim
x→∞
from which (b) follows.
32. By part (b) of Exercise 31,
x
P
Z>t+
x
t
lim P Z > t + Z ≥ t = lim
t→∞
t→∞
t
P (Z ≥ t)
1
x
exp
−
t+
x √
t
t+
2π
t
1
2
√ e−t /2
t 2π
= lim
t→∞
2
2
x2
t2
exp
−
x
−
= e−x .
t→∞ t 2 + x
2t 2
= lim
33. Let X be the amount of soft drink in a random bottle. We are given that P (X < 15.5) = 0.07
15.5 − µ
and P (X > 16.3) = 0.10. These imply that
= 0.07 and and
σ
0.90. Using Tables 1 and 2 of the appendix, we obtain
16.3 − µ
σ
⎧ 15.5 − µ
⎪
= −1.48
⎪
⎨
σ
16.3 − µ
⎪
⎪
= 1.28.
⎩
σ
Solving these two equations in two unknowns, we obtain µ = 15.93 and σ = 0.29.
34. Let X be the height of a randomly selected skeleton from group 1. Then
185 − 172
= P (Z > 1.44) = 0.0749.
P (X > 185) = P Z >
9
=
Section 7.3
Exponential Random Variables
139
Now suppose that the skeleton’s of the second group belong to the family of the first group.
The probability of finding three or more skeleton’s with heights above 185 centimeters is
5
i=3
5
(0.0749)i (0.9251)5−i = 0.0037.
i
Since the chance of this event is very low, it is reasonable to assume that the second group is
not part of the first one. However, we must be careful that in reality, this observation is not
sufficient to make a judgment. In the lack of other information, if a decision is to be made
solely based on this observation, then we must reject the hypothesis that the second group is
part of the first one.
35. For t ∈ (0, ∞), let A be the region whose points have a (positive) distance t or less from the
given tree. The area of A is π t 2 . Let X be the distance from the given tree to its nearest tree.
We have that
e−λπ t (λπ t 2 )0
2
P (X > t) = P (no trees in A) =
= e−λπ t .
0!
2
Now by Remark 6.4,
E(X) =
∞
P (X > t) dt =
0
∞
e−λπ t dt.
2
0
√
Letting u =
2λπ t, we obtain
1 1
E(X) = √ √
λ 2π
∞
1
1 1
= √ .
du = √
λ2
2 λ
e−u
2 /2
0
36. Note that dy = xds; so
I =
2
=
∞
0 ∞
0
7.3
∞
e
0
=
∞
0
−(x 2 +x 2 s 2 )/2
x ds dx =
1
1
e
du ds =
2
2
0
∞ π
1
ds = arctan s
= .
0
1 + s2
2
∞
−u(1+s 2 )/2
∞
0
∞
0
∞
0
−
e−x
2 (1+s 2 )/2
x dx ds
(let u = x 2 )
∞
2
−u(1+s 2 )/2
e
ds
0
1 + s2
EXPONENTIAL RANDOM VARIABLES
1. Let X be the time until the next customer arrives; X is exponential with parameter λ = 3.
Hence P (X > x) = e−λx , and P (X > 3) = e−9 = 0.0001234.
140
Chapter 7
Special Continuous Distributions
2. Let m be the median of an exponential random variable with rate λ. Then P (X > m) = 1/2;
thus e−λm = 1/2 or m =
ln 2
.
λ
3. For −∞ < y < ∞,
−y
P (Y ≤ y) = P (− ln X ≤ y) = P X ≥ e−y = e−e .
Thus g(y), the probability density function of Y is given by
g(y) =
−y
−y
d
P (Y ≤ y) = e−y · e−e = e−y − e .
dy
4. Let X be the time between the first and second heart attacks. We are given that P (X ≤ 5) =
1/2. Since exponential is memoryless, the probability that a person who had one heart attack
five years ago will not have another one during the next five years is still P (X > 5) which is
1 − P (X ≤ 5) = 1/2.
5. (a) Suppose that the next customer arrives in X minutes. By the memoryless property, the
desired probability is
1
P X<
= 1 − e−5(1/30) = 0.1535.
30
(b) Let Y be the time between the arrival times of the 10th and 11th customers; Y is exponential
with λ = 5. So the answer is
1
P Y ≤
= 1 − e−5(1/30) = 0.1535.
30
6.
1 2
P |X − E(X)| ≥ 2σX = P X − ≥
λ
λ
2
1
2
1
+P X− ≤−
=P X− ≥
λ
λ
λ
λ
1
3
+P X ≤−
=P X≥
λ
λ
−λ(3/λ)
−3
=e
+ 0 = e = 0.049787.
7. (a) P (X > t) = e−λt .
(b) P (t ≤ X ≤ s) = 1 − e−λs − 1 − e−λt = e−λt − e−λs .
8. The number of documents typed by the secretary on a given eight-hour working day is Poisson
with parameter λ = 8. So the answer is
∞
e−8 8 i
=1−
i!
i=12
11
i=0
e−8 8 i
= 1 − 0.888 = 0.112.
i!
Section 7.3
9. The answer is
Exponential Random Variables
141
1
E 350 − 40N (12) = 350 − 40
· 12 = 323.33.
18
10. Mr. Jones makes his phone calls when either A or B is finished his call. At that time the
remaining phone call of A or B, whichever is not finished, and the duration of the call of
Mr. Jones both have the same distribution due to the memoryless property of the exponential
distribution. Hence, by symmetry, the probability that Mr. Jones finishes his call sooner than
the other one is 1/2.
11. Let N(t) be the number of change-of-states occurring in [0, t]. Let X1 be the time until the
machine breaks down for the first time. Let X2 be the time it will take to repair the machine,
X3 be the time since the machine was fixed until it breaks down again, and so on. Clearly, X1 ,
X2 , . . . are the times between consecutive change of states. Since {X1 , X2 , . . . } is a sequence
of independent
and identically
distributed exponential random variables with mean 1/λ, by
Remark 7.2, N(t) : t ≥ 0 is a Poisson process with rate λ. Therefore, N(t) is a Poisson
random variable with parameter λt.
12. The probability mass function of L is given by
n = 1, 2, 3, . . . .
P (L = n) = (1 − p)n−1 p,
Hence
P (L > n) = (1 − p)n ,
n = 0, 1, 2, . . . .
Therefore,
P (T ≤ x) = P (L ≤ 1000x) = 1 − P (L > 1000x) = 1 − (1 − p)1000x
= 1 − e1000x ln(1−p) = 1 − e−x[−1000 ln(1−p)] ,
x > 0.
This shows that T is exponential with parameter λ = −1000 ln(1 − p).
∞
13. (a) We must have
ce−|x| dx = 1; thus
−∞
c=
1
∞
−∞
e−|x| dx
=
2
1
∞
e−x dx
=
1
.
2
0
∞
1 2n+1 −|x|
e
dx = 0, because the integrand is an odd function.
x
2
−∞
∞
∞
1 2n −|x|
2n
E(X ) =
dx =
x 2n e−x dx,
x e
−∞ 2
0
∞
because the integrand is an even function. We now use induction to prove that
x n e−x dx =
(b) E(X2n+1 ) =
0
n!. For n = 1, the integral is the expected value of an exponential random variable with
142
Chapter 7
Special Continuous Distributions
parameter 1; so it equals to 1 = 1!. Assume that the identity is valid for n − 1. Using
integration by parts, we show it for n.
∞
∞ ∞
n −x
n −x
x e dx = − − x e
+
nx n−1 e−x dx = 0 + n(n − 1)! = n!.
0
0
0
Hence E(X2n ) = (2n)!.
n+1
14. P [X] = n = P (n ≤ X < n + 1) =
n+1
n
λe−λx dx = −e−λx
= e−λ 1 − e−λ . This
n
n
is the probability mass function of a geometric random variable with parameter p = 1 − e−λ .
15. Let that G(t) = P (X > t) = 1 − F (t). By the memoryless property of X,
P (X > s + t | X > t) = P (X > s),
for all s ≥ 0 and t ≥ 0. This implies that
P (X > s + t) = P (X > s)P (X > t),
or
G(s + t) = G(s)G(t),
t ≥ 0, s ≥ 0.
Now for arbitrary positive integers n and m, (24) gives that
1 1
1 1
1 2
2
=G
+
=G
,
G
= G
G
n
n n
n
n
n
3
2 1
2 1
1 2 1
1
G
G
=G
+
=G
G
= G
=G
n
n n
n
n
n
n
n
..
.
1 m
m
= G
.
G
n
n
Also
1 n
1
1 1
+ + ··· +
G(1) = G
= G
n.
n
+n n ,n terms
yields
1
1/n
= G(1)
.
G
n
(24)
3
,
(25)
Hence
m/n
.
G(m/n) = G(1)
(26)
Section 7.3
Exponential Random Variables
143
Now we show that G(1) > 0. If not, G(1) = 0 and by (25), G(1/n) = 0 for all positive
integer n. This and right continuity of G imply that
1
P (X ≤ 0) = F (0) = 1 − G(0) = 1 − G lim
n→∞ n
1
= 1 − 0 = 1,
= 1 − lim G
n→∞
n
which is a contradiction to the
given
fact that X is a positive random variable. Thus G(1) > 0
and we can define λ = − ln G(1) . This gives
G(1) = e−λ ,
and by (26),
G(m/n) = e−λ(m/n) .
Thus far, we have proved that for any positive rational t,
G(t) = e−λt .
(27)
To prove the same relation for a positive irrational number t, recall from calculus that for each
1
1
positive integer n, there exists a rational number tn in t, t +
. Since t < tn < t + ,
n
n
limn→∞ tn exists and is t. On the other hand because F is right continuous, G = 1 − F is also
right continuous and so
G(t) = lim G(tn ).
n→∞
But since tn is rational, (27) implies that, G(tn ) = e−λtn . Hence
G(t) = lim e−λtn = e−λt .
n→∞
Thus F (t) = 1 − e−λt for all t, and X is exponential.
Remark: If X is memoryless, then P (X ≤ 0) = 0. To see this, note that P (X > s + t | X >
t) = P (X > s) implies P (X ≤ s + t | X > t) = P (X ≤ s). Letting s = t = 0, we get
P (X ≤ 0 | X > 0) = P (X ≤ 0). But P (X ≤ 0 | X > 0) = 0; therefore P (X ≤ 0) = 0.
This shows that the memoryless property cannot be defined for random variables possessing
nonpositive values with positive probability.
144
Chapter 7
7.4
GAMMA DISTRIBUTIONS
Special Continuous Distributions
1. Let f be the probability density function of a gamma random variable with parameters r and
λ. Then
f (x) =
λr x r−1 e−λx
.
(r)
Therefore,
f (x) =
λr+1 r−2 −λx
r −1
λr
x−
− λe−λx x r−1 + e−λx (r − 1)x r−2 = −
x e
.
(r)
(r)
λ
This relation implies that the function f is increasing if x < (r − 1)/λ, it is decreasing if
x > (r − 1)/λ, and f (x) = 0 if x = (r − 1)/λ. Therefore, x = (r − 1)/λ is a maximum
of the function f . Moreover, since f has only one root, the point x = (r − 1)/λ is the only
maximum of f .
2. We have that
0
t
=
0
=
0
(λe−λx )(λx)r−1
dx
(let u = cx)
(r)
λe−λu/c (λu/c)r−1
(1/c) du
(r)
(λ/c)e−λu/c (λu/c)r−1
du.
(r)
t/c
P (cX ≤ t) = P (X ≤ t/c) =
t
This shows that cX is gamma with parameters r and λ/c.
3. Let N(t) be the number of babies born at or prior to t. N(t) : t ≥ 0 is a Poisson process
with λ = 12. Let X be the time it takes before the next three babies are born. The random
variable X is gamma with parameters 3 and 12. The desired probability is
∞
∞
12e−12x (12x)2
P (X ≥ 7/24) =
dx = 864
x 2 e−12x dx.
(3)
7/24
7/24
Applying integration by parts twice, we get
1
1
1 −12x
x 2 e−12x dx = − x 2 e−12x − xe−12x −
e
+ c.
12
72
864
Thus
7
1
1
1 −12x ∞
P X≥
= 864 − x 2 e−12x − xe−12x −
e
= 0.3208.
7/24
24
12
72
864
Remark: A simpler way to do this problem is to avoid gamma random variables and use the
properties of Poisson processes:
i
2
2
7
7
e−(7/24)12 (7/24)12
P N
≤2 =
=i =
= 0.3208.
P N
24
24
i!
i=0
i=0
Section 7.4
Gamma Distributions
145
4.
∞
−∞
f (x) dx =
0
∞
λe−λx (λx)r−1
λr
dx =
(r)
(r)
∞
e−λx x r−1 dx.
0
Let t = λx; then dt = λdx, so
∞
λr
t r−1
f (x) dx =
e−t · r−1
(r) 0
λ
−∞
∞
1
=
e−t t r−1 dt
(r) 0
∞
1
dx
λ
1
=
(r) = 1.
(r)
·
5. Let X be the time until the restaurant starts to make profit; X is a gamma random variable with
parameters 31 and 12. Thus E(X) = 31/12; that is, two hours and 35 minutes.
6. By the method of Example 5.17, the number of defective light bulbs produced is a Poisson
process at the rate of (200)(0.015) = 3 per hour. Therefore, X, the time until 25 defective
light bulbs are produced is gamma with parameters λ = 3 and r = 25. Hence
E(X) =
r
25
=
= 8.33.
λ
3
That is, it will take, on average, 8 hours and 20 minutes to fill up the can.
7.
1
2
=
∞
t −1/2 e−t dt.
0
Making the substitution t = y 2 /2, we get
√ ∞
2
2
e
dy =
e−y /2 dy
2
2 −∞
0
∞
√
√
1
2
= π·√
e−y /2 dy = π .
2π −∞
1
√
= 2
∞
−y 2 /2
146
Chapter 7
Special Continuous Distributions
Hence
3
2
5
2
7
2
=
1 1
1 √
= · π,
2 2
2
=
3 3
3 1 √
= · · π,
2 2
2 2
=
5 5
5 3 1 √
= · · · π,
2 2
2 2 2
..
.
n+
2n + 1
1
2n − 1 2n − 3
7 5 3 1 √
=
=
·
··· · · · · π
2
2
2
2
2 2 2 2
=
22n
√
(2n)!
π
(2n) · · · 6 · 4 · 2
√
√
(2n)! π
(2n)! π
= n n
= n
.
2 · 2 · n!
4 · n!
8. (a) Let F be the probability distribution function of Y . For t ≤ 0, F (t) = P (Z 2 ≤ t) = 0.
For t > 0,
√
√
F (t) = P (Y ≤ t) = P Z 2 ≤ t = P − t ≤ Z ≤ t
=
√
t −
−
√
t =
√
t − 1−
√
√
t =2
t − 1.
Let f be the probability density function of Y . For t ≤ 0, f (t) = 0. For t > 0,
1 −t/2 1 −1/2
e
t
2
,
= 2
(1/2)
√
1
1
1
e−t/2
t = √ · √ e−t/2 = √
t
2π
2π t
√
where by the previous exercise, π = (1/2). This shows that Y is gamma with parameters
λ = 1/2 and r = 1/2.
1
f (t) = F (t) = 2 · √
2 t
(b) Since (X − µ)/σ is standard normal, by part (a), W is gamma with parameters λ = 1/2
and r = 1/2.
9. The following solution is an intuitive one. A rigorous mathematical solution would have to
consider the sum of two random variables, each being the minimum of n exponential random
Section 7.5
Beta Distributions
147
variables; so it would require material from joint distributions. However, the intuitive solution
has its own merits and it is important for students to understand it.
Let the time Howard enters the bank be the origin and let N (t) be the number of customers
served by time t. As long asall of the servers
are busy, due to the memoryless property of
the exponential distribution, N(t) : t ≥ 0 is a Poisson process with rate nλ. This follows
because if one server serves at the rate λ, n servers will serve at the rate nλ. For the Poisson
process N(t) : t ≥ 0 , every time a customer is served and leaves, an “event” has occurred.
Therefore, again because of the memoryless property, the service time of the person ahead
of Howard begins when the first “event” occurs and Howard’s service time begins when the
second “event” occurs. Therefore, Howard’s
waiting
time in the queue is the time of the
second event of the Poisson process N(t), t ≥ 0 . This period, as we know, has a gamma
distribution with parameters 2 and nλ.
10. Since the lengths of the characters are independent of each other and identically distributed,
for any two intervals 1 and 2 with the same length, the probability that n characters are
emitted during 1 is equal to the probability that n characters are emitted in 2 . Moreover, for
s > 0, the number of characters being emitted during (t, t + s] is independent of the number of
characters that
have beenemitted in [0, t]. Clearly, characters are not emitted simultaneously.
Therefore, N(t) : t ≥ 0 is stationary, possesses independent increments, and is orderly. So
it is a Poisson process. By Exercise 11, Section 7.3, the time
until the first
character is emitted
is exponential with parameter λ = −1000 ln(1 − p). Thus N (t) : t ≥ 0 is a Poisson process
with parameter λ = −1000 ln(1 − p). Knowing this, we have that the time until the message
is emitted, that is, the time until the kth character is emitted is gamma with parameters k and
λ = −1000 ln(1 − p).
7.5
BETA DISTRIBUTIONS
1. Yes, it is a probability density function of a beta random variable with parameters α = 2 and
β = 3. Note that
1
4!
=
= 12. We have
B(2, 3)
1! 2!
E(X) =
2
,
5
VarX =
6
1
= .
6(52 )
25
2. No, it is not because, for α = 3 and β = 5, we have
7!
1
=
= 105 = 120.
B(3, 5)
2! 4!
3. Let α = 5 and β = 6. Then f is the probability density function of a beta random variable
with parameters 5 and 6 for
c=
1
10!
=
= 1260.
B(5, 6)
4! 5!
148
Chapter 7
Special Continuous Distributions
For this value of c,
E(X) =
5
,
11
VarX =
30
5
=
.
2
12(11 )
242
4. The answer is
1
1
x 19 (1 − x)12 dx
B(20,
13)
0.60
1
32!
=
x 19 (1 − x)12 dx = 0.538.
19! 12! 0.60
P (p ≥ 0.60) =
5. Let X be the proportion of resistors the procurement office purchases from this vendor. We
know that X is beta. Let α and β be the parameters of the density function of X. Then
⎧
α
1
⎪
⎪
⎪
⎨α + β = 3
⎪
⎪
⎪
⎩
1
αβ
= .
(α + β + 1)(α + β)2
18
Solving this system of 2 equations in 2 unknowns, we obtain α = 1 and β = 2. The desired
probability is
P (X ≥ 7/12) =
1
7/12
1
x 1−1 (1 − x)2−1 dx = 2
B(1, 2)
1
7/12
(1 − x) dx =
50
≈ 0.17.
288
6. Let X be the median of the fractions for the 13 sections of the course; X is a beta random
variable with parameters 7 and 7. Let Y be a binomial random variable with parameters 13
and 0.40. By Theorem 7.2,
P (X ≤ 0.40) = P (Y ≥ 7).
Therefore,
6
13
(0.40)i (0.60)13−i = 0.771156.
i
P (X ≥ 0.40) = P (Y ≤ 6) =
i=0
7. Let Y be a binomial random variable with parameters 25 and 0.25; by Theorem 7.2,
P (X ≤ 0.25) = P (Y ≥ 5).
Therefore,
4
P (X ≥ 0.25) = P (Y < 5) =
i=0
25
(0.25)i (0.75)25−i = 0.214.
i
Section 7.5
Beta Distributions
149
8. (a) Clearly,
E(Y ) = a + (b − a)E(X) = a + (b − a)
Var(X) = (b − a)2 Var(X) =
(b)
α
,
α+β
(b − a)2 αβ
.
(α + β + 1)(α + β)2
Note that 0 < X < 1 implies that a < Y < b. Let a < t < b; then
t −a
P (Y ≤ t) = P a + (b − a)X ≤ t = P X ≤
b−a
(t−a)/(b−a)
1
=
x α−1 (1 − x)β−1 dx.
B(α, β)
0
Let y = (b − a)x + a; we have
t
1
1 y − a α−1
y − a β−1
P (Y ≤ t) =
·
1−
dy
b−a
b−a
a B(α, β) b − a
t
1
1 y − a α−1 b − y β−1
=
dy.
·
b−a
a b − a B(α, β) b − a
This shows that the probability density function of Y is
1
1 y − a α−1 b − y
f (y) =
·
b − a B(α, β) b − a
b−a
(c)
β−1
,
a < y < b.
Note that a = 2, b = 6. Hence
3
1 4! y − 2 6 − y 2
·
P (Y < 3) =
dy
4
4
2 4 1! 2!
3
67
3 67
3
·
=
≈ 0.26.
(y − 2)(6 − y)2 dy =
=
64 2
64 12
256
9. Suppose that
f (x) =
1
x α−1 (1 − x)β−1 ,
B(α, β)
0 < x < 1,
is symmetric about a point a. Then f (a − x) = f (a + x). That is, for 0 < x < min(a, 1 − a),
(a − x)α−1 (1 − a + x)β−1 = (a + x)α−1 (1 − a − x)β−1 .
(28)
Since α and β are not necessarily integers, for (a −x)α−1 and (1−a −x)β−1 to be well-defined,
we need to restrict ourselves to the range 0 < x < min(a, 1 − a). Now, if a < 1 − a, then,
by continuity, (28) is valid for x = a. Substituting a for x in (28), we obtain
(2a)α−1 (1 − 2a)β−1 = 0.
150
Chapter 7
Special Continuous Distributions
Since a = 0, this implies that a = 1/2. If 1 − a < a, then, by continuity, (28) is valid for
x = 1 − a. Substituting 1 − a for x in (28), we obtain
(2a − 1)α−1 (2 − 2a)β−1 = 0.
Since a = 1, this implies that a = 1/2. Therefore, in either case a = 1/2. In (28), substituting
a = 1/2, and taking x = 1/4, say, we get
(1/4)α−1 (3/4)β−1 = (3/4)α−1 (1/4)β−1 .
This gives 3β−α = 0, which can only hold for α = β. Therefore, only beta density functions
with α = β are symmetric, and they are symmetric about a = 1/2.
2t
dt, we have
(1 + t 2 )2
∞
β−1
2t
·
dt = 2
t 2α−1 (1 + t 2 )−(α+β) dt.
(1 + t 2 )2
0
10. t = 0 gives x = 0; t = ∞ gives x = 1. Since dx =
B(α, β) =
∞
0
t2
1 + t2
α−1
1
1 + t2
11. We have that
1
B(α, β) =
x α−1 (1 − x)β−1 dx.
0
Let x = cos θ to obtain
2
π/2
B(α, β) = 2
(cos θ)2α−1 (sin θ)2β−1 dθ.
0
Now
∞
(α) =
t α−1 e−t dt.
0
Use the substitution t = y to obtain
2
(α) = 2
∞
y 2α−1 e−y dy.
2
0
This implies that
(α) (β) = 4
0
∞
∞
x 2α−1 y 2β−1 e−(x
2 +y 2 )
dxdy.
0
Now we evaluate this double integral by means of a change of variables to polar coordinates:
y = r sin θ , x = r cos θ ; we obtain
∞ π/2
2
(α) (β) = 4
r 2(α+β)−1 (cos θ)2α−1 (sin θ)2β−1 e−r dθdr
0
0
∞
∞
2
= 2B(α, β)
r 2(α+β)−1 e−r dr = B(α, β)
uα+β−1 e−u du (let u = r 2 )
0
= B(α, β) (α + β).
0
Section 7.5
Thus
B(α, β) =
Beta Distributions
151
(α) (β)
.
(α + β)
12. We will show that E(X2 ) = n/(n − 2). Since E(X2 ) < ∞, by Remark 6.6, E(X) < ∞.
Since E(X) exists and xf (x) is an odd function, we have
∞
E(X) =
xf (x) dx = 0.
−∞
Consequently,
2
Var(X) = E(X2 ) − E(X) =
n
.
n−2
Therefore, all we need to find is E(X2 ). By Theorem 6.3,
n + 1
∞
x 2 −(n+1)/2
2
2
n
x2 1 +
dx.
E(X ) = √
n
−∞
nπ
2
√
Substituting x = ( n )t in this integral yields
n + 1
∞
√
2
2
n
E(X ) = √
(nt 2 )(1 + t 2 )−(n+1)/2 n dt
−∞
nπ
2
n + 1
∞
2
n · 2n
=√
t 2 (1 + t 2 )−(n+1)/2 dt.
0
π
2
By the previous two exercises,
3 n − 2
∞
3 n − 2
2
2
.
=
t 2 (1 + t 2 )−(n+1)/2 dt = B ,
2
n + 1
2
2
0
2
Therefore,
3 n − 2
3 n − 2
n + 1
n
2
2
2
2
n2
.
=
E(X2 ) = √
n + 1
√
n ·n·
π·
π
2
2
2
√
By the solution to Exercise 7, Section 7.4, (1/2) = π . Using the identity (r +1) = r (r),
we have
√
3
1 1
π
=
=
;
2
2 2
2
n
n − 2
n − 2 n − 2
=
+1 =
.
2
2
2
2
152
Chapter 7
Special Continuous Distributions
Consequently,
√
n−2
π
2
2
E(X2 ) =
√
n − 2 n − 2
π·
2
2
n
7.6
=
n
.
n−2
SURVIVAL ANALYSIS AND HAZARD FUNCTIONS
1. Let X be the lifetime of the electrical component, F be its probability distribution function,
and λ(t) be its failure rate. For some constants α and β, we are given that
λ(t) = αt + β.
Since λ(48) = 0.10 and λ(72) = 0.15,
48α + β = 0.10
72α + β = 0.15.
Solving this system of two equations in two unknowns gives α = 1/480 and β = 0. Hence
λ(t) = t/480. By (7.6), for t > 0,
t u
2
P (X > t) = F̄ (t) = exp −
du = e−t /960 .
0 480
Let f be the probability density function of X. This also gives
f (t) = −
d
t −t 2 /960
F̄ (t) =
e
.
dt
480
The answer to part (a) is
P (X > 30) = e−900/960 = e−0.9375 = 0.392.
The exact value for part (b) is
P (30 < X < 31)
P (X > 30)
31
0.02411
1
2
(t/480)e−t /960 dt =
= 0.0615.
=
0.392
0.392 30
P (X < 31 | X > 30) =
Note that for small t , λ(t)t is approximately the probability that the component fails
within t hours after t, given that it has not yet failed by time t. Letting t = 1, for t = 30,
λ(t)t ≈ 0.0625 which is relatively close to the exact value of 0.0615. This is interesting
because t = 1 is not that small, and one may not expect close approximations anyway.
Chapter 7
Review Problems
153
2. Let F̄ be the survival function of a Weibull random variable. We have
F̄ (t) =
∞
αx α−1 e−x dx.
α
t
Letting u = x α , we have du = αx α−1 dx. Thus
∞
∞
α
−u
−u
e du = −e α = e−t .
F̄ (t) =
t
tα
Therefore,
αt α−1 e−t
= αt α−1 ·
λ(t) =
e−t α
λ(t) = 1, for α = 1; so the Weibull in this case is exponential with parameter 1. Clearly, for
α < 1, λ (t) < 0; so λ(t) is decreasing. For α > 1, λ (t) > 0; so λ(t) is increasing. Note that
for α = 2, the failure rate is the straight line λ(t) = 2t.
α
REVIEW PROBLEMS FOR CHAPTER 7
1.
30 − 25
5
=
.
37 − 25
12
2. Let X be the weight of a randomly selected women from this community. The desired quantity
is
170 − 130
P Z>
P (X > 170)
20
=
P (X > 170 | X > 140) =
P (X > 140)
140 − 130
P Z>
20
=
1 − (2)
1 − 0.9772
P (Z > 2)
=
=
= 0.074.
P (Z > 0.5)
1 − (0.5)
1 − 0.6915
3. Let X be the number of times the digit
√X is binomial with parameters n = 1000
√ 5 is generated;
and p = 1/10. Thus np = 100 and np(1 − p) =
and making correction for continuity,
90 = 9.49. Using normal approximation
93.5 − 100
P (X ≤ 93.5) = P Z ≤
= P (Z ≤ −0.68) = 1 −
9.49
4. The given relation implies that
1 − e−2λ = 2 (1 − e−3λ ) − (1 − e−2λ ) .
(0.68) = 0.248.
154
Chapter 7
Special Continuous Distributions
This is equivalent to
3e−2λ − 2e−3λ − 1 = 0,
or, equivalently,
2
e−λ − 1 2e−λ + 1 = 0.
The only root of this equation is λ = 0 which is not acceptable. Therefore, it is not possible
that X satisfy the given relation.
5. Let X be the lifetime of a random light bulb. Then
P (X < 1700) = 1 − e−(1/1700)·1700 = 1 − e−1 .
The desired probability is
1 − P (none fails) − P (one fails)
19
20
20
−1 0 −1 20
=1−
(1 − e ) (e ) −
1 − e−1 e−1 = 0.999999927.
0
1
6. Note that limx→0 x ln x = 0; so
E(− ln X) =
1
(− ln x) dx = x − x ln x
1
0
0
= 1.
7. Let X be the diameter of the randomly chosen disk in inches. We are given that X ∼ N(4, 1).
We want to find the distribution function of 2.5X; we have
1
P (2.5X ≤ x) = P (X ≤ x/2.5) = √
2π
x/2.5
e−(t−4)
2 /2
dt.
−∞
8. If α < 0, then α + β < β; therefore,
P (α ≤ X ≤ α + β) = P (0 ≤ X ≤ α + β) ≤ P (0 ≤ X ≤ β).
If α > 0, then e−λα < 1. Thus
P (α ≤ X ≤ α + β) = 1 − e−λ(α+β) − 1 − e−λα
= e−λα 1 − e−λβ < 1 − e−λβ = P (0 ≤ X ≤ β).
9. We are given that 1/λ = 1.25; so λ = 0.8. Let X be the time it takes for a random student to
complete the test. Since P (X > 1) = e−(0.8)1 = e−0.8 , the desired probability is
10
1 − e−0.8 = 1 − e−8 = 0.99966.
Chapter 7
10. Note that
f (x) = ke−[x−(3/2)]
2 +17/4
Review Problems
155
= ke17/4 · e−[x−(3/2)] .
2
Comparing this with the probability density function
of a normal random variable with mean
√
2
17/4
3/2, we see that σ = 1/2 and ke
= 1/(σ 2π ). Therefore,
k=
1
1
√ e−17/4 = e−17/4 .
π
σ 2π
11. Let X be the grade of a randomly selected student.
90 − 72
=1−
P (X ≥ 90) = P Z ≥
7
(2.57) = 0.0051.
Similarly,
P (80 ≤ X < 90) = P (1.14 ≤ Z < 2.57) = 0.122,
P (70 ≤ X < 80) = P (−0.29 ≤ Z < 1.14) = 0.487,
P (60 ≤ X < 70) = P (−1.71 ≤ Z < −0.29) = 0.3423,
P (X < 60) = P (Z < −1.71) = 0.0436.
Therefore, approximately 0.51% will get A, 12.2% will get B, 48.7% will get C, 34.23% D,
and 4.36% F.
12. Since E(X) = 1/λ,
P X > E(X) = e−λ(1/λ) = e−1 = 0.36788.
13. Round off error to the nearest integer is uniform over (−0.5, 0.5); round off error to the nearest
1st decimal place is uniform over (−0.05, 0.05); round off error to the nearest 2nd decimal
place is uniform over (−0.005, 0.005), and so on. In general, round off error to the nearest k
decimal places is uniform over (−5/10k+1 , 5/10k+1 ).
14. We want to find the smallest a for which P (X ≤ a) ≥ 0.90. This implies
a − 175
≥ 0.90.
P Z≤
22
Using Table 1 of the appendix, we see that (a − 175)/22 = 1.29 or a = 203.38.
15. Let X be the breaking strength of the yarn under consideration. Clearly,
100 − 95
=1−
P (X ≥ 100) = P Z ≥
11
(0.45) = 0.33.
So the desired probability is
10
10
(0.33)1 (0.67)9 = 0.89.
1−
(0.33)0 (0.67)10 −
1
0
156
Chapter 7
Special Continuous Distributions
16. Let X be the time until the 91st call is received. X is a gamma random variable with parameters
r = 91 and λ = 23. The desired probability is
∞
23e−23x (23x)91−1
P (X ≥ 4) =
dx
(91)
4
4
23e−23x (23x)91−1
dx
=1−
90!
0
2391 4 90 −23x
x e
dx = 1 − 0.55542 = 0.44458.
=1−
90! 0
17. Clearly,
Now
E(X) =
(1 − θ) + (1 + θ)
= 1,
2
Var(X) =
θ2
(1 + θ − 1 + θ)2
= .
12
3
2 θ 2
E X2 − E(X) =
3
implies that
θ2
+ 1,
E X2 =
3
which yeilds 3E(X2 ) − 1 = θ 2 , or, equivalently, E(3X2 − 1) = θ 2 . Therefore, one choice for
g(X) is g(X) = 3X2 − 1.
18. Let α and β be the parameters of the density function of X/. Solving the following two
equations in two unknowns,
E(X/) =
Var(X/) =
α
3
= ,
α+β
7
αβ
3
= ,
2
(α + β + 1)(α + β)
98
we obtain α = 3 and β = 4. Therefore, X/ is beta with parameters 3 and 4. The desired
probability is
1/3
1
P (/7 < X < /3) = P (1/7 < X/ < 1/3) =
x 2 (1 − x)3 dx
1/7 B(3, 4)
1/3
x 2 (1 − x)3 dx = 0.278.
= 60
1/7
Chapter 8
Bivariate Distributions
8.1
JOINT DISTRIBUTIONS OF TWO RANDOM VARIABLES
2
2
k(x/y) = 1 implies that k = 2/9.
(b) pX (x) = 2y=1 (2x)/(9y) = x/3, x = 1, 2.
1. (a)
x=1
y=1
pY (y) =
2
x=1
(2x)/(9y) = 2/(3y),
p(2, 1)
pY (1)
2
2 x
x·
=
9 y
x=1
(c) P (X > 1 | Y = 1) =
2
(d) E(X) =
y=1
2. (a)
3
2
x=1
y=1
(b) pX (x) =
pY (y) =
2
4/9
= .
2/3
3
5
;
3
2
2
y·
E(Y ) =
y=1 x=1
2x
4
= .
9 y
3
c(x + y) = 1 implies that c = 1/21.
2
y=1 (1/21)(x
3
x=1 (1/21)(x
(c) P (X ≥ 2 | Y = 1) =
3
=
y = 1, 2.
2
(d) E(X) =
x=1 y=1
+ y) = (2x + 3)/21. x = 1, 2, 3.
+ y) = (6 + 3y)/21. y = 1, 2.
p(2, 1) + p(3, 1)
7/21
7
=
= .
pY (1)
9/21
9
46
1
x(x + y) = ;
21
21
3
2
E(Y ) =
x=1 y=1
11
1
y(x + y) = .
21
7
3. (a) k(1 + 1 + 1 + 9 + 4 + 9) = 1 implies that k = 1/25.
(b)
pX (1) = p(1, 1) + p(1, 3) = 12/25,
pX (2) = p(2, 3) = 13/25;
pY (1) = p(1, 1) = 2/25,
pY (3) = p(1, 3) + p(2, 3) = 23/25.
158
Chapter 8
Bivariate Distributions
Therefore,
pX (x) =
(c) E(X) = 1 ·
⎧
⎨12/25
if x = 1
⎩13/25
if x = 2,
13
38
12
+2·
= ;
25
25
25
pY (y) =
E(Y ) = 1 ·
⎧
⎨2/25
if y = 1
⎩23/25
if y = 3.
2
23
71
+3·
= .
25
25
25
4. P (X > Y ) = p(1, 0) + p(2, 0) + p(2, 1) = 2/5,
P (X + Y ≤ 2) = p(1, 0) + p(1, 1) + p(2, 0) = 7/25,
P (X + Y = 2) = p(1, 1) + p(2, 0) = 6/25.
5. Let X be the number of sheep stolen; let Y be the number of goats stolen. Let p(x, y) be the
joint probability mass function of X and Y . Then, for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ x + y ≤ 4,
7 8
5
x y 4−x−y
;
p(x, y) =
20
4
p(x, y) = 0, for other values of x and y.
6. The following table gives p(x, y), the joint probability mass function of X and Y ; pX (x), the
marginal probability mass function of X; and pY (y), the marginal probability mass function
of Y .
y
x
2
3
4
5
6
7
8
9
10
11
12
pY (y)
0
1/36
0
1/36
0
1/36
0
1/36
0
1/36
0
1/36
6/36
1
0
2/36
0
2/36
0
2/36
0
2/36
0
2/36
0
10/36
2
0
0
2/36
0
2/36
0
2/36
0
2/36
0
0
8/36
3
0
0
0
2/36
0
2/36
0
2/36
0
0
0
6/36
4
0
0
0
0
2/36
0
2/36
0
0
0
0
4/36
7. p(1, 1) = 0, p(1, 0) = 0.30, p(0, 1) = 0.50, p(0, 0) = 0.20.
5
0
0
0
0
0
2/36
0
0
0
0
0
2/36
pX (x)
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
Section 8.1
Joint Distributions of Two Random Variables
159
8. (a) For 0 ≤ x ≤ 7, 0 ≤ y ≤ 7, 0 ≤ x + y ≤ 7,
p(x, y) =
13 13
26
x
y
7−x−y
.
52
7
For all other values of x and y, p(x, y) = 0.
(b) P (X ≥ Y ) =
3
7−y
y=0
x=y
p(x, y) = 0.61107.
x
9. (a) fX (x) =
2 dy = 2x,
0 ≤ x ≤ 1;
fY (y) =
0
1
2 dx = 2(1 − y),
0 ≤ y ≤ 1.
y
1
(b) E(X) =
0
1
E(Y ) =
1
xfX (x) dx =
1
yfY (y) dy =
0
x(2x) dx = 2/3;
0
2y(1 − y) dy = 1/3.
0
1
(c) P X <
=
2
1/2
0
P (X < 2Y ) =
0
1
fX (x) dx =
1/2
0
x
2 dy dx =
x/2
P (X = Y ) = 0.
x
10. (a) fX (x) =
8xy dy = 4x 3 ,
1
2x dx = ,
4
1
,
2
0 ≤ x ≤ 1,
0
1
fY (y) =
8xy dx = 4y(1 − y 2 ),
0 ≤ y ≤ 1.
y
(b) E(X) =
1
0
E(Y ) =
1
11. fX (x) =
0
2
1 −x
ye dy = e−x ,
2
x · 4x 3 dx = 4/5;
0
yfY (y) dy =
0
1
xfX (x) dx =
1
y · 4y(1 − y 2 ) dy = 8/15.
0
fY (y) =
x > 0;
0
∞
1
1 −x
ye dx = y,
2
2
0 < y < 2.
12. Let R = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Since area(R) = 1, P (X + Y ≤ 1/2) is the area
of the region (x, y) ∈ R : x + y ≤ 1/2 which is 1/8. Similarly, P (X − Y ≤ 1/2) is the
160
Chapter 8
Bivariate Distributions
area of the region (x, y) ∈ R : x − y≤ 1/2 which is 7/8. P (X 2 + Y 2 ≤ 1) is the area of
2
2
the region (x,
y) ∈ R : x + y ≤ 1 which is π/4. P (XY ≤ 1/4) is the sum of the area
of the region (x, y) : 0 ≤ x ≤ 1/4, 0 ≤ y ≤ 1 which is 1/4 and the area of the region under
the curve y = 1/(4x) from 1/4 to 1. (Draw a figure.) Therefore,
P (XY ≤ 1/4) =
13. (a) The area of R is
0
1
(b) fX (x) =
x
f (x, y) dy =
x2
√
fY (y) =
y
E(Y ) =
0
6
0
if (x, y) ∈ R
elsewhere.
√
y
0 < x < 1;
√
6 dx = 6( y − y),
0 < y < 1.
y
1
xfX (x) dx =
0
1/4
1
dx ≈ 0.597.
4x
6 dy = 6x(1 − x),
x2
y
(c) E(X) =
1
x
f (x, y) dx =
1
(x − x 2 ) dx = ; so
6
f (x, y) =
1
+
4
1
6x 2 (1 − x) dx = 1/2;
0
1
yfY (y) dy =
1
√
6y( y − y) dy = 2/5.
0
14. Let X and Y be the minutes past 11:30
that the man and his fiancée arrive at the
lobby,
respectively. We have that X and Y are uniformly distributed over (0, 30). Let
S = (x, y) : 0 ≤ x ≤ 30, 0 ≤ y ≤ 30 , and R = (x, y) ∈ S : y ≤ x − 12 or y ≥ x + 12 .
The desired probability is the area of R divided by the area of S: 324/900 = 0.36. (Draw a
figure.)
A.M.
15. Let
X and Y be two randomly selected points from the interval (0, ). We are interested in
E |X − Y | . Since the joint probability density function of X and Y is
⎧
⎪
⎨1
2
f (x, y) =
⎪
⎩0
0 < x < , 0 < y <
elsewhere,
Section 8.1
E |X − Y | =
Joint Distributions of Two Random Variables
161
1
|x − y| 2 dx dy
0
0
y
1
1
= 2
(y − x) dx dy + 2
(x − y) dx dy
0
0
0
y
= + = .
6 6
3
16. The problem is equivalent to the following: Two random numbers X and Y are selected at
random
and independently from (0,). What is the probability that |X − Y | < X? Let
S = (x, y) : 0 < x < , 0 < y < and
R = (x, y) ∈ S : |x − y| < x = (x, y) ∈ S : y < 2x .
The desired probability is the area of R which is 3 2 /4 divided by 2 . So the answer is 3/4.
(Draw a figure.)
17. Let S = (x, y) : 0 < x < 1, 0 < y < 1 and R = (x, y) ∈ S : y ≤ x and x 2 + y 2 ≤ 1 .
The desired probability is the area of R which is π/8 divided by the area of S which is 1. So
the answer is π/8.
18. We prove this for the case in which X and Y are continuous random variables with joint
probability density function f . For discrete random variables the proof is similar. The relation
P (X ≤ Y ) = 1, implies that f (x, y) = 0 if x > y. Hence by Theorem 8.2,
∞ ∞
xf (x, y) dx dy
E(X) =
−∞ −∞
∞ y
=
xf (x, y) dx dy
−∞
−∞
∞
y
≤
yf (x, y) dx dy
−∞ −∞
∞ ∞
=
yf (x, y) dx dy = E(Y ).
−∞
−∞
19. Let H be the distribution
function of a random variable with probability density function h.
x
That is, let H (x) =
h(y) dy. Then
−∞
P (X ≥ Y ) =
∞
−∞
x
−∞
h(x)h(y) dy dx =
∞
x
h(x)
−∞
h(y) dy dx
−∞
2 ∞
1
1
1
=
h(x)H (x) dx = H (x)
= (12 − 02 ) = .
2
2
2
−∞
−∞
∞
20. Since 0 ≤ 2G(x) − 1 ≤ 1, 0 ≤ 2H (y) − 1 ≤ 1, and −1 ≤ α ≤ 1, we have that
−1 ≤ α 2G(x) − 1 2H (y) − 1 ≤ 1.
162
Chapter 8
Bivariate Distributions
So
0 ≤ 1 + α 2G(x) − 1 2H (y) − 1 ≤ 2.
This and g(x) ≥ 0, h(y) ≥ 0 imply that f (x, y) ≥ 0. To prove that f is a joint probability
∞
∞
−∞
−∞
density function, it remains to show that
∞
−∞
f (x, y) dx dy = 1.
∞
f (x, y) dx dy
∞ ∞
∞ ∞
=
g(x)h(y) dx dy + α
g(x)h(y) 2G(x) − 1 2H (y) − 1 dx dy
−∞ −∞
−∞ −∞
∞
∞
=1+α
h(y) 2H (y) − 1 dy
g(x) 2G(x) − 1 dx
−∞
−∞
−∞
2 ∞ 1
2 ∞
1
= 1 + α · 0 · 0 = 1.
= 1 + α 2H (y) − 1
2G(x) − 1
−∞ 4
−∞
4
Now we calculate the marginals.
∞
fX (x) =
g(x)h(y) 1 + α 2G(x) − 1 2H (y) − 1 dy
−∞
∞
∞
=
g(x)h(y) dy + α
g(x)h(y) 2G(x) − 1 2H (y) − 1 dy
−∞
−∞
∞
∞
= g(x)
h(y) dy + αg(x) 2G(x) − 1
h(y) 2H (y) − 1 dy
−∞
1
−∞
2 ∞
2H (y) − 1
= g(x) + αg(x) 2G(x) − 1
−∞
4
= g(x) + αg(x) 2G(x) − 1 · 0 = g(x) + 0 = g(x).
Similarly, fY (y) = h(y).
21. Orient the circle counterclockwise and let X be the length of the arc N M and Y be length of
the arc NL. Let R be the radius of the circle; clearly, 0 ≤ X ≤ 2π R and 0 ≤ Y ≤ 2π R.
The angle MN L is acute if and only if |Y − X| < π R. Therefore, the sample space of this
experiment is
S = (x, y) : 0 ≤ x ≤ 2π R, 0 ≤ y ≤ 2π R
and the desired event is
E = (x, y) ∈ S : |y − x| < πR .
The probability that MN L is acute is the area of E which is 3π 2 R 2 divided by the area of S
which is 4π 2 R 2 ; that is, 3/4.
22. Let
S = (x, y) ∈ R2 : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 ,
B = (x, y) ∈ S : 0.5 < x + y < 1.5 ,
A = (x, y) ∈ S : 0 < x + y < 0.5 ,
C = (x, y) ∈ S : x + y > 1.5 .
Section 8.1
Joint Distributions of Two Random Variables
The probability that the integer nearest to x + y is 0 is
integer nearest to x + y is 1 is
x + y is 2 is
163
1
area(A)
= , The probability that the
area (S)
8
area(B)
3
= , and the probability that the nearest integer to
area(S)
4
area (C)
1
= .
area(S)
8
4
=4
3
ways we can select three of X, a − X, Y , and b − Y . If X, a − X, and Y are selected, a
triangular pen is possible to make if and only if X < (a − X) + Y , a − X < X + Y , and
Y < X + (a − X). The probability of this event is the area of
(x, y) ∈ R2 : 0 < x < a, 0 < y < b, 2x − y < a, 2x + y > a, y < a
23. Let X be a random number from (0, a) and Y be a random number from (0, b). In
which is a 2 /2 divided by the area of
S = (x, y) ∈ R2 : 0 < x < a, 0 < y < b
which is ab: (a 2 /2)/ab = a/(2b). Similarly, for each of the other three 3-combinations of
X, a − x, Y , and b − Y also the probability that the three segments can be used to form a
triangular pen is a/(2b). Thus the desired probability is
1 a
1 a
1 a
a
1 a
·
+ ·
+ ·
+ ·
=
.
4 2b 4 2b 4 2b 4 2b
2b
24. Let X and Y be the two points that are placed on the segment. Let E be the event that the length
of none of the three parts exceeds the given value α. Clearly, P (E | X < Y ) = P (E | Y < X)
and P (X < Y ) = P (Y < X) = 1/2. Therefore,
P (E) = P (E | X < Y )P (X < Y ) + P (E | Y < X)P (Y < X)
1
1
= P (E | X < Y ) + P (E | X < Y ) = P (E | X < Y ).
2
2
This shows that for calculation of P (E), we may reduce the sample space to the case where
X < Y . The reduced sample space is
S = (x, y) : x < y, 0 < x < , 0 < y < .
The desired probability is the area of
R = (x, y) ∈ S : x < α, y − x < α, y > − α
divided by area(S) = 2 /2. But
⎧
(3α − )2
⎪
⎪
⎪
⎨
2
area(R) =
2
⎪
3 2
α
⎪
⎪
−
1−
⎩
2
2
≤α≤
3
2
if ≤ α ≤ .
2
if
2
164
Chapter 8
Bivariate Distributions
Hence the desired probability is
⎧ 3α
2
⎪
⎪
⎨ −1
P (E) =
⎪
α
⎪
⎩1 − 3 1 −
≤α≤
3
2
if ≤ α ≤ .
2
if
2
25. R is the square bounded by the lines x + y = 1, −x + y = 1, −x − y = 1, and x − y = 1; its
area is 2. To find the probability density function of X, the x-coordinate of the point selected
at random from R, first we calculate P (X ≤ t), ∀t. For −1 ≤ t < 0, P (X ≤ t) is the area of
the triangle bound by the lines −x + y = 1, −x − y = 1, and x = t which is (1 + t)2 divided
by area(R) = 2. (Draw a figure.) For 0 ≤ t < 1, P (X ≤ t) is the area inside R to the left of
the line x = t which is 2 − (1 − t)2 divided by area(R) = 2. Therefore,
⎧
⎪
0
⎪
⎪
⎪
⎪
⎪
2
⎪
⎪ (1 + t)
⎪
⎨
2
P (X ≤ t) =
⎪
2
−
(1 − t)2
⎪
⎪
⎪
⎪
⎪
2
⎪
⎪
⎪
⎩1
and hence
⎧
⎪
⎨1 + t
d
P (X ≤ t) = 1 − t
⎪
dt
⎩
0
t < −1
−1 ≤ t < 0
0≤t <1
t ≥ 1,
−1 ≤ t < 0
0≤t <1
otherwise.
This shows that fX (t), the probability density function of X is given by fX (t) = 1 − |t|,
−1 ≤ t ≤ 1; 0, elsewhere.
26. Clearly,
P (Z ≤ z) =
f (x, y) dx dy.
{(x,y) : y/x≤z}
Now for x > 0, y/x ≤ z if and only if y ≤ xz; for x < 0, y/x ≤ z if and only if y ≥ xz.
Therefore, integration region is
(x, y) : x < 0, y ≥ xz ∪ (x, y) : x > 0, y ≤ xz .
Thus
P (Z ≤ z) =
0
−∞
∞
xz
f (x, y) dy dx +
∞
xz
f (x, y) dy dx.
0
−∞
Section 8.1
Joint Distributions of Two Random Variables
Using the substitution y = tx, we get
0 −∞
P (Z ≤ z) =
xf (x, tx) dt dx +
=
=
=
−∞
0
−∞
0
−∞
z
−∞
∞
−∞
z
−∞
−∞
∞
z
−xf (x, tx) dt dx +
−∞
∞ z
xf (x, tx) dt dx
0
−∞
∞ z
0
−∞
∞
|x|f (x, tx) dt dx +
|x|f (x, tx) dt dx =
z
xf (x, tx) dt dx
0
z
165
z
−∞
−∞
|x|f (x, tx) du dx
|x|f (x, tx) dx dt.
Differentiating with respect to z, Fundamental Theorem of Calculus implies that,
∞
d
|x|f (x, xz) dx.
P (Z ≤ z) =
fZ (z) =
dz
−∞
27. Note that there are exactly n such closed semicircular disks because the probability that the
diameter through Pi contains any other point Pj is 0. (Draw a figure.) Let E be the event that
all the points are contained in a closed semicircular disk. Let Ei be the event that the points
are all in Di . Clearly, E = ∪ni=1 Ei . Since there is at most one Di , 1 ≤ i ≤ n, that contains all
the Pi ’s, the events E1 , E2 , . . . , En are mutually exclusive. Hence
n
n
1 n−1
n
1 n−1
Ei =
P (Ei ) =
=n
,
P (E) = P
2
2
i=1
i=1
i=1
where the next-to-the-last equality follows because P (Ei ) is the probability that P1 , P2 ,
. . . , Pi−1 , Pi+1 , . . . , Pn fall inside Di . The probability that any of these falls inside Di is
(area of Di )/(area of the disk) = 1/2 independently of the others. Hence the probability that
all of them fall inside Di is (1/2)n−1 .
28. We have that
(α + β + γ ) 1−x α−1 β−1
fX (x) =
x y (1 − x − y)γ −1 dy
(α) (β) (γ ) 0
1−x
1
α−1
x
y β−1 (1 − x − y)γ −1 dy.
=
B(α, β + γ )B(β, γ )
0
Let z = y/(1 − x); then dy = (1 − x) dz, and
1−x
β−1
γ −1
β+γ −1
y (1 − x − y)
dy = (1 − x)
0
1
zβ−1 (1 − z)γ −1 dz = (1 − x)β+γ −1 B(β, γ ).
0
So
1
x α−1 (1 − x)β+γ −1 B(β, γ )
B(α, β + γ )B(β, γ )
1
x α−1 (1 − x)β+γ −1 .
=
B(α, β + γ )
fX (x) =
166
Chapter 8
Bivariate Distributions
This shows that X is beta with parameters (α, β + γ ). A similar argument shows that Y is
beta with parameters (β, γ + α).
29. It is straightforward to check that f (x, y) ≥ 0, f is continuous and
∞
−∞
∞
−∞
f (x, y) dx dy = 1.
∂F
Therefore, f is a continuous probability density function. We will show that
does not
∂x
∂F
exist at (0, 0). Similarly, one can show that
does not exist at any point on the y-axis. Note
∂x
that for small x > 0,
F (x, 0) − F (0, 0) = P (X ≤ x , Y ≤ 0) − P (X ≤ 0 , Y ≤ 0)
0 x
f (x, y) dx dy.
= P (0 ≤ X ≤ x , Y ≤ 0) =
−∞
0
Now, from the definition of f (x, y), we must have x < (1/2)ey or, equivalently,
y > ln(2x). Thus, for small x > 0,
x
0
x
(1 − 2xe−y ) dx dy = (x)2 − (x) ln(2x) +
F (x, 0) − F (0, 0) =
.
2
ln(2x) 0
This implies that
F (x, 0) − F (0, 0)
1
= lim x − ln(2x) −
= ∞,
x→0+
x→0+
x
2
lim
showing that
8.2
∂F
does not exist at (0, 0).
∂x
INDEPENDENT RANDOM VARIABLES
1. Note that pX (x) = (1/25)(3x 2 + 5), pY (y) = (1/25)(2y 2 + 5). Now pX (1) = 8/25,
pY (0) = 5/25, and p(1, 0) = 1/25. Since p(1, 0) = pX (1)pY (0), X and Y are dependent.
2. Note that
1
p(1, 1) = ,
7
1 2
3
+ = ,
7 7
7
6
1 5
pY (1) = p(1, 1) + p(2, 1) = + = .
7 7
7
pX (1) = p(1, 1) + p(1, 2) =
Since p(1, 1) = pX (1)pY (1), X and Y are dependent.
Section 8.2
Independent Random Variables
167
3. By the independence of X and Y ,
P (X = 1, Y = 3) = P (X = 1)P (Y = 3) =
12 12
·
2 3 2 3
3
=
4
.
81
P (X + Y = 3) = P (X = 1, Y = 2) + P (X = 2, Y = 1)
4
12 12 2 12 2 12
·
= .
+
·
=
2 3 2 3
2 3
2 3
27
4. No, they are not independent because, for example, P (X = 0 | Y = 8) = 1 but
39
8
P (X = 0) = = 0.08175 = 1,
52
8
showing that P (X = 0 | Y = 8) = P (X = 0).
7 1
2 2
5. The answer is
2 1 5
2
8 1
·
2 2
2 1 6
2
= 0.0179.
6. We have that
P max(X, Y ) ≤ t = P (X ≤ t, Y ≤ t) = P (X ≤ t)P (Y ≤ t) = F (t)G(t).
P min(X, Y ) ≤ t = 1 − P min(X, Y ) > t
= 1 − P (X > t, Y > t) = 1 − P (X > t)P (Y > t)
= 1 − 1 − F (t) 1 − G(t) = F (t) + G(t) − F (t)G(t).
7. Let X and Y be the number of heads obtained by Adam and Andrew, respectively. The desired
probability is
n
n
P (X = i, Y = i) =
i=0
P (X = i)P (Y = i)
i=0
n
n 1 i 1 n−i n 1 i 1
·
i 2
i 2
2
2
i=0
1 2n n n2 1 2n 2n
=
=
,
2
2
i
n
i=0
=
where the last equality follows by Example 2.28.
n−i
168
Chapter 8
Bivariate Distributions
An Intuitive Solution: Let Z be the number of tails obtained by Andrew. The desired probability is
n
n
n
P (X = i, Y = i) =
i=0
P (X = i, Z = i) =
i=0
P (X = i, Y = n − i)
i=0
= P (Adam and Andrew get a total of n heads)
1 2n 2n
= P ( n heads in 2n flips of a fair coin) =
.
2
n
8. For i, j ∈ 0, 1, 2, 3 , the sum of the numbers in the ith row is pX (i) and the sum of the
numbers in the j th row is pY (j ). We have that
pX (0) = 0.41,
pX (1) = 0.44,
pX (2) = 0.14,
pX (3) = 0.01;
pY (0) = 0.41,
pY (1) = 0.44,
pY (2) = 0.14,
pY (3) = 0.01.
Since for all x, y ∈ 0, 1, 2, 3 , p(x, y) = pX (x)pY (y), X and Y are independent.
9. They are not independent because
x
fX (x) =
2 dy = 2x,
0 ≤ x ≤ 1;
0
1
fY (y) =
2 dx = 2(1 − y),
0 ≤ y ≤ 1;
y
and so f (x, y) = fX (x)fY (y).
10. Let X and Y be the amount of cholesterol in the first and in the second sandwiches, respectively.
Since X and Y are continuous random variables, P (X = Y ) = 0 regardless of what the
probability density functions of X and Y are.
11. We have that
fX (x) =
0
fY (y) =
0
∞
∞
x 2 e−x(y+1) dy = xe−x ,
x 2 e−x(y+1) dx =
x ≥ 0;
2
,
(y + 1)3
y ≥ 0,
where the second integral is calculated by applying integration by parts twice. Now since
f (x, y) = fX (x)fY (y), X and Y are not independent.
Section 8.2
Independent Random Variables
169
12. Clearly,
E(XY ) =
1
1
0
1
(xy)(8xy) dy dx =
0
x
1
1
x
4
8y 2 dy x 2 dx = ,
9
1
8
x(8xy) dy dx = ,
E(X) =
15
0
x
1 1
4
y(8xy) dy dx = .
E(Y ) =
5
0
x
So E(XY ) = E(X)E(Y ).
13. Since
f (x, y) = e−x · 2e−2y = fX (x)fY (y),
X and Y are independent exponential random variables with parameters 1 and 2, respectively.
Therefore,
1
E(X 2 Y ) = E(X2 )E(Y ) = 2 · = 1.
2
14. The joint probability density function of X and Y is given by
f (x, y) =
e−(x+y)
0
x > 0, y > 0
elsewhere.
Let G be the probability distribution function, and g be the probability density function of
X/Y . For t > 0,
X
≤ t = P (X ≤ tY )
Y
∞ ty
=
e−(x+y) dx dy =
G(t) = P
0
0
t
.
1+t
Therefore, for t > 0,
g(t) = G (t) =
1
.
(1 + t)2
Note that G (t) = 0 for t < 0; G (0) does not exist.
15. Let F and f be the probability distribution and probability density functions of max(X, Y ),
respectively. Clearly,
F (t) = P max(X, Y ) ≤ t = P (X ≤ t, Y ≤ t) = (1 − e−t )2 ,
Thus
f (t) = F (t) = 2e−t (1 − e−t ) = 2e−t − 2e−2t .
t ≥ 0.
170
Chapter 8
Bivariate Distributions
Hence
E max(X, Y ) = 2
∞
te−t dt −
0
∞
2te−2t dt = 2 −
0
1
3
= .
2
2
∞
te−t dt is the expected value of an exponential random variable with parameter
∞
1, thus it is 1. Also,
2te−2t dt is the expected value of an exponential random variable
Note that
0
0
with parameter 2, thus it is 1/2.
16. Let F and f be the probability distribution and probability density functions of max(X, Y ).
For −1 < t < 1,
t + 1
F (t) = P max(X, Y ) ≤ t = P (X ≤ t, Y ≤ t) = P (X ≤ t)P (Y ≤ t) =
2
Thus
t +1
, −1 < t < 1.
f (t) = F (t) =
2
Therefore,
1
1
t +1
dt = .
t
E(X) =
2
3
−1
2
.
17. Let F and f be the probability distribution and probability density functions of XY , respectively. Clearly, for t ≤ 0, F (t) = 0 and for t ≥ 1, F (t) = 1. For 0 < t < 1,
1 1
F (t) = P (XY ≤ t) = 1 − P (XY > t) = 1 −
dy dx = t − t ln t.
t
Hence
f (t) = F (t) =
− ln t
0
t/x
0 0, y > 0. So the desired probability is
∞ ∞
2 −(2y)/11 1 −x/6
11
dy e
dx = .
e
P (Y > X) =
11
6
23
0
x
21. If IA and IB are independent, then
P (IA = 1, IB = 1) = P (IA = 1)P (IB = 1).
This is equivalent to P (AB) = P (A)P (B) which shows that A and B are independent.
c On
c
thecother
hand, if {A, B} is an independent set, so are the following: A, B , A , B , and
A , B c . Therefore,
P (AB) = P (A)P (B),
P (AB c ) = P (A)P (B c ),
P (Ac B) = P (Ac )P (B),
P (Ac B c ) = P (Ac )P (B c ).
These relations, respectively, imply that
P (IA = 1, IB = 1) = P (IA = 1)P (IB = 1),
P (IA = 1, IB = 0) = P (IA = 1)P (IB = 0),
P (IA = 0, IB = 1) = P (IA = 0)P (IB = 1),
P (IA = 0, IB = 0) = P (IA = 0)P (IB = 0).
These four relations show that IA and IB are independent random variables.
22. The joint probability density function of B and C is
⎧ 2 2
⎪
⎨ 9b c
676
f (b, c) =
⎪
⎩0
1 < b < 3, 1 < c < 3
otherwise.
For X2 +BX+C to have two real roots we must have B 2 −4C > 0, or, equivalently, B 2 > 4C.
Let
E = (b, c) : 1 < b < 3, 1 < c < 3, b2 > 4c ;
172
Chapter 8
Bivariate Distributions
the desired probability is
9b2 c2
db dc =
676
E
3
2
b2 /4
1
9b2 c2
dc db ≈ 0.12.
676
(Draw a figure to verify the region of integration.)
23. Note that
fX (x) =
−∞
fY (y) =
∞
∞
−∞
g(x)h(y) dy = g(x)
∞
h(y) dy,
−∞
g(x)h(y) dx = h(y)
Now
fX (x)fY (y) = g(x)h(y)
−∞
∞
∞
g(x) dx
−∞
∞
h(y)g(x) dy dx
−∞
= f (x, y)
g(x) dx.
−∞
h(y) dy
= f (x, y)
∞
∞
∞
−∞
−∞
∞
−∞
f (x, y) dy dx = f (x, y).
This relation shows that X and Y are independent.
24. Let G and
g be the probability distribution and probability density functions of
max(X, Y ) min(X, Y ). Then G(t) = 0 if t < 1. For t ≥ 1,
G(t) = P
max(X, Y )
min(X, Y )
≤ t = P max(X, Y ) ≤ t min(X, Y )
= P X ≤ t min(X, Y ), Y ≤ t min(X, Y )
X
Y
= P min(X, Y ) ≥ , min(X, Y ) ≥
t
t
X
X
Y
Y
=P X≥ , Y ≥ , X≥ , Y ≥
t
t
t
t
X
Y
X
=P Y ≥ , X≥
=P
≤ Y ≤ tX .
t
t
t
This quantity is the area of the region
(x, y) : 0 < x < 1, 0 < y < 1,
x
≤ y ≤ tx
t
Section 8.2
Independent Random Variables
173
which is equal to (t − 1)/t. Hence
G(t) =
and therefore,
⎧
⎪
⎨0
t <1
t −1
⎪
⎩
t
t ≥ 1,
⎧
⎪
⎨1
2
g(t) = G (t) = t
⎪
⎩0
t ≥1
elsewhere.
25. Let F be the distribution function of X/(X + Y ). Since X/(X + Y ) ∈ (0, 1), we have that
F (t) =
0
1
t <0
t ≥ 1.
For 0 ≤ t < 1,
∞ ∞
X
1−t
2
P
≤t =P Y ≥
X =λ
e−λx e−λy dy dx
X+Y
t
0
[(1−t)x]/t
∞
∞
−λx −[λ(1−t)x]/t
=λ
e e
dx = λ
e−λx/t dt = t.
0
Therefore,
0
⎧
⎪
⎨0
F (t) = t
⎪
⎩
1
t <0
0≤t <1
t ≥ 1.
This shows that X/(X + Y ) is uniform over (0, 1).
26. The fact that if X and Y are both normal with mean 0 and equal variance implies that f (x, y) is
circularly symmetrical is straightforward. We prove the converse; suppose that f is circularly
symmetrical, then there exists a function ϕ so that
)
fX (x)fY (y) = ϕ x 2 + y 2 .
Differentiating this relation with respect to x and using
fY (y) =
yields
)
fX (x)fY (y)
= ϕ x 2 + y 2 /fX (x)
fX (x)
)
x2 + y2
f (x)
= X
.
)
)
xfX (x)
ϕ x2 + y2 x2 + y2
ϕ
174
Chapter 8
Bivariate Distributions
)
Now the right side of this equation
is
a
function
of
x
while
its
left
side
is
a
function
of
x2 + y2.
This implies that fX (x)/ xfX (x) is constant. To prove this, we show that for any given x1
and x2 ,
fX (x1 )
f (x2 )
= X
.
x1 fX (x1 )
x2 fX (x2 )
Let y1 = x2 and y2 = x1 ; then x12 + y12 = x22 + y22 and we have
/
/
ϕ x12 + y12
ϕ x22 + y22
f (x2 )
= /
.
= /
= X
/
/
x1 fX (x1 )
x2 fX (x2 )
ϕ x12 + y12
x12 + y12
ϕ x22 + y22
x22 + y22
fX (x1 )
We have shown that for some constant k,
fX (x)
= k.
xfX (x)
Therefore,
fX (x)
1
= kx and hence ln fX (x) = kx 2 + c, or
fX (x)
2
fX (x) = e(1/2)kx
where α = e . Now since
c
then fX (x) = αe−x
2 /(2σ 2 )
2 +c
2
= αe(1/2)kx ,
)
2
αe(1/2)kx dx = 1, we have that k < 0. Let σ = −1/k;
−∞
∞
√
2
2
and
αe−x /(2σ ) dx = 1 implies that α = 1/(σ 2π ). So
∞
−∞
1
2
2
fX (x) = √ e−x /(2σ ) , showing that X ∼ N(0, σ 2 ). The fact that Y ∼ N(0, σ 2 ) is
σ 2π
proved similarly.
8.3
CONDITIONAL DISTRIBUTIONS
2
1. pY (y) =
p(x, y) =
x=1
pX|Y (x|y) =
1
(2y 2 + 5). Thus
25
p(x, y)
(1/25)(x 2 + y 2 )
x2 + y2
=
=
x = 1, 2, y = 0, 1, 2,
pY (y)
(1/25)(2y 2 + 5)
2y 2 + 5
P (X = 2 | Y = 1) = pX|Y (2|1) = 5/7,
2
E(X|Y = 1) =
2
xpX|Y (x|1) =
x=1
x
x=1
x2 + 1
12
=
.
7
7
Section 8.3
2. Since
Conditional Distributions
175
y
fY (y) =
2 dx = 2y, 0 < y < 1,
0
we have that
2
1
f (x, y)
=
= , 0 < x < y, 0 < y < 1.
fY (y)
2y
y
fX|Y (x|y) =
3. Let X be the number of flips of the coin until the sixth head is obtained. Let Y be the number
of flips of the coin until the third head is obtained. Let Z be the number of additional flips
of the coin after the third head occurs until the sixth head occurs; Z is a negative binomial
random variable with parameters 3 and 1/2. By the independence of the trials,
x − 6 1 3 1 x−8
pX|Y (x|5) = P (Z = x − 5) =
2
2
2
x − 6 1 x−5
=
, x = 8, 9, 10, . . . .
2
2
4. Note that
3
3 x 2 + (9/16)
1
fX|Y x
=
= (48x 2 + 27).
4
(27/16) + 1
43
Therefore,
P
1
4
1
3
=
Y =
2
4
|y|,
|y|
fY |X (y|x) =
(1/2)e−x
x
(1/2)e−x dy
=
1
, −x < y < x.
2x
−x
(c) By part (b), given X = x, Y is a uniform random variable over (−x, x). Therefore,
E(Y |X = x) = 0 and
2
x − (−x)
x2
Var(Y |X = x) =
= .
3
12
11. Let f (x, y) be the joint probability density function of X and Y . Since
fX|Y (x|y) =
⎧
⎪
⎨
3
1
=
2y
20 + (2y)/3 − 20
⎪
⎩
0
20 < x < 20 +
otherwise,
and
fY (y) =
⎧
⎨1/30
0 < y < 30
⎩0
elsewhere,
2y
3
178
Chapter 8
Bivariate Distributions
we have that
⎧
1
⎪
⎨
f (x, y) = fX|Y (x|y)fY (y) = 20y
⎪
⎩0
20 < x < 20 +
2y
, 0 < y < 30
3
elsewhere.
12. Let X be the first arrival time. Clearly,
0
P X ≤ x | N(t) = 1 =
1
if x < 0
if x ≥ t.
For 0 ≤ x < t,
P X ≤ x, N (t) = 1
P N(x) = 1, N (t − x) = 0
P X ≤ x | N(t) = 1 =
=
P N(t) = 1
P N(t) = 1
0
e−λx (λx)1 e−λ(t−x) λ(t − x)
·
P N(x) = 1 P N(t − x) = 0
x
1!
0!
=
= ,
=
−λt
1
t
P N(t) = 1
e (λt)
1!
where the third equality follows from the independence of the random variables N(x) and
N(t − x) (recall that Poisson processes possess independent increments). We have shown that
⎧
⎪
0
if x < 0
⎪
⎪
⎨
P X ≤ x | N (t) = 1 = x/t if 0 ≤ x < t
⎪
⎪
⎪
⎩
1
if x ≥ t.
This shows that the conditional distribution of X given N(t) = 1 is uniform on (0, 1).
13. For x ≤ y, the fact that the conditional distribution of X given Y = y is hypergeometric
follows from the following:
P (X = x | Y = y) =
P (X = x, Y = y)
P (X = x)P (Y − X = y − x)
=
P (Y = y)
P (Y = y)
n − m y−x
m n−m
m x
m−x
(n−m)−(y−x)
·
p (1 − p)
p (1 − p)
y−x
x
y−x
x
=
.
=
n y
n
p (1 − p)n−y
y
y
Section 8.3
Conditional Distributions
179
It must be clear that the conditional distribution of Y given that X = x is binomial with
parameters n − m and p. That is,
n − m y−x
P (Y = y | X = x) =
p (1 − p)n−m−y+x ,
y−x
y = x, x + 1, . . . , n − m + x.
14. Let f (x, y) be the joint probability density function of X and Y . By the solution to Exercise 25,
Section 8.1,
f (x, y) =
⎧
⎨1/2
|x| + |y| ≤ 1
⎩0
elsewhere,
and
fY (y) = 1 − |y|, −1 ≤ y ≤ 1.
Hence
1
1/2
, −1 + |y| ≤ x ≤ 1 − |y|, −1 ≤ y ≤ 1.
=
1 − |y|
2 1 − |y|
fX|Y (x|y) =
15. Let λ be the parameter of N(t) : t ≥ 0 . The fact that for s < t, the conditional distribution of
N(s) given N(t) = n is binomial with parameters n and p = s/t, follows from the following
relations for i ≤ n.
P N(s) = i, N (t) = n
P N(s) = i | N(t) = n =
P N (t) = n
P N (s) = i P N (t) − N (s) = n − i
P N(s) = i, N(t) − N (s) = n − i
=
=
P N(t) = n
P N(t) = n
P N(s) = i P N(t − s) = n − i
=
=
P N(t) = n
=
n s
i
t
i
1−
s
t
n−i
e−λs (λs)i e−λ(t−s) λ(t − s)
·
i!
(n − i)!
e−λt (λt)n
n!
n−i
,
where the third equality follows since Poisson processes possess independent increments and
the fourth equality follows since Poisson processes are stationary.
180
Chapter 8
Bivariate Distributions
For i ≥ k,
P N(t) = i | N(s) = k = P N(t) − N(s) = i − k | N(s) = k
= P N(t) − N (s) = i − k = P N (t − s) = i − k
i−k
e−λ(t−s) λ(t − s)
=
(i − k)!
shows that the conditional distribution of N(t) given N (s) = k is Poisson with parameter
λ(t − s).
16. Let p(x, y) be the joint probability mass function of X and Y . Clearly,
12
pY (5) =
and
⎧
11
⎪
⎪
⎪
⎪
⎪
⎨ 13
p(x, 5) = 0
⎪
⎪
⎪ 11
⎪
⎪
⎩
13
x−1
4
4
13
1 12
13 13
1
,
13
4−x
x−6
1 12
13 13
1
13
x<5
x=5
1
13
x > 5.
Using these, we have that
∞
∞
xpX|Y (x|5) =
E(X | Y = 5) =
x=1
4
=
x=1
x
x=1
1 11
x
11 12
= 0.72932 +
x
∞
+
x
x=6
11
4
12
11
4
1
13
1
12
13
11 4 1
= 0.702932 +
12
13
= 0.72932 +
p(x, 5)
pY (5)
11
4
12
1 12
13 13
12
∞
(y + 6)
y=0
∞
y
y=0
12
13
x−6
y
13
y
∞
+6
y=0
12
y
13
12/13
1
= 13.412.
+
6
(1/13)2
1 − (12/13)
Remark: In successive draws of cards from an ordinary deck of 52 cards, one at a time,
randomly, and with replacement, the expected value of the number of draws until the first ace
is 1/(1/13) = 13. This exercise shows that knowing the first king occurred on the fifth trial
will increase, on the average, the number of trials until the first ace 0.412 draws.
Section 8.3
Conditional Distributions
181
17. Let X be the number of blue chips in the first 9 draws and Y be the number of blue chips drawn
altogether. We have that
9
E(X | Y = 10) =
x
x=0
p(x, 10)
pY (10)
9 12
9
=
x
x
22
x=1
9
12
·
10 − x 22
22
18 12 10 10 8
10 22
22
x 10 9−x
9
9
9
x 10 − x
=
x
18
x=1
10
=
10−x 10 x−1
22
9 × 10
= 5,
18
where the last sum is (9 × 10)/18 because it is the expected value of a hypergeometric random
variable with N = 18, D = 9, and n = 10.
18. Clearly,
1
fX (x) =
n(n − 1)(y − x)n−2 dy = n(1 − x)n−1 .
x
Thus
f (x, y)
n(n − 1)(y − x)n−2
(n − 1)(y − x)n−2
=
=
.
fX (x)
n(1 − x)n−1
(1 − x)n−1
fY |X (y|x) =
Therefore,
E(Y | X = x) =
1
y
x
(n − 1)(y − x)n−2
n−1
dy =
n−1
(1 − x)
(1 − x)n−1
1
y(y − x)n−2 dy.
x
But
1
y(y − x)
n−2
dy =
x
1
(y − x + x)(y − x)n−2 dy
x
=
x
=
Thus
E(Y | X = x) =
1
(y − x)n−1 dy +
1
x(y − x)n−2 dy
x
(1 − x)n x(1 − x)n−1
+
.
n
n−1
n−1
n−1 1
(1 − x) + x =
+ x.
n
n
n
182
Chapter 8
Bivariate Distributions
19. (a) The area of the triangle is 1/2. So
f (x, y) =
1−y
(b) fY (y) =
2
0
if x ≥ 0, y ≥ 0, x + y ≤ 1
elsewhere.
2 dx = 2(1 − y), 0 < y < 1. Therefore,
0
fX|Y (x|y) =
1
2
=
, 0 ≤ x ≤ 1 − y, 0 ≤ y < 1.
2(1 − y)
1−y
(c) By part (b), given that Y = y, X is a uniform random variable over (0, 1 − y). Thus
E(X | Y = y) = (1 − y)/2, 0 < y < 1.
20. Clearly,
x
pX (x) =
y=0
1
1
= 2
2
e y! (x − y)!
e x!
x
y=0
x!
e−2
=
y! (x − y)!
x!
x
e−2 · 2x
=
,
y
x!
y=0
x
x
is the number of subsets of a set with x
where the last equality follows since y=0
y
elements and hence is equal to 2x . Therefore, pX (x) is Poisson with parameter 2 and so
x
p(x, y)
x −x
pY |X (y|x) =
=
2 .
pX (x)
y
This yields
x −x
E(Y | X = x) =
y
2 =
y
y=0
x
x 1
y
y 2
y=0
x
y 1 x−y
2
=
x
,
2
where the last equality follows because the last sum is the expected value of a binomial random
variable with parameters x and 1/2.
21. Let X be the lifetime of the dead battery. We want to calculate E(X | X < s). Since X is a
continuous random variable, this is the same as E(X | X ≤ s). To find this quantity, let
FX|X≤s (t) = P (X ≤ t | X ≤ s),
and fX|X≤s (t) = FX|X≤s
(t). Then
E(X | X ≤ s) =
∞
tfX|X≤s (t) dt.
0
Section 8.4 Transformations of Two Random Variables
Now
P (X ≤ t, X ≤ s)
FX|X≤s (t) = P (X ≤ t | X ≤ s) =
P (X ≤ s)
⎧
P
(X
≤
t)
⎨
if t < s
= P (X ≤ s)
⎩
1
if t ≥ s.
Differentiating FX|X≤s (t) with respect to t, we obtain
⎧
⎨ f (t) if t < s
fX|X≤s (t) = F (s)
⎩
0
otherwise.
This yields
E(X | X ≤ s) =
8.4
1
F (s)
s
tf (t) dt.
0
TRANSFORMATIONS OF TWO RANDOM VARIABLES
1. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) =
1
0
0 < x < 1, 0 < y < 1
elswhere.
The system of two equations in two unknowns
−2 ln x = u
−2 ln y = v
defines a one-to-one transformation of
R = (x, y) : 0 < x < 1, 0 < y < 1
onto the region
Q = (u, v) : u > 0, v > 0 .
It has the unique solution x = e−u/2 , y = e−v/2 . Hence
1 −u/2
− e
0
1
2
J=
= e−(u+v)/2 = 0.
4
1
−v/2
0
−
e
2
By Theorem 8.8, g(u, v), the joint probability density function of U and V is
1
1
g(u, v) = f e−u/2 , e−v/2 e−(u+v)/2 = e−(u+v)/2 , u > 0, v > 0.
4
4
183
184
Chapter 8
Bivariate Distributions
2. Let f (x, y) be the joint probability density function of X and Y . Clearly,
f (x, y) = f1 (x)f2 (y),
x > 0, y > 0.
Let V = X and g(u, v) be the joint probability density functions of U and V . The probability
density function of U is gU (u), its marginal density function. The system of two equations in
two unknowns
x/y = u
x=v
defines a one-to-one transformation of
R = (x, y) : x > 0, y > 0
onto the region
Q = (u, v) : u > 0, v > 0 .
It has the unique solution x = v, y = v/u. Hence
0
1
v
J=
= 2 = 0.
v 1 u
− 2
u
u
By Theorem 8.8,
v
v v
v v
v
= 2 f1 (v)f2
g(u, v) = f v, 2 = 2 f v,
u u
u
u
u
u
Therefore,
gU (u) =
0
∞
v
v
f
(v)f
dv,
1
2
u2
u
u > 0, v > 0.
u > 0.
3. Let g(r, θ ) be the joint probability density function of R and . We will show that g(r, θ ) =
gR (r)g (θ). This proves the surprising result that R and are independent. Let f (x, y) be
the joint probability density function of X and Y . Clearly,
f (x, y) =
1 −(x 2 +y 2 )/2
,
e
2π
−∞ < x < ∞, −∞ < y < ∞.
Let R be the entire xy-plane excluded the set of points on the x-axis with x ≥ 0. This causes
no problems since
P (Y = 0, X ≥ 0) = P (Y = 0)P (X ≥ 0) = 0.
The system of two equations in two unknowns
⎧)
⎨ x2 + y2 = r
⎩ arctan y = θ
x
Section 8.4 Transformations of Two Random Variables
185
defines a one-to-one transformation of R onto the region
Q = (r, θ ) : r > 0, 0 < θ < 2π .
It has the unique solution
x = r cos θ
y = r sin θ.
cos θ
J =
sin θ
Hence
−r sin θ
= r = 0.
r cos θ
By Therorem 8.8, g(r, θ ) is given by
g(r, θ ) = f (r cos θ, r sin θ)|r| =
Now
2π
gR (r) =
0
and
1 −r 2 /2
re
2π
0 < θ < 2π, r > 0.
1 −r 2 /2
2
dθ = re−r /2 ,
re
2π
r > 0,
∞
1
1 −r 2 /2
re
, 0 < θ < 2π.
dr =
2π
2π
0
Therefore, g(r, θ ) = gR (r)g (θ ), showing that R and are independent random variables.
The formula for g (θ) indicates that is a uniform random variable over the interval (0, 2π ).
The probability density function obtained for R is called Rayleigh.
g (θ) =
4. Method 1: By the convolution theorem (Theorem 8.9), g, the probability density function of
the sum of X and Y , the two random points selected from (0, 1) is given by
∞
g(t) =
f1 (x)f2 (t − x) dx,
−∞
where f1 and f2 are, respectively, the probability density functions of X and Y . Since
f1 (x) = f2 (x) =
1
0
x ∈ (0, 1)
elsewhere,
the integrand, f1 (x)f2 (t − x) is nonzero if 0 < x < 1 and t − 1 < x < t. This shows that for
t < 0 and t ≥ 2, g(t) = 0. For 0 ≤ t < 1, t − 1 < 0; thus
t
g(t) =
dx = t.
0
For 1 ≤ t < 2, 0 < t − 1 < 1; therefore,
1
dx = 1 − (t − 1) = 2 − t.
g(t) =
t−1
186
Chapter 8
Bivariate Distributions
So
⎧
⎪
⎨t
g(t) = 2 − t
⎪
⎩
0
if 0 ≤ t < 1
if 1 ≤ t < 2
otherwise.
Method 2: Note that the sample space of the experiment of choosing two random numbers
from (0, 1) is
S = (x, y) ∈ R2 : 0 < x < 1, 0 < y < 1 .
So, for 0 ≤ t < 1, P (X + Y ≤ t) is the area of the region
(x, y) ∈ S : 0 < x ≤ t, 0 < y ≤ t, x + y ≤ t
divided by the area of S: t 2 /2. For 1 ≤ t < 2, P (X + Y ≤ t) is the area of
S − (x, y) ∈ S : t − 1 ≤ x < 1, t − 1 ≤ y < 1, x + y > t
(2 − t)2
divided by the area of S: 1 −
. (Draw figures to verify these regions.) Let G be the
2
probability distribution function of X + Y . We have shown that
⎧
⎪
0
t <0
⎪
⎪
⎪
⎪
⎪
2
⎪
t
⎪
⎪
0≤t <1
⎨
2
G(t) =
⎪
(2 − t)2
⎪
⎪
1≤t <2
1−
⎪
⎪
⎪
2
⎪
⎪
⎪
⎩1
t ≥ 2.
Therefore,
⎧
⎪
⎨t
g(t) = G (t) = 2 − t
⎪
⎩
0
0≤t <1
1≤t <2
otherwise.
5. (a) Clearly, pX (x) = 1/3 for x = −1, 0, 1 and pY (y) = 1/3 for y = −1, 0, 1. Since
⎧
⎪
1/9 z = −2, +2
⎪
⎪
⎨
P (X + Y = z) = 2/9 z = −1, +1
⎪
⎪
⎪
⎩3/9 z = 0,
the relation
pX (x)pY (z − x)
P (X + Y = z) =
x
is easily seen to be true.
Section 8.4 Transformations of Two Random Variables
187
(b) p(x, y) = pX (x)pY (y) for all possible values x and y of X and Y if and only if (1/9)+c =
1/9 and (1/9) − c = 1/9; that is, if and only if c = 0.
6. Let h(x, y) be the joint probability density function of X and Y . Then
⎧ 1
⎪
⎪
⎨ x2y2
h(x, y) =
⎪
⎪
⎩
0
x ≥ 1, y ≥ 1
elsewhere.
Consider the system of two equations in two unknowns
x/y = u
(29)
xy = v.
This system has the unique solution
√
uv
√
y = v/u.
x=
We have that
(30)
x ≥ 1 ⇐⇒
√
uv ≥ 1
y ≥ 1 ⇐⇒
√
v/u ≥ 1 ⇐⇒
⇐⇒ u ≥
1
,
v
v ≥ u.
1
Clearly, x ≥ 1, y ≥ 1 imply that v = xy ≥ 1, so > 0. Therefore, the system of equations
v
(29) defines a one-to-one transformation of
R = (x, y) : x ≥ 1, y ≥ 1
onto the region
By (30),
1
Q = (u, v) : 0 < ≤ u ≤ v .
v
(
1 v
2 u
J = √
v
− √
2u u
(
u
v
= 1 = 0.
1 2u
√
2 uv
1
2
Hence, by Theorem 8.8, g(u, v), the joint probability density function of U and V is given by
(
√
v
1
1
g(u, v) = h
uv,
, 0 < ≤ u ≤ v.
|J| =
2
u
2uv
v
188
Chapter 8
Bivariate Distributions
7. Let h be the joint probability density function of X and Y . Clearly,
h(x, y) =
e−(x+y)
x > 0, y > 0
0
elsewhere.
Consider the system of two equations in two unknowns
x+y =u
(31)
ex = v.
This system has the unique solution
x = ln v
(32)
y = u − ln v.
We have that
x > 0 ⇐⇒
ln v > 0
⇐⇒
v > 1,
y > 0 ⇐⇒ u − ln v > 0 ⇐⇒ eu > v.
Therefore, the system of equations (31) defines a one-to-one transformation of
R = (x, y) : x > 0, y > 0
onto the region
Q = (u, v) : u > 0, 1 < v < eu .
By (32),
1
0
v
1
J=
= − = 0.
v
1 − 1
v
Hence, by Theorem 8.8, g(u, v), the joint probability density function of U and V is given by
g(u, v) = h(ln v, u − ln v)|J| =
1 −u
e ,
v
u > 0, 1 < v < eu .
8. Let U = X + Y and V = X − Y . Let g(u, v) be the joint probability density function of U
and V . We will show that g(u, v) = gU (u)gV (v). To do so, let f (x, y) be the joint probability
density function of X and Y . Then
f (x, y) =
1 −(x 2 +y 2 )/2
e
,
2π
−∞ < x < ∞, −∞ < y < ∞.
The system of two equations in two unkowns
x+y =u
x−y =v
Section 8.4 Transformations of Two Random Variables
189
defines a one-to-one correspondence from the entire xy-plane onto the entire uv-plane. It has
the unique solution
⎧
u+v
⎪
⎨x =
2
u
−
v
⎪
⎩y =
.
2
Hence
1/2 1/2
= − 1 = 0.
J =
2
1/2 −1/2
By Theorem 8.8,
u + v u − v
,
|J|
g(u, v) = f
2
2
⎡
u+v 2
=
⎢
1
exp ⎢
⎣−
4π
2
+
u − v
2
2
2
⎤
⎥
⎥ = 1 e−(u2 +v2 )/4 ,
⎦ 4π
−∞ < u, v < ∞.
This gives
∞
1
1 −u2 /4 ∞ −v2 /4
−(u2 +v 2 )/4
gU (u) =
e
dv =
e
e
dv
4π −∞
4π
−∞
∞
1
1
1
2
2
2
= √ e−u /4
√ e−v /4 dv = √ e−u /4 , −∞ < u < ∞,
2 π
2
π
2
π
−∞
1
2
√ e−v /4 is the probability density function of
2 π
a normal random variable with mean 0 and variance 2. Thus its integral over the interval
(−∞, ∞) is 1. Similarly,
where the last equality follows because
1
2
gV (v) = √ e−v /2 ,
2 π
−∞ < v < ∞.
Since g(u, v) = gU (u)gV (v), U and V are independent normal random variables each with
mean 0 and variance 2.
9. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) =
λr1 +r2 x r1 −1 y r2 −1 e−λ(x+y)
,
(r1 ) (r2 )
Consider the system of two equations in two unknowns
⎧
⎨x +y = u
x
⎩
= v.
x+y
x > 0, y > 0.
(33)
190
Chapter 8
Bivariate Distributions
Clearly, (33) implies that u > 0 and v > 0. This system has the unique solution
x = uv
(34)
y = u − uv.
We have that
x > 0 ⇐⇒
uv > 0
⇐⇒ u > 0 and v > 0,
y > 0 ⇐⇒ u − uv > 0 ⇐⇒
v < 1.
Therefore, the system of equations (33) defines a one-to-one transformation of
R = (x, y) : x > 0, y > 0
onto the region
By (34),
Q = (u, v) : u > 0, 0 < v < 1 .
v
u
= −u = 0.
J =
1 − v −u
Hence by Thereom 8.8, the joint probability density function of U and V is given by
g(u, v) = f (uv, u − uv)|J| =
λr1 +r2 ur1 +r2 −1 e−λu v r1 −1 (1 − v)r2 −1
(r1 ) (r2 )
u > 0, 0 < v < 1.
Note that
g(u, v) =
=
λe−λu (λu)r1 +r2 −1
·
(r1 + r2 )
(r1 + r2 ) r1 −1
(1 − v)r2 −1
v
(r1 ) (r2 )
λe−λu (λu)r1 +r2 −1
1
·
v r1 −1 (1 − v)r2 −1 ,
(r1 + r2 )
B(r1 , r2 )
u > 0, 0 < v < 1.
This shows that
g(u, v) = gU (u)gV (v).
That is, U and V are independent. Furthermore, it shows that gU (u) is the probability density
function of a gamma random variable with parameter r1 + r2 and λ; gV (v) is the probability
density function of a beta random variable with parameters r1 and r2 .
10. Let f be the joint probability density function of X and Y . Clearly,
f (x, y) = λ2 e−λ(x+y) ,
x > 0, y > 0.
The system of two equations in two unknowns
x+y =u
x/y = v
Chapter 8
Review Problems
191
defines a one-to-one transformation of
R = (x, y) : x > 0, y > 0
onto the region
Q = (u, v) : u > 0, v > 0 .
It has he unique solution x = uv/(1 + v), y = u/(1 + v). Hence
v
u
2
1 + v
(1 + v)
u
J=
= 0.
=−
1
(1 + v)2
u
1 + v − (1 + v)2
By Theorem 8.8, g(u, v), the joint probability density function of U and V is
uv
u
λ2 u
g(u, v) = f
e−λu , u > 0, v > 0.
,
|J| =
1+v 1+v
(1 + v)2
This shows that g(u, v) = gU (u)gV (v), where
gU (u) = λ2 ue−λu ,
u > 0,
and
1
, v > 0.
(1 + v)2
Therefore, U = X + Y and V = X/Y are independent random variables.
gV (v) =
REVIEW PROBLEMS FOR CHAPTER 8
1. (a) We have that
P (XY ≤ 6) = p(1, 2) + p(1, 4) + p(1, 6) + p(2, 2) + p(3, 2)
= 0.05 + 0.14 + 0.10 + 0.25 + 0.15 = 0.69.
(b) First we calculate pX (x) and pY (y), the marginal probability mass functions of X and Y .
They are given by the following table.
x
y
1
2
3
pY (y)
2
4
6
0.05
0.14
0.10
0.25
0.10
0.02
0.15
0.17
0.02
0.45
0.41
0.14
pX (x)
0.29
0.37
0.34
192
Chapter 8
Bivariate Distributions
Therefore,
E(X) = 1(0.29) + 2(0.37) + 3(0.34) = 2.05;
E(Y ) = 2(0.45) + 4(0.41) + 6(0.14) = 3.38.
2. (a) and (b) p(x, y), the joint probability mass function of X and Y , and pX (x) and pY (y), the
marginal probability mass functions of X and Y are given by the following table.
y
x
2
3
4
5
6
7
8
9
10
11
12
pY (y)
(c) E(X) =
15
x=2
1
1/36
0
0
0
0
0
0
0
0
0
0
1/36
xpX (x) = 7;
2
0
2/36
1/36
0
0
0
0
0
0
0
0
3/36
3
4
5
6
pX (x)
0
0
0
0
1/36
0
0
0
0
2/36
2/36
0
0
0
3/36
2/36 2/36
0
0
4/36
1/36 2/36 2/36
0
5/36
0
2/36 2/36 2/36
6/36
0
1/36 2/36 2/36
5/36
0
0
2/36 2/36
4/36
0
0
1/36 2/36
3/36
0
0
0
2/36
2/36
0
0
0
1/36
1/36
5/36 7/36 9/36 11/36
E(Y ) = 6y=1 ypY (y) = 161/36 ≈ 4.47.
3. Let X be the number of spades and Y be the number of hearts in the random bridge hand. The
desired probability mass function is
13 13
26
x
4
9−x
52
13
26
13
p(x, 4)
x
9−x
pX|Y (x|4) =
, 0 ≤ x ≤ 9.
=
=
39
pY (4)
13 39
9
4
9
52
13
4. The set of possible values of X and Y , both, is 0, 1, 2, 3 . Let p(x, y) be their joint probability
mass function; then
13 13
26
x
y
3−x−y
p(x, y) =
, 0 ≤ x, y, x + y ≤ 3.
52
3
Chapter 8
13
13
x
6−x
5. Reducing the sample space, the answer is
,
26
6
2
6. (a)
0
x
0
Review Problems
193
0 ≤ x ≤ 6.
c
dy dx = 1 ⇒ c = 1/2.
x
x
(b) fX (x) =
0
fY (y) =
y
2
1
1
dy = ,
2x
2
0 0, y > 0.
∂x ∂y
Therefore, by symmetry,
P (X > 2Y ) + P (Y > 2X) = 2P (X > 2Y ) = 2
∞
0
12. We have that
1−x
fX (x) =
0
∞
2y
2
2
2
4xye−x e−y dx dy = .
5
3
3
3(x + y) dy = − x 2 + , 0 < x < 1,
2
2
By symmetry,
3
3
fY (y) = − y 2 + , 0 < y < 1.
2
2
Therefore,
P (X + Y > 1/2) =
1/2
0
=
1−x
3(x + y) dy dx +
1
1/2
(1/2)−x
1−x
3(x + y) dy dx
0
9
5
29
+
= .
64 16
64
13. Since
fX|Y (x|y) =
f (x, y)
e−y
= 1,
= *1
−y dx
fY (y)
e
0
we have that
E(X | Y = y) =
n
0
1
x n · 1 dx =
0 < x < 1, y > 0,
1
,
n+1
n ≥ 1.
Chapter 8
Review Problems
14. Let p(x, y) be the joint probability mass function of X and Y . We have that
10 1
p(x, y) =
x
4
10
=
x
1
15.
0
1
x 3 10−x
4
15 1
y
4
15 1
·
4
y
x+y 3 25−x−y
4
y 3 15−y
4
, 0 ≤ x ≤ 10, 0 ≤ y ≤ 15.
cx(1 − x) dy dx = 1 ⇒ c = 12. Clearly,
x
1
fX (x) =
12x(1 − x) dy = 12x(1 − x)2 ,
0 < x < 1,
x
y
fY (y) =
12x(1 − x) dx = 6y 2 − 4y 3 ,
0 < y < 1.
0
Since f (x, y) = fX (x)fY (y), X and Y are not independent.
16. The area of the region bounded by y = x 2 − 1 and y = 1 − x 2 is
1
−1
1−x 2
x 2 −1
8
dy dx = .
3
Therefore f (x, y), the joint probability density function of X and Y is given by
f (x, y) =
Clearly,
3/8
x 2 − 1 < y < 1 − x 2 , −1 < x < 1
0
elsewhere.
fX (x) =
1−x 2
x 2 −1
3
3
dy = (1 − x 2 ), −1 < x < 1.
8
4
To find fY (y), note that for −1 < y < 0,
fY (y) =
and, for 0 ≤ y < 1,
1+y
√
− 1+y
fY (y) =
√
√
1−y
√
− 1−y
3
3)
1+y
dx =
8
4
3
3)
dx =
1 − y.
8
4
195
196
Chapter 8
Bivariate Distributions
⎧ )
3
⎪
⎪
1+y
⎪
⎪
4
⎪
⎪
⎨
fY (y) = 3 )
1−y
⎪
⎪
⎪4
⎪
⎪
⎪
⎩0
So
−1 < y < 0
0≤y<1
otherwise.
Since f (x, y) = fX (x)fY (y), X and Y are not independent.
17. Let f (x, y) be the joint probability density function of X and Y , G be the probability distribution function of X/Y , and g be the probability density function of X/Y . We have that
f (x, y) =
1/2
0 < x < 1, 0 < y < 2
0
otherwise.
Clearly, P (X/Y ≤ t) = 0 if t < 0. For 0 ≤ t < 1/2,
P
For t ≥ 1/2,
P
X
Y
X
Y
≤t =
2
0
≥t =
0
1
ty
0
2
x/t
1
dx dy = t.
2
1
1
dy dx = 1 − .
2
4t
(Draw appropriate figures to verify the limits of these integrals.) Therefore,
⎧
⎪
0
t <0
⎪
⎪
⎪
⎪
⎨
1
0≤t <
G(t) = t
2
⎪
⎪
⎪
⎪
⎪
⎩1 − 1 t ≥ 1 .
4t
2
This gives
⎧
⎪
0
⎪
⎪
⎪
⎪
⎨
g(t) = G (t) = 1
⎪
⎪
⎪
⎪
⎪
⎩ 1
4t 2
t <0
0≤t <
t≥
1
2
1
.
2
18. No, because G(∞, ∞) = F (∞) + F (∞) = 2 = 1.
19. The problem is equivalent to the following: Two points X and Y are selected independently
and at random from the interval (0, ). What is the probability that the length of at least one
Chapter 8
Review Problems
197
interval is less than /20? The solution to this problem is as follows:
P min(X, Y − X, − Y ) <
X < Y P (X < Y )
20
+ P min(Y, X − Y, − X) <
X > Y P (X > Y )
20
= 2P min(X, Y − X, − Y ) <
X < Y P (X < Y )
20
1
= 2P min(X, Y − X, − Y ) <
X x ;
that is,
17 17
×
2
20
20
= 0.7225.
÷
2
2
Therefore, the desired probability is 1 − 0.7225 = 0.2775.
20. Let p(x, y) be the joint probability mass function of X and Y .
p(x, y) = P (X = x, Y = y) = (0.90)x−1 (0.10)(0.90)y−1 (0.10) = (0.90)x+y−2 (0.10)2 .
21. We have that
fX (x) =
x
−x
dy = 2x, 0 < x < 1,
⎧ 1
⎪
⎪
⎪
dx = 1 + y
⎪
⎨ −y
fY (y) =
1
⎪
⎪
⎪
⎪
dx = 1 − y
⎩
y
−1 < y < 0
0 < y < 1,
198
Chapter 8
Bivariate Distributions
fX|Y (x|y) =
⎧
1
⎪
⎪
⎪
⎨1 + y
−y < x < 1, −1 < y < 0
1
⎪
⎪
⎪
⎩1 − y
y < x < 1, 0 < y < 1,
and
fY |X (y|x) =
Thus
E(Y | X = x) =
and
1
, −x < y < x.
2x
x
−x
y
dy = 0 = 0 · x + 0,
2x
⎧ 1
x
1−y
⎪
⎪
dx =
, −1 < y < 0
⎪
⎪
⎨ −y 1 + y
2
E(X | Y = y) = 1
1+y
x
⎪
⎪
⎪
dx =
, 0 < y < 1.
⎪
⎩ y 1−y
2
22. We present the solution given by Merryfield, Viet, and Watson, in the August–September 1997
issue of the American Mathematical Monthly. Let f be the joint probability density function
of X and Y .
b
b
E(WA ) =
WA (x, y)f (x, y) dxdy,
a
a
b
E(WB ) =
b
WB (x, y)f (x, y) dxdy.
a
a
Let U = Y , V = X, h1 (x, y) = y and h2 (x, y) = x. Then the system of equations
y=u
x=v
has the unique solution x = v, y = u, and
0 1
= −1 = 0.
J=
1 0
Applying the change of variables formula for multiple intergrals, we obtain
b
b
E(WA ) =
b
WA (x, y)f (x, y) dxdy =
a
a
b
=
WA (v, u)f (v, u) dudv.
a
a
WA (v, u)f (v, u)|J| dudv
a
b
b
a
Chapter 8
Review Problems
199
Since the distribution of the money in each player’s wallet is the same, the joint distributions
of (X, Y ) and (Y, X) have the same probability density function f satisfying f (x, y) =
f (y, x). Observing that WA (Y, X) = WB (X, Y ), we have that WA (v, u) = WB (u, v). This
and f (v, u) = f (u, v) imply that
b
E(WA ) =
b
WB (u, v)f (u, v) dudv = E(WB ).
a
a
On the other hand, WA (X, Y ) = −WB (X, Y ) implies that E(WA ) = −E(WB ). Thus
E(WA ) = −E(WA ), implying that E(WA ) = E(WB ) = 0.
Chapter 9
Multivariate Distributions
9.1
JOINT DISTRIBUTIONS OF n > 2 RANDOM VARIABLES
1. Let p(h, d, c, s) be the joint probability mass function of the number of hearts, diamonds,
clubs, and spades selected. We have
13 13 13 13
h
d
c
s
, h + d + c + s = 13, 0 ≤ h, d, c, s ≤ 13.
p(h, d, c, s) =
52
13
2. Let p(a, h, n, w) be the joint probability mass function of A, H , N , and W . Clearly,
8 7 3 20
a h n w
p(a, h, n, w) =
,
38
12
a + h + n + w = 12, 0 ≤ a ≤ 8, 0 ≤ h ≤ 7, 0 ≤ n ≤ 3, 0 ≤ w ≤ 12.
The marginal probability mass function of A is given by
8
30
a 12 − a
, 0 ≤ a ≤ 8.
pA (a) =
38
12
3. (a) The desired joint marginal probability mass functions are given by
2
pX,Y (x, y) =
z=1
5
pY,Z (y, z) =
x=4
3
pX,Z (x, z) =
y=1
xy
xyz
=
, x = 4, 5, y = 1, 2, 3.
162
54
yz
xyz
=
, y = 1, 2, 3, z = 1, 2.
162
18
xyz
xz
=
, x = 4, 5, z = 1, 2.
162
27
Section 9.1
3
2
3
(b) E(Y Z) =
Joint Distributions of n > 2 Random Variables
2
yzpY,Z (y, z) =
y=1 z=1
y=1 z=1
201
(yz)2
35
= .
18
9
4. (a) The desired marginal joint probability mass functions are given by
fX,Y (x, y) =
∞
6e−x−y−z dz = 6e−x−2y , 0 < x < y < ∞.
y
z
fX,Z (x, z) =
6e−x−y−z dy = 6e−x−z (e−x − e−z ), 0 < x < z < ∞.
x
fY,Z (y, z) =
y
6e−x−y−z dx = 6e−y−z (1 − e−y ), 0 < y < z < ∞.
0
(b) E(X) =
1/3.
0
∞
∞
xfX,Y (x, y) dy dx =
∞
0
x
∞
6xe−x−2y dy dx =
∞
3xe−3x dx =
0
x
5. They are not independent because P (X1 = 1, X2 = 1, X3 = 0) = 1/4, whereas
P (X1 = 1)P (X2 = 1)P (X3 = 0) = 1/8.
6. Note that
∞
∞
x 2 e−x(1+y+z) dy dz
∞
∞
2 −x
−xz
−xy
e
e
dy dz = e−x , x > 0,
=x e
fX (x) =
0
0
0
fY (y) =
∞
0
∞
2 −x(1+y+z)
x e
0
0
dz dx =
1
, y > 0,
(1 + y)2
and similarly,
fZ (z) =
Also
fX,Y (x, y) =
∞
1
, z > 0.
(1 + z)2
x 2 e−x(1+y+z) dz = xe−x(1+y) , y > 0.
0
Since
f (x, y, z) = fX (x)fY (y)fZ (z),
X, Y , and Z are not independent. Since fX,Y (x, y) = fX (x)fY (y), X, Y , and Z are not
pairwise independent either.
202
Chapter 9
Multivariate Distributions
7. (a) The marginal probability distribution functions of X, Y , and Z are, respectively, given by
FX (x) = F (x, ∞, ∞) = 1 − e−λ1 x , x > 0,
FY (y) = F (∞, y, ∞) = 1 − e−λ2 y , y > 0,
FZ (z) = F (∞, ∞, z) = 1 − e−λ3 z , z > 0.
Since F (x, y, z) = FX (x)FY (y)FZ (z), the random variables X, Y , and Z are independent.
(b) From part (a) it is clear that X, Y , and Z are independent exponential random variables
with parameters λ1 , λ2 , and λ3 , respectively. Hence their joint probability density functions is
given by
f (x, y, z) = λ1 λ2 λ3 e−λ1 x−λ2 y−λ3 z .
(c) The desired probability is calculated as follows:
∞ ∞ ∞
P (X < Y < Z) =
f (x, y, z) dz dy dx
0
x
y
∞
∞
∞
−λ1 x
−λ2 y
−λ3 z
e
e
e
dz dy dx
= λ1 λ2 λ3
0
x
y
λ1 λ2
=
.
(λ2 + λ3 )(λ1 + λ2 + λ3 )
8. (a) Clearly f (x, y, z) ≥ 0 for the given domain. Since
1
0
0
x
y
0
ln x
dz
−
xy
dy dx = 1,
f is a joint probability density function.
y
ln x
ln x
−
dz = −
, 0 ≤ y ≤ x ≤ 1.
xy
x
0
1 y
1
ln x
dz dx = (ln y)2 , 0 ≤ y ≤ 1.
−
fY (y) =
xy
2
y
0
(b) fX,Y (x, y) =
9. For 1 ≤ i ≤ n, let Xi be the distance of the ith point selected at random from the origin. For
r < R, the desired probability is
P (X1 ≥ r, X2 ≥ r, . . . , Xn ≥ r) = P (X1 ≥ r)P (X2 ≥ r) · · · P (Xn ≥ r)
π R2 − π r 2 n
r2 n
=
1
−
.
=
π R2
R2
For r ≥ R, the desired probability is 0.
10. The sphere inscribed
in the
cube has radius a and is centered at the origin. Hence the desired
probability is (4/3)πa 3 /(8a 3 ) = π/6.
Section 9.1
Joint Distributions of n > 2 Random Variables
203
11. Yes, it is because f ≥ 0 and
∞
0
∞
x1
=
x2
∞
∞
0
∞
···
∞
x1
∞
···
x2
∞
= ··· =
e−xn dxn dxn−1 · · · dx1
xn−1
e−xn−1 dxn−1 · · · dx1
xn−2
∞
e
0
∞
−x2
dx2 dx1 =
∞
e−x1 dx1 = 1.
0
x1
12. Let f (x1 , x2 , x3 ) be the joint probability density function of X1 , X2 , and X3 , the lifetimes of
the original, the second, and the third transistors, respectively. We have that
1
1
1
1 −(x1 +x2 +x3 )/5
e
.
f (x1 , x2 , x3 ) = e−x1 /5 · e−x2 /5 · e−x3 /5 =
5
5
5
125
Now
15
15−x1
15−x1 −x2
1 −(x1 +x2 +x3 )/5
dx3 dx2 dx1
e
125
0
0
0
15 15−x1
1 −3
1 −(x1 +x2 )/5
=
− e
e
dx2 dx1
25
25
0
0
15
1 −3
1 −x1 /5 4 −3
=
− e + e x1 dx1
e
5
5
25
0
P (X1 + X2 + X3 < 15) =
=1−
17 −3
e = 0.5768.
2
Therefore, the desired probability is P (X1 + X2 + X3 ≥ 15) = 1 − 0.5768 = 0.4232.
13. Let F be the distribution function of X. We have that
F (t) = P (X ≤ t) = 1 − P (X > t) = 1 − P (X1 > t, X2 > t, . . . , Xn > t)
= 1 − P (X1 > t)P (X2 > t) · · · P (Xn > t) = 1 − e−λ1 t e−λ2 t · · · e−λn t
= 1 − e−(λ1 +λ2 +···+λn )t ,
t > 0.
Thus X is exponential with parameter λ1 + λ2 + · · · + λn .
14. Let Y be the number of functioning components of the system. The random variable Y is
binomial with parameters n and p. The reliability of this system is given by
n
r = P (X = 1) = P (Y ≥ k) =
i=k
n i
p (1 − p)n−i .
i
204
Chapter 9
Multivariate Distributions
15. Let Xi be the lifetime of the ith part. The time until the item fails is the random variable
min(X1 , X2 , . . . , Xn ) which by the solution to Exercise 13 is exponentially distributed with
parameter nλ. Thus the average life of the item is 1/(nλ).
16. Let X1 , X2 , . . . be the lifetimes of the transistors selected at random. Clearly,
N = min n : Xn > s .
Note that
P XN ≤ t | N = n = P Xn ≤ t | X1 ≤ s, X2 ≤ s, . . . , Xn−1 ≤ s, Xn > s).
This shows that for s ≥ t, P XN ≤ t | N = n = 0. For s < t,
P (s < Xn ≤ t, X1 ≤ s, X2 ≤ s, . . . , Xn−1 ≤ s)
P XN ≤ t | N = n =
P (X1 ≤ s, X2 ≤ s, . . . , Xn−1 ≤ s, Xn > s)
=
P (s < Xn ≤ t)P (X1 ≤ s)P (X2 ≤ s) · · · P (Xn−1 ≤ s)
P (X1 ≤ s)P (X2 ≤ s) · · · P (Xn−1 ≤ s)P (Xn > s)
=
P (s < Xn ≤ t)
F (t) − F (s)
=
.
P (Xn > s)
1 − F (s)
This relation shows that the probability distribution function of XN given N = n does not
depend on n. Therefore, XN and N are independent.
17. Clearly,
X = X1 1 − (1 − X2 )(1 − X3 ) 1 − (1 − X4 )(1 − X5 X6 ) X7
= X1 X7 X2 X4 + X3 X4 − X2 X3 X4 + X2 X5 X6 + X3 X5 X6
− X2 X3 X5 X6 − X2 X4 X5 X6 − X3 X4 X5 X6 + X2 X3 X4 X5 X6 .
The reliability of this system is
r = p1 p7 p2 p4 + p3 p4 − p2 p3 p4 + p2 p5 p6 + p3 p5 p6
− p2 p3 p5 p6 − p2 p4 p5 p6 − p3 p4 p5 p6 + p2 p3 p4 p5 p6 .
18. Let G and F be the distribution functions of max1≤i≤n Xi and min1≤i≤n Xi , respectively. Let
g and f be their probability density functions, respectively. For 0 ≤ t < 1,
G(t) = P (X1 ≤ t, X2 ≤ t, . . . , Xn ≤ t)
= P (X1 ≤ t)P (X2 ≤ t) · · · P (Xn ≤ t) = t n .
Section 9.1
Joint Distributions of n > 2 Random Variables
⎧
⎪
⎨0
G(t) = t n
⎪
⎩
1
So
t <0
0≤t <1
t ≥ 1.
Therefore,
nt n−1
0
g(t) = G (t) =
This gives
E max Xi =
1≤i≤n
Similarly, for 0 ≤ t < 1,
F (t) = P
1
0 t
1≤i≤n
= 1 − P (X1 > t)P (X2 > t) · · · P (Xn > t)
0 ≤ t < 1.
= 1 − (1 − t)n ,
Hence
⎧
⎪
⎨0
F (t) = 1 − (1 − t)n
⎪
⎩
1
t <0
0≤t <1
t ≥ 1,
n(1 − t)n−1
0
0 t
= 1 − P (X1 > t, X2 > t, . . . , Xn > t)
= 1 − P (X1 > t)P (X2 > t) · · · P (Xn > t)
n
= 1 − 1 − F (t) .
205
206
Chapter 9
Multivariate Distributions
20. We have that
Thus
x
P (Yn > x) = P min(X1 , X2 , . . . , Xn ) >
n
x
x
x
= P X1 > , X2 > , . . . , Xn >
n
n
n
x
x
x
= P X1 >
P X2 >
· · · P Xn >
n
n
n
n
x
= 1−
.
n
x n
lim P (Yn > x) = lim 1 −
= e−x , x > 0.
n→∞
n→∞
n
21. We have that
P (X < Y < Z) =
∞
∞
∞
h(x)h(y)h(z) dz dy dx
−∞
∞
x
y
h(x)h(y) 1 − H (y) dy dx
−∞ x
∞
2 ∞
1
=
h(x) − 1 − H (y)
dx
2
−∞
x
2
1 ∞
=
h(x) 1 − H (x) dx
2 −∞
3 ∞
1
1
1
=
− 1 − H (x)
= .
2
3
6
−∞
=
∞
22. Noting that Xi2 = Xi , 1 ≤ i ≤ 5, we have
X = max{X2 X5 , X2 X3 X4 , X1 X4 , X1 X3 X5 }
= 1 − (1 − X2 X5 )(1 − X2 X3 X4 )(1 − X1 X4 )(1 − X1 X3 X5 )
= X2 X5 + X1 X4 + X1 X3 X5 + X2 X3 X4 − X1 X2 X3 X4 − X1 X2 X3 X5
− X1 X2 X4 X5 − X1 X3 X4 X5 − X2 X3 X4 X5 + 2X1 X2 X3 X4 X5 .
Therefore, whenever the system is turned on for water to flow from A to B, water reaches B
with probability r given by,
r = P (X = 1) = E(X) = p2 p5 + p1 p4 + p1 p3 p5 + p2 p3 p4 − p1 p2 p3 p4
− p1 p2 p3 p5 − p1 p2 p4 p5 − p1 p3 p4 p5 − p2 p3 p4 p5 + 2p1 p2 p3 p4 p5 .
23. Clearly, B = (1 × 1)/2 and h = 1. So the volume of the pyramid is (1/3)Bh = 1/6.
Therefore, the joint probability density function of X, Y , and Z is
f (x, y, z) =
6
0
(x, y, z) ∈ V
otherwise.
Section 9.1
Thus
fX (x) =
1−x
Joint Distributions of n > 2 Random Variables
1−x−y
6 dz
0
207
dy = 3(1 − x)2 ,
0 < x < 1.
0
Similarly, fY (y) = 3(1 − y)2 , 0 < y < 1, and fZ (z) = 3(1 − z)2 , 0 < z < 1. Since
f (x, y, z) = fX (x)fY (y)fZ (z),
X, Y , and Z are not independent.
24. The probability that Ax 2 +Bx+C = 0 has real roots is equal to the probability that B 2 −4AC ≥
0. To calculate this quantity, we will first evaluate the distribution functions of B 2 and −4AC
and then use the convolution theorem to find the distribution function of B 2 − 4AC.
⎧
⎪
0
if t < 0
⎪
⎨√
FB 2 (t) = P (B 2 ≤ t) =
t if 0 ≤ t < 1
⎪
⎪
⎩
1
if t ≥ 1,
⎧
1
⎨ √
if 0 < t < 1
fB 2 (t) = F 2 (t) = 2 t
B
⎩
0
otherwise,
and
⎧
⎪
0
⎪
⎪
⎨
t
F−4AC (t) = P (−4AC ≤ t) = P AC ≥ −
⎪
4
⎪
⎪
⎩
1
if t < −4
if −4 ≤ t < 0
if t ≥ 0.
Now A and C are random numbers from (0, 1); hence (A, C) is a random
point from
the
square (0, 1) × (0, 1) in the ac-plane. Therefore, P (AC ≥ −t/4) = P C ≥ −t/(4A) is the
t
area of the shaded region bounded by a = 1, c = 1, c = −
of Figure 1.
4a
c
1
-t/4
0
Figure 1
-t/4
1
a
The shaded region of Exercise 24.
208
Chapter 9
Multivariate Distributions
Thus, for −4 ≤ t < 0,
F−4AC (t) =
1
−t/4
1
−t/(4a)
dc da = 1 +
t
t t
− ln − .
4 4
4
Therefore,
F−4AC (t) = P (−4AC ≤ t) =
⎧
⎪
0
⎪
⎪
⎨
if t < −4
t
t
t
1 + − ln −
⎪
4
4
4
⎪
⎪
⎩
1
if −4 ≤ t < 0
if t > 0.
Applying convolution theorem, we obtain
P B 2 − 4AC ≥ 0 = 1 − P B 2 − 4AC < 0
∞
=1−
F−4AC (0 − x)fB 2 (x)dx
−∞
1
=1−
1−
0
Letting y =
x
x x
1
+ ln
√ dx.
4
4 4 2 x
√
1
x/2, we get dy = √ dx. So
4 x
P B 2 − 4AC ≥ 0 = 1 −
1/2
(1 − y 2 + y 2 ln y 2 )2dy
0
=1−
=2
1/2
2dy + 2
0
1/2
1/2
(y 2 − y 2 ln y 2 )dy
0
(y 2 − y 2 ln y 2 )dy.
0
Now by integration by parts (u = ln y 2 , dv = y 2 dy),
1
2
y 2 ln y 2 dy = y 3 ln y 2 − y 3 .
3
9
Thus
1/2
1
10 3 2 3
5
y − y ln y 2
+ ln 2 ≈ 0.25.
=
P B 2 − 4AC ≥ 0 =
0
9
3
36 6
25. The following solution by Scott Harrington, Duke University, Durham, NC, was given in The
College Mathematics Journal, September 1993.
Let V be the set of points (A, B, C) ∈ [0, 1]3 such that f (x) = x 3 +Ax 2 +Bx +C = 0
has all real roots. The probability that all of the roots are real is the volume of V .
Section 9.1
Joint Distributions of n > 2 Random Variables
The function is cubic, so it either has one real root and two complex roots or
three real roots. Since the coefficient of x 3 is positive, limx→−∞ f (x) = −∞ and
limx→+∞ f (x) = +∞. The number of real roots of the graph of f (x) depends on
the nature of the critical points of the function f .
f (x) = 3x 2 + 2Ax + B = 0,
with roots
1
1) 2
x =− A±
A − 3B.
3
3
√
1
1
Let D = A2 − 3B, x1 = − (A + D), and x2 = − (A − D). If A2 < 3B then the
3
3
critical points are imaginary, so the graph of f (x) is strictly increasing and there
must be exactly one real root. Thus we may assume A2 ≥ 3B.
In
multiplicities, the local maximum
order for
there to be three real
roots, counting
x1 , f (x1 ) and local minimum x2 , f (x2 ) must satisfy f (x1 ) ≥ 0 and f (x2 ) ≤ 0;
that is,
f (x1 ) = −
1 3
(A + 3A2 D + 3AD 2 + D 3 )
27
1
1
+ A(A2 + 2AD + D 2 ) − B(A + D) + C ≥ 0,
9
3
f (x2 ) = −
1 3
(A − 3A2 D + 3AD 2 − D 3 )
27
1
1
+ A(A2 − 2AD + D 2 ) − B(A − D) + C ≤ 0.
9
3
Simplifying produces two half-spaces:
1
− 2A3 + 9AB − 2(A2 − 3B)3/2 ,
27
1
− 2A3 + 9AB + 2(A2 − 3B)3/2 ,
C≤
27
C≥
(constraint surface 1);
(constraint surface 2).
1
These two surfaces intersect at the curve given parametrically by A = t, B = t 2
3
1 3
t . Note that all points in the intersection of these two half-spaces
and C =
27
1
satisfy B ≤ A2 . Surface 2 intersects the plane C = 0 at the A-axis, but surface 1
3
1
intersects the plane C = 0 at the curve B = A2 , which is a quadratic curve in the
4
plane C = 0 located between the A-axis and the upper limit B = 13 A2 . Therefore, V
is the region above the plane C = 0 and constraint surface 1, and below constraint
surface 2. The volume of V is the volume V2 under surface 2 minus the volume V1
under surface 1. Now
1 (1/3)a 2
1
− 2a 3 + 9ab − 2(a 2 − 3b)3/2 db da
V1 =
a=0 b=(1/4)a 2 27
209
210
Chapter 9
Multivariate Distributions
=
1
0
1
1
(1/3)a 2
1
9
4
− 2a 3 b + ab2 + (a 2 − 3b)5/2
da
27
2
15
b=(1/4)a 2
1
7 5
7
·
a da =
, and
27
160
25,
920
0
1 (1/3)a 2
1
V2 =
− 2a 3 + 9ab + 2(a 2 − 3b)3/2 db da
27
a=0 b=0
=
=
0
(1/3)a 2
1
9
4
1
1
1 5
− 2a 3 b + ab2 − (a 2 − 3b)5/2
a da =
.
da =
27
2
15
270
1620
0
b=0
Thus
V = V2 − V 1 =
9.2
7
1
1
−
=
.
1, 620 25, 920
2, 880
ORDER STATISTICS
1. By Theorem 9.5, we have that
f3 (x) =
2
4!
f (x) F (x) 1 − F (x) ,
2! 1!
where
f (x) =
⎧
⎨1
0 x) dx.
0
Now
P X(n) > x = 1 − P X(n) ≤ x
n
= 1 − P (X1 ≤ x, X2 ≤ x, . . . , Xn ≤ x) = 1 − F (x) .
So
E X(n) =
∞
n
1 − F (x)
dx.
0
5. To find P X(i) = k , 0 ≤ k ≤ n, note that
P X(i) = k = 1 − P X(i) < k − P X(i) > k .
Let N be the number of Xj ’s that are less than k. Then N is a binomial random variable with
parameters m and
k−1
p1 =
l=0
n l
p (1 − p)n−l .
l
(35)
Let L be the number of Xj ’s that are greater than k. Then L is a binomial random variable
with parameters m and
n l
p (1 − p)n−l .
l
l=k+1
n
p2 =
(36)
212
Chapter 9
Multivariate Distributions
Clearly,
m
P X(i) < k = P (N ≥ i) =
j =i
and
P X(i)
m j
p1 (1 − p1 )m−j ,
j
m j
p2 (1 − p2 )m−j .
> k = P (L ≥ m − i + 1) =
j
j =m−i+1
m
Thus, for 0 ≤ k ≤ n,
P X(i) = k = 1 −
m
j =i
m
m j
m j
m−j
−
p1 (1 − p1 )
p2 (1 − p2 )m−j ,
j
j
j =m−i+1
where p1 and p2 are given by (35) and (36).
6. By Theorem 9.6, the joint probability density function of X(1) and X(n) is given by
n−2
,
f1n (x, y) = n(n − 1)f (x)f (y) F (y) − F (x)
Therefore,
x < y.
X(1) + X(n)
G(t) = P
≤ t = P X(1) + X(n) ≤ 2t
2
t 2t−x
n−2
n(n − 1)f (x)f (y) F (y) − F (x)
dy dx
=
−∞
=n
t
−∞
x
n−1
F (2t − x) − F (x)
f (x) dx.
7. By Theorem 9.5, f1 (x), the probability density function of X(1) is given by
f1 (x) =
1−1 −λx 2−1
2!
= 2λe−2λx ,
λe−λx 1 − e−λx
e
(1 − 1)! (2 − 1)!
x ≥ 0.
By Theorem 9.6, f12 (x, y), the joint probability density function of X(1) and X(2) is given by
2!
λe−λx λe−λy
(1 − 1)! (2 − 1 − 1)! (2 − 2)!
1−1 −λx
2−1−1
= 1 − e−λx
e
− e−λy
= 2λ2 e−λ(x+y) ,
f12 (x, y) =
0 ≤ x < y < ∞.
Let U = X(1) and V = X(2) − X(1) . We will show that g(u, v), the joint probability density
function of U and V satisfy g(u, v) = gU (u)gV (v). This proves that U and V are independent.
To find g(u, v), note that the system of two equations in two unknowns
x=u
y−x =v
Section 9.2
Order Statistics
213
defines a one-to-one transformation of
R = (x, y) : 0 ≤ x < y < ∞
onto the region
Q = (u, v) : u ≥ 0, v > 0 .
It has the unique solution x = u, y = u + v. Hence
1 0
= 1 = 0.
J =
1 1
By Thereom 8.8,
g(u, v) = f12 (u, u + v)|J| = 2λ2 e−λ(u+2v) ,
u ≥ 0, v > 0.
Since
g(u, v) = gU (u)gV (v),
where
gU (u) = 2λe−2λu ,
and
gV (v) = λe−λv ,
u ≥ 0,
v > 0,
we have that U and V are independent. Furthermore, U is exponential with parameter 2λ and
V is exponential with parameter λ.
8. Let f12 (x, y) be the joint probability density function of X(1) and X(2). By Theorem 9.6,
1
1
2
2
2
2
f12 (x, y) = 2! f (x)f (y) = 2 · √ e−x /2σ · √ e−y /2σ
σ 2π
σ 2π
1 −x 2 /2σ 2 −y 2 /2σ 2
= 2 e
·e
, −∞ < x < y < ∞.
σ π
Therefore,
∞
y
1 −x 2 /2σ 2 −y 2 /2σ 2
·e
dx dy
e
σ 2π
−∞ −∞
y
∞
1
−y 2 /2σ 2
−x 2 /2σ 2
e
xe
dx dy
= 2
σ π −∞
−∞
∞
1
2
2
2
2
= 2
e−y /2σ · (−σ 2 )e−y /2σ dy
σ π −∞
1 ∞ −y 2 /σ 2
=−
e
dy
π −∞
E X(1) =
x·
214
Chapter 9
Multivariate Distributions
√
1
1
·σ π · σ √
π
√ · 2π
2
√
1
σ
= − · σ π · 1 = −√ .
π
π
=−
∞
y2
√
2
e 2(σ/ 2) dy
−
−∞
9. (a) By Theorem 9.6, the joint probability density function of X(1) and X(n) is given by
n−2
n(n − 1)f (x)f (y) F (y) − F (x)
f1n (x, y) =
0
x 0 .
It has the unique solution x = v − r, y = v. Hence
−1 1
= −1 = 0.
J =
0 1
By Theorem 8.8, g(u, v) is given by
g(r, v) = f1n (v − r, v)|J|
n−2
,
= n(n − 1)f (v − r)f (v) F (v) − F (v − r)
This implies
gR (r) =
∞
−∞
−∞ < v < ∞, r > 0.
n−2
n(n − 1)f (v − r)f (v) F (v) − F (v − r)
dv,
r > 0.
(37)
Section 9.3
Multinomial Distributions
215
(b) The probability density function of n random numbers from (0, 1) is obtained by letting
⎧
⎨1
0 t) = 1 − P (X1 > t, X2 > t, . . . , Xn > t)
⎧
0
t <1
⎪
⎪
⎪
⎪
n
⎪
⎪
⎪
1 − 65
1≤t <2
⎪
⎪
⎪
⎪
⎪
4 n
⎪
⎪
2≤t <3
⎪1 − 6
⎪
⎨
n
3 n
= 1 − P (X1 > t) = 1 − 6
3≤t <4
⎪
⎪
⎪
n
⎪
⎪
4≤t <5
1 − 26
⎪
⎪
⎪
⎪
⎪
⎪
1 n
⎪
5≤t <6
⎪
⎪1 − 6
⎪
⎪
⎩
1
t ≥ 6.
The probability mass function of X is
p(x) = P (X = x) =
7−x
6
n
−
6−x
6
n
, x = 1, 2, 3, 4, 5, 6.
3. Let D1 , D2 , . . . , Dn be the distances of the points selected from the origin. Let D =
min(D1 , D2 , . . . , Dn ). The desired probability is
n
n
P (D ≥ r) = P (D1 ≥ r, D2 ≥ r, . . . , Dn ≥ r) = P (D1 ≥ r) = 1 − P (D1 < r)
n
(4/3)πr 3
π r 3 n
= 1−
=
1
−
.
8a 3
6 a
1
1
4. (a) c
0
0
0
1
(x + y + 2z) dz dy dx = 1 ⇒ c = 1/2.
Chapter 9
Review Problems
219
(b) We have that
1
1
1
P
X< ,Y < ,Z<
1
1
1
3
2
4
=
P X< Y < ,Z<
1
1
3
2
4
P Y < ,Z<
2
4
1/3 1/2 1/4
1
(x + y + 2z) dz dy dx
2
1/36
2
0
0
0
=
= .
=
1
1/2 1/4
1/8
9
1
(x + y + 2z) dz dy dx
2
0
0
0
5. The joint probability mass function of thenumber of times each face appears is multinomial.
Hence the desired probability is
18!
(3!)6
1
6
18
= 0.00135.
6. Using the multinomial distribution, the answer is
7!
(0.4)3 (0.35)2 (0.25)2 = 0.1029.
3! 2! 2!
≤ i ≤ n, let Xi be the lifetime of the ith component.
Then
min(X1 , X2 , . . . , Xn ) is the lifetime of the system. Let F̄ (t) be the survival function of
the system. By the independence of the lifetimes of the components, for all t > 0,
7. For 1
F̄ (t) = P min(X1 , X2 , . . . , Xn ) > t = P (X1 > t, X2 > t, . . . , Xn > t)
= P (X1 > t)P (X2 > t) · · · P (Xn > t) = F¯1 (t)F¯2 (t) · · · F¯n (t).
≤ i ≤ n, let Xi be the lifetime of the ith component.
Then
max(X1 , X2 , . . . , Xn ) is the lifetime of the system. Let F̄ (t) be the survival function of
the system. By the independence of the lifetimes of the components, for all t > 0,
8. For 1
F̄ (t) = P max(X1 , X2 , . . . , Xn ) > t
= 1 − P max(X1 , X2 , . . . , Xn ) ≤ t
= 1 − P (X1 ≤ t, X2 ≤ t, . . . , Xn ≤ t)
= 1 − P (X1 ≤ t)P (X2 ≤ t) · · · P (Xn ≤ t)
= 1 − F1 (t)F2 (t) · · · Fn (t).
9. The problem is equivalent to the following: Two points X and Y are selected independently
and at random from the interval (0, ). What is the probability that the length of at least one
220
Chapter 9
Multivariate Distributions
interval is less than /20? The solution to this problem is as follows:
P min(X, Y − X, − Y ) <
X < Y P (X < Y )
20
+ P min(Y, X − Y, − X) <
X > Y P (X > Y )
20
= 2P min(X, Y − X, − Y ) <
X < Y P (X < Y )
20
1
= 2P min(X, Y − X, − Y ) <
X x ;
that is,
17 17
×
2
20
20
= 0.7225.
÷
2
2
Therefore, the desired probability is 1 − 0.7225 = 0.2775.
10. Let f13 (x, y) be the joint probability density function of X(1) and X(3) . By Theorem 9.6,
f13 (x, y) = 6(y − x),
0 < x < y < 1.
X(1) + X(3)
and V = X(1) . Using Theorem 8.8, we will find g(u, v), the joint
Let U =
2
probability density function of U and V . The probability density function of the midrange of
these three random variables is gU (u). The system of two equations in two unknowns
⎧
⎨x+y =u
2
⎩x = v
Chapter 9
Review Problems
defines a one-to-one transformation of
R = (x, y) : 0 < x < y < 1
onto the region
v+1
<1
Q = (u, v) : 0 < v < u <
2
that has the unique solution
x=v
y = 2u − v.
0 1
= −2 = 0;
J =
2 −1
Hence
therefore,
g(u, v) = f13 (v, 2u − v)|J| = 24(u − v),
To find gU (u), draw the region Q to see that
⎧ u
⎪
⎪
24(u − v) dv
⎪
⎪
⎨ 0
gU (u) =
u
⎪
⎪
⎪
⎪
⎩
24(u − v) dv
0 x, Y > y) dx dy.
0
0
18. Clearly N > i if and only if
X1 ≥ X2 ≥ X3 ≥ · · · ≥ Xi .
Hence for i ≥ 2,
1
P (N > i) = P X1 ≥ X2 ≥ X3 ≥ · · · ≥ Xi−1 ≥ Xi =
i!
because Xi ’s are independent and identically distributed. So, by Theorem 10.2,
∞
∞
P (N ≥ i) =
E(N ) =
i=1
∞
=1+1+
i=2
1
=
i!
∞
P (N > i) = P (N > 0) + P (N > 1) +
i=0
∞
i=0
i=2
1
i!
1
= e.
i!
19. If the first red chip is drawn on or before the 10th draw, let N be the number of chips before
the first red chip. Otherwise, let N = 10. Clearly,
1 i1
1 i+1
P (N = i) =
=
, 0 ≤ i ≤ 9;
2
2
2
P (N = 10) =
1
2
10
.
The desired quantity is
1
9
E(10 − N) =
(10 − i)
i=0
2
i+1
+ (10 − 10) ·
1
2
10
≈ 9.001.
20. Clearly, if for some λ ∈ R, X = λY , Cauchy-Schwarz’s inequality becomes equality. We
show that the converse of this is also true. Suppose that for random variables X and Y ,
)
E(XY ) = E(X2 )E(Y 2 ).
Then
2
4 E(XY ) − 4E(X2 )E(Y 2 ) = 0.
Section 10.2
Covariance
227
Now the left side of this equation is the discriminant of the quadratic equation
E(Y 2 )λ2 − 2 E(XY ) λ + E(X2 ) = 0.
Hence this quadratic equation has exactly one root. On the other hand,
E(Y 2 )λ2 − 2 E(XY ) λ + E(X2 ) = E (X − λY )2 .
So the equation
E (X − λY )2 = 0
has a unique solution. That is, there exists a unique number λ1 ∈ R such that
E (X − λ1 Y )2 = 0.
Since the expected value of a positive random variable is positive, this implies that with
probability 1, X − λ1 Y = 0 or X = λ1 Y.
10.2
COVARIANCE
1. Since X and Y are independent random variables, Cov(X, Y ) = 0.
4
3
2.
E(X) =
x=1 y=3
E(Y ) =
3
4
x=1
3
17
1 2
x (x + y) = ;
70
7
4
E(XY ) =
x=1 y=3
1
y=3 70
xy(x + y) =
124
;
35
1 2
43
x y(x + y) = .
70
5
Therefore,
Cov(X, Y ) = E(XY ) − E(X)E(Y ) =
1
43 17 124
−
·
=−
.
5
7 35
245
3. Intuitively, E(X) is the average of 1, 2, . . . , 6 which is 7/2; E(Y ) is (7/2)(1/2) = 7/4. To
show these, note that
6
6
xpX (x) =
E(X) =
x=1
x(1/6) = 7/2.
x=1
228
Chapter 10
More Expectations and Variances
By the table constructed for p(x, y) in Example 8.2,
E(Y ) = 0 ·
63
120
99
64
29
8
1
7
+1·
+2·
+3·
+4·
+5·
+6·
= .
384
384
384
384
384
384
384
4
By the same table,
6
6
E(XY ) =
xyp(x, y) = 91/12.
x=1 y=0
Therefore,
Cov(X, Y ) = E(XY ) − E(X)E(Y ) =
91 7 7
35
− · =
> 0.
12 2 4
24
This shows that X and Y are positively correlated. The higher the outcome from rolling the
die, the higher the number of tails obtained—a fact consistent with our intuition.
4. Let X be the number of sheep stolen; let Y be the number of goats stolen. Let p(x, y) be the
joint probability mass function of X and Y . Then, for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ x + y ≤ 4,
7 8
5
x y 4−x−y
p(x, y) =
;
20
4
p(x, y) = 0, for other values of x and y. Clearly, X is a hypergeometric random variable with
parameters n = 4, D = 7, and N = 20. Therefore,
E(X) =
nD
28
7
=
= .
N
20
5
Y is a hypergeometric random variable with parameters n = 4, D = 8, and N = 20. Therefore,
E(Y ) =
nD
32
8
=
= .
N
20
5
Since
4
4−x
E(XY ) =
xyp(x, y) =
x=0 y=0
168
,
95
we have
Cov(X, Y ) = E(XY ) − E(X)E(Y ) =
168 7 8
224
− · =−
< 0.
95
5 5
475
Therefore, X and Y are negatively correlated as expected.
Section 10.2
Covariance
229
5. Since Y = n − X,
E(XY ) = E(nX − X2 ) = nE(X) − E(X2 ) = nE(X) − Var(X) + E(X)2
= n · np − np(1 − p) + n2 p2 = n(n − 1)p(1 − p),
and
Cov(X, Y ) = E(XY ) − E(X)E(Y ) = n(n − 1)p(1 − p) − np · n(1 − p) = −np(1 − p).
This confirms the (obvious) fact that X and Y are negatively correlated.
6. Both (a) and (b) are straightforward results of relation (10.6).
7. Since Cov(X, Y ) = 0, we have
Cov(X, Y + Z) = Cov(X, Y ) + Cov(X, Z) = Cov(X, Z).
8. By relation (10.6),
Cov(X + Y, X − Y ) = E(X2 − Y 2 ) − E(X + Y )E(X − Y )
2
2
= E(X2 ) − E(Y 2 ) − E(X) + E(Y ) = Var(X) − Var(Y ).
9. In Theorem 10.4, let a = 1 and b = −1.
10. (a) This is an immediate result of Exercise 8 above.
(b) By relation (10.6),
Cov(X, XY ) = E(X2 Y ) − E(X)E(XY )
2
= E(X2 )E(Y ) − E(X) E(Y ) = E(Y )Var(X).
11. The probability density function of is given by
⎧
⎪
⎨ 1
f (θ ) = 2π
⎪
⎩0
if θ ∈ [0, 2π]
otherwise.
Therefore,
E(XY ) =
2π
0
E(Y ) =
1
dθ = 0,
sin θ cos θ
2π
2π
cos θ
0
1
dθ = 0.
2π
Thus Cov(X, Y ) = E(XY ) − E(X)E(Y ) = 0.
E(X) =
2π
sin θ
0
1
dθ = 0,
2π
230
Chapter 10
More Expectations and Variances
12. The joint probability density function of X and Y is given by
⎧
⎪
⎨1
f (x, y) = π
⎪
⎩0
x2 + y2 ≤ 1
elsewhere.
X and Y are dependent because, for example,
1
1
P 0 P (A), ⇐⇒
P (B)
P (A | B) > P (A). The proof that IA and IB are positively correlated if and only if P (B|A) >
P (B) follows by symmetry.
This shows that Cov(IA , IB ) > 0 ⇐⇒ P (AB) > P (A)P (B) ⇐⇒
234
Chapter 10
More Expectations and Variances
23. By Exercise 6,
Cov(aX + bY, cZ + dW ) = a Cov(X, cZ + dW ) + b Cov(Y, cZ + dW )
= ac Cov(X, Z) + ad Cov(X, W ) + bc Cov(Y, Z) + bd Cov(Y, W ).
24. By Exercise 6 and an induction on n,
n
m
Cov
n
bj Yj =
ai Xi ,
j =1
i=1
ai Cov Xi ,
m
bj Yj .
j =1
i=1
By Exercise 6 and an induction on m,
m
m
bj Yj =
Cov Xi ,
j =1
bj Cov(Xi , Yj ).
j =1
The desired identity follows from these two identities.
25. For 1 ≤ i ≤ n, let Xi = 1 if the outcome of the ith throw is 1; let Xi = 0, otherwise. For
1 ≤ j ≤ n, let Yj = 1 if the outcome of the j th throw is 6; let Yj = 0, otherwise. Clearly,
Cov(Xi , Yj ) = 0 if i = j . By Exercise 24,
Cov
n
n
j =1
i
n
=
n
n
Yj =
Xi ,
n
Cov(Xi , Yj ) =
j =1 i=1
Cov(Xi , Yi )
i=1
E(Xi Yi ) − E(Xi )E(Yi ) =
i=1
n
0−
i=1
1 1
n
·
=− .
6 6
36
As expected, in n throws of a fair die, the number of ones and the number of sixes are negatively
correlated.
26. Let Sn =
n
i=1
ai Xi , µi = E(Xi ); then
n
E(Sn ) =
n
Sn − E(Sn ) =
ai µi ,
i=1
ai (Xi − µi ).
i=1
Thus
Var(Sn ) = E
2
n
ai (Xi − µi )
i=1
n
=
ai2 E (Xi − µi )2 + 2
ai aj E (Xi − µi )(Xj − µj )
i 2.
The time that the admissions office has to wait before doubling its student recruitment efforts
is SN+1 = X1 + X2 + · · · + XN +1 . Therefore,
E(SN+1 ) = E E(SN +1 | N ) =
∞
E(SN +1 | N = i)P (N = i).
i=0
Now, for i ≥ 0,
i+1
E(SN+1 | N = i) = E(X1 + X2 + · · · + Xi+1 | N = i) =
E(Xj | N = i)
j =1
i
E(Xj | Xj ≤ 2) + E(Xi+1 | Xi+1 > 2),
=
j =1
where by Remark 8.1,
1
E(Xj | Xj ≤ 2) =
F (2)
E(Xi+1 | Xi+1
2
tf (t) dt,
0
1
> 2) =
1 − F (2)
∞
tf (t) dt,
2
Section 10.4
Conditioning on Random Variables
245
F and f being the probability distribution and density functions of Xi ’s, respectively. That is,
for t ≥ 0, F (t) = 1 − e−5t , f (t) = 5e−5t . Thus, for 1 ≤ j ≤ i,
2
1
1 −5t 2
−5t
E(Xj | Xj ≤ 2) =
e
5t
e
dt
=
(1.0000454)
−
t
−
0
1 − e−10 0
5
= (1.0000454)(0.19999) = 0.1999092
and, for j = i + 1,
E(Xi+1 | Xi+1 > 2) =
Thus, for i ≥ 0,
1
e−10
∞
5t e−5t dt = e10
−t −
2
1 −5t ∞
= 2.2.
e
2
5
E(SN+1 | N = i) = (0.1999092)i + 2.2.
To find P (N = i), note that for i ≥ 0,
P (N = i) = P (X1 ≤ 2, X2 ≤ 2, . . . , Xi ≤ 2, Xi+1 > 2)
i
= F (2) 1 − F (2) = (0.9999546)i (0.0000454).
Putting all these together, we obtain
∞
E(SN+1 ) =
E(SN+1 | N = i)P (N = i)
i=0
∞
=
(0.1999092)i + 2.2 (0.9999546)i (0.0000454)
i=0
∞
∞
i(0.9999546)i + (0.00009988)
= (0.00000908)
i=0
(0.9999546)i
i=0
0.9999546
1
= (0.00000908) ·
+ (0.00009988) ·
2
(1 − 0.9999546)
1 − 0.9999546
= 4407.286,
∞ i
∞ i
= r/(1 − r)2 , and
=
where the next to last equality follows from
i=1 ir
i=0 r
1/(1 − r), |r| < 1. Since an academic year is 9 months long, and contains approximately
180 business days, the admission officers should not be concerned about this rule at all. It
will take 4,407.286 business days, on average, until there is a lapse of two days between two
consecutive applications.
14. Let Xi be the number of calls until Steven has not missed Adam in exactly i consecutive calls.
We have that
Xi−1 + 1
E Xi | Xi−1 =
Xi−1 + 1 + E(Xi )
with probability p
with probability 1 − p.
246
Chapter 10
More Expectations and Variances
Therefore,
E(Xi ) = E E(Xi | Xi−1 ) = E(Xi−1 ) + 1 p + E(Xi−1 ) + 1 + E(Xi ) (1 − p).
Solving this equation for E(Xi ), we obtain
E(Xi ) =
1
1 + E(Xi−1 ) .
p
Now X1 is a geometric random variable with parameter p. So E(X1 ) = 1/p. Thus
1
1
1
E(X2 ) =
1 + E(X1 ) =
1+
,
p
p
p
1
1
1
1
E(X3 ) =
1 + E(X2 ) =
1+ + 2 ,
p
p
p p
..
.
E(Xk ) =
1 (1/p k ) − 1
1
1
1 − pk
1
1
1 + + 2 + · · · + k−1 = ·
= k
.
p
p p
p
p (1/p) − 1
p (1 − p)
15. Let N be the number of games to be played until Emily wins two of the most recent three
games. Let X be the number of games to be played until Emily wins a game for the first time.
The random variable X is geometric with parameter 0.35. Hence E(X) = 1/0.35. First, we
find the random variable E(N | X) in terms of X. Then we obtain E(N ) by calculating the
expected value of E(N | X). Let W be the event that Emily wins the (X + 1)st game as well.
Let LW be the event that Emily loses the (X + 1)st game but wins the (X + 2)nd game. Let
LL be the event that Emily loses both the (X + 1)st and the (X + 2)nd games. Given X = x,
we have
E(N | X = x) = (x + 1)P (W ) + (x + 2)P (LW ) + (x + 2) + E(N) P (LL).
So
E(N | X = x) = (x + 1)(0.35) + (x + 2)(0.65)(0.35) + (x + 2) + E(N) (0.65)2 .
This gives
E(N | X = x) = x + (0.4225)E(N ) + 1.65.
Therefore,
E(N | X) = X + (0.4225)E(N ) + 1.65.
Hence
E(N ) = E E(N | X) = E(X) + (0.4225)E(N ) + 1.65 =
Solving this for E(N ) gives E(N) = 7.805.
1
+ (0.4225)E(N ) + 1.65.
0.35
Section 10.4
Conditioning on Random Variables
247
16. Since hemophilia is a sex-linked disease, and John is phenotypically normal, John is H .
Therefore, no matter what Kim’s genotype is, none of the daughters has hemophilia. Whether
a boy has hemophilia or not depends solely on the genotype of Kim. Let X be the number
of the boys who have hemophilia. To find, E(X), the expected number of the boys who have
hemophilia, let
⎧
⎪
0 if Kim is hh
⎪
⎨
Z = 1 if Kim is H h
⎪
⎪
⎩
2 if Kim is H H .
Then
E(X) = E E(X | Z)
= E(X | Z = 0)P (Z = 0) + E(X | Z = 1)P (Z = 1) + E(X | Z = 2)P (Z = 2)
= 4(0.02)(0.02) + 4(1/2) 2(0.98)(0.02) + 0 0.98)(0.98) = 0.08.
Therefore, on average, 0.08 of the boys and hence 0.08 of the children are expected to have
hemophilia.
17. Let X be the number of bags inspected until an unacceptable bag is found. Let Kn be the number
of consequent bags inspected until n consecutive acceptable bags are found. The number of
bags inspected in one inspection cycle is X + Km . We are interested in E(X + Km ) =
E(X) + E(K
X is a geometric random variable with parameter α(1 − p). So
m ). Clearly,
E(X) = 1/ α(1 − p) . To find E(Km ), note that ∀n,
E(Kn ) = E E(Kn | Kn−1 ) .
Now
E(Kn | Kn−1 = i) = (i + 1)p + i + 1 + E(Kn ) (1 − p)
= (i + 1) + (1 − p)E(Kn ).
(41)
To derive this relation, we noted the following. It took i inspections to find n − 1 consecutive
acceptable bags. If the next bag inspected is also acceptable, we have the n consecutive
acceptable bags required in i + 1 inspections. This occurs with probability p. However, if
the next bag inspected is unacceptable, then, onthe average, we need an additional E(Kn )
inspections a total of i + 1 + E(Kn ) inspections until we get n consecutive acceptable bags
of cinnamon. This happens with probability 1 − p.
From (41), we have
E(Kn | Kn−1 ) = (Kn−1 + 1) + (1 − p)E(Kn ).
Finding the expected values of both sides of this relation gives
E(Kn ) = E(Kn−1 ) + 1 + (1 − p)E(Kn ).
248
Chapter 10
More Expectations and Variances
Solving for E(Kn ), we obtain
1
E(Kn−1 )
+
.
p
p
E(Kn ) =
Noting that E(K1 ) = 1/p and solving recursively, we find that
E(Kn ) =
1
1
1
+ 2 + ··· + n.
p p
p
Therefore, the desired quantity is
E(X + Km ) = E(X) + E(Km )
=
1
1
1
1
+
1 + + · · · + m−1
α(1 − p) p
p
p
1
=
m
−1
1
1
(1 − α)pm + α
p
+ ·
.
=
1
α(1 − p) p
αpm (1 − p)
−1
p
18. For 0 < t ≤ 1, let N(t) be the number of batteries changed by time t. Let X be the lifetime
of the initial battery used; X is a uniform random variable over the interval (0, 1). Therefore,
fX , the probability density function of X, is given by
fX (x) =
1
0
if 0 < x < 1
otherwise.
We are interested in K(t) = E N (t) . Clearly,
∞
E N(t) | X = x fX (x) dx
E N(t) = E E N(t) | X =
0
t
t
E N (t − x) dx
1 + E N(t − x) dx = t +
=
0
0
t
K(u) du,
=t+
0
where the last
* t equality follows from the substitution u = t − x. Differentiating both sides of
K(t) = t + 0 K(u) du with respect to t, we obtain K (t) = 1 + K(t) which is equivalent to
K (t)
= 1.
1 + K(t)
Thus, for some constant c,
ln 1 + K(t) = t + c,
Section 10.4
Conditioning on Random Variables
249
or,
1 + K(t) = et+c .
The initial condition K(0) = E N (0) = 0 yields ec = 1; so
K(t) = et − 1.
On average, after 950 hours of operation, K(0.95) = 1.586 batteries are used.
19. Since E(X|Y ) is a function of Y , by Example 10.23,
E(XZ) = E E(XZ|Y ) = E E XE(X|Y )|Y
= E E(X|Y )E(X|Y ) = E(Z 2 ).
Therefore,
2
E X − E(X|Y )
= E (X − Z)2
= E(X2 − 2ZX + Z 2 ) = E(X2 ) − 2E(Z 2 ) + E(Z 2 )
= E(X2 ) − E(Z 2 ) = E(X2 ) − E E(X|Y )2 .
20. Let Z = E(X|Y ); then
Var(X|Y ) = E (X − Z)2 |Y
= E(X2 − 2XZ + Z 2 |Y )
= E(X2 |Y ) − 2E(XZ|Y ) + E(Z 2 |Y ).
Since E(X|Y ) is a function of Y , by Example 10.23,
E(XZ|Y ) = E XE(X|Y )|Y = E(X|Y )E(X|Y ) = Z 2 .
Also
E(Z 2 |Y ) = E E(X|Y )2 |Y = E(X|Y )2 = Z 2
since, in general, E f (Y )|Y = f (Y ): if Y = y, then E f (Y )|Y is defined to be
E f (Y )|Y = y = E f (y)|Y = y = f (y).
Therefore,
Var(X|Y ) = E(X2 |Y ) − 2Z 2 + Z 2 = E(X2 |Y ) − E(X|Y )2 .
21. By the definition of variance,
Var
N
Xi = E
i=1
2
N
Xi
i=1
− E
2
N
Xi
i=1
,
(42)
250
Chapter 10
More Expectations and Variances
where by Wald’s equation,
2
N
Xi
E
2
2
2
= E(X)E(N) = E(N) · E(X) .
(43)
i=1
Now since N is independent of {X1 , X2 , . . . },
2
N
Xi
E
=E E
N
N
2
Xi
i=1
i=1
∞
=
N
E
n=1
n
E
2
N = n P (N = n)
Xi
n=1
i=1
∞
=
N = n P (N = n)
i=1
∞
=
2
Xi
n
2
E
n=1
Xi
P (N = n).
i=1
Thus
2
N
E
Xi
∞
=
Xi2 + 2
E
n=1
i=1
n
i=1
Xi Xj P (N = n)
i 0 and ρ 2 = ρ
σY
1
=
σX
2
and
ρ
σX
1
= .
σY
2
σY
σX
1
·ρ
= . Therefore ρ = 1/2.
σX
σY
4
6. We use Theorem 8.8 to find the joint probability density function of X and Y . The joint
probability density function of Z and W is given by
1
1
exp − z2 + w2 .
2π
2
)
Let h1 (z, w) = σ1 z + µ1 and h2 (z, w) = σ2 ρz + 1 − ρ 2 w + µ2 . The system of equations
⎧
⎨σ1 z + µ1 = x
⎩σ ρz + )1 − ρ 2 w + µ = y
2
2
f (z, w) =
defines a one-to-one transformation of R2 in the zw-plane onto R2 in the xy-plane. It has a
unique solution
z=
x − µ1
,
σ1
w=)
1
1−ρ2
y − µ2 ρ(x − µ1 )
−
σ2
σ1
for z and w in terms of x and y. Moreover,
1
0
σ
1
1
=
J=
= 0.
)
ρ
1
σ1 σ2 1 − ρ 2
− )
)
σ1 1 − ρ 2
σ2 1 − ρ 2
Hence, by Theorem 8.8, the joint probability density function of X and Y is given by
x − µ
1
1
x − µ1
y − µ2
1
f
, )
−ρ
.
)
σ1
σ2
σ1
σ1 σ2 1 − ρ 2
1−ρ2
1
1
Noting that f (z, w) =
exp − z2 + w2 . Straightforward calculations will result in
2π
2
(10.24), showing that the joint probability density function of X and Y is bivariate normal.
254
Chapter 10
More Expectations and Variances
7. Using Theorem 8.8, it is straightforward to show that the joint probability density function of
X + Y and X − Y is bivariate normal. Since
ρ(X + Y, X − Y ) =
Var(X) − Var(Y )
Cov(X + Y, X − Y )
=
= 0,
σX+Y · σX−Y
σX+Y · σX−Y
X + Y and X − Y are uncorrelated. But for bivariate normal, uncorrelated and independence
are equivalent. So X + Y and X − Y are independent.
REVIEW PROBLEMS FOR CHAPTER 10
1. Number the last 10 graduates who will walk on the stage 1 through 10. Let Xi = 1 if the ith
graduate receives his or her own diploma; 0, otherwise. The number of graduates who will
receive their own diploma is X = X1 + X2 + · · · + Xn . Since
1
1
1
+0· 1−
= ,
n
n
n
E(Xi ) = 1 ·
we have
E(X) = E(X1 ) + E(X2 ) + · · · + E(Xn ) = n ·
2. Since
2
E(X) =
1
and
2
E(X 3 ) =
1
= 1.
n
5
(2x 2 − 2x) dx = ,
3
(2x 4 − 2x 3 ) dx =
1
49
,
10
we have that
E(X3 + 2X − 7) =
3. Since
E(X 2 ) =
and
1
E(XY ) =
3
we have that E(X 2 + 2XY ) =
1
3
0
1
0
1
2
37
49 10
+
−7=
.
10
3
30
(3x 5 + x 3 y) dy dx =
0
2
(3x 4 y + x 2 y 2 ) dy dx =
0
1 188
511
+
=
.
2 135
270
1
,
2
94
,
135
Chapter 10
Review Problems
255
4. Let X1 , X2 , . . . , Xn be geometric random variables with parameters 1, (n − 1)/n, (n − 2)/n,
. . . , 1/n, respectively. The desired quantity is
n
n
+
+ ··· + n
E(X1 + X2 + · · · + Xn ) = 1 +
n−1 n−2
1
1
1
+
+ · · · + + 1 = 1 + nan−1 .
=1+n
n−1 n−2
2
5. Let X be the number of tosses until 4 consecutive sixes. Let Y be the number of tosses until
the first non-six outcome is obtained. We have
∞
E(X) = E E(X|Y ) =
E(X | Y = i)P (Y = i)
i=1
∞
4
=
E(X | Y = i)P (Y = i) +
i=1
4
=
E(X | Y = i)P (Y = i)
i=5
i + E(X)
i=1
1
i−1 5
6
6
∞
+
4
i=5
1
i−1 5
6
6
.
This equation reduces to
5
1 5
1 2 5
E(X) = 1 + E(X) + 2 + E(X) · + 3 + E(X)
6
6 6
6
6
5 (1/6)4
1 3 5
+4
.
+ 4 + E(X)
6
6
6 1 − (1/6)
Solving this equation for E(X), we obtain E(X) = 1554.
6. f (x, y, z) = (2x)(2y)(2z), 0 < x < 1, 0 < y < 1, 0 < z < 1. Since 2x, 0 < x < 1
is a probability density function, 2y, 0 < y < 1 is a probability density function, and 2z,
0 < z < 1 is also a probability density function, these three functions are fX (x), fY (y), and
fZ (z), respectively. Therefore, f (x, y, z) = fX (x)fY (y)fZ (z) showing that X, Y , and Z are
independent. Thus
ρ(X, Y ) = ρ(Y, Z) = ρ(X, Z) = 0.
7. Since Cov(X, Y ) = σX σY ρ(X, Y ) = 2,
Var(3X − 5Y + 7) = Var(3X − 5Y ) = 9Var(X) + 25Var(Y ) − 15Cov(X, Y )
= 9 + 225 − 30 = 204.
8. Clearly,
pX (1) = p(1, 1) + p(1, 3) = 12/25,
pX (2) = p(2, 3) = 13/25;
pY (1) = p(1, 1) = 2/25,
pY (3) = p(1, 3) + p(2, 3) = 23/25.
256
Chapter 10
More Expectations and Variances
Therefore,
pX (x) =
⎧
⎨12/25
if x = 1
⎩13/25
if x = 2,
pY (y) =
⎧
⎨2/25
if y = 1
⎩23/25
if y = 3.
These yield
12
13
38
+2·
= ;
25
25
25
2
23
71
E(Y ) = 1 ·
+3·
= ;
25
25
25
1
1
1
22
E(XY ) = (1)(1) (12 + 12 ) + (1)(3) (12 + 32 ) + (2)(3) (22 + 32 ) =
.
25
25
25
5
E(X) = 1 ·
Thus
Cov(X, Y ) = E(XY ) − E(X)E(Y ) =
22 38 71
52
−
·
=
.
5
25 25
625
9. In Exercise 6, Section 8.1, we calculated p(x, y), pX (x), and pY (y). The results of that
exercise yield
12
E(X) =
xpX (x) = 7;
x=2
5
E(Y ) =
ypY (y) = 35/18;
y=0
12
5
E(XY ) =
xyp(x, y) = 245/18.
x=2 y=0
Therefore,
Cov(X, Y ) = E(XY ) − E(X)E(Y ) = (245/18) − 7(35/18) = 0.
This shows that X and Y are uncorrelated. Note that X and Y are not independent as the
following shows.
1/36 = p(2, 0) = pX (2)pY (0) = (1/36)(6/36) = 1/216.
10. Let p be the probability mass function of |X −Y |, q be the probability mass function of X +Y ,
and r be the probability mass function of |X2 − Y 2 |. We have
x
p(x)
0
726/1296
1
520/1296
2
50/1296,
Chapter 10
0
625/1296
x
q(x)
x
r(x)
1
500/1296
0
726/1296
2
150/1296
1
500/1296
3
20/1296
3
20/1296
Review Problems
257
4
1/1296,
4
50/1296.
Using these we obtain
760
E |X2 − Y 2 | =
,
1296
720
,
E |X − Y |2 =
1296
σX+Y = 0.831.
620
E |X − Y | =
,
1296
E (X + Y )2 = 1,
E(X + Y ) =
864
,
1296
σ|X−Y | = 0.572,
Therefore,
Cov |X − Y |, X + Y
ρ |X − Y |, |X + Y | =
σ|X−Y | · σX+Y
2
E |X − Y 2 | − E |X − Y | E(X + Y )
=
σ|X−Y | · σX+Y
=
(760/1296) − (620/1296)(864/1296)
= 0.563.
(0.831)(0.572)
11. One way to solve this problem is to note that the desired probability is the area of the region
under the curve y = sin x from x = 0 to x = π/2 divided by the area of the rectangle
[0, π/2] × [0, 1]. Hence it is
π/2
sin x dx
2
0
= .
π/2
π
A second way to find this probability is to note that (X, Y ) lies below the curve y = sin x if
and only if Y < sin X. Noting that f , the probability density function of X is given by
⎧
⎪
⎨ 2 if 0 < x < π
2
f (x) = π
⎪
⎩0 otherwise,
and conditioning on X, we obtain
π/2
π/2
sin x − 0 2
P (Y < sin X | X = x)f (x)dx =
· dx
P (Y < sin X) =
1−0
π
0
0
π/2
2
2
= − cos x = .
π
π
0
258
Chapter 10
More Expectations and Variances
12. (a) Clearly,
x
fX (x) =
e−x dy = xe−x ,
0 < x < ∞,
0
∞
fY (y) =
e−x dx = e−y ,
0 < y < ∞.
y
(b) we have that
∞
E(X) =
x 2 e−x = 2,
E(Y ) =
0
E X2 =
∞
ye−y = 1,
0
∞
E Y2 =
x 3 e−x dx = 6,
0
∞
y 2 e−y = 2.
0
Therefore, Var(X) = 2 and Var(Y ) = 1. Also
∞
E(XY ) =
0
∞
e−x dx dy = 3.
y
Thus
1
E(XY ) − E(X)E(Y )
3−2
=√ .
=√
σX σY
2·1
2
13. Let h(α, β) = E (Y − α − βX)2 . Then
h(α, β) = E Y 2 + α 2 + β 2 E X2 − 2αE(Y ) − 2βE(XY ) + 2αβE(X).
ρ(X, Y ) =
Setting
∂h
∂h
= 0 and
= 0, we obtain
∂α
∂β
α + E(X)β = E(Y )
E(X)α + E X2 β = E(XY ).
Solving this system of two equations in two unknowns, we obtain
Cov(X, Y )
ρσX σY
σY
=
=ρ ,
σX
σX2
σX2
σY
α = µY − ρ µX .
σX
β=
Therefore, Y = µY + ρ
14. We have that
∞
E(X) =
σY
(X − µX ).
σX
∞
xye
0
0
−y(1+x)
dy dx =
0
∞
x
1+x
0
∞
(1 + x)ye
−y(1+x)
dy
dx.
Chapter 10
Review Problems
259
*∞
Now 0 (1 + x)ye−y(1+x) dy is the expected value of an exponential random variable with
parameter 1 + x, so it is 1/(1 + x). Letting u = 1 + x, we have
∞
∞
x
u−1
E(X) =
dx =
du
2
(1 + x)
u2
0
1
∞
∞
∞
1
1
− 1 = ∞.
=
du −
du
=
ln
u
u
u2
1
1
1
(b) To find E(X|Y ), note that
∞
E(X | Y = y) =
xfX|Y (x|y) dx =
0
where
fY (y) =
∞
ye
x
0
−y(1+x)
dx = e
−y
0
∞
∞
f (x, y)
dx,
fY (y)
ye−yx dx = e−y .
0
*∞
Note that 0 ye−yx dx = 1 because ye−yx is the probability density function of an exponential
random variable with parameter 1. So
∞
∞
ye−y e−yx
1
E(X | Y = y) =
x
dx
=
xye−xy dx = ,
−y
e
y
0
0
where the last equality holds because the last integral is the expected value of an exponential
random variable with parameter y. Since ∀y > 0, E(X | Y = y) = 1/y, E(X|Y ) = 1/Y.
15. Let X and Y denote the number of minutes past 10:00 A.M. that bus A and bus B arrive at
the station, respectively. X is uniformly distributed over (0, 30). Given that X = x, Y is
uniformly distributed over (0, x). Let f (x, y) be the joint probability density function of X
and Y . We calculate E(Y ) by conditioning on X:
∞
30
x 1
30
E(Y ) = E E(Y |X) =
E(Y | X = x)fX (x) dx =
·
dx =
.
2
30
4
−∞
0
Thus the expected arrival time of bus B is 7.5 minutes past 10:00 A.M.
16. To find the distribution function of
N
i=1
∞
N
Xi ≤ t =
P
Xi , note that
n=1
∞
i=1
N
P
=
i=1
n
P
n=1
∞
=
Xi ≤ t N = n P (N = n)
i=1
n
Xi ≤ t P (N = n),
P
n=1
Xi ≤ t N = n P (N = n)
i=1
260
Chapter 10
More Expectations and Variances
where the last inequality follows since N is independent of X1 , X2 , X3 , . . . . Now ni=1 Xi
is a gamma random variable with parameters n and λ. Thus
(λx)n−1
dx (1 − p)n−1 p
Xi ≤ t =
λe
(n
−
1)!
0
n=1
n−1
∞ t
λ(1 − p)x
−λx
dx
λpe
=
(n − 1)!
n=1 0
n−1
t
∞
λ(1 − p)x
−λx
=
λpe
dx
(n − 1)!
0
n=1
t
λpe−λx eλ(1−p)x dx
=
0 t
λpe−λpx dx = 1 − e−λpt .
=
∞
N
P
i=1
t
−λx
0
This shows that
N
i=1
Xi is exponential with parameter λp.
17. Let
X1 , X2 , . . . , Xi , . . . , X20 be geometric random variables with parameters 1, 19/20, . . . ,
20 − (i − 1) /20, . . . , 1/20. The desired quantity is
20
20
E(Xi ) =
Xi =
E
i=1
20
i=1
i=1
20
= 71.9548.
20 − (i − 1)
Chapter 11
S ums of I ndependent
R andom Variables
and L imit Theorems
11.1
MOMENT-GENERATING FUNCTIONS
5
1. MX (t) = E etX =
etx p(x) =
x=1
2. (a) For t = 0,
MX (t) = E e
1 t
e + e2t + e3t + e4t + e5t .
5
tX
=
3
−1
1 tx
1 e3t − e−t
e dx =
,
4
4
t
whereas for t = 0, MX (0) = 1. Thus
⎧ 3t
−t
⎪
⎨1 e − e
t
MX (t) = 4
⎪
⎩
1
if t = 0
if t = 0.
2
3 − (−1)
−1 + 3
4
Since X is uniform over (−1, 3), E(X) =
= 1 and Var(X) =
= .
2
12
3
(b) By the definition of derivative,
MX (h) − MX (0)
1 e3h − e−h
= lim
−1
h→0
h→0 h
h
4h
E(X) = MX (0) = lim
e3h − e−h − 4h
3e3h + e−h − 4
9e3h − e−h
=
lim
=
lim
= 1,
h→0
h→0
h→0
4h2
8h
8
= lim
where the fifth and sixth equalities follow from L’Hôpital’s rule.
262
Chapter 11
Sums of Independent Random Variables and Limit Theorems
3. Note that
MX (t) = E etX =
∞
etx · 2
1
x=1
x
3
∞
∞
etx · e−x ln 3 = 2
=2
x=1
ex(t−ln 3) .
x=1
Restricting the domain of MX (t) to the set t : t < ln 3 and using the geometric series
theorem, we get
et−ln 3
2et
MX (t) = 2
.
=
3 − et
1 − et−ln 3
(Note that e− ln 3 = 1/3.) Differentiating MX (t), we obtain
6et
MX (t) =
3 − et
2 ,
which gives E(X) = MX (0) = 3/2.
4. For t = 0, MX (0) = 1. For t = 0, using integration by parts, we obtain
MX (t) =
1
2xetx dx =
0
2et
2
2et
− 2 + 2.
t
t
t
5. (a) For t = 0, MX (0) = 1. For t = 0,
MX (t) =
1
e · 6x(1 − x) dx = 6
tx
0
xe dx − 6
t
−
t
e
1
+ 2
2
t
t
1
x 2 etx dx
tx
0
et
=6
1
0
et
12(1 − et ) 6(1 + et )
2et
2et
2
−6
− 2 + 3 − 3 =
+
.
t
t
t
t
t3
t2
(b) By the definition of derivative,
12(1 − et ) 6(1 + et )
+
−1
MX (t) − MX (0)
t3
t2
= lim
E(X) = MX (0) = lim
t→0
t→0
t
t
12(1 − et ) + 6t (1 + et ) − t 3
1
= ,
4
t→0
t
2
= lim
where the last equality is calculated by applying L’Hôpital’s rule four times.
6. Let A be the set of possible values of X. Clearly, MX (t) =
x∈A
etx p(x), where p(x) is the
Section 11.1
Moment-Generating Functions
263
probability mass function of X. Therefore,
MX (t) =
xetx p(x),
x∈A
MX (t)
=
x 2 etx p(x),
x∈A
..
.
MX(n) (t) =
x n etx p(x).
x∈A
Therefore,
MX(n) (0) =
x n p(x) = E(Xn ).
x∈A
7. (a) By definition,
MX (t) = E e
tX
∞
=
e
x=0
(b) From
and
tx e
∞
λ
(λet )x
= e−λ
= e−λ exp(λet ) = exp λ(et − 1) .
x!
x!
x=0
−λ x
MX (t) = λet exp λ(et − 1)
2
MX (t) = λet exp λ(et − 1) + λet exp λ(et − 1) ,
we obtain E(X) = MX (0) = λ and E(X2 ) = MX (0) = λ2 + λ. Therefore,
Var(X) = (λ2 + λ) − λ2 = λ.
8. The probability density function of X is given by
⎧
⎪
⎨
f (x) =
1
b−a
⎪
⎩0
if a < x < b
otherwise.
Therefore, for t = 0,
MX (t) = E etX =
a
b
1 etb − eta
1 tx
e dx =
,
b−a
b−a
t
whereas for t = 0, MX (0) = 1. Thus
⎧
etb − eta
⎨ 1
MX (t) = b − a
t
⎩
1
if t = 0
if t = 0.
264
Chapter 11
Sums of Independent Random Variables and Limit Theorems
9. The probability mass function of a geometric random variable X, p(x) with parameter p is
given by
p(x) = pq x−1 ,
Thus
q = 1 − p,
∞
MX (t) =
pq x−1 etx =
x=1
∞
p
q
∞
x
qet .
x=1
Now by the geometric series theorem, x=1 qe converges to qet / 1 − qet if qet < 1
or, equivalently, if t < − ln q. Restricting the domain of MX (t) to the set {t : t < − ln q}, we
obtain
∞
t x
p
p
pet
qet
qe = ·
=
.
MX (t) =
q x=1
q 1 − qet
1 − qet
Now
MX (t) =
pet
(1 − qet )2
t x
x = 1, 2, 3, . . . .
MX (t) =
and
Therefore,
p
1
= .
(1 − q)2
p
E(X) = MX (0) =
and
E(X2 ) = MX (0) =
Thus
pet + pqe2t
.
(1 − qet )3
p(1 + q)
1+q
=
.
3
(1 − q)
p2
2 1 + q
1
q
Var(X) = E(X2 ) − E(X) =
− 2 = 2.
2
p
p
p
10. Let X be a discrete random variable with the probability mass function p(x) = x/21, x =
1, 2, 3, 4, 5, 6. The moment-generating function of X is the given function.
11. X is a discrete random variable with the set of possible values {1, 3, 4, 5} and probability mass
function
x
p(x)
1
5/15
3
4/15
4
2/15
5
4/15.
12. We have that
M2X+1 (t) = E e(2X+1)t = et E e2tX = et MX (2t) =
13. Note that
MX (t) =
24
,
(2 − t)4
MX (t) =
et
,
1 − 2t
96
.
(2 − t)5
Therefore,
24
3
= ,
16
2
and hence Var(X) = 3 − (9/4) = 3/4.
E(X) = MX (0) =
t<
E(X2 ) = MX (0) =
96
= 3,
32
1
.
2
Section 11.1
Moment-Generating Functions
265
14. Since for odd r’s, MX(r) (t) = (et − e−t )/6 and for even r’s, MX(r) (t) = (et + e−t )/6, we have
that E(Xr ) = 0 if r is odd and E(Xr ) = 1/3 if r is even.
15. For a random variable X, we must have MX (0) = 1. Since t/(1 − t) is 0 at 0, it cannot be a
moment-generating function.
16. (a) The distribution of X is binomial with parameters 7 and 1/4.
(b) The distribution of X is geometric with parameter 1/2.
(c) The distribution of X is gamma with parameters r and 2.
(d) The distribution of X is Poisson with parameter λ = 3.
17. Since
1
2 4
,
3
3
X is a binomial random variable with parameters 4 and 1/3; therefore,
2
4 1 i 2 4−i
8
P (X ≤ 2) =
= .
3
9
i 3
i=0
MX (t) =
18. By relation (11.2),
∞
MX (t) =
n=0
et +
2n n
t =
n!
∞
n=0
(2t)n
= e2t .
n!
This shows that X = 2 with probability 1.
19. We know that for t = 0,
MX (t) =
et − 1
et − 1
=
.
t (1 − 0)
t
Therefore, for t = 0,
MaX+b (t) = E et (aX+b) = ebt E eatX = ebt MX (at)
= ebt ·
eat − 1
e(a+b)t − ebt
,
=
at
(a + b) − b t
which is the moment-generating function of a uniform random variable over (b, a + b).
20. Let µn = E(Z n ); then
∞
MX (t) =
Now et =
∞
n=0 (t
n=0
n
MXn (0) n
t =
n!
∞
n=0
µn n
t .
n!
/n!). Therefore,
e
t 2 /2
∞
=
n=0
(t 2 /2)n
=
n!
∞
n=0
t 2n
=
2n n!
∞
n=0
(2n)! t 2n
.
2n n! (2n)!
(44)
266
Chapter 11
Sums of Independent Random Variables and Limit Theorems
(2n)!
comparing this relation with (44), we obtain E(Z 2n+1 ) = 0, ∀n ≥ 0 and E(Z 2n ) = n ,
2 n!
∀n ≥ 1.
21. By definition,
MX (t) =
λr
(r)
∞
etx x r−1 e−λx dx =
0
λr
(r)
∞
e(t−λ)x x r−1 dx.
0
This integral converges if t < λ. Therefore, if we restrict the range of MX (t) to t < λ, by the
substitution u = (λ − t)x, we obtain
λr
MX (t) =
(r)
∞
0
λ
λr
(r)
e−u ur−1
du
=
·
=
(λ − t)r
(r) (λ − t)r
λ−t
r
.
Now MX (t) = rλr (λ − t)−r−1 ; thus E(X) = MX (0) = r/λ. Also
MX (t) = r(r + 1)λr (λ − t)−r−2 ;
therefore, E(X2 ) = MX (0) = r(r + 1) /λ2 , and hence
Var(X) =
r(r + 1) r
−
λ2
λ
2
=
r
.
λ2
22. (a) Let F be the distribution function of X. We have that
P (−X ≤ t) = P (X ≥ −t) =
∞
f (x) dx.
−t
Letting u = −x and noting that f (−u) = f (u), we obtain
P (−X ≤ t) =
−∞
f (−u) (− du) =
t
t
−∞
f (u) du = F (t).
This shows that the distribution function of −X is also F .
(b) Clearly,
∞
MX (−t) =
e−tx f (x) dx.
−∞
Letting u = −x, we get
MX (−t) =
∞
−∞
e f (−u) du =
tu
∞
−∞
etu f (u) du = MX (t).
A second way to explain this is to note that MX (−t) is the moment-generating function of
−X. Since X and −X are identically distributed, we must have that MX (t) = MX (−t).
Section 11.1
23. Note that
MX (t) = E etX =
∞
x=1
Moment-Generating Functions
6 tx
6
e = 2
2
2
π x
π
∞
x=1
267
etx
.
x2
Now by the ratio test,
et (x+1) /(x + 1)2
x2
=
lim
et = et
x→∞
x→∞ x 2 + 2x + 1
etx /x 2
lim
∞ etx
x=1 2 diverges on (0, ∞) and thus on no interval
x
of the form (−δ, δ), δ > 0, MX (t) exists.
which is > 1 for t ∈ (0, ∞). Therefore,
24. For t < 1/2, (11.2) implies that
∞
MX (t) =
n=0
E(Xn ) n
t =
n!
∞
∞
1
(n + 1)(2t) =
2
n=0
1 d
·
2 dt
n=0
1/2 2
1
=
.
=
(1 − 2t)2
(1/2) − t
=
1 d
2 dt
∞
n
(2t)n+1 =
n=0
d
(2t)n+1
dt
1 d
1
(2t)n − 1 = ·
−1
2 dt 1 − 2t
n=0
∞
We see that for t < 1/2, MX (t) exists; furthermore, it is the moment-generating function of a
gamma random variable with parameters r = 2 and λ = 1/2.
25. (a) At the end of the first period, with probability 1, the investment will grow to
A+A
X
X
=A 1+
;
k
k
at the end of the second period, with probability 1, it will grow to
X
X
X
X 2
A 1+
+A 1+
·
=A 1+
;
k
k
k
k
(b)
X n
and, in general, at the end of the nth period, with probability 1, it will grow to A 1+
.
k
Dividing a year into k equal periods allows the banks to compound interest quarterly,
monthly, or daily. If we increase k, we can compound interest every minute, second,
or even fraction of a second. For an infinitesimal ε > 0, suppose that the interest
is compounded at the end of each period of length ε. If ε → 0, then the interest is
compounded continuously. Since a year is 1/ε periods, each of length ε, the interest
rate per period of length ε is the random variable X/(1/ε) = εX. Suppose that at time
t, the investment has grown to A(t). Then at t + ε, with probability 1, the investment
will be
A(t + ε) = A(t) + A(t) · εX.
268
Chapter 11
Sums of Independent Random Variables and Limit Theorems
This implies that
P
A(t + ε) − A(t)
= XA(t) = 1.
ε
Letting ε → 0, yields
A(t + ε) − A(t)
P lim
= XA(t) = 1
ε→0
ε
or, equivalently, with probability 1,
A (t) = XA(t).
(c)
Part (b) implies that, with probability 1,
A (t)
= X.
A(t)
Integrating both sides of this equation, we obtain that, with probability 1,
ln[A(t)] = tX + C,
or
A(t) = etX+c .
Considering the fact that A(0) = A, this equation yields A = ec . Therefore, with
probability 1,
A(t) = etX · ec = AetX .
This shows that if the interest rate is compounded continuously, then an initial investment
of A dollars will grow, in t years, with probability 1, to the random variable AetX , whose
expected value is
E(AetX ) = AE(etX ) = AMX (t).
We have shown the following:
If money is invested in a bank at an annual rate X, where X is a random
variable, and if the bank compounds interest continuously, then, on average, the money will grow by a factor of MX (t), the moment-generating
function of the interest rate.
26. Since Xi and Xj are binomial with parameters (n, pi ) and (n, pj ),
E(Xi ) = npi ,
)
σXi = npi (1 − pi ),
E(Xj ) = npj ,
)
σXj = npj (1 − pj ).
Section 11.2
Sums of Independent Random Variables
To find E(Xi Xj ), note that
M(t1 , t2 ) = E et1 Xi +t2 Xj
n
n−xi
et1 xi +t2 xj P (Xi = xi , Xj = xj )
=
xi =0 xj =0
n
n−xi
et1 xi +t2 xj ·
=
xi =0 xj =0
n!
x
pixi pj j (1 − pi − pj )n−xi −xj
xi ! xj ! (n − xi − xj )!
n−xi
t1 xi t2 xj
n!
e pi
e pj (1 − pi − pj )n−xi −xj
x
!
x
!
(n
−
x
−
x
)!
i
j
xi =0 xj =0 i j
n
t1
= pi e + pj et2 + 1 − pi − pj ,
n
=
where the last equality follows from multinomial expansion (Theorem 2.6). Therefore,
n−2
∂ 2M
(t1 , t2 ) = n(n − 1)pi pj et1 et2 pi et1 + pj et2 + 1 − pi − pj
,
∂t1 ∂t2
and so
E(Xi Xj ) =
Thus
11.2
∂ 2M
(0, 0) = n(n − 1)pi pj .
∂t1 ∂t2
(
pi pj
n(n − 1)pi pj − (npi )(npj )
ρ(Xi , Xj ) = √
.
=−
)
(1 − pi )(1 − pj )
npi (1 − pi ) · npj (1 − pj )
SUMS OF INDEPENDENT RANDOM VARIABLES
1. MαX (t) = E etαX = MX (tα) = exp αµt + (1/2)α 2 σ 2 t 2 .
2. Since
MX1 +X2 +···+Xn (t) = MX1 (t)MX2 (t) · · · MXn (t) =
n
pet
,
1 − (1 − p)et
X1 + X2 + · · · + Xn is negative binomial with parameters (n, p).
3. Since
MX1 +X2 +···+Xn (t) = MX1 (t)MX2 (t) · · · MXn (t) =
X1 + X2 + · · · + Xn is gamma with parameters n and λ.
λ
λ−t
n
,
269
270
Chapter 11
Sums of Independent Random Variables and Limit Theorems
4. For 1 ≤ i ≤ n, let Xi be negative binomial with parameters ri and p. We have that
M X1 +X2 +···+Xn (t) = MX1 (t)MX2 (t) · · · MXn (t)
=
r1
r2
rn
pet
pet
pet
·
·
·
1 − (1 − p)et
1 − (1 − p)et
1 − (1 − p)et
=
r1 +r2 +···+rn
pet
.
1 − (1 − p)et
Thus X1 + X2 + · · · + Xr is negative binomial with parameters r1 + r2 + · · · + rn and p.
5. Since
MX1 +X2 +···+Xn (t) = MX1 (t)MX2 (t) · · · MXn (t)
λ r1 λ r2
λ
=
···
λ−t
λ−t
λ−t
λ r1 +r2 +···+rn
=
,
λ−t
rn
X1 + X2 + · · · + Xn is gamma with parameters r1 + r2 + · · · + rn and λ.
6. By Theorem 11.4, the total number of underfilled bottles is binomial with parameters 180 and
0.15. Therefore, the desired probability is
180
(0.15)27 (0.85)153 = 0.083.
27
7. For j < i, P (X = i | X + Y = j ) = 0. For j ≥ i,
P (X = i | X + Y = j ) =
P (X = i)P (Y = j − i)
P (X = i, Y = j − i)
=
P (X + Y = j )
P (X + Y = j )
n i
n
m
m
n−i
j −i
m−(j −i)
p (1 − p) ·
p (1 − p)
i
i j −i
j −i
.
=
=
n+m j
n+m
n+m−j
p (1 − p)
j
j
Interpretation: Given that in n + m trials exactly j successes have occurred, the probability
mass function of the number of successes in the first n trials is hypergeometric. This should
be intuitively clear.
Section 11.2
Sums of Independent Random Variables
271
8. Since X + Y + Z is Poisson with parameter λ1 + λ2 + λ3 and X + Z is Poisson with parameter
λ1 + λ3 , we have that
P (Y = y | X + Y + Z = t) =
P (Y = y, X + Z = t − y)
P (X + Y + Z = t)
e−λ2 λ2 e−(λ1 +λ3 ) (λ1 + λ3 )t−y
·
y!
(t − y)!
y
=
=
e−(λ1 +λ2 +λ3 ) (λ1 + λ2 + λ3 )t
t!
t
λ2
y λ1 + λ2 + λ3
y
λ1 + λ3
λ1 + λ2 + λ3
t−y
.
9. Let X be the remaining calling time of the person in the booth. Let Y be the calling time of the
person ahead of Mr. Watkins. By the memoryless property of exponential, X is exponential
with parameter 1/8. Since Y is also exponential with parameter 1/8, assuming that X and Y
are independent, the waiting time of Mr. Watkins, X + Y , is gamma with parameters 2 and
1/8. Therefore,
∞
5
1 −x/8
P (X + Y ≥ 12) =
xe
dx = e−3/2 = 0.558.
2
12 64
10. By Theorem 11.7, X + Y ∼ N(5, 9), X − Y ∼ N (−3, 9), and 3X + 4Y ∼ N (19, 130). Thus
X + Y − 5
0−5
>
= 1 − (−1.67) = (1.67) = 0.9525,
3
3
X − Y + 3
2+3
<
= (1.67) = 0.9525,
P (X − Y < 2) = P
3
3
P (X + Y > 0) = P
and
P (3X + 4Y > 20) = P
3X + 4Y − 19
20 − 19
> √
=1−
√
130
130
(0.9) = 0.4641.
11. Theorem 11.7 implies that X̄ ∼ N(110, 1.6), where X̄ is the average of the IQ’s of the
randomly selected students. Therefore,
112 − 110
X̄ − 110
≥ √
=1−
P ( X̄ ≥ 112) = P
√
1.6
1.6
(1.58) = 0.0571.
12. Let X̄1 be the average of the accounts selected at store 1 and X̄2 be the average of the accounts
selected at store 2. We have that
900
= N(90, 90)
X̄1 ∼ N 90,
10
and
500
2500
X̄2 ∼ N 100,
= N 100,
.
15
3
272
Chapter 11
Sums of Independent Random Variables and Limit Theorems
770
Therefore, X̄1 − X̄2 ∼ N − 10,
and so
3
X̄1 − X̄2 + 10
0 + 10
P ( X̄1 > X̄2 ) = P ( X̄1 − X̄2 > 0) = P
>√
√
770/3
770/3
= 1 − (0.62) = 0.2676.
13. By Exercise 6, Section 10.5, X and Y are sums of independent standard normal random
variables. Hence αX + βY is a linear combination of independent standard normal random
variables. Thus, by Theorem 11.7, αX + βY is normal.
14. By Exercise 13, X − Y is normal; its mean is 71 − 60 = 11, its variance is
Var(X − Y ) = Var(X) + Var(Y ) − 2Cov(X, Y )
= Var(X) + Var(Y ) − 2ρ(X, Y )σX σY
= 9 + (2.7)2 − 2(0.45)(3)(2.7) = 9.
Therefore,
P (X − Y ≥ 8) = P
X − Y − 11
8 − 11
≥
=1−
3
3
(−1) =
(1) = 0.8413.
15. Let X̄ be the average of the weights of the 12 randomly selected athletes. Let X1 , X2 , . . . ,
X12 be the weights of these athletes. Since
625
252
= N 225,
,
X̄ ∼ N 225,
12
12
we have that
2700
= P ( X̄ ≤ 225)
P (X1 + X2 + · · · + X12 ≤ 2700) = P X̄ ≤
12
225 − 225
X̄ − 225
≤ √
=
=P √
625/12
625/12
(0) =
1
.
2
16. Let X̄1 and X̄2 be the averages of the final grades of the probability and calculus courses
Dr. Olwell teaches, respectively. We have that
418
X̄1 ∼ N 65,
= N (65, 19) and
22
448
X̄2 ∼ N 72,
= N(72, 16).
28
Therefore, X̄1 − X̄2 ∼ N(−7, 35) and hence the desired probability is
P |X̄1 − X̄2 | ≥ 2 = P ( X̄1 − X̄2 ≥ 2) + P ( X̄1 − X̄2 ≤ −2)
2+7
−2 + 7
X̄1 − X̄2 + 7
X̄1 − X̄2 + 7
≥ √
≤ √
=P
+P
√
√
35
35
35
35
=1−
(1.52) +
(0.85) = 1 − 0.9352 + 0.8023 = 0.8671.
Section 11.2
Sums of Independent Random Variables
273
17. Let X and Y be the lifetimes of the mufflers of the first and second cars, respectively.
(a) To calculate the desired probability, P (|X − Y | ≥ 1.5), note that by symmetry,
P |X − Y | ≥ 1.5 = 2P (X − Y ≥ 1.5).
Now X − Y ∼ N(0, 2), hence
1.5 − 0
X−Y −0
=2 1−
≥ √
P |X − Y | ≥ 1.5 = 2P
√
2
2
(1.06) = 0.289.
(b) Let Z be the lifetime of the first muffler the family buys. By symmetry, the desired
probability is
2P (Y > X + Z) = 2P (Y − X − Z > 0).
Now Y − X − Z ∼ N(−3, 3). Hence
Y −X−Z+3
0+3
2P (Y − X − Z > 0) = 2P
> √
=2 1−
√
3
3
(1.73) = 0.0836.
18. Let n be the maximum number of passengers who can use the elevator and X1 , X2 , . . . , Xn
be the weights of n random passengers. We must have
P (X1 + X2 + · · · Xn > 3000) < 0.0003
or, equivalently,
P (X1 + X2 + · · · + Xn ≤ 3000) > 0.9997.
Let X̄ be the mean of the weights of the n random passengers. We must have
3000
> 0.9997.
P X̄ ≤
n
625
, we must have
Since X̄ ∼ N 155,
n
P
or
X̄ − 155
(3000/n) − 155
> 0.9997,
√ ≤
√
25/ n
25/ n
√
3000
155 n
> 0.9997.
√ −
25
25 n
Using Table 2 of the Appendix, this gives
√
3000
155 n
≥ 3.49
√ −
25
25 n
or, equivalently,
√
155n + 87.25 n − 3000 ≤ 0.
274
Chapter 11
Sums of Independent Random Variables and Limit Theorems
√
Since the roots of the quadratic equation 155n + 87.25 n − 3000 = 0 are (approximately)
√
√
n = 4.127 and n = −4.69, the inequality is valid if and only if
√
√
n + 4.69
n − 4.127 ≤ 0.
√
√
But n + 4.69 > 0, so the inequality is valid if and only if n − 4.127 ≤ 0 or n ≤ 17.032.
Therefore the answer is n = 17.
19. By Remark 9.3, the marginal joint probability mass function of X1 , X2 , . . . , Xk is multinomial
with parameters n and (p1 , p2 , . . . , pk , 1 − p1 − p2 − · · · − pk ). Thus, letting p = p1 + p2 +
· · · + pk and x = x1 + x2 + · · · + xk , we have that
p(x1 , x2 , . . . , xk ) =
n!
p x1 px2 · · · pkxk (1 − p)n−x .
x1 ! x2 ! · · · xk ! (n − x)! 1 2
This gives
P (X1 + X2 + · · · + Xk = i)
=
x1 +x2 +···+xk =i
n!
p1x1 p2x2 · · · pkxk (1 − p)n−i
x1 ! x2 ! · · · xk ! (n − i)!
n!
i!
(1 − p)n−i
p1x1 p2x2 · · · pkxk
i! (n − i)!
x
!
x
!
·
·
·
x
!
k
x1 +x2 +···+xk =i 1 2
n
=
(1 − p)n−i (p1 + p2 + · · · + pk )i
i
n i
=
p (1 − p)n−i .
i
=
This shows that X1 + X2 + · · · + Xk is binomial with parameters n and p = p1 + p2 + · · · + pk .
20. First note that if Y1 and Y2 are two exponential random variables each with rate λ, min(Y1 , Y2 )
is exponential with rate 2λ. Now let A1 , A2 , . . . , A11 be the customers in the line ahead of
Kim. Due to the memoryless property of exponential random variables, X1 , the time until
A1 ’s turn to make a call is exponential with rate 2(1/3) = 2/3. The time until A2 ’s turn to
call is X1 + X2 , where X2 is exponential with rate 2(1/3) = 2/3. Continuing this argument
and considering the fact that Kim is the 12th person waiting in the line, we have that the time
until Kim’s turn to make a phone call is X1 + X2 + · · · + X12 , where {X1 , X2 , . . . , X12 }
is an independent and identically distributed sequence of exponential random variables each
with rate 2/3. Hence the distribution of the waiting time of Kim is gamma with parameters
(12, 2/3). Her expected waiting time is 12(2/3) = 18.
11.3
MARKOV AND CHEBYSHEV INEQUALITIES
1. Let X be the lifetime (in months) of a randomly selected dollar bill. We are given that
E(X) = 22. By Markov inequality,
Section 11.3
Markov and Chebyshev Inequalities
275
22
= 0.37.
60
This shows that at most 37% of the one-dollar bills last 60 or more months; that is, at least
five years.
P (X ≥ 60) ≤
2. We have that P (X ≥ 2) = 2/5. Hence, by Markov’s inequality,
E(X)
2
= P (X ≥ 2) ≤
.
5
2
This gives E(X) ≥ 4/5.
E(X)
5
=
= 0.4545.
11
11
σ2
42 − 25
=
= 0.472.
(b) P (X ≥ 11) = P (X − 5 ≥ 6) ≤ P |X − 5| ≥ 6 ≤
36
36
3. (a) P (X ≥ 11) ≤
4. Let X be the lifetime of the randomly selected light bulb; we have
2500
P (X ≤ 700) ≤ P |X − 800| ≥ 100 ≤
= 0.25.
10, 000
5. Let X be the number of accidents that will occur tomorrow. Then
(a) P (X ≥ 5) ≤
2
= 0.4.
5
4
(b) P (X ≥ 5) = 1 −
i=0
e−2 2i
= 0.053.
i!
2
(c) P (X ≥ 5) = P (X − 2 ≥ 3) ≤ P |X − 2| ≥ 3 ≤ = 0.222
9
6. Let X be the IQ of a randomly selected student from this campus; we have
15
P (X > 140) ≤ P |X − 110| > 30 ≤
= 0.017.
900
Therefore, less than 1.7% of these students have an IQ above 140.
7. Let X be the waiting period from the time Helen orders the book until she receives it. We want
to find a so that P (X < a) ≥ 0.95 or, equivalently, P (X ≥ a) ≤ 0.05. But
P (X ≥ a) = P (X − 7 ≥ a − 7) ≤ P |X − 7| ≥ a − 7 ≤
4
.
(a − 7)2
So we should determine the value of a for which 4/(a − 7)2 ≤ 0.05; it is easily seen that
a ≥ 15.9 or a = 16. Therefore, Helen should order the book 16 days earlier.
276
Chapter 11
Sums of Independent Random Variables and Limit Theorems
8. By Markov’s inequality, P (X ≥ 2µ) ≤
µ
1
= .
2µ
2
9. P (X > 2µ) = P (X − µ > µ) ≤ P |X − µ| ≥ µ ≤
µ
1
= .
µ2
µ
10. We have that
P (38 < X̄ < 46) = P (−4 < X̄ − 42 < 4) = P |X̄ − 42| < 4
= 1 − P |X̄ − 42| ≥ 4 .
By (11.3),
P |X̄ − 42| ≥ 4 ≤
Hence
P (38 < X̄ < 46) ≥ 1 −
3
60
=
.
16(25)
20
17
3
=
= 0.85.
20
20
11. For i = 1, 2, . . . , n, let Xi be the IQ of the ith student selected at random. We want to find n,
so that
X1 + X2 + · · · + Xn
P −3<
− µ < 3 ≥ 0.92
n
or, equivalently,
P (|X̄ − µ| ≥ 3) ≤ 0.08.
Since E(Xi ) = µ and Var(Xi ) = 150, by (11.3),
P (|X̄ − µ| ≥ 3) ≤
150
.
32 · n
Therefore, all we need to do is to find n for which 150/(9n) ≤ 0.08. This gives n ≥
150/[9(0.08)] = 208.33. Thus the psychologist should choose a sample of size 209.
12. Let X1 , X2 , . . . , Xn be the random sample, µ be the expected value of the distribution, and σ 2
be the variance of the distribution. We want to find n so that
P (|X̄ − µ| < 2σ ) ≥ 0.98
or, equivalently,
P (|X̄ − µ| ≥ 2σ ) < 0.02.
By (11.3),
1
σ2
=
.
(2σ )2 · n
4n
Therefore, all we need to do is to make sure that 1/(4n) ≤ 0.02. This gives n ≥ 12.5. So a
sample of size 13 gives a mean which is within 2 standard deviations from the expected value
with a probability of at least 0.98.
P (|X̄ − µ| ≥ 2σ ) ≤
Section 11.3
Markov and Chebyshev Inequalities
277
13. Call a random observation success, if the operator is busy. Call it failure, if he is free. In
(11.5), let ε = 0.05 and α = 0.04; we have
n≥
1
= 2500.
4(0.05)2 (0.04)
Therefore, at least 2500 independent observations should be made to ensure that (1/n) ni=1
estimates p, the proportion of time that the airline operator is busy, with a maximum error of
0.05 with probability 0.96 or higher.
14. By (11.5),
1
= 1666.67.
4(0.05)2 (0.06)
Therefore, it suffices to flip the coin n = 1667 times independently.
E (X − µ)2n
2n
2n
15. P |X − µ| ≥ α = P |X − µ| ≥ α ≤
.
α 2n
E ekX
kX
kt
16. By Markov’s inequality, P (X > t) = P e > e ≤
.
ekt
n≥
17. By the Corollary of Cauchy-Schwarz Inequality (Theorem 10.3),
2
E(X − Y )
≤ E (X − Y )2 = 0.
This gives that E(X − Y ) = 0. Therefore,
2
Var(X − Y ) = E (X − Y )2 − E(X − Y ) = 0.
We have shown that X −Y is a random variable with mean 0 and variance 0; by Example 11.16,
P (X − Y = 0) = 1. So with probability 1, X = Y .
18. If Y = X with probability 1, Theorem 10.5 implies that ρ(X, Y ) = 1. Suppose that ρ(X, Y ) =
1; we show that X=Y with probability 1.) Note that E(X) = E(Y ) = (n + 1)/2, Var(X) =
Var(Y ) = (n2 − 1)/12, and σX = σY = (n2 − 1)/12. These and
1 = ρ(X, Y ) =
E(XY ) − E(X)E(Y )
σX σY
imply that E(XY ) = (2n2 + 3n + 1)/6. Therefore,
E (X − Y )2 = E(X2 − 2XY + Y 2 ) = E(X2 ) + E(Y 2 ) − 2E(XY )
2
2
= Var(X) + E(X) + Var(Y ) + E(Y ) − 2E(XY )
n2 − 1 n + 1 2 n2 − 1 n + 1 2 2n2 + 3n + 1
+
+
= 0.
+
−
=
12
2
12
2
3
E (X − Y )2 = 0 implies that with probability 1, X=Y (see Exercise 17 above).
278
Chapter 11
Sums of Independent Random Variables and Limit Theorems
19. By Markov’s inequality,
E etX
tX
1
1
P X ≥ ln α = P (tX ≥ ln α) = P e ≥ α ≤
= MX (t).
t
α
α
20. Using gamma function introduced in Section 7.4,
1
E(X) =
n!
E(X2 ) =
Hence
σX2
1
n!
∞
x n+1 e−x dx =
(n + 1)!
(n + 2)
=
= n + 1,
n!
n!
x n+2 e−x dx =
(n + 2)!
(n + 3)
=
= (n + 1)(n + 2).
n!
n!
0
∞
0
= (n + 1)(n + 2) − (n + 1)2 = n + 1. Now
P (0 < X < 2n + 2) = 1 − P (X ≥ 2n + 2),
and by Chebyshev’s inequality,
P (X ≥ 2n + 2) = P X − (n + 1) ≥ n + 1 ≤ P X − (n + 1) ≥ n + 1
n+1
1
≤
=
.
(n + 1)2
n+1
Therefore,
P (0 < X < 2n + 1) ≥ 1 −
11.4
1
n
=
.
n+1
n+1
LAWS OF LARGE NUMBERS
1. Since
E(Xi ) =
0
1
1
x · 4x(1 − x) dx = ,
3
by the strong law of large numbers,
X1 + X2 + · · · + Xn
1
P lim
=
= 1.
n→∞
n
3
2. If X1 > M with probability 1, then X2 > M with probability 1 since X1 and X2 are identically
distributed. Therefore, X1 + X2 > 2M > M with probability 1. This argument shows that
{X1 > M} ⊆ {X1 + X2 > M} ⊆ {X1 + X2 + X3 > M} ⊆ · · · .
Therefore, by the continuity of probability function (Theorem 1.8),
lim P (X1 + X2 + · · · + Xn > M) = P lim X1 + X2 + · · · + Xn > M .
n→∞
n→∞
Section 11.4
Laws of Large Numbers
279
By this relation, it suffices to show that ∀M > 0,
lim X1 + X2 + · · · + Xn > M
(45)
n→∞
with probability 1. Let S be the sample space over which Xi ’s are defined. Let µ = E(Xi );
we are given that µ > 0. By the central limit theorem,
X1 + X2 + · · · Xn
P lim
= µ = 1.
n→∞
n
Therefore, letting
X1 (ω) + X2 (ω) + · · · Xn (ω)
V = ω ∈ S : lim
=µ ,
n→∞
n
we have that P (V ) = 1. To establish (45), it is sufficient to show that ∀ω ∈ V ,
lim X1 (ω) + X2 (ω) + · · · Xn (ω) = ∞.
(46)
n→∞
To do so, applying the definition of limit to
lim
n→∞
X1 (ω) + X2 (ω) + · · · Xn (ω)
= µ,
n
we have that for ε = µ/2, there exists a positive integer N (depending on ω) such that ∀n > N,
X (ω) + X (ω) + · · · X (ω)
µ
2
n
1
− µ < ε =
n
2
or, equivalently,
−
µ
X1 (ω) + X2 (ω) + · · · Xn (ω)
µ
<
−µ< .
2
n
2
This yields
µ
X1 (ω) + X2 (ω) + · · · Xn (ω)
> .
n
2
Thus, for all n > N,
X1 (ω) + X2 (ω) + · · · Xn (ω) >
nµ
,
2
which establishes (46).
3. For 0 < ε < 1,
P |Yn − 0| > ε = 1 − P |Yn − 0| ≤ ε = 1 − P (X ≤ n) = 1 −
f (x) dx.
0
Therefore,
lim P |Yn − 0| > ε = 1 −
n→∞
showing that Yn converges to 0 in probability.
0
∞
n
f (x) dx = 1 − 1 = 0,
280
Chapter 11
Sums of Independent Random Variables and Limit Theorems
4. By the strong law of large numbers, Sn /n converges to µ almost surely. Therefore, Sn /n
converges to µ in probability and hence
Sn
lim P n(µ − ε) ≤ Sn ≤ n(µ + ε) = lim P µ − ε ≤
≤µ+ε
n→∞
n→∞
n
S
n
= lim P − µ ≤ ε
n→∞
n
S
n
= 1 − lim P − µ > ε = 1 − 0 = 1.
n→∞
n
5. Suppose that the bank will never be empty of customers again. We will show a contradiction.
Let Un = T1 + T2 + · · · + Tn . Then Un is the time the nth new customer arrives. Let Wi be the
service time of the ith new customer served. Clearly, W1 , W2 , W3 , . . . are independent and
identically distributed random variables with E(Wi ) = 1/µ. Let Zn = T1 +W1 +W2 +· · ·+Wn .
Since the bank will never be empty of customers, Zn is the departure time of the nth new
customer served. By the strong law of large numbers,
lim
n→∞
1
Un
=
n
λ
and
T
W1 + W2 + · · · + Wn
Zn
1
= lim
+
n→∞ n
n→∞ n
n
T1
W1 + W2 + · · · + Wn
1
1
+ lim
=0+ = .
= lim
n→∞ n
n→∞
n
µ
µ
lim
Clearly, the bank will never remain empty of customers again if and only if ∀n,
Un+1 < Zn .
This implies that
Un+1
Zn
<
n
n
or, equivalently,
Zn
n + 1 Un+1
·
<
.
n
n+1
n
Thus
lim
n→∞
n + 1 Un+1
Zn
·
≤ lim
n→∞
n
n+1
n
(47)
n+1
Un+1
Zn
1
1
= 1, and with probability 1, lim
=
and lim
= , (47)
n→∞
n→∞ n + 1
n→∞ n
n
λ
µ
1
1
implies that ≤
or λ ≥ µ. This is a contradiction to the fact that λ < µ. Hence, with
λ
µ
probability 1, eventually, for some period, the bank will be empty of customers again.
Since lim
Section 11.4
Laws of Large Numbers
281
6. Suppose that the bank will never be empty of customers again. We will show a contradiction.
Let Un = T1 + T2 + · · · + Tn . Then Un is the time the nth new customer arrives. Let R be the
sum of the remaining service time of the customer being served and the sums of the service
times of the m customers present in the queue at t = 0. Let Zn = R + S1 + S2 + · · · + Sn .
Since the bank will never be empty of customers, and customers are served on a first-come,
first-served basis, we have that U1 < R and hence Zn is the departure time of the nth new
customer. By the strong law of large numbers,
Un
1
=
n→∞ n
λ
lim
and
R S + S + · · · + S
Zn
1
2
n
= lim
+
n→∞ n
n→∞ n
n
1
1
R
S1 + S2 + · · · + Sn
+ lim
=0+ = .
= lim
n→∞ n
n→∞
n
µ
µ
lim
Clearly, the bank will never remain empty of customers if and only if ∀n,
Un+1 < Zn .
This implies that
Un+1
Zn
<
n
n
or, equivalently,
n + 1 Un+1
Zn
·
<
.
n
n+1
n
Thus
lim
n→∞
n + 1 Un+1
Zn
·
≤ lim
n
n + 1 n→∞ n
(48)
n+1
Un+1
Zn
1
1
= 1, and with probability 1, lim
=
and lim
= , (48)
n→∞
n→∞
n
n+1
λ
n
µ
1
1
implies that ≤
or λ ≥ µ. This is a contradiction to the fact that λ < µ. Hence, with
λ
µ
probability 1, eventually, for some period, the bank will be empty of customers.
7. Xn converges to 0 in probability because for every ε > 0, P |Xn − 0| ≥ ε is the probability
i i + 1
that the random point selected from [0, 1] is in k , k . Now n → ∞ implies that 2k → ∞
2
2
i i + 1
and the length of the interval k , k
→ 0, Therefore, limn→∞ P |Xn − 0| ≥ ε = 0.
2
2
However, Xn does not converge at any point because for all positive natural number N, there
are always m > N and n > N, such that Xm = 0 and Xn = 1 making it impossible for
|Xn − Xm | to be less than a given 0 < ε < 1.
Since lim
n→∞
282
11.5
Chapter 11
Sums of Independent Random Variables and Limit Theorems
CENTRAL LIMIT THEOREM
1. Let X1 , X2 , . . . , X150 be the random points selected from the interval (0,
√ 1). For 1 ≤ i ≤ 150,
Xi is uniform over (0, 1). Therefore, E(Xi ) = µ = 0.5 and σXi = 1/ 12. We have
X1 + X2 + · · · + X150
P 0.48 <
< 0.52 = P (72 < X1 + X2 + · · · + X150 < 78)
150
72 − (150)(0.5)
X1 + X2 + · · · + X150 − (150)(0.5)
78 − (150)(0.5)
=P √
< √
√
√ <
√
√
150 1/ 12
150 1/ 12
150 1/ 12
≈
(0.85) −
(−0.85) = 2 (0.85) − 1 = 2(0.8023) − 1 = 0.6046.
2. For 1 ≤ i ≤ 35, let Xi be the score of the ith student selected at random. By the central limit
theorem
X1 + X2 + · · · + X35
P (460 < X̄ < 540) = P 460 <
< 540
35
= P (16100 < X1 + X2 + · · · + X35 < 18900)
X1 + X2 + · · · + X35 − 35(500)
18900 − 35(500)
16100 − 35(500)
<
<
=P
√
√
√
100 35
100 35
100 35
X1 + X2 + · · · + X35 − 35(500)
< 2.37
= P − 2.37 <
√
100 35
=
(2.37) −
(−2.37) = 0.9911 − 0.0089 = 0.9822.
3. We have that
µ=
3
1
E(X2 ) =
1
3
5
56
1
x x+
dx =
= 2.07,
9
2
27
125
1 2
5
x x+
dx =
,
9
2
27
)
σX = (125/27) − (56/27)2 = 0.57.
The desired probability is
X1 + X2 + · · · + X24
P (2 < X̄ < 2.15) = P 2 <
< 2.15
24
= P (48 < X1 + X2 + · · · + X24 < 51.6)
Section 11.5
X1 + X2 + · · · + X24 − 24(2.07)
51.6 − 24(2.07)
48 − 24(2.07)
<
<
=P
√
√
√
0.57 24
0.57 24
0.57 24
≈
(0.69) −
283
Central Limit Theorem
(−0.60) = 0.7549 − 0.2743 = 0.4806.
4. Let X1 , X2 , . . . , Xn be the sample. Since f is an even function, for 1 ≤ i ≤ n,
∞
1 −|x|
xe
dx = 0
2
−∞
∞
∞
1 2 −|x|
2
E(Xi ) =
x e
dx =
x 2 e−x dx = 2
−∞ 2
0
√
√
σXi = 2 − 0 = 2.
E(Xi ) =
By the central limit theorem,
X + X + · · · + X
1
2
n
>0
P (X̄ > 0) = P
n
X + X + · · · + X − n(0)
1
2
n
>0 =1−
=P
√ √
2 n
(0) = 0.5.
√
5. Let µ = E(Xi ) and σ = σXi . Clearly, E(Sn ) = nµ and σSn = σ n; thus, by the central limit
theorem,
√
√
P E(Sn ) − σSn ≤ Sn ≤ E(Sn ) + σSn = P nµ − σ n ≤ Sn ≤ nµ + σ n
Sn − nµ
=P −1≤
≤ 1 ≈ (1) − (−1) = 2 (1) − 1 = 0.6826.
√
σ n
6. For 1 ≤ i ≤ 300, let Xi be the amount of the ith expenditure minus Jim’s
/ ith record; Xi is approximately uniform over (−1/2, 1/2). Hence E(Xi ) = 0 and σXi =
√
1/(2 3). The desired probability is
2
(1/2) − (−1/2) /12 =
P (−10 < X1 + X2 + · · · + X300 < 10)
−10 − 300(0)
X1 + X2 + · · · + X300 − 300(0)
10 − 300(0)
=P √
<√
√
√ <
√
√
300 1/(2 3)
300 1/(2 3)
300 1/(2 3)
≈
(2) −
(−2) = 0.9772 − 0.0228 = 0.9544.
7. Note that actual value is a nebulous concept. In this exercise, like everywhere else, we are
using it to mean the average of a very large number of measurements. Let Xi be the error in
284
Chapter 11
Sums of Independent Random Variables and Limit Theorems
√
the ith measurement; µ = E(Xi ) = 0, σ = σXi = 1/ 3. Hence
X1 + X2 + · · · + X50
P − 0.25 <
< 0.25
50
= P (−12.5 < X1 + X2 + · · · + X50 < 12.5)
−12.5
X1 + X2 + · · · + X50
12.5
= P √ √ <
< √ √
√ √
1/ 3 50
1/ 3 50
1/ 3 50
≈
(3.06) −
(−3.06) = 2 (3.06) − 1 = 0.9778.
8. For 1 ≤ i ≤ 300, let Xi = 2, if the ith employee attends with his or her spouse; let Xi = 1,
if the ith employee attends alone; let Xi = 0,
if the ith employee does not attend. To find the
desired quantity, the probability of the event 300
i=1 Xi ≥ 320, note that
µ = E(Xi ) = 2 ·
1
1
1
+ 1 · + 0 · = 1,
3
3
3
1
1
1
5
+1· +0· = ,
3
3
3
3
(
2
5
2
= −1= ,
σXi =
.
3
3
3
E(Xi2 ) = 4 ·
σX2 i
Thus
300
P
i=1
300 X − 300
320 − 300
i=1 i
Xi ≥ 320 = P
≥√ √
≈1−
√ √
2/3 300
2/3 300
9. Direct calculations show that
µ=
6
xf (x) dx = 2/ ln(3/2) = 4.93,
4
E(X2 ) =
6
x 2 f (x) dx = 10/ ln(3/2)
4
!
σX =
We want to find n so that
10
4
−
= 0.577.
ln(3/2) [ln(3/2)]2
P |X̄ − µ| ≤ 0.07 ≥ 0.98
or, equivalently,
P (−0.07 ≤ X̄ − µ ≤ 0.07) ≥ 0.98.
(1.41) = 0.0793.
Section 11.5
Central Limit Theorem
285
Since
X1 + X2 + · · · + Xn
− µ < 0.07
P − 0.07 ≤
n
= P (−0.07n ≤ X1 + X2 + · · · + Xn − nµ ≤ 0.07n)
−0.07n
X1 + X2 + · · · + Xn − nµ
0.07n
=P
≤
√ ≤
√
√
0.577 n
0.577 n
0.577 n
√
√
√
− 0.12 n = 2 0.12 n − 1,
≈ 0.12 n −
all we need to do is to find n so that
2
√
0.12 n − 1 ≥ 0.98,
√
√
or
0.12 n ≥ 0.99. By Table 2 of the appendix, this is satisfied if 0.12 n ≥ 2.33, or
n ≥ 377.007. Therefore, for all sample sizes of 378 or larger, the sample mean is within
±0.07 of the µ.
10. Let
Xi =
0.125
with probability 1/2
−0.125 with probability 1/2.
The change in the stock price, per share, after 60 days is X1 + X2 + · · · + X60 . Clearly,
E(Xi ) = 0 and σXi = 0.125. To find the distribution of X1 + X2 + · · · + X60 , note that for
all t,
60
60
t
t
i=1 Xi − 60(0)
P
.
Xi ≤ t = P
≤
≈
√
√
0.968
0.125 60
0.125 60
i=1
This relation implies that
P
X + X + · · · + X
1
2
60
≤t ≈
0.968
(t).
So (X1 + X2 + · · · + X60 )/0.968 is approximately standard normal and hence
X1 + X2 + · · · + X60 ∼ N (0, 0.9682 ).
Since the most likely value of a normal random variable with mean 0 is 0, the change in the
stock price after 60 days is most likely 0 and hence the most likely value of the holdings of
this investor after 60 days is 50,000.
11. Let X1 be the number of tosses until the first tails. Let X2 be the number of additional tosses
until the second tails; X3 be the number of tosses after the second tails until the third tails,
and so on. Clearly, Xi ’s are independent geometric random variables, each with parameter
286
Chapter 11
Sums of Independent Random Variables and Limit Theorems
1/2. To√
find the desired probability, P (X1 + X2 + · · · + X50 ≥ 75), note that E(Xi ) = 2 and
)
1 − (1/2)
= 2 1/2. Therefore,
σXi =
1/2
P (X1 + X2 + · · · + X50 ≥ 75)
X1 + X2 + · · · + X50 − 50(2)
75 − 50(2)
=P
≥√
√
√
√
50 · 2 1/2
50 · 2 1/2
≈ 1 − (−2.5) = (2.5) = 0.9938.
12. By Exercise 8, Section 7.4, for each i, i ≥ 1, the random variable Xi2 is gamma with parameters
λ = 1/2 and r = 1/2. Therefore,
µ = E(Xi2 ) =
r
=1
λ
and
σ 2 = Var(Xi2 ) =
r
= 2.
λ2
Therefore, by central limit theorem,
S − n
√
n
≤1
lim P Sn ≤ n + 2n = lim P √
n→∞
n→∞
2n
S − nµ
n
= lim P
≤1 =
√
n→∞
σ n
13. Let Yn =
n
i=1
(1) = 0.8413.
Xi ; Yn is Poisson with rate n. On the one hand,
n
P (Yn ≤ n) =
k=0
e−n nk
1
= n
k!
e
n
k=0
nk
,
k!
and on the other hand,
lim P (Yn ≤ n) = lim P
n→∞
n→∞
n
Xi ≤ n
i=1
n
= lim P
n→∞
So
1
n→∞ en
Xi − n
n−n
≤ √
√
n
n
i=1
∞
lim
k=0
nk
1
= .
k!
2
=
(0) =
1
.
2
Chapter 11
Review Problems
287
REVIEW PROBLEMS FOR CHAPTER 11
1. X̄, the average wage
√ of a sample of 10 employees is normal with mean $27000 and standard
deviation $4900/ 10 = $1549.52. Therefore, the desired probability is
30, 000 − 27000
X̄ − 27000
≥
= 1 − (1.94) = 0.0262.
P ( X̄ ≥ 30, 000) = P
1549.52
1549.52
2. MX (t) is the moment-generating function of a binomial random variable with parameters 10
2 1
20
× =
and
3 3
9
10
10 2 i 1
P (X ≥ 8) =
i
3
3
i=8
and 2/3. Therefore, Var(X) = 10 ×
10−i
= 0.299.
3. MX (t) is the moment-generating function of a discrete random variable X with P (X = 1) =
1/6, P (X = 2) = 1/3, and P (X = 3) = 1/2. Therefore, F , the distribution function of X is
given by
⎧
0
t <1
⎪
⎪
⎪
⎪
⎪
⎪
⎨1/6 1 ≤ t < 2
F (x) =
⎪
⎪
1/2 2 ≤ t < 3
⎪
⎪
⎪
⎪
⎩
1
t ≥ 3.
4. MX (t) is the moment-generating function of a normal random variable with mean 1 and
variance 4.
5. X is a uniform random variable over the interval (−1/2, 1/2).
6. X is a Poisson random variable with parameter λ = 1/2. Therefore,
P (X > 0) = 1 − P (X = 0) = 1 − e−1/2 = 0.393.
7. Note that
MX(n) (t) =
(−1)n+1 (n + 1)!
.
(1 − t)n+2
Therefore, E(Xn ) = MX(n) (0) = (−1)n+1 (n + 1)!.
8. Let X̄ be the average of the heights of 10 randomly selected men and Ȳ be the average
40
heights of 6 randomly selected women. Theorem 10.7 implies that X̄ ∼ N 173,
and
10
20
22
Ȳ ∼ N 160,
; thus X̄ − Ȳ ∼ N 13,
. Therefore,
6
3
5 − 13
X̄ − Ȳ − 13
≥√
= (2.95) = 0.9984.
P ( X̄ − Ȳ ≥ 5) = P
√
22/3
22/3
288
Chapter 11
Sums of Independent Random Variables and Limit Theorems
9. By definition,
E e
tX
=
∞
−∞
1
=
2
1 −|x| tx
e e dx =
2
0
1
e(1+t)x dx +
2
−∞
0
1 x tx
e · e dx +
2
−∞
∞
∞
0
1 −x tx
e · e dx
2
ex(t−1) dx.
0
Now for these integrals to exist, we must restrict the domain of the moment-generating function
of X to {t ∈ R : − 1 < t < 1}. In this domain,
0
∞
1
1
e(1+t)x
ex(t−1)
+
−∞
0
2(1 + t)
2(t − 1)
1
1
1
+
=
.
=
2(1 + t) 2(1 − t)
1 − t2
MX (t) = E etX =
10. (a) By the law of total probability (Theorem 3.4),
n
P (X + Y = n | X = i)P (X = i)
P (X + Y = n) =
i=0
n
=
n
P (X + Y = n, X = i) =
i=0
n
=
P (Y = n − i, X = i)
i=0
P (X = i)P (Y = n − i).
i=0
(b) By part (a),
n
P (X + Y = n) =
=
1
e−λ λi e−µ µn−i
·
= e−(λ+µ) · ·
i!
(n − i)!
n!
i=0
−(λ+µ)
e
n
i=0
n i n−i
λµ
i
(λ + µ)
,
n!
n
where the last equality follows from the binomial expansion (Theorem 2.5).
11. We have
X1 + X2 + · · · + X28
P 0.95 <
< 1.05 = P (26.6 < X1 + X2 + · · · + X28 < 29.4)
28
X1 + X2 + · · · + X28 − 28(1)
29.4 − 28
26.6 − 28
<
<
=P
√
√
√
2 28
2 28
2 28
≈
(0.13) −
(−0.13) = 0.5517 − 0.4483 = 0.1034.
Chapter 11
Review Problems
289
12. In (11.5), let ε = 0.01 and α = 0.06; we have
n≥
1
= 41, 666.67.
4(0.01)2 (0.06)
Therefore, at least 41667 patients should participate in the trial.
13. By (11.4),
p(1 − p)
1
P |8
p − p| < 0.05 ≥ 1 −
≥1−
= 0.98,
2
(0.05) 5000
4(0.05)2 5000
since p(1 − p) ≤ 1/4 implies that −p(1 − p) ≥ −1/4.
14. For i = 1, 2, 3, . . . , n, let Xi be the IQ of the ith student of the sample. We want to determine
n so that
X1 + X2 + · · · + Xn
P − 0.2 <
− µ < .2 ≥ 0.98.
n
Since E(Xi ) = µ and Var(Xi ) = 170, by the central limit theorem,
n
n
Xi
P − 0.2 < i=1
− µ < 0.2 = P − (0.2)n <
Xi − nµ < (0.2)n
n
i=1
−(0.2)n
<
=P √
170n
n
Xi − nµ
√
170n
i=1
(0.2)n
<√
170n
0.2√n
−(0.2)n
≈
=2
− 1 ≥ 0.98.
√
√
170n
170
√ √
n/ 170) ≥ 0.98. From Table 2 of the
Therefore, we should determine
n
so
that
(0.2
√ √
Appendix, we find (0.2) n/ 170 = 2.33, which implies that n = 23072.8250; therefore,
the psychologist should choose a sample of size 23073.
(0.2)n
−
√
170n
15. Let Xi be the amount chopped off on the ith charge in dollars. Let X be the actual amount
Ed has charged to his credit card this month minus the amount his record shows. Clearly,
X = X1 + X2 + · · · + X20 , and for 1 ≤ i ≤ 20, Xi is uniform over (0, 1). Thus E(Xi ) = 1/2
and Var(Xi ) = 1/12 and hence E(X) = 20/2 = 10 and Var(X) = 20/12 = 5/3. Therefore,
by Chebyshev’s inequality,
P (X > 15) = P (X − 10 > 5) ≤ P |X − 10| > 5
5/3
= P |X − E(X)| > 5 ≤
= 0.0667.
25
16. P (X ≥ 45) ≤ P |X − 0| ≥ 45 ≤ 152 /452 = 1/9.
290
Chapter 11
Sums of Independent Random Variables and Limit Theorems
17. Suppose that the ith randomly selected book is Xi centimeters thick. The desired probability
is
P (X1 + X2 + · · · + X31
X1 + X2 + · · · + X31 − 3(31)
87 − 3(31)
≤ 87) = P
≤
√
√
1 31
1 31
87 − 93
= (−1.08) = 1 − 0.8599 = 0.1401.
≈
√
31
18. For 1 ≤ i ≤ 20, let Xi denote the outcome of the ith roll. We have
6
E(Xi ) =
i·
i=1
6
7
1
= ,
6
2
E(Xi2 ) =
i2 ·
i=1
91
1
= .
6
6
35
91 49
−
=
, and hence
6
4
12
20
20
65 − 70
75 − 70
i=1 Xi − 70
P 65 ≤
Xi ≤ 75 = P √
√ ≤√
√ ≤√
√
35/12 · 20
35/12 · 20
35/12 · 20
i=1
Thus Var(Xi ) =
≈
(0.65) −
19. By Markov’s inequality, P (X ≥ nµ) ≤
20. Let X =
26
i=1
(−0.65) = 2 (0.65) − 1 = 0.4844.
1
µ
= . So nP (X ≥ nµ) ≤ 1.
nµ
n
Xi. We have that
E(Xi ) = 26/51 = 0.5098,
E(Xi2 ) = E(Xi ) = 0.5098,
Var(Xi ) = 0.5098 − (0.5098)2 = 0.2499,
E(Xi Xj ) = P (Xi = 1, Xj = 1) = P (Xi = 1)P (Xj = 1 | Xi = 1) =
26 25
·
= 0.2601,
51 49
and
Cov(Xi , Xj ) = E(Xi Xj ) − E(Xi )E(Xj ) = 0.2601 − (0.5098)2 = 0.0002.
Thus E(X) = 26(0.5098) = 13.2548 and
26
Var(X) =
Var(Xi ) + 2
Cov(Xi , Xj )
i 15.
The time Linda has to wait before being able to cross the street is 0 if N = 0 (i.e., X1 > 15),
and is SN = X1 + X2 + · · · + XN , otherwise. Therefore,
E(SN ) = E E(SN | N) =
∞
E(SN | N = i)P (N = i)
i=0
∞
E(SN | N = i)P (N = i),
=
i=1
292
Chapter 12
Stochastic Processes
where the last equality follows since for N = 0, we have that SN = 0. Now
i
E(SN | N = i) = E(X1 + X2 + · · · + Xi | N = i) =
E(Xj | N = i)
j =1
i
E(Xj | Xj ≤ 15),
=
j =1
where by Remark 8.1,
E(Xj | Xj ≤ 15) =
1
F (15)
15
tf (t) dt;
0
F and f being the probability distribution and density functions of Xi ’s, respectively. That is,
for t ≥ 0, F (t) = 1 − e−t/7 , f (t) = (1/7)e−t/7 . Thus
15
15
1
t −t/7
−t/7
E(Xj | Xj ≤ 15) =
dt = (1.1329) − (t + 7)e
e
1 − e−15/7 0 7
0
= (1.1329)(4.41898) = 5.00631.
This gives E(SN | N = i) = 5.00631i. To find P (N = i), note that for i ≥ 1,
P (N = i) = P (X1 ≤ 15, X2 ≤ 15, . . . , Xi ≤ 15, Xi+1 > 15)
i
= F (15) 1 − F (15) = (0.8827)i (0.1173).
Putting all these together, we obtain
∞
E(SN ) =
∞
E(SN | N = i)P (N = i) =
i=1
(5.00631i)(0.8827)i (0.1173)
i=1
∞
i(0.8827)i = (0.5872) ·
= (0.5872)
i=1
0.8827
= 37.6707,
(1 − 0.8827)2
i
2
where the next to last equality follows from ∞
i=1 ir = r/(1 − r) , |r| < 1. Therefore, on
average, Linda has to wait approximately 38 seconds before she can cross the street.
4. Label the time point 9:00 A.M. as t = 0. Then
to 1:00 P.M. Let N(t) be
t = 4 corresponds
the number of fish caught at or prior to t; N(t) : t ≥ 0 is a Poisson process with rate 2.
Let X1 , X2 , . . . , X6 be six uniformly distributed independent random variables over [0, 4].
By theorem 12.4, given that N(4) = 6, the time that the fisherman caught the first fish is
Y = min(X1 , X2 , . . . , X6 ). Therefore, the desired probability is
P (Y < 1) = 1 − P (Y ≥ 1) = 1 − P min(X1 , X2 , . . . , X6 ) ≥ 1
= 1 − P (X1 ≥ 1, X2 ≥ 1, . . . , X6 ≥ 1)
= 1 − P (X1 ≥ 1)P (X2 ≥ 1) · · · P (X6 ≥ 1) = 1 −
3
4
6
= 0.822.
Section 12.2
More on Poisson Processes
293
5. Let S1 , S2 , and S3 be the number of meters of wire manufactured, after the inspector left,
until the first, second, and third fractures appeared, respectively. By Theorem 12.4, given that
N(200) = 3, the joint probability density function of S1 , S2 , and S3 is
fS1 ,S2 ,S3 |N(200) (t1 , t2 , t3 | 3) =
3!
,
8, 000, 000
0 < t1 < t2 < t3 < 200.
Using this, the probability we are interested in, is given by the following triple integral:
80 140 200
3!
dt3 dt2 dt1
P (S1 + 60 < S2 , S2 + 60 < S3 ) =
0
t1 +60 t2 +60 8, 000, 000
80 140
3!
=
(140 − t2 ) dt2 dt1
8, 000, 000 0
t1 +60
80
6
1
=
3200 − 80t1 + t12 dt1
8, 000, 000 0
2
80
1 3
6
=
t1 − 40t12 + 3200t1
0
8, 000, 000 6
=
8
= 0.064.
125
6. By (12.8), the conditional probability density function of Sk , given that N(t) = n, is
fSk |N(t) (x|n) =
n!
1x
·
(n − k)! (k − 1)! t t
Therefore,
E Sk | N(t) = n =
0
t
k−1
1−
n!
1x
x·
(n − k)! (k − 1)!
t t
Letting x/t = u, we have (1/t) dx = du. Thus
E Sk | N(t) = n =
n!
t
(n − k)! (k − 1)!
1
x
t
n−k
0 ≤ x ≤ t.
1−
x
t
k−1
,
n−k
dx.
uk (1 − u)n−k du.
0
What we want to show follows from the following relations discussed in Section 7.5:
1
(k + 1) (n − k + 1)
k! (n − k)!
uk (1 − u)n−k du = B(k + 1, n − k + 1) =
=
.
(n + 2)
(n + 1)!
0
7. Let T be the time until the next arrival, and let S be the time until the next departure. By
the memoryless property of exponential random variables, T and S are exponential random
variables with parameters λ and µ, respectively. They are independent by the definition of an
M/M/1 queue. Thus
P (A) = P (T > t and S > T ) = P (T > t)P (S > t) = e−λt · e−µt = e−(λ+µ)t ,
294
Chapter 12
Stochastic Processes
P (B) = P (S > T ) =
∞
P (S > T | T = u)λe−λu du
0
∞
=
P (S > u | T = u)λe
−λu
du =
0
∞
P (S > u)λe−λu du
0
∞
=λ
e−µu · eλu du =
0
A similar calculation shows that
λ
.
λ+µ
P (AB) = P (S > T > t) =
∞
P (S > T | T = u)λe−λu du
t
∞
=
e−µu · λe−λu du =
t
λ
e−(λ+µ)t = P (A)P (B).
λ+µ
8. (a) Let X be the number of customers arriving to the queue during a service period S. Then
∞
P (X = n) =
0
n
λ µ
=
n!
P (X = n | S = t)µe−µt dt =
0
∞
n −(λ+µ)t
t e
0
n
λ µ
dt =
n! (λ + µ)
∞
e−λt (λt)n −µt
µe
dt
n!
∞
t n (λ + µ)e−(λ+µ)t dt.
0
Note that (λ + µ)e−(λ+µ)t is the probability density function of an exponential random
variable Z with parameter λ + µ. Hence
P (X = n) =
λn µ
E(Z n ).
n! (λ + µ)
By Example 11.4,
E(Z n ) =
n!
.
(λ + µ)n
Therefore,
P (X = n) =
λn µ
λ
=
1−
n+1
(λ + µ)
λ+µ
n
µ
,
λ+µ
n ≥ 0.
This is the probability mass function of a geometric random variable with parameter
µ/(λ + µ).
(b)
Due to the memoryless property of exponential random variables, the remaining service
time of the customer being served is also exponential with parameter µ. Hence we want
to find the number of new customers arriving during a period, which is the sum of n + 1
independent exponential random variables. Since during each of these service times the
number of new arrivals is geometric with parameter µ/(λ + µ), during the entire period
under consideration, the distribution of the total number of new customers arriving is the
sum of n + 1 independent geometric random variables each with parameter µ/(λ + µ),
which is negative binomial with parameters n + 1 and µ/(λ + µ).
Section 12.2
More on Poisson Processes
295
9. It is straightforward to check that M(t)
is stationary,
orderly, and possesses independent
increments. Clearly, M(0) = 0. Thus M(t) : t ≥ 0 is a Poisson process. To find its rate,
note that, for 0 ≤ k < ∞,
∞
P M(t) = k =
P M(t) = k | N(t) = n P N (t) = n
n=k
n k
e−λt (λt)n
p (1 − p)n−k ·
=
k
n!
n=k
∞
e−λt pk
=
k! (1 − p)k
∞
n=k
n
λt (1 − p)
(n − k)!
k
e−λt pk
=
·
λt
(1
−
p)
k! (1 − p)k
∞
n=k
n−k
λt (1 − p)
(n − k)!
e−λt pk
(λpt)k −λpt
.
(λt)k eλt (1−p) =
e
k!
k!
This shows that the parameter of M(t) : t ≥ 0 is λp.
10. Note that P Vi = min(V1 , V2 , . . . , Vk ) is the probability that the first shock occurring to
the system is of type i. Suppose that the first shock occurs to the system at time u. If we
label the time point u as t = 0, then from that point on, by stationarity and the independentincrements property, probabilistically, the behavior of these Poisson processes is identical to
the system considered prior to u. So the probability that the second shock is of type i is
identical
to the probability that the
first shock is of type i, and so on. Hence they are all equal
to P Vi = min(V1 , V2 , . . . , Vk ) . To find this probability, note that, for 1 ≤ j ≤ k, Vj ’s,
are independent exponential random variables, and the probability density function of Vj is
λj e−λj t . Thus P (Vj > u) = e−λj u . By conditioning on Vi , we have
P Vi = min(V1 , . . . , Vk )
∞
=
P min(V1 , . . . , Vk ) = Vi | Vi = u λi e−λi u du
=
0
= λi
∞
P min(V1 , . . . , Vk ) = u | Vi = u e−λi u du
0
= λi
∞
P (V1 ≥ u, . . . , Vi−1 ≥ u, Vi+1 ≥ u, . . . , Vk ≥ u | Vi = u)e−λi u du
0
= λi
∞
P (V1 ≥ u, . . . , Vi−1 ≥ u, Vi+1 ≥ u, . . . , Vk ≥ u)e−λi u du
0
= λi
0
∞
P (V1 ≥ u) · · · P (Vi−1 ≥ u)P (Vi+1 ≥ u) · · · P (Vk ≥ u)e−λi u du
296
Chapter 12
Stochastic Processes
= λi
∞
e−λ1 u · · · e−λi−1 u · e−λi+1 u · · · e−λk u · e−λi u du
0
= λi
∞
e
−(λ1 +···+λk )u
0
12.3
du = λi
∞
e−λu du =
0
λi
.
λ
MARKOV CHAINS
1. {Xn : n = 1, 2, . . . } is not a Markov chain. For example, P (X4 = 1) depends on all the values
of X1 , X2 , and X3 , and not just X3 . That is, whether or not the fourth person selected is female
depends on the genders of all three persons selected prior to the fourth and not only on the
gender of the third person selected.
2. For j ≥ 0,
∞
P (Xn = j ) =
∞
P (Xn = j | X0 = i)P (X0 = i) =
i=0
pijn p(i),
i=0
where pijn is the ij th entry of the matrix P n .
3. The transition probability matrix of this Markov chain is
⎛
⎞
0 1/2 0
0
0 1/2
⎜1/2 0 1/2 0
0
0 ⎟
⎜
⎟
⎜ 0 1/2 0 1/2 0
0 ⎟
⎜
⎟.
P =⎜
0 1/2 0 1/2 0 ⎟
⎜ 0
⎟
⎝ 0
0
0 1/2 0 1/2⎠
1/2 0
0
0 1/2 0
By calculating P 4 and P 5 , we will find that, (a) the probability that in 4 transitions the Markov
4
chain returns to 1 is P11
= 3/8; (b) the probability that, in 5 transitions, the Markov chain
enters 2 or 6 is
11 11
11
5
5
p12
+ p16
=
+
= .
32 32
16
4. Solution 1:
Starting at 0, the process eventually enters 1 or 2 with equal probabilities. Since
2 is absorbing, “never entering 1” is equivalent to eventually entering 2 directly from 0. The
probability of that is 1/2.
Solution 2: Let Z be the number of transitions until the first visit to 1. Note that state 2 is
absorbing. If the process enters 2, it will always remain there. Hence Z = n if and only if the
Section 12.3
Markov Chains
297
first n − 1 transitions are from 0 to 0, and the nth transition is from 0 to 1, implying that
1 n−1 1
,
n = 1, 2, . . . .
P (Z = n) =
2
4
The probability that the process ever enters 1 is
∞
P (Z < ∞) =
1
n−1 1
2
4
n=1
=
1/4
1
= .
1 − (1/2)
2
Therefore, the probability that the process never enters 1 is 1 − (1/2) = 1/2.
5. (a) By the Markovian property, given the present, the future is independent of the past.
Thus the probability that tomorrow Emmett will not take the train to work is, simply,
p21 + p23 = 1/2 + 1/6 = 2/3.
(b)
The desired probability is
p21 p11 + p21 p13 + p23 p31 + p23 p33 = 1/4.
6. Let Xn denote the number of balls in urn I after n transfers. The stochastic process {Xn : n =
0, 1, . . . } is a Markov chain with state space {0, 1, . . . , 5} and transition probability matrix
⎛
⎞
0
1
0
0
0
0
⎜1/5 0 4/5 0
0
0 ⎟
⎟
⎜
⎟
⎜ 0 2/5 0 3/5 0
0
⎟.
P =⎜
⎜ 0
0 3/5 0 2/5 0 ⎟
⎜
⎟
⎝ 0
0
0 4/5 0 1/5⎠
0
0
0
0
1
0
Direct calculations show that
⎛
241
⎜ 3125
⎜
⎜
⎜
⎜ 0
⎜
⎜
⎜
⎜
⎜ 1022
⎜
⎜ 15625
⎜
P (6) = P 6 = ⎜
⎜
⎜
⎜ 0
⎜
⎜
⎜
⎜ 168
⎜
⎜ 3125
⎜
⎜
⎝
0
0
2044
3125
0
168
625
5293
15625
0
9492
15625
0
0
9857
15625
0
4746
15625
4746
15625
0
9857
15625
0
0
9492
15625
0
5293
15625
168
625
0
2044
3125
0
⎞
0
⎟
⎟
⎟
168 ⎟
⎟
⎟
3125 ⎟
⎟
⎟
⎟
0 ⎟
⎟
⎟
⎟.
1022 ⎟
⎟
⎟
15625 ⎟
⎟
⎟
⎟
0 ⎟
⎟
⎟
⎟
241 ⎠
3125
298
Chapter 12
Stochastic Processes
Hence, by Theorem 12.5,
P (X6 = 4) = 0 ·
168
1
2 4746
3
4 5293
5
+
·0+
·
+
·0+
·
+
· 0 = 0.1308.
625 15
15 15625 15
15 15625 15
7. By drawing a transition graph, it is readily seen that this Markov chain consists of the recurrent
classes {0, 3} and {2, 4} and the transient class {1}.
8. Let Zn be the outcome of the nth toss. Then
Xn+1 = max(Xn , Zn+1 )
shows that {Xn : n = 1, 2, . . . } is a Markov chain. Its state space is {1, 2, . . . , 6}, and its
transition probability matrix is given by
⎛
⎞
1/6 1/6 1/6 1/6 1/6 1/6
⎜ 0 2/6 1/6 1/6 1/6 1/6⎟
⎜
⎟
⎜ 0
⎟
0
3/6
1/6
1/6
1/6
⎟.
P =⎜
⎜ 0
0
0 4/6 1/6 1/6⎟
⎜
⎟
⎝ 0
0
0
0 5/6 1/6⎠
0
0
0
0
0
1
It is readily seen that no two states communicate with each other. Therefore, we have six
classes of which {1}, {2}, {3}, {4}, {5}, are transient, and {6} is recurrent (in fact, absorbing).
9. This can be achieved more easily by drawing a transition graph. An example of a desired
matrix is as follows:
⎛
0
⎜1
⎜
⎜0
⎜
⎜0
⎜
⎜0
⎜
⎜0
⎜
⎝0
0
⎞
0 1/2 0 1/2 0
0
0
0 0
0
0
0
0
0 ⎟
⎟
1 0
0
0
0
0
0 ⎟
⎟
0 1/3 2/3 0
0
0
0 ⎟
⎟.
0 0
0
0 2/5 0 3/5⎟
⎟
0 0
0 1/2 0 1/2 0 ⎟
⎟
0 0
0
0 3/5 0 2/5⎠
0 0
0 1/3 0 2/3 0
10. For 1 ≤ i ≤ 7, starting from state i, let xi be the probability that the Markov chain will
eventually be absorbed into state 4. We are interested in x6 . Applying the law of total
Section 12.3
Markov Chains
299
probability repeatedly, we obtain the following system of linear equations:
⎧
⎪
x1 = (0.3)x1 + (0.7)x2
⎪
⎪
⎪
⎪
⎪
⎪
x2 = (0.3)x1 + (0.2)x2 + (0.5)x3
⎪
⎪
⎪
⎪
⎪
⎪
x = (0.6)x4 + (0.4)x5
⎪
⎨ 3
x4 = 1
⎪
⎪
⎪
⎪
x5 = x3
⎪
⎪
⎪
⎪
⎪
⎪
x6 = (0.1)x1 + (0.3)x2 + (0.1)x3 + (0.2)x5 + (0.2)x6 + (0.1)x7
⎪
⎪
⎪
⎪
⎩x = 0.
7
Solving this system of equations, we obtain
⎧
⎪
x = x2 = x3 = x4 = x5 = 1
⎪
⎨ 1
x6 = 0.875
⎪
⎪
⎩
x7 = 0.
Therefore, the probability is 0.875 that, starting from state 6, the Markov chain will eventually
be absorbed into state 4.
11. Let π1 , π2 , and π3 be the long-run probabilities that the sportsman devotes to horseback riding,
sailing, and scuba diving, respectively. Then, by Theorem 12.7, π1 , π2 , and π3 are obtained
from solving the system of equations.
⎛ ⎞ ⎛
⎞⎛ ⎞
π1
0.20 0.32 0.60
π1
⎝π2 ⎠ = ⎝0.30 0.15 0.13⎠ ⎝π2 ⎠
0.50 0.53 0.27
π3
π3
along with π1 + π2 + π3 = 1. The matrix equation above gives us the following system of
equations
⎧
⎪
⎨π1 = 0.20π1 + 0.32π2 + 0.60π3
π2 = 0.30π1 + 0.15π2 + 0.13π3
⎪
⎩
π3 = 0.50π1 + 0.53π2 + 0.27π3 .
By choosing any two of these equations along with π1 + π2 + π3 = 1, we obtain a system of
three equations in three unknowns. Solving that system yields π1 = 0.38856, π2 = 0.200056,
and π3 = 0.411383. Hence the long-run probability that on a randomly selected vacation day
the sportsman sails is approximately 0.20.
12. For n ≥ 1, let
Xn =
1
0
if the nth fish caught is trout
if the nth fish caught is not trout.
300
Chapter 12
Stochastic Processes
Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1} and transition probability
matrix
10/11 1/11
8/9
1/9
Let π0 be the fraction of fish in the lake that are not trout, and π1 be the fraction of fish in the
lake that are trout. Then, by Theorem 12.7, π0 and π1 satisfy
π0
10/11 8/9
π0
=
,
1/11 1/9
π1
π1
which gives us the following system of equations
⎧
⎨π0 = (10/11)π0 + (8/9)π1
⎩π = (1/11)π + (1/9)π .
1
0
1
By choosing any one of these equations along with the relation π0 + π1 = 1, we obtain a
system of two equations in two unknown. Solving that system yields π0 = 88/97 ≈ 0.907
and π1 = 9/97 ≈ 0.093. Therefore, approximately 9.3% of the fish in the lake are trout.
13. Let
⎧
⎪
⎨1
Xn = 2
⎪
⎩
3
if the nth card is drawn by player I
if the nth card is drawn by player II
if the nth card is drawn by player III.
{Xn : n = 1, 2, . . . } is a Markov chain with probability transition matrix
⎛
⎞
48/52 4/52
0
39/52 13/52⎠ .
P =⎝ 0
12/52
0
40/52
Let π1 , π2 , and π3 be the proportion of cards drawn by players I, II, and III, respectively. π1 ,
π2 , and π3 are obtained from
⎛ ⎞ ⎛
⎞⎛ ⎞
π1
12/13 0
3/13
π1
⎝π2 ⎠ = ⎝ 1/13 3/4
⎠
⎝
π2 ⎠
0
0
1/4 10/13
π3
π3
and π1 + π2 + π3 = 1, which gives π1 = 39/64 ≈ 0.61, π2 = 12/64 ≈ 0.19, and π3 =
13/64 ≈ 0.20.
14. For 1 ≤ i ≤ 9, let πi be the probability that the mouse is in cell i, 1 ≤ i ≤ 9, at a random time
Section 12.3
Markov Chains
301
in the future. Then πi ’s satisfy
⎞⎛ ⎞
⎛ ⎞ ⎛
0 1/3 0 1/3 0
0
0
0
0
π1
π1
⎟
⎜
⎜π2 ⎟ ⎜1/2 0 1/2 0 1/4 0
0
0
0 ⎟ ⎜π2 ⎟
⎟
⎜ ⎟ ⎜
⎜ ⎟
⎜π3 ⎟ ⎜ 0 1/3 0
0
0 1/3 0
0
0 ⎟
⎟ ⎜π3 ⎟
⎜ ⎟ ⎜
⎜ ⎟
⎜π4 ⎟ ⎜1/2 0
0
0 1/4 0 1/2 0
0 ⎟
⎟ ⎜π4 ⎟
⎜ ⎟ ⎜
⎜π5 ⎟ = ⎜ 0 1/3 0 1/3 0 1/3 0 1/3 0 ⎟ ⎜π5 ⎟ .
⎟⎜ ⎟
⎜ ⎟ ⎜
⎜ ⎟
⎜π6 ⎟ ⎜ 0
0 1/2 0 1/4 0
0
0 1/2⎟
⎟ ⎜π6 ⎟
⎜ ⎟ ⎜
⎟
⎟
⎜π7 ⎟ ⎜ 0
0
0 1/3 0
0
0 1/3 0 ⎟ ⎜
⎜π7 ⎟
⎜ ⎟ ⎜
⎝π8 ⎠ ⎝ 0
0
0
0 1/4 0 1/2 0 1/2⎠ ⎝π8 ⎠
0
0
0
0
0 1/3 0 1/3 0
π9
π9
9
Solving this system of equations along with i=1 π1 , we obtain
π1 = π3 = π7 = π9 = 1/12,
π2 = π4 = π6 = π8 = 1/8,
π5 = 1/6.
15. Let Xn denote the number of balls in urn I after n transfers. The stochastic process {Xn : n =
0, 1, . . . } is a Markov chain with state space {0, 1, . . . , 5} and transition probability matrix
⎛
⎞
0
1
0
0
0
0
⎜1/5 0 4/5 0
0
0 ⎟
⎜
⎟
⎜ 0 2/5 0 3/5 0
0 ⎟
⎜
⎟.
P =⎜
⎟
0
0
3/5
0
2/5
0
⎜
⎟
⎝ 0
0
0 4/5 0 1/5⎠
0
0
0
0
1
0
Clearly, {Xn : n = 0, 1, . . . } is an irreducible recurrent Markov chain; since it is finite-state,
it is positive recurrent. However, {Xn : n = 0, 1, . . . } is not aperiodic, and the period of each
state is 2. Hence the limiting probabilities do not exist. For 0 ≤ i ≤ 5, let πi be the fraction of
time urn I contains i balls. Then with this interpretation, πi ’s satisfy the following equations
⎛ ⎞ ⎛
⎞⎛ ⎞
π0
0 1/5 0
0
0 0
π0
⎜π1 ⎟ ⎜1 0 2/5 0
⎟ ⎜ π1 ⎟
0
0
⎜ ⎟ ⎜
⎟⎜ ⎟
⎜π2 ⎟ ⎜0 4/5 0 3/5 0 0⎟ ⎜π2 ⎟
⎜ ⎟=⎜
⎟⎜ ⎟
⎜π3 ⎟ ⎜0 0 3/5 0 4/5 0⎟ ⎜π3 ⎟ ,
⎜ ⎟ ⎜
⎟⎜ ⎟
⎝π4 ⎠ ⎝0 0
0 2/5 0 1⎠ ⎝π4 ⎠
π5
5
i=0
0
0
0
0
1/5 0
πi = 1. Solving these equations, we obtain
π0 = π5 = 1/31,
π1 = π4 = 5/31,
π2 = π3 = 10/31.
π5
302
Chapter 12
Stochastic Processes
Therefore, the fraction of time an urn is empty is π0 + π5 = 2/31. Hence the expected number
of balls transferred between two consecutive times that an urn becomes empty is 31/2 = 15.5.
16. Solution 1: Let Xn be the number of balls in urn I immediately before the nth game begins.
Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , 7} and transition
probability matrix
⎛
⎞
3/4 1/4 0
0
0
0
0
0
⎜1/4 1/2 1/4 0
0
0
0
0 ⎟
⎜
⎟
⎜ 0 1/4 1/2 1/4 0
0
0
0 ⎟
⎜
⎟
⎜ 0
⎟
0
1/4
1/2
1/4
0
0
0
⎟.
P =⎜
⎜ 0
0
0 1/4 1/2 1/4 0
0 ⎟
⎜
⎟
⎜ 0
0
0
0 1/4 1/2 1/4 0 ⎟
⎜
⎟
⎝ 0
0
0
0
0 1/4 1/2 1/4⎠
0
0
0
0
0
0 1/4 3/4
Since the transition probability matrix is doubly stochastic; that is, the sum of each column
is also 1, for i = 0, 1, . . . , 7, πi , the long-run probability that the number of balls in urn I
immediately before a game begins is 1/8 (see Example 12.35). This implies that the long-run
probability mass function of the number of balls in urn I or II is 1/8 for i = 0, 1, . . . , 7.
Solution 2: Let Xn be the number of balls in the urn selected at step 1 of the nth game. Then
{Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , 7} and transition probability
matrix
⎞
⎛
1/2 0
0
0
0
0
0 1/2
⎜1/4 1/4 0
0
0
0 1/4 1/4⎟
⎜
⎟
⎜ 0 1/4 1/4 0
0 1/4 1/4 0 ⎟
⎟
⎜
⎜ 0
0 1/4 1/4 1/4 1/4 0
0 ⎟
⎜
⎟.
P =⎜
⎟
0
0
0
1/2
1/2
0
0
0
⎜
⎟
⎜ 0
0 1/4 1/4 1/4 1/4 0
0 ⎟
⎜
⎟
⎝ 0 1/4 1/4 0
0 1/4 1/4 0 ⎠
1/4 1/4
0
0
0
0
1/4 1/4
Since the transition probability matrix is doubly stochastic; that is, the sum of each column
is also 1, for i = 0, 1, . . . , 7, πi , the long-run probability that the number of balls in the
urn selected at step 1 of a game is 1/8 (see Example 12.35). This implies that the long-run
probability mass function of the number of balls in urn I or II is 1/8 for i = 0, 1, . . . , 7.
17. For i ≥ 0, state i is directly accessible from 0. On the other hand, i is accessible from i + 1.
These two facts make it possible for all states to communicate with each other. Therefore, the
Markov chain has only one class. Since 0 is recurrent and aperiodic (note that p00 > 0 makes
0 aperiodic), all states are recurrent and aperiodic. Let πk be the long-run probability that a
Section 12.3
Markov Chains
303
computer selected at the end of a semester will last at least k additional semesters. Solving
⎛ ⎞ ⎛
⎞⎛ ⎞
π0
p1 1 0 0 . . .
π0
⎜π1 ⎟ ⎜p2 0 1 0 . . .⎟ ⎜π1 ⎟
⎜ ⎟ ⎜
⎟⎜ ⎟
⎜π2 ⎟ = ⎜p3 0 0 1 . . .⎟ ⎜π2 ⎟
⎝ ⎠ ⎝
⎠⎝ ⎠
..
..
..
.
.
.
along with
∞
i=0
πi = 1, we obtain
π0 =
πk =
1+
1
,
(1
−
p
1 − p2 − · · · − pi )
i=1
∞
1 − p1 − p2 − · · · − pk
∞
,
1 + i=1 (1 − p1 − p2 − · · · − pi )
k ≥ 1.
18. Let DN denote the state at which the last movie Mr. Gorfin watched was not a drama, but
the one before that was a drama. Define DD, ND, and NN similarly, and label the states
DD, DN, ND, and NN by 0, 1, 2, and 3, respectively. Let Xn = 0 if the nth and (n − 1)st
movies Mr. Gorfin watched were both dramas. Define Xn = 1, 2, and 3 similarly. Then
{Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, 2, 3} and transition probability
matrix
⎛
⎞
7/8 1/8 0
0
⎜ 0
0 1/2 1/2⎟
⎟.
P =⎜
⎝1/2 1/2 0
0 ⎠
0
0 1/8 7/8
(a)
If the first two movies Mr. Gorfin watched last weekend were dramas, the probability
2
2
that the fourth one is a drama is p00
+ p02
. Since
⎛
⎞
49/64 7/64 1/16 1/16
⎜ 1/4
1/4 1/16 7/16 ⎟
⎟,
P2 = ⎜
⎝ 7/16 1/16 1/4
1/4 ⎠
1/16 1/16 7/64 49/64
the desired probability is (49/64) + (1/16) = 53/64.
(b)
Let π0 denote the long-run probability that Mr. Gorfin watches two dramas in a row.
Define π1 , π2 , and π3 similarly. We have that,
⎛ ⎞ ⎛
⎞⎛ ⎞
π0
7/8 0 1/2 0
π0
⎜π1 ⎟ ⎜1/8 0 1/2 0 ⎟ ⎜π1 ⎟
⎜ ⎟=⎜
⎟⎜ ⎟
⎝π2 ⎠ ⎝ 0 1/2 0 1/8⎠ ⎝π2 ⎠ .
0 1/2 0 7/8
π3
π3
Solving this system along with π0 + π1 + π2 + π3 = 1, we obtain π0 = 2/5, π1 = 1/10,
π2 = 1/10, and π3 = 2/5. Hence the probability that Mr. Gorfin watches two dramas
in a row is 2/5.
304
Chapter 12
Stochastic Processes
19. Clearly,
Xn+1 =
0
1 + Xn
if the (n + 1)st outcome is 6
otherwise.
This relation shows that {Xn : n = 1, 2, . . . } is a Markov chain. Its transition probability
matrix is given by
⎞
⎛
1/6 5/6 0
0
0 ...
⎜1/6 0 5/6 0
0 . . .⎟
⎟
⎜
⎜1/6 0
0 5/6 0 . . .⎟
P =⎜
⎟.
⎟
⎜1/6 0
0
0
5/6
.
.
.
⎠
⎝
..
.
It is readily seen that all states communicate with 0. Therefore, by transitivity of the communication property, all states communicate with each other. Therefore, the Markov chain is
irreducible. Clearly, 0 is recurrent. Since p00 > 0, it is aperiodic as well. Hence all states
are recurrent and aperiodic. On the other hand, starting at 0, the expected number of transitions until the process returns to 0 is 6. This is because the number of tosses until the next
6 obtained is a geometric random variable with probability of success p = 1/6, and hence
expected value 1/p = 6. Therefore, 0, and hence all other states are positive recurrent. Next,
a simple probabilistic argument shows that,
5 i1
πi =
, i = 0, 1, 2, . . . .
6
6
This can also be shown by solving the following system of equations:
⎞⎛ ⎞
⎧⎛ ⎞ ⎛
⎪
π
1/6
1/6
1/6
1/6
.
.
.
π0
0
⎪⎜ ⎟ ⎜
⎪
⎟⎜ ⎟
⎪
⎪
⎟ ⎜
⎜ ⎟
0
0 . . .⎟
⎪
⎪⎜
⎜π1 ⎟ ⎜5/6 0
⎟ ⎜π1 ⎟
⎪
⎪
⎟
⎟
⎜
⎟
⎨⎜
0 . . .⎟ ⎜
⎜π2 ⎟ = ⎜ 0 5/6 0
⎜π2 ⎟
⎜ ⎟ ⎜
⎟⎜ ⎟
⎜π3 ⎟ ⎜ 0
0 5/6 0 . . .⎟ ⎜π3 ⎟
⎪
⎪
⎝ ⎠ ⎝
⎠⎝ ⎠
⎪
⎪
.
..
.
⎪
.
.
⎪
⎪
.
.
.
⎪
⎪
⎩
π0 + π1 + π2 + · · · = 1.
20. (a) Let
Xn =
1
0
if Alberto wins the nth game
if Alberto loses the nth game.
Then {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1}. Its transition
1−p p
probability matrix is P =
p 1 − p . Using induction, we will now show that
⎛
P (n)
1 1
+ (1 − 2p)n
⎜
2
2
n
=P =⎜
⎝1 1
− (1 − 2p)n
2 2
⎞
1 1
− (1 − 2p)n ⎟
2 2
⎟.
⎠
1 1
n
+ (1 − 2p)
2 2
Section 12.3
Markov Chains
305
Clearly, for n = 1, P (1) = P . Suppose that
⎛
P (n)
1 1
n
⎜ 2 + 2 (1 − 2p)
⎜
=⎜
⎝1 1
− (1 − 2p)n
2 2
⎞
1 1
− (1 − 2p)n ⎟
2 2
⎟
⎟.
⎠
1 1
n
+ (1 − 2p)
2 2
We will show that
⎛
P n+1
1 1
n+1
⎜ 2 + 2 (1 − 2p)
⎜
=⎜
⎝1 1
− (1 − 2p)n+1
2 2
⎞
1 1
− (1 − 2p)n+1 ⎟
2 2
⎟
⎟.
⎠
1 1
+ (1 − 2p)n+1
2 2
To do so, note that
P (n+1) =
p00 p01
p10 p11
n
n
n
n
n
n
p00
p00 p00
p01
+ p01 p10
p00 p01
+ p01 p11
=
.
n
n
n
n
n
n
p10
p11
p10 p00
+ p11 p10
p10 p01
+ p11 p11
Thus
1 1
− (1 − 2p)n + (1 − p)
2 2
1
1
= p + (1 − p) + (1 − 2p)n − p + (1 − p) =
2
2
n+1
n
n
p11
= p10 p01
+ p11 p11
=p
1 1
+ (1 − 2p)n
2 2
1 1
+ (1 − 2p)n+1 .
2 2
1 1
n+1
This establishes what we wanted to show. The proof that p00
= + (1 − 2p)n+1 is
2 2
identical to what we just showed. We have
n+1
n+1
P01
= 1 − P00
=1−
1 1
1 1
+ (1 − 2p)n = − (1 − 2p)n .
2 2
2 2
Similarly,
n+1
n+1
p10
= 1 − p11
=
(b)
1 1
− (1 − 2p)n .
2 2
Let π0 and π1 be the long-run probabilities that Alberto loses and wins a game, respectively. Then
π0
1−p
p
π0
=
,
p
1−p
π1
π1
and π0 + π1 = 1 imply that π0 = π1 = 1/2. Therefore, the expected number of games
Alberto will play between two consecutive wins is 1/π1 = 2.
306
Chapter 12
Stochastic Processes
21. For each j ≥ 0, limn→∞ pijn exists and is independent of i if the following system of equations,
in π0 , π1 , . . . , have a unique solution.
⎧⎛ ⎞ ⎛
⎪
π
1−p 1−p
0
0
0
⎪
⎪⎜ 0 ⎟ ⎜
⎪
⎪
⎜π1 ⎟ ⎜ p
0
1−p
0
0
⎪
⎪
⎟ ⎜
⎪⎜
⎪
⎟
⎜
⎨⎜
p
0
1−p
0
⎜π2 ⎟ = ⎜ 0
⎜ ⎟ ⎜
⎜π3 ⎟ ⎜ 0
0
p
0
1−p
⎪
⎪
⎝ ⎠ ⎝
⎪
⎪
.
.
⎪
..
..
⎪
⎪
⎪
⎪
⎩
π0 + π1 + π2 + · · · = 1.
From the matrix equation, we obtain
p
i
π0 ,
πi =
1−p
0
0
0
0
⎞⎛ ⎞
...
π0
⎟⎜ ⎟
⎜ ⎟
. . .⎟
⎟ ⎜π1 ⎟
⎟
⎟
. . .⎟ ⎜
⎜π2 ⎟
⎟⎜ ⎟
. . .⎟ ⎜π3 ⎟
⎠⎝ ⎠
..
.
i = 0, 1, . . . .
p
i
to
1−p
i=0
converge. Hence we must have p < 1 − p, or p < 1/2. Therefore, for p < 1/2, this
irreducible, aperiodic Markov chain which is positively recurrent has limiting probabilities.
Note that, for p < 1/2,
∞
i
p
π0
=1
1−p
i=0
For these quantities to satisfy
yields π0 = 1 −
∞
∞
i=0 πi = 1, we need the geometric series
p
. Thus the limiting probabilities are
1−p
p
i
p
1−
πi =
,
i = 0, 1, 2, . . . .
1−p
1−p
22. Let Yn be Carl’s fortune after the nth game. Let Xn be Stan’s fortune after the nth game. Let
Zn = Yn − Xn . The {Zn : n = 0, 1, . . . } is a random walk with state space {0, ±2, ±4, . . . }.
We have that Z0 = 0, and at each step either the process moves two units to the right with
probability 0.46 or two units to the left with probability 0.54. Let A be the event that, starting
at 0, the random walk will eventually enter 2; P (A) is the desired quantity. By the law of total
probability,
P (A) = P (A | Z1 = 2)P (Z1 = 2) + P (A | Z1 = −2)P (Z1 = −2)
2
= 1 · (0.46) + P (A) · (0.54).
2
To show that P (A | Z1 = −2) = P (A) , let E be the event of, starting from −2, eventually
entering 0. It should be clear that P (E) = P (A). By independence of E and A, we have
2
P (A | Z = −2) = P (EA) = P (E)P (A) = P (A) .
Section 12.3
Markov Chains
307
We have shown that P (A), the quantity we are interested in, satisfies
2
(0.54) P (A) − P (A) + 0.46 = 0.
This is a quadratic equation in P (A). Solving it gives P (A) = 23/27 ≈ 0.85.
23. We will use induction on m. For m = 1, the relation is, simply, the Markovian property, which
is true. Suppose that the relation is valid for m − 1. We will show that it is also valid for m.
We have
P (Xn+m = j | X0 = i0 , X1 = i1 , . . . , Xn = in )
=
P (Xn+m = j | X0 = i0 , . . . , Xn = in , Xn+m−1 = i)
i∈S
P (Xn+m−1 = i | X0 = i0 , . . . , Xn = in )
=
P (Xn+m = j | Xn+m−1 = i)P (Xn+m−1 = i | Xn = in )
i∈S
P (Xn+m = j | Xn+m−1 = i, Xn = in )P (Xn+m−1 = i | Xn = in )
=
i∈S
= P (Xn+m = j | Xn = in ),
where the following relations are valid from the definition of Markov chain: given the present
state, the process is independent of the past.
P (Xn+m = j | X0 = i0 , . . . , Xn = in , Xn+m−1 = i) = P (Xn+m = j | Xn+m−1 = i),
P (Xn+m = j | Xn+m−1 = i) = P (Xn+m = j | Xn+m−1 = i, Xn = in ).
2n+1
24. Let (0, 0), the origin, be denoted by O. It should be clear that, for all n ≥ 0, POO
= 0. Now,
for n ≥ 1, let Z1 , Z2 , Z3 , and Z4 be the number of transitions to the right, left, up, and down,
respectively. The joint probability mass function of Z1 , Z2 , Z3 , and Z4 is multinomial. We
have
n
2n
POO
=
P (Z1 = i, Z2 = i, Z3 = n − i, Z4 = n − i)
i=0
n
=
i=0
i=0
1
(2n)!
n!
n!
·
·
n! n! i! (n − i)! i! (n − i)! 4
1
2n
n
=
=
1 i1 i1
(2n)!
i! i! (n − i)! (n − i)! 4
4
4
4
2n
n
n
i=0
2
n
.
i
n−i 1 n−i
4
2n
308
Chapter 12
Stochastic Processes
n
By Example 2.28,
i=0
2
1
n
2n
n
=
. Thus POO
=
i
4
n
(Stirling’s formula),
1 2n 2n2 1
=
n
4
4
∞
n
POO
Therefore,
1
Since
π
n=1
∞
n=1
2n
2
2n
n
. Now, by Theorem 2.7
√
4π n (2n)2n e−2n 2
(2n)! 2
1
1
·
∼ 2n · √
=
.
n! n!
4
πn
( 2π n · nn · e−n )2
∞
∞
1 2n 2n 2
1
is convergent.
=
is convergent if and only if
4
πn
n
n=1
n=1
2n
1
n
is divergent, ∞
n=1 POO is divergent, showing that the state (0, 0) is recurrent.
n
25. Clearly, P (Xn+1 = 1 | Xn = 0) = 1. For i ≥ 1, given Xn = i, either Xn+1 = i + 1 in which
case we say that a transition to the right has occurred, or Xn+1 = i − 1 in which case we say
that a transition to the left has occurred. For i ≥ 1, given Xn = i, when the nth transition
occurs, let S be the remaining service time of the customer being served or the service time
of a new customer, whichever applies. Let T be the time from the nth transition until the next
arrival. By the memoryless property of exponential random variables, S and T are exponential
random variables with parameters µ and λ, respectively. For i ≥ 1,
∞
P (Xn+1 = i + 1 | Xn = i) = P (T < S) =
P (S > T | T = t)λe−λt dt
=
0
0
∞
P (S > t)λe−λt dt =
∞
e−µt · λe−λt dt =
0
λ
.
λ+µ
Therefore,
P (Xn+1 = i − 1 | Xn = i) = P (T > S) = 1 −
µ
λ
=
.
λ+µ
λ+µ
These calculations show that knowing Xn , the next transition does not depend on the values of
Xj for j < n. Therefore, {Xn : n = 1, 2, . . . } is a Markov chain, and its transition probability
matrix is given by
⎞
⎛
0
1
0
0
0
...
⎟
⎜ µ
λ
⎟
⎜
0
0
0
.
.
.
⎟
⎜λ + µ
λ+µ
⎟
⎜
λ
µ
⎟
⎜
0
0
.
.
.
0
⎟.
⎜
P =⎜
⎟
λ+µ
λ+µ
⎟
⎜
λ
µ
⎜ 0
0
. . .⎟
0
⎟
⎜
λ+µ
λ+µ
⎠
⎝
..
.
Since all states are accessible from each other, this Markov chain is irreducible. Starting from
0, for the Markov chain to return to 0, it needs to make as many transitions to the left as it
Section 12.3
Markov Chains
309
n
makes to the right. Therefore, P00
> 0 only for positive even integers. Since the greatest
common divisor of such integers is 2, the period of 0, and hence the period of all other states
is 2.
26. The ij th element of P Q is the product of the ith row of P with the jth column of Q. Thus
pi qj . To show that the sum of each row of P Q is 1, we will now calculate the sum
it is
of the elements of the ith row of P Q, which is
pi qj =
j
qj = 1 and
Note that
j
pi qj =
j
qj =
pi
j
pi = 1.
pi = 1 since the sum of the elements of the th row of Q and
the sum of the elements of the ith row of P are 1.
27. If state j is accessible from state i, there is a path
i = i1 , i2 , i3 , . . . , in = j
from i to j . If n ≤ K, we are done. If n > K, by the pigeonhole principle, there must exist k
and (k < ) so that ik = i . Now the path
i = i1 , i2 , . . . , ik , ik+1 , . . . , i , i+1 , . . . , in = j
can be reduced to
i = i1 , i2 , . . . , ik , i+1 , . . . , in = j
which is still a path from i to j but in fewer steps. Repeating this procedure, we can eliminate
all of the states that appear more than once from the path and yet reach from i to j with a
positive probability. After all such eliminations are made, we obtain a path
i = i1 , im1 , im2 , . . . , in = j
in which the states i1 , im1 , im2 , . . . , in are distinct states. Since there are K states altogether,
this path has at most K states.
n
28. Let I = {n ≥ 1 : piin > 0} and J = {n ≥ 1 : pjj
> 0}. Then d(i), the period of i, is the
greatest common divisor of the elements of I , and d(j ), the period of j , is the greatest common
divisor of the elements of J . If d(i) = d(j ), then one of d(i) and d(j ) is smaller than the
other one. We will prove the theorem for the case in which d(j ) < d(i). The proof for the
case in which d(i) < d(j ) follows by symmetry. Suppose that for positive integers n and m,
k
pijn > 0 and pjmi > 0. Let k ∈ J ; then pjj
> 0. We have
piin+m ≥ pijn pjmi > 0,
310
Chapter 12
Stochastic Processes
and
k m
pj i > 0.
piin+k+m ≥ pijn pjj
By these inequalities, we have that d(i) divides n + m and n + k + m. Hence it divides
(n + k + m) − (n + m) = k. We have shown that, if k ∈ J , then d(i) divides k. This means
that d(i) divides all members of J . It contradicts the facts that d(j ) is the greatest common
divisor of J and d(j ) < d(i). Therefore, we must have d(i) = d(j ).
29. The stochastic process {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, . . . , k−1}.
For 0 ≤ i ≤ k − 2, a transition is only possible from state i to 0 or i + 1. The only transition
from k − 1 is to 0. Let Z be the number of weeks it takes Liz to play again with Bob from the
time they last played. The event Z > i occurs if and only if Liz has not played with Bob since
i Sundays ago, and the earliest she will play with him is next Sunday. Now the probability is
i/k that Liz will play with Bob if last time they played was i Sundays ago; hence
i
P (Z > i) = 1 − ,
k
i = 1, 2, . . . , k − 1.
Using this fact, for 0 ≤ i ≤ k − 2, we obtain
pi(i+1) = P (Xn+1 = i + 1 | Xn = i) =
P (Xn = i, Xn+1 = i + 1)
P (Xn = i)
i+1
1−
P (Z > i + 1)
k = k − i − 1,
=
=
i
P (Z > i)
k−i
1−
k
pi0 = P (Xn+1 = 0 | Xn = i) = 1 −
k−i−1
1
=
,
k−i
k−i
p(k−1)0 = P (Xn+1 = 0 | Xn = k − 1) = 1.
Hence the transition probability matrix of {Xn : n = 1, 2, . . . } is given by
Section 12.3
⎛ 1
⎜ k
⎜
⎜
⎜ 1
⎜
⎜k − 1
⎜
⎜
⎜ 1
⎜
⎜
⎜k − 2
⎜
⎜
P =⎜
⎜ 1
⎜k − 3
⎜
⎜
⎜
⎜ ..
⎜ .
⎜
⎜
⎜
⎜ 1
⎜
⎜ 2
⎝
1−
311
⎞
1
k
1−
0
1
Markov Chains
0
0
0
...
0
1
k−1
0
0
...
0
1
k−2
0
...
0
1
k−3
...
0
1−
0
0
0
0
0
0
0
0
0
...
0
0
0
0
0
...
0
1−
0
⎟
⎟
⎟
⎟
0⎟
⎟
⎟
⎟
⎟
⎟
0⎟
⎟
⎟
⎟
⎟.
0⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
1⎟
⎟
2⎟
⎠
0
It should be clear that the Markov chain under consideration is irreducible, aperiodic, and
positively recurrent. For 0 ≤ i ≤ k − 1, let πi be the long-run probability that Liz says no to
Bob for i consecutive weeks. π0 , π1 , . . . , πk−1 are obtained from solving the following matrix
equation along with k−1
i=0 πi = 1.
⎛ 1
⎜ k
⎛
⎞ ⎜
⎜
π0
⎜
1
⎜
⎟ ⎜
1−
⎜
⎟ ⎜
k
⎜
⎜ π1 ⎟ ⎜
⎜
⎟ ⎜
⎜
⎟ ⎜
⎜
⎟
0
⎜ π2 ⎟ ⎜
⎜
⎟ ⎜
⎜
⎜
⎟ ⎜
⎜
⎟
⎜ π3 ⎟ ⎜
⎜
⎟=⎜
0
⎜
⎟ ⎜
⎜
⎟ ⎜
⎜
⎜ .. ⎟ ⎜
⎜ . ⎟ ⎜
⎜
⎟ ⎜ 0
⎜
⎟ ⎜
⎜
⎟
⎜πk−2 ⎟ ⎜
⎜
⎟ ⎜
⎝
⎠ ⎜
⎜ ..
⎜ .
πk−1
⎜
⎜
⎝
0
⎞
1
k−1
1
k−2
1
k−3
...
1
2
1
0
0
0
...
0
0
1
k−1
0
0
...
0
0
1
k−2
0
...
0
0
1
k−3
...
0
0
0
...
1
2
0
1−
0
1−
0
0
0
0
1−
The matrix equation gives
πi =
k−i
π0 ,
k
i = 1, 2, . . . , k − 1.
⎟
⎟⎛
⎞
⎟
⎟ π0
⎟⎜
⎟
⎟⎜
⎟
⎟⎜
⎟ ⎜ π1 ⎟
⎟
⎟⎜
⎟
⎟⎜
⎟
⎟⎜
π
⎟⎜ 2 ⎟
⎟
⎟⎜
⎟
⎟⎜
⎟
⎟⎜
π
⎟⎜ 3 ⎟
⎟.
⎟⎜
⎟
⎟⎜
⎟⎜ . ⎟
⎟⎜ . ⎟
⎟⎜ . ⎟
⎟
⎟⎜
⎟
⎟⎜
⎟
⎟⎜
⎟ ⎜πk−2 ⎟
⎟
⎟⎝
⎠
⎟
⎟
π
⎟
k−1
⎟
⎠
312
Chapter 12
Using
k−1
i=0
Stochastic Processes
πi = 1, we obtain
k−1
π0
i=0
k−i
=1
k
or, equivalently,
π0
k
k−1
i = 1.
k−1
k−
i=0
i=0
This implies that
π0 2 (k − 1)k
k −
= 1,
k
2
which gives π0 = 2/(k + 1). Hence
πi =
2(k − i)
,
k(k + 1)
i = 0, 1, 2, . . . , k − 1.
30. Let Xi be the amount of money player A has after i games. Clearly, X0 = a and {Xn : n =
0, 1, . . . } is a Markov chain with state space {0, 1, . . . , a, a + 1, . . . , a + b}. For 0 ≤ i ≤
a + b, let mi = E(T | X0 = i). Let F be the event that A wins the first game. Then, for
1 ≤ i ≤ a + b − 1,
E(T | X0 = i) = E(T | X0 = i, F )P (F | X0 = i) + E(T | X0 = i, F c )P (F c | X0 = i).
This gives
1
1
mi = (1 + mi+1 ) + (1 + mi−1 ) ,
2
2
1 ≤ i ≤ a + b − 1,
or, equivalently,
2mi = 2 + mi+1 + mi−1 ,
1 ≤ i ≤ a + b − 1.
Now rewrite this relation as
mi+1 − mi = −2 + mi − mi−1 ,
and, for 1 ≤ i ≤ a + b, let
1 ≤ i ≤ a + b − 1,
yi = mi − mi−1 .
Then
yi+1 = −2 + yi ,
and, for 1 ≤ i ≤ a + b,
1 ≤ i ≤ a + b − 1,
mi = y1 + y2 + · · · + yi .
Clearly, m0 = 0, ma+b = 0, y1 = m1 , and
y2 = −2 + y1 = −2 + m1 ,
y3 = −2 + y2 = −2 + (−2 + m1 ) = −4 + m1
..
.
yi = −2(i − 1) + m1 ,
1 ≤ i ≤ a + b.
Section 12.3
Markov Chains
313
Hence, for 1 ≤ i ≤ a + b,
mi = y1 + y2 + · · · + yi
= im1 − 2 1 + 2 + · · · + (i − 1)
= im1 − i(i − 1) = i(m1 − i + 1).
This and ma+b = 0 imply that
(a + b)(m1 − a − b + 1) = 0,
or m1 = a + b − 1. Therefore,
mi = i(a + b − i),
and hence the desired quantity is
E(T | X0 = a) = ma = ab.
31. Let q be a positive solution of the equation x =
∞
i=0
show that ∀n ≥ 0, P (Xn = 0) ≤ q. This implies that
αi x i . Then q =
∞
i=0
αi q i . We will
p = lim P (Xn = 0) ≤ q.
n→∞
To establish that P (Xn = 0) ≤ q, we use induction. For n = 0, P (X0 = 0) = 0 ≤ q is
trivially true. Suppose that P (Xn = 0) ≤ q. We have
∞
P (Xn+1 = 0) =
P (Xn+1 = 0 | X1 = i)P (X1 = i).
i=0
It should be clear that
i
P (Xn+1 = 0 | X1 = i) = P (Xn = 0 | X0 = 1) .
However, since P (X0 = 1) = 1,
P (Xn = 0 | X0 = 1) = P (Xn = 0).
Therefore,
i
P (Xn+1 = 0 | X1 = i) = P (Xn = 0) .
Thus
∞
P (Xn+1 = 0) =
i
∞
P (Xn = 0) P (X1 = i) ≤
i=0
This establishes the theorem.
q i αi = q.
i=0
314
Chapter 12
Stochastic Processes
32. Multiplying P successively, we obtain
1
13
9 1
1
=
+ ,
13 13
13
9 1
9 2 1
1
+
+ ,
=
13
13
13 13
13
p12 =
2
p12
3
p12
and in general,
n
p12
=
1 9
13 13
1
=
·
13
n−1
+
9
13
9
1−
13
1−
9
13
n
=
n−2
+ ··· + 1
9
1
1−
4
13
n
.
n
= 1/4.
Hence the desired probability is limn→∞ p12
33. We will use induction. Let n = 1; then, for 1 + j − i to be nonnegative, we must have
1+j −i
i − 1 ≤ j . For the inequality
≤ 1 to be valid, we must have j ≤ i + 1. Therefore,
2
i − 1 ≤ j ≤ i + 1. But, for j = i, 1 + j − i is not even. Therefore, if 1 + j − i is an even
1+j −i
nonnegative integer satisfying
≤ 1, we must have j = i − 1 or j = i + 1. For
2
j = i − 1,
1+i−1−i
n+j −i
=
=0
2
2
Hence
and
n−j +i
1−i+1+i
=
= 1.
2
2
1 0
p (1 − p)1 ,
P (X1 = i − 1 | X0 = i) = 1 − p =
0
showing that the relation is valid. For j = i + 1,
1+i+1−i
n+j −i
=
=1
2
2
Hence
and
n−j +i
1−i−1+i
=
= 0.
2
2
1 1
p (1 − p)0 ,
P (X1 = i + 1 | X0 = i) = p =
1
showing that the relation is valid in this case as well. Since, for a simple random walk, the
only possible transitions from i are to states i + 1 and i − 1, in all other cases
P (X1 = j | X0 = i) = 0.
Section 12.4
Continuous-Time Markov Chains
315
We have established the theorem for n = 1. Now suppose that it is true for n. We will show
it for n + 1 by conditioning on Xn :
P (Xn+1 = j | X0 = i) = P (Xn+1 = j | X0 = i, Xn = j − 1)P (Xn = j − 1 | X0 = i)
+ P (Xn+1 = j | X0 = i, Xn = j + 1)P (Xn = j + 1 | X0 = i)
= P (Xn+1 = j | Xn = j − 1)P (Xn = j − 1 | X0 = i)
+ P (Xn+1 = j | Xn = j + 1)P (Xn = j + 1 | X0 = i)
n
= p · n + j − 1 − i p (n+j −1−i)/2 (1 − p)(n−j +1+i)/2
2
n
+ (1 − p) n + j + 1 − i p (n+j +1−i)/2 (1 − p)(n−j −1+i)/2
2
n
n
(n+1+j −i)/2
=
(1 − p)(n+1−j +i)/2
+
n−1+j −i
n+1+j −i p
2
2
n+1
= n + 1 + j − i p (n+1+j −i)/2 (1 − p)(n+1−j +i)/2 .
2
12.4
CONTINUOUS-TIME MARKOV CHAINS
1. By Chapman-Kolmogorov equations,
∞
pij (t + h) − pij (t) =
pik (h)pkj (t) − pij (t)
k=0
pik (h)pkj (t) + pii (h)pij (t) − pij (t)
=
k=i
pik (h)pkj (t) + pij (t) pii (h) − 1 .
=
k=i
Thus
pij (t + h) − pij (t)
=
h
k=i
1 − pii (h)
pik (h)
pkj (t) − pij (t)
.
h
h
Letting h → 0, by (12.13) and (12.14), we have
pij (t) =
qik pkj (t) − νi pij (t).
k=i
316
Chapter 12
Stochastic Processes
2. Clearly, X(t) : t ≥ is a continuous-time Markov chain. Its balance equations are as follows:
Input rate to
=
Output rate from
f
µπ0
=
λπf
0
λπf + µπ1 + µπ2 + µπ3
=
µπ0 + λπ0
1
λπ0
=
λπ1 + µπ1
2
λπ1
=
λπ2 + µπ2
3
λπ2
=
µπ3 .
State
Solving these equations along with
π f + π0 + π1 + π2 + π3 = 1
we obtain
πf =
π1 =
π3 =
µ2
,
λ(λ + µ)
λµ
,
(λ + µ)2
λ
3
π0 =
µ
,
λ+µ
π2 =
λ2 µ
,
(λ + µ)3
.
λ+µ
3. The fact that X(t) : t ≥ 0 is a continuous-time Markov chain should be clear. The balance
equations are
State
Input rate to
=
Output rate from
(0, 0)
µπ(1,0) + λπ(0,1)
=
λπ(0,0) + µπ(0,0)
(n, 0)
µπ(n+1,0) + λπ(n−1,0)
=
λπ(n,0) + µπ(n,0) ,
λπ(0,m+1) + µπ(0,m−1)
=
n≥1
m ≥ 1.
4. Let X(t) be the number of customers in the system at time t. Then the process X(t) : t ≥ 0
is a birth and death process with λn = λ, n ≥ 0, and µn = nµ, n ≥ 1. To find π0 , the
probability that the system is empty, we will first calculate the sum in (12.18). We have
(0, m)
∞
n=1
λ0 λ1 · · · λn−1
=
µ1 µ2 · · · µn
∞
n=1
λn
=
n! µn
∞
n=1
1 λ
n! µ
λπ(0,m) + µπ(0,m)
n
∞
= −1 +
Hence, by (12.18),
π0 =
1
= e−λ/µ .
1 − 1 + eλ/µ
n=0
1 λ
n! µ
n
= −1 + eλ/µ .
Section 12.4
Continuous-Time Markov Chains
317
By (12.17),
πn =
λn π0
(λ/µ)n e−λ/µ
,
=
n!µn
n!
n = 0, 1, 2, . . . .
This shows that the long-run number of customers in such an M/M/∞ queueing system is
Poisson with parameter λ/µ. The average number of customers in the system is, therefore,
λ/µ.
5. Let X(t) be the number of operators busy serving customers at time t. Clearly, X(t) : t ≥ 0
is a finite-state birth and death process with state space {0, 1, . . . , c}, birth rates λn = λ,
n = 0, 1, . . . , c, and death rates µn = nµ, n = 0, 1, . . . , c. Let π0 be the proportion of
time that all operators are free. Let πc be the proportion of time all of them are busy serving
customers.
(a)
πc is the desired quantity. By (12.22),
1
π0 =
c
1+
n=1
n
λ
n! µn
=
1
1 λ
c
n=0
.
n
n! µ
By (12.21),
1
(λ/µ)c
c!
πc =
.
c 1
n
(λ/µ)
n=0
n!
This formula is called Erlang’s loss formula.
(b)
We want to find the smallest c for which
1/c!
≤ 0.004.
n=0 (1/n!)
c
For c = 5, the left side is 0.00306748. For c = 4, it is 0.01538462. Therefore, the
airline must hire at least five operators to reduce the probability of losing a call to a
number less than 0.004.
6. No, it is not because it is possible for the process to enter state 0 directly from state 2. In a
birth and death process, from a state i, transitions are only possible to the states i − 1 and i + 1.
7. For n ≥ 0, let Hn be the time, starting from n, until the process enters state n + 1 for the first
time. Clearly, E(H0 ) = 1/λ and, by Lemma 12.2,
E(Hn ) =
1
+ E(Hn−1 ),
λ
n ≥ 1.
318
Chapter 12
Stochastic Processes
Hence
1
,
λ
1
E(H1 ) = +
λ
1
E(H2 ) = +
λ
E(H0 ) =
1
2
= ,
λ
λ
2
3
= .
λ
λ
Continuing this process, we obtain,
E(Hn ) =
n+1
,
λ
n ≥ 0.
The desired quantity is
j −1
j −1
E(Hn ) =
n=i
n=i
n+1
1
= (i + 1) + (i + 2) + · · · + j
λ
λ
1
(1 + 2 + · · · + j ) − (1 + 2 + · · · + i)
λ
1 j (j + 1) i(i + 1) j (j + 1) − i(i + 1)
=
−
=
.
λ
2
2
2λ
=
8. Suppose that a birth occurs each time that an out-of-order machine is repairedand begins to
operate, and a death occurs each time that a machine breaks down. The fact that X(t) : t ≥ 0
is a birth and death process with state space {0, 1, . . . , m} should be clear. The birth and death
rates are
λn =
kλ
n = 0, 1, . . . , m − k
(m − n)λ n = m − k + 1, m − k + 2, . . . , m,
µn =
nµ
n = 0, 1, . . . , m.
9. The Birth rates are
λ0 = λ
λn = αn λ, n ≥ 1.
The death rates are
µ0 = 0
µn = µ + (n − 1)γ ,
n ≥ 1.
10. Let X(t) be the population size at time t. Then X(t) : t ≥ 0 is a birth and death process
with birth rates λn = nλ + γ , n ≥ 0, and death rates µn = nµ, n ≥ 1. For i ≥ 0, let Hi
Section 12.4
Continuous-Time Markov Chains
319
be the time, starting from i, until the population size reaches i + 1 for the first time. We are
interested in E(H0 ) + E(H1 ) + E(H2 ). Note that, by Lemma 12.2,
E(Hi ) =
1
µi
+ E(Hi−1 ),
λi
λi
i ≥ 1.
Since E(H0 ) = 1/γ ,
E(H1 ) =
1
µ
1
µ+γ
+
· =
,
λ+γ
λ+γ γ
γ (λ + γ )
E(H2 ) =
2µ
µ+γ
γ (λ + γ ) + 2µ(µ + γ )
1
+
·
=
.
2λ + γ
2λ + γ γ (λ + γ )
γ (λ + γ )(2λ + γ )
and
Thus the desired quantity is
E(H0 ) + E(H1 ) + E(H2 ) =
(λ + γ )(2λ + γ ) + (µ + γ )(2λ + 2µ + γ ) + γ (λ + γ )
.
γ (λ + γ )(2λ + γ )
11. Let X(t) be the number of deaths in the time interval
[0, t]. Since there are no births, by
Remark 7.2, it should be clear that X(t) : t ≥ 0 is a Poisson process with rate µ as long as
the population is not extinct. Therefore, for 0 < j ≤ i,
pij (t) =
e−µt (µt)i−j
.
(i − j )!
Clearly, p00 (t) = 1. For i > 0, j = 0, we have
i
pi0 (t) = 1 −
i
pij (t) = 1 −
j =1
j =1
e−µt (µt)i−j
=1−
(i − j )!
1
j =i
e−µt (µt)i−j
.
(i − j )!
Letting k = i − j yields
i−1
pi0 (t) = 1 −
k=0
e−µt (µt)k
=
k!
∞
k=i
e−µt (µt)k
.
k!
12. Suppose that a birth occurs whenever a physician takes a break, and a death occurs whenever
he or she becomes available
to answerpatients’ calls. Let X(t) be the number of physicians
on break at time t. Then X(t) : t ≥ 0 is a birth and death process with state space {0, 1, 2}.
Clearly, X(t) = 0 if at t both of the physicians are available to answer patients’ calls, X(t) = 1
if at t only one of the physicians is available to answer patients’ calls, and X(t) = 2 if at t
none of the physicians is available to answer patients’ calls. We have that
λ0 = 2λ,
λ1 = λ,
λ2 = 0,
320
Chapter 12
Stochastic Processes
µ0 = 0,
µ1 = µ,
µ2 = 2µ.
Therefore,
ν0 = 2λ,
ν1 = λ + µ,
ν2 = 2µ.
Also,
p01 = p21 = 1,
p02 = p20 = 0,
p10 =
µ
,
λ+µ
p12 =
λ
.
λ+µ
Therefore,
q01 = ν0 p01 = 2λ,
q10 = ν1 p10 = µ,
q12 = ν1 p12 = λ,
q21 = ν2 p21 = 2µ,
q02 = q20 = 0.
Substituting these quantities in the Kolmogorov backward equations
pij (t) =
qik pkj (t) − νi pij (t),
k=i
we obtain
p00
(t) = 2λp10 (t) − 2λp00 (t)
(t) = 2λp11 (t) − 2λp01 (t)
p01
(t) = 2λp12 (t) − 2λp02 (t)
p02
(t) = λp20 (t) + µp00 (t) − (λ + µ)p10 (t)
p10
(t) = λp21 (t) + µp01 (t) − (λ + µ)p11 (t)
p11
(t) = λp22 (t) + µp02 (t) − (λ + µ)p12 (t)
p12
(t) = 2µp10 (t) − 2µp20 (t)
p20
(t) = 2µp11 (t) − 2µp21 (t)
p21
(t) = 2µp12 (t) − 2µp22 (t).
p22
13. Let X(t) be the number of customers in the system at time t. Then X(t) : n ≥ 0 is a birth
and death process with λn = λ, for n ≥ 0, and
µn =
nµ
n = 0, 1, . . . , c
cµ
n > c.
By (12.21), for n = 1, 2, . . . c,
πn =
λn
1 λ
π
=
0
n! µn
n! µ
n
π0 ;
for n > c,
πn =
λn
λn
cc λ
π
=
π
=
0
0
c! µc (cµ)n−c
c! cn−c µn
c! cµ
n
π0 =
cc n
ρ π0 .
c!
Section 12.4
Noting that
c
n=0
πn +
∞
n=c+1
c
n=0
∞
n=c+1
1 λ
n! µ
n
+ π0
c
n=0
1 λ
n! µ
n
cc
+
c!
cc
c!
∞
ρ n = 1.
n=c+1
ρ c+1
. Therefore,
1−ρ
ρn =
1
π0 =
321
πn = 1, we have
π0
Since ρ < 1, we have
Continuous-Time Markov Chains
=
∞
ρ
c! (1 − ρ)
n
c! (1 − ρ)
c
1 λ n
n=0
n=c+1
n! µ
.
+c ρ
c c+1
14. Let s, t > 0. If j < i, then pij (s + t) = 0, and
∞
∞
i−1
pik (s)pkj (t) =
k=0
pik (s)pkj (t) +
k=0
pik (s)pkj (t) = 0,
k=i
since pik (s) = 0 if k < i, and pkj (t) = 0 if k ≥ i > j. Therefore, for j < i, the ChapmanKolmogorov equations are valid. Now suppose that j > i. Then
∞
j
pik (s)pkj (t) =
k=0
pik (s)pkj (t)
k=i
j
=
k=i
=
=
e−λs (λs)k−i e−λt (λt)j −k
·
(k − i)!
(j − k)!
e−λ(t+s)
(j − i)!
k=i
−λ(t+s) j −i
e
(j − i)!
e−λ(t+s)
=
(j − i)!
=
j
(j − i)!
(λs)k−i (λt)j −k
k − i)! (j − k)!
=0
(j − i)!
(λs) (λt)(j −i)−
! (j − i − )!
j −i
=0
j −i
(λs) (λt)(j −i)−
−λ(t+s)
e
(λs + λt)j −i
(j − i)!
where the last equality follows by Theorem 2.5, the binomial expansion. Since
j −i
e−λ(t+s)
= pij (s + t),
λ(t + s)
(j − i)!
we have shown that the Chapman-Kolmogorov equations are satisfied.
322
Chapter 12
Stochastic Processes
15. Let X(t) be the number of particles
in the shower
t units of time after the cosmic particle enters
the earth’s atmosphere. Clearly, X(t) : t ≥ 0 is a continuous-time
Markov chain with state
space {1, 2, . . . } and νi = iλ, i ≥ 1. In
fact,
X(t)
:
t
≥
0
is
a
pure
birth process, but that
fact will not help us solve this exercise. Clearly, for i ≥ 1, j ≥ 1,
pij =
1
0
if j = i + 1
if j = i + 1.
qij =
νi
0
if j = i + 1
if j = i + 1.
Hence
We are interested in finding p1n (t). This is the desired probability. For n = 1, p11 (t) is the
probability that the cosmic particle does not collide with any air particles during the first t
units of time in the earth’s atmosphere. Since the time it takes the particle to collide with
another particle is exponential with parameter λ, we have p11 (t) = e−λt . For n ≥ 2, by the
Kolmogorov’s forward equation,
p1n
(t) =
qkn p1k (t) − νn p1n (t)
k=n
= q(n−1)n p1(n−1) (t) − νn p1n (t)
= νn−1 p1(n−1) (t) − νn p1n (t).
Therefore,
(t) = (n − 1)λp1(n−1) (t) − nλp1n (t).
p1n
For n = 2, this gives
or, equivalently,
(49)
(t) = λp11 (t) − 2λp12 (t)
p12
(t) = λe−λt − 2λp12 (t).
p12
Solving this first order linear differential equation with boundary condition p12 (0) = 0, we
obtain
p12 (t) = e−λt (1 − e−λt ).
For n = 3, by (49),
or, equivalently,
(t) = 2λp12 (t) − 3λp13 (t)
p13
(t) = 2λe−λt (1 − e−λt ) − 3λp13 (t).
p13
Solving this first order linear differential equation with boundary condition p13 (0) = 0 yields
p13 (t) = e−λt (1 − e−λt )2 .
Continuing this process, and using induction, we obtain that
p1n (t) = e−λt (1 − e−λt )n−1
n ≥ 1.
Section 12.4
Continuous-Time Markov Chains
323
16. It is straightforward to see that
π(i,j ) =
λ
µ1
i
1−
λ λ
µ1 µ2
j
1−
λ
,
µ2
i, j ≥ 0,
satisfy the following balance equations for the tandem queueing system under consideration.
Hence, by Example 12.43, π(i,j ) is the product of an M/M/1 system having i customers in
the system, and another M/M/1 queueing system having j customers in the system. This
establishes what we wanted to show.
State
(0, 0)
(i, 0), i ≥ 1
(0, j ), j ≥ 1
(i, j ), i, j ≥ 1
Input rate to
=
Output rate from
µ2 π(0,1)
µ2 π(i,1) + λπ(i−1,0)
µ2 π(0,j +1) + µ1 π(1,j −1)
µ2 π(i,j +1) + µ1 π(i+1,j −1) + λπ(i−1,j )
=
=
=
=
λπ(0,0)
λπ(i,0) + µ1 π(i,0)
λπ(0,j ) + µ2 π(0,j )
λπ(i,j ) + µ1 π(i,j ) + µ2 π(i,j ) .
17. Clearly, X(t) : t ≥ 0 is a birth and death process with birth rates λi = iλ, i ≥ 0, and death
rates µi = iµ + γ , i > 0; µ0 = 0. For some m ≥ 1, suppose that X(t) = m. Then, for
infinitesimal values of h, by (12.5), the population at t +h is m+1 with probability mλh+o(h),
it is m − 1 with probability (mµ + γ )h + o(h), and it is still m with probability
1 − mλh − o(h) − (mµ + γ )h − o(h) = 1 − (mλ + mµ + γ )h + o(h).
Therefore,
E X(t + h) | X(t) = m = (m + 1) mλh + o(h) + (m − 1) (mµ + γ )h + o(h)
+ m 1 − (mλ + mµ + γ )h + o(h)
= m + m(λ − µ) − γ h + o(h).
This relation implies that
E X(t + h) | X(t) = X(t) + (λ − µ)X(t) − γ h + o(h).
Equating the expected values of both sides, and noting that
E E X(t + h) | X(t) = E X(t + h) ,
we obtain
E X(t + h) = E X(t) + h(λ − µ)E X(t) − γ h + o(h).
For simplicity, let g(t) = E X(t) . We have shown that
g(t + h) = g(t) + h(λ − µ)g(t) − γ h + o(h)
324
Chapter 12
Stochastic Processes
or, equivalently,
g(t + h) − g(t)
o(h)
= (λ − µ)g(t) − γ +
.
h
h
As h → 0, this gives
g (t) = (λ − µ)g(t) − γ .
If λ = µ, then g (t) = −γ . So g(t) = −γ t + c. Since g(0) = n, we must have c = n, or
g(t) = −γ t + n. If λ = µ, to solve the first order linear differential equation,
g (t) = (λ − µ)g(t) − γ ,
let f (t) = (λ − µ)g(t) − γ . Then
1
f (t) = f (t),
λ−µ
or
f (t)
= λ − µ.
f (t)
This yields
ln |f (t)| = (λ − µ)t + c,
or
f (t) = e(λ−µ)t+c = Ke(λ−µ)t ,
where K = ec . Thus
g(t) =
K (λ−µ)t
γ
e
+
.
λ−µ
λ−µ
Now g(0) = n implies that K = n(γ − µ) − γ . Thus
g(t) = E X(t) = ne(λ−µ)t +
γ
1 − e(λ−µ)t .
λ−µ
18. For n ≥ 0, let En be the event that, starting from state n, eventually extinction will occur. Let
αn = P (En ). Clearly, α0 = 1. We will show that αn = 1, for all n. For n ≥ 1, starting from
n, let Zn be the state to which the process will move. Then Zn is a discrete random variable
with set of possible values {n − 1, n + 1}. Conditioning on Zn yields
P (En ) = P (En | Zn = n − 1)P (Zn = n − 1) + P (En | Zn = n + 1)P (Zn = n + 1).
Hence
αn = αn−1 ·
µn
λn
+ αn+1 ·
,
λn + µn
λn + µn
n ≥ 1,
or, equivalently,
λn (αn+1 − αn ) = µn (αn − αn−1 ),
n ≥ 1.
Section 12.4
Continuous-Time Markov Chains
325
For n ≥ 0, let yn = αn+1 − αn . We have
λn yn = µn yn−1 ,
or
yn =
µn
yn−1 ,
λn
n ≥ 1,
n ≥ 1.
Therefore,
µ1
y0
λ1
µ2
µ1 µ2
y2 =
y1 =
y0
λ2
λ1 λ2
y1 =
..
.
yn =
µ1 µ2 · · · µn
y0 .
λ1 λ2 · · · λn
n ≥ 1.
On the other hand, by yn = αn+1 − αn , n ≥ 0,
α1 = α0 + y0 = 1 + y0
α2 = α1 + y1 = 1 + y0 + y1
..
.
αn+1 = 1 + y0 + y1 + · · · + yn .
Hence
n
αn+1 = 1 + y0 +
yk
k=1
n
= 1 + y0 + y0
k=1
= 1 + y0 1 +
n
k=1
µ1 µ2 · · · µk
λ1 λ2 · · · λk
µ1 µ2 · · · µk
λ1 λ2 · · · λk
= 1 + (α1 − 1) 1 +
n
k=1
∞
n
µ1 µ2 · · · µk
.
λ1 λ2 · · · λk
µ1 µ2 · · · µk
µ1 µ2 · · · µk
= ∞, the sequence
increases without bound. For
λ1 λ2 · · · λk
λ1 λ2 · · · λk
k=1
k=1
αn ’s to exist, this requires that α1 = 1, which in turn implies that αn+1 = 1, for n ≥ 1.
Since
326
12.5
Chapter 12
Stochastic Processes
BROWNIAN MOTION
1. (a) By the independent-increments property of Brownian motions, the desired probability
is
P − 1/2 < Z(10) < 1/2 | Z(5) = 0
= P − 1/2 < Z(10) − Z(5) < 1/2 | Z(5) = 0
= P − 1/2 < Z(10) − Z(5) < 1/2 .
Since Z(10) − Z(5) is normal with mean 0 and variance (10 − 5)σ 2 = 45, letting
Z ∼ N(0, 1), we have
P − 1/2 < Z(10) − Z(5) < 1/2
−0.5 − 0
0.5 − 0
=P
ε = P |X(t)| > εt
= P X(t) > εt + P X(t) < −εt
εt
εt
=P Z > √ +P Z <− √
σ t
σ t
√
√
ε t
ε t
+P Z <−
=P Z>
σ
σ
√
√
− ε t/σ
= 1 − ε t/σ +
√
√
√
= 1 − ε t/σ + 1 − ε t/σ = 2 − 2 ε t/σ .
This implies that
lim P
t→0
whereas
lim P
t→∞
|X(t)|
t
|X(t)|
t
> ε = 2 − 1 = 1.
> ε = 2 − 2 = 0,
4. Let F be the probability distribution function of 1/Y 2 . Let Z ∼ N (0, 1). We have
√
√
F (t) = P 1/Y 2 ≤ t = P Y 2 ≥ 1/t = P Y ≥ 1/ t + P Y ≤ −1/ t
α
α
=P Z ≥ √ +P Z ≤− √
σ t
σ t
α
α
α
− √ =2 1−
=1−
√ +
√ ,
σ t
σ t
σ t
which, by (12.35), is also the distribution function of Tα .
5. Clearly, P (T < x) = 0 if x ≤ t. For x > t, by Theorem 12.10,
2
P (T < x) = P at least one zero in (t, x) = arccos
π
(
t
.
x
Let F be the distribution function of T . We have shown that
⎧
⎪
x≤t
⎨0
(
F (x) = 2
t
⎪
⎩ arccos
x ≥ t.
π
x
6. Rewrite X(t1 ) + X(t2 ) as X(t1 ) + X(t2 ) = 2X(t1 ) + X(t2 ) − X(t1 ). Now 2X(t1 ) and X(t2 ) −
X(t1 ) are independent
variables.
By Theorem 11.7, 2X(t1 ) ∼ N(0, 4σ 2 t1 ). Since
random
X(t2 ) − X(t1 ) ∼ N 0, σ 2 (t2 − t1 ) , applying Theorem 11.7 once more implies that
2X(t1 ) + X(t2 ) − X(t1 ) ∼ N 0, 4σ 2 t1 + σ 2 (t2 − t1 ) .
328
Chapter 12
Stochastic Processes
Hence X(t1 ) + X(t2 ) ∼ N(0, 3σ 2 t1 + σ 2 t2 ).
7. Let f (x, y) be the joint probability density function of X(t) and X(t +u). Let fX(t+u)|X(t) (y|a)
be the conditional probability density function of X(t + u) given that X(t) = a. Let fX(t) (x)
be the probability density function of X(t). We know that X(t) is normal with mean 0 and
variance σ 2 t. The formula for f (x, y) is given by (12.28). Using these, we obtain
1
1 a 2 (y − a)2
+
exp
−
√
2σ 2 t
u
2σ 2 π tu
f (a, y)
fX(t+u)|X(t) (y|a) =
=
2
fX(t) (a)
1
a
exp −
√
2σ 2 t
σ 2π t
1
1
2
= √
.
(y
−
a)
exp −
2σ 2 u
σ 2π u
This shows that the conditional probability density function of X(t + u) given that X(t) = a
is normal with mean a and variance σ 2 u. Hence
E X(t + u) | X(t) = a = a.
This implies that
E X(t + u) | X(t) = X(t).
8. By Example 10.23,
E X(t)X(t + u) | X(t) = X(t)E X(t + u) | X(t) .
By Exercise 7 above,
E X(t + u) | X(t) = X(t).
Hence
E X(t)X(t + u) = E E X(t)X(t + u) | X(t)
= E X(t)E X(t + u) | X(t)
= E X(t) · X(t) = E X(t)2
2
= Var X(t) + E X(t) = σ 2 t + 0 = σ 2 t.
9. For t > 0, the probability density function of Z(t) is
φt (x) =
1
x2
.
exp −
√
2σ 2 t
σ 2π t
Section 12.5
Brownian Motion
329
Therefore,
E V (t) = E |Z(t)| =
∞
=2
∞
−∞
|x|φt (x) dx
0
Making the change of variable u =
E V (t) = σ
(
2t
π
∞
xφt (x) dx = 2
0
x
√ yields
σ t
(
∞
ue
0
x
2
2
e−x /(2σ t) dx.
√
σ 2π t
−u2 /2
du = σ
∞
2t
2
− e−u /2
=σ
0
π
(
2t
.
π
2
2σ 2 t
Var V (t) = E V (t)2 − E V (t) = E Z(t)2 −
π
2σ 2 t
2
= σ 2t 1 −
,
= σ 2t −
π
π
since
2
E Z(t)2 = Var Z(t) + E Z(t) = σ 2 t + 0 = σ 2 t.
To find P V (t) ≤ z | V (0) = z0 , note that, by (12.27),
P V (t) ≤ z | V (0) = z0 = P |Z(t)| ≤ z | V (0) = z0
= P − z ≤ Z(t) ≤ z | V (0) = z0
z
1
2
2
e−(u−z0 ) /(2σ t) du.
=
√
−z σ 2π t
Letting U ∼ N(z0 , σ 2 t) and Z ∼ N(0, 1), this implies that
P V (t) ≤ z | V (0) = z0 = P (−z ≤ U ≤ z)
−z − z
z − z0
0
=P
√ ≤z≤ √
σ t
σ t
z − z
−z − z
0
0
=
−
√
√
σ t
σ t
z + z
z − z
0
0
=
+
− 1.
√
√
σ t
σ t
10. Clearly, D(t) =
)
X(t)2 + Y (t)2 + Z(t)2 . Since X(t), Y (t), and Z(t) are independent and
330
Chapter 12
Stochastic Processes
identically distributed normal random variables with mean 0 and variance σ 2 t, we have
∞ ∞ ∞)
1
1
2
2
2
2
E D(t) =
x 2 + y 2 + z2 · √
e−x /(2σ t) · √
e−y /(2σ t)
σ 2π t
σ 2π t
−∞ −∞ −∞
1
=
√
3
2π σ t 2π t
∞
−∞
∞
−∞
1
2
2
· √
e−z /(2σ t) dx dy dz
σ 2π t
∞
−∞
)
2
2
2
2
x 2 + y 2 + z2 · e−(x +y +z )/(2σ t) dx dy dz.
We now make a change of variables to spherical coordinates: x = ρ sin φ cos θ, y =
ρ sin φ sin θ, z = ρ cos φ, ρ 2 = x 2 + y 2 + z2 , dx dy dz = ρ 2 sin φ dρ dφ dθ, 0 ≤ ρ < ∞,
0 ≤ φ ≤ π, and 0 ≤ θ ≤ 2π. We obtain
2π π ∞
1
2
2
ρe−ρ /(2σ t) · ρ 2 sin φ dρ dφ, dθ
E D(t) =
√
2π σ 3 t 2π t 0
0
0
2π π ∞
1
2
2
ρ 3 e−ρ /(2σ t) dρ sin φ dφ dθ
=
√
2π σ 3 t 2π t 0
0
0
2π π
∞
1
2
2
sin φ dφ dθ
− σ 2 t (ρ 2 + 2σ 2 t)e−ρ /(2σ t)
=
√
0
2π σ 3 t 2π t 0
0
(
2π π
1
2t
4 2
=
.
sin φ dφ dθ = 2σ
· 2σ t
√
π
2π σ 3 t 2π t
0
0
11. Noting that
√
5.29 = 2.3, we have
V (t) = 95e−2t+2.3W (t) ,
where W (t) : t ≥ 0 is a standard Brownian motion. Hence W (t) ∼ N (0, t). The desired
probability is
P V (0.75) < 80 = P 95e−2(0.75)+2.3W (0.75) < 80
= P e2.3W (0.75) < 3.774 = P W (0.75) < 0.577
W (0.75) − 0
0.577
=P
<√
= P (Z < 0.67) = (0.67) = 0.7486.
√
0.75
0.75
Chapter 12
Review Problems
331
REVIEW PROBLEMS FOR CHAPTER 12
1. Label
the time point 10:00 as t = 0. We are given that N (180) = 10 and are interested in
P S10 ≥ 160 | N(180) = 10 . Let X1 , X2 , . . . , X10 be 10 independent random variables uniformly distributed over the interval [0, 180]. Let Y = max(X1 , . . . , X10 ). By Theorem 12.4,
P S10 > 160 | N(180) = 10 = P (Y > 160) = 1 − P (Y ≤ 160)
= 1 − P max(X1 , . . . , X10 ) ≤ 160
= 1 − P (X1 ≤ 160)P (X2 ≤ 160) · · · P (X10 ≤ 160)
160 10
=1−
= 0.692.
180
2. For all positive integer n, we have that
P
2n
1 0
=
0 1
and P
2n+1
0 1
=
.
1 0
Therefore, {Xn : n = 0, 1, . . . } is not regular.
3. By drawing a transition graph, it can be readily seen that, if states 0, 1, 2, 3, and 4 are renamed
0, 4, 2, 1, and 3, respectively, then the transition probability matrix P 1 will change to P 2 .
4. Let Z be the number of transitions until the first visit to 1. Clearly, Z is a geometric random
variable with parameter p = 3/5. Hence its expected value is 1/p = 5/3.
5. By drawing a transition graph, it is readily seen that this Markov chain consists of two recurrent
classes {3, 5} and {4}, and two transient classes {1} and {2}.
6. We have that
Xn+1 =
Xn
1 + Xn
if the (n + 1)st outcome is not 6
if the (n + 1)st outcome is 6.
This shows that {Xn : n = 1, 2, . . . } is a Markov chain with state space {0, 1, 2, . . . }. Its
transition probability matrix is given by
⎛
⎞
5/6 1/6 0
0
0 ...
⎜ 0 5/6 1/6 0
0 . . .⎟
⎜
⎟
⎜ 0
0 5/6 1/6 0 . . .⎟
P =⎜
⎟.
⎜ 0
0
0 5/6 1/6 . . .⎟
⎝
⎠
..
.
All states are transient; no two states communicate with each other. Therefore, we have
infinitely many classes; namely, {0}, {1}, {2}, . . . , and each one of them is transient.
332
Chapter 12
Stochastic Processes
7. The desired probability is
p11 p11 + p11 p12 + p12 p22 + p12 p21 + p21 p11 + p21 p12 + p22 p21 + p22 p22
= (0.20)2 + (0.20)(0.30) + (0.30)(0.15) + (0.30)(0.32)
+ (0.32)(0.20) + (0.32)(0.30) + (0.15)(0.32) + (0.15)2 = 0.4715.
8. The following is an example of such a transition probability matrix:
⎛
⎞
0 0 1 0 0
0 0 0
⎜1 0 0 0 0
0 0 0⎟
⎜
⎟
⎜0 0 0 1 0
⎟
0
0
0
⎜
⎟
⎜0 1/2 0 0 1/2 0 0 0⎟
⎟
P =⎜
⎜0 0 0 0 1/3 2/3 0 0⎟ .
⎜
⎟
⎜0 0 0 0 0
0 1 0⎟
⎜
⎟
⎝0 0 0 0 0
0 0 1⎠
0 0 0 0 0
1 0 0
9. For n ≥ 1, let
Xn =
1
if the nth golfball produced is defective
0
if the nth golfball produced is good.
Then {X
n : n = 1, 2, . . . } is a Markov chain with state space {0, 1} and transition probability
15/18 3/18
matrix 11/12 1/12 . Let π0 be the fraction of golfballs produced that are good, and π1 be
the fraction of the balls produced that are defective. Then, by Theorem 12.7, π0 and π1 satisfy
π0
15/18 11/12
π0
=
,
3/18 1/12
π1
π1
which gives us the following system of equations
⎧
⎨π0 = (15/18)π0 + (11/12)π1
⎩π = (3/18)π + (1/12)π .
1
0
1
By choosing any one of these equations along with the relation π0 + π1 = 1, we obtain a
system of two equations in two unknowns. Solving that system yields
π0 =
11
≈ 0.85
13
and π1 =
2
≈ 0.15.
13
Therefore, approximately 15% of the golfballs produced have no logos.
10. Let
⎧
⎪
⎨1
Xn = 2
⎪
⎩
3
if the nth ball is drawn by Carmela
if the nth ball is drawn by Daniela
if the nth ball is drawn by Lucrezia.
Chapter 12
Review Problems
333
The process {Xn : n = 1, 2, . . . } is an irreducible, aperiodic, positive recurrent Markov chain
with transition probability matrix
⎛
⎞
7/31 11/31 13/31
P = ⎝7/31 11/31 13/31⎠ .
7/31 11/31 13/31
Let π1 , π2 , and π3 be the long-run proportion of balls drawn by Carmela, Daniela, and Lucrezia,
respectively. Intuitively, it should be clear that these quantities are 7/31, 11/31, and 13/31,
respectively. However, that can be seen also by solving the following matrix equation along
with π0 + π1 + π3 = 1.
⎛ ⎞ ⎛
⎞⎛ ⎞
π1
7/31 7/31 7/31
π1
⎝π2 ⎠ = ⎝11/31 11/31 11/31⎠ ⎝π2 ⎠ .
13/31 13/31 13/31
π3
π3
11. Let π1 and π2 be the long-run probabilities that Francesco devotes to playing golf and playing
tennis, respectively. Then, by Theorem 12.7, π1 and π2 are obtained from solving the system
of equations
π1
0.30 0.58
π1
=
0.70 0.42
π2
π2
along with π1 + π2 = 1. The matrix equation above gives the following system of equations:
π1 = 0.30π1 + 0.58π2
π2 = 0.70π1 + 0.42π2 .
By choosing any one of these equations along with the relation π1 + π2 = 1, we obtain
a system of two equations in two unknowns. Solving that system yields π1 = 0.453125
and π2 = 0.546875. Therefore, the long-run probability that, on a randomly selected day,
Francesco plays tennis is approximately 0.55.
12. Suppose that a train leaves the station at t = 0. Let X1 be the time until the first passenger
arrives at the station after t = 0. Let X2 be the additional time it will take until a train arrives
at the station, X3 be the time after that until a passenger arrives, and so on. Clearly, X1 ,
X2 , . . . are the times between consecutive change of states. By the memoryless property
of exponential random variables, {X1 , X2 , . . . } is a sequence of independent andidentically
distributed
exponential random variables with mean 1/λ. Hence, by Remark 7.2, N(t) : t ≥
0 is a Poisson process with rate λ. Therefore, N(t) is a Poisson random variable with
parameter λt.
13. Let X(t) be the number of components working at time t. Clearly, X(t) : t ≥ 0 is a
continuous-time Markov chain with state space {0, 1, 2}. Let π0 , π1 , and π2 be the long-run
proportion
oftime the process is in states 0, 1, and 2, respectively. The balance equations for
X(t) : t ≥ 0 are as follows:
334
Chapter 12
Stochastic Processes
State
Input rate to
=
Output rate from
0
λπ1
=
µπ0
1
2λπ2 + µπ0
=
µπ1 + λπ1
2
µπ1
=
2λπ2
µ
µ2
π0 and π2 = 2 π0 . Using π0 + π1 + π2 = 1 yields
λ
2λ
2λ2
π0 = 2
.
2λ + 2λµ + µ2
From these equations, we obtain π1 =
Hence the desired probability is
1 − π0 =
µ(2λ + µ)
.
+ 2λµ + µ2
2λ2
14. Suppose that every time an out-of-order machine is repaired and is ready to operate a birth
occurs.
Suppose
that a death occurs every time that a machine breaks down. The fact that
X(t) : t ≥ 0 is a birth and death process should be clear. The birth and death rates are
⎧
⎪
kλ
n = 0, 1, . . . , m + s − k
⎪
⎨
λn = (m + s − n)λ n = m + s − k + 1, m + s − k + 2, . . . , m + s
⎪
⎪
⎩
0
n ≥ m + s;
⎧
⎪
nµ n = 0, 1, . . . , m
⎪
⎨
µn = mµ n = m + 1, m + 2, . . . , m + s
⎪
⎪
⎩
0
n > m + s.
15. Let X(t) be the number of machines operating at time t. For 0 ≤ i ≤ m, let πi be the long-run
proportion of time that there are exactly i machines operating. Suppose that a birth occurs
each time that an out-of-order machine is repaired
and begins
to operate, and a death occurs
each time that a machine breaks down. Then X(t) : t ≥ 0 is a birth and death process with
state space {0, 1, . . . , m}, and birth and death rates, respectively, given by λi = (m − i)λ and
µi = iµ for i = 0, 1, . . . , m. To find π0 , first we will calculate the following sum:
m
m
(mλ) (m − 1)λ (m − 2)λ · · · (m − i + 1)λ
λ0 λ1 · · · λi−1
=
µ1 µ2 · · · µi
µ(2µ)(3µ) · · · (iµ)
i=1
i=1
m
=
i=1
λi
=
i! µi
m Pi
m
= −1 +
i=0
m
i=1
m λ
µ
i
i
m λ i m−i
λ
1
= −1 + 1 +
i
µ
µ
m
,
Chapter 12
Review Problems
335
where m Pi is the number of i-element permutations of a set containing m objects. Hence, by
(12.22),
m
λ −m λ + µ −m µ
π0 = 1 +
=
=
.
µ
µ
λ+µ
By (12.21),
i
λ0 λ1 · · · λi−1
m Pi λ
π0 =
π0
µ1 µ2 · · · µi
i! µi
m
i µ
m λ i µ
m λ i µ
=
=
µ
λ+µ
µ
λ+µ
λ+µ
i
i
i
m−i
m
λ
λ
=
1−
, 0 ≤ i ≤ m.
λ+µ
i
λ+µ
πi =
m−i
Therefore, in steady-state, the number of machines that are operating is binomial with parameters m and λ/(λ + µ).
16. Let X(t) be the number
of cars at the
center, either being inspected or waiting to be inspected,
at time t. Clearly, X(t) : t ≥ 0 is a birth and death process with rates λn = λ/(n + 1),
n ≥ 0, and µn = µ, n ≥ 1. Since
∞
n=1
λ0 λ1 · · · λn−1
=
µ1 µ2 · · · µn
∞
λ·
n=1
λ λ
λ
· ···
∞
1 λ
2 3
n
=
−1
+
µn
n! µ
n=0
n
= eλ/µ − 1.
By (12.18), π0 = e−λ/µ . Hence, by (12.17),
πn =
λ·
λ
λ λ
· ···
2 3
n −λ/µ (λ/µ)n e−λ/µ
,
e
=
µn
n!
n ≥ 0.
Therefore, the long-run probability that there are n cars at the center for inspection is Poisson
with rate λ/µ.
17. Let X(t) be the population size at time t. Then X(t) : t ≥ 0 is a birth and death process with
birth rates λn = nλ, n ≥ 1, and death rates µn = nµ, n ≥ 0. For i ≥ 0, let Hi be the time,
starting from i, until the population size reaches i + 1 for the first time. We are interested in
4
i=1 E(Hi ). Note that, by Lemma 12.2,
E(Hi ) =
1
µi
+ E(Hi−1 ),
λi
λi
Since E(H0 ) = 1/λ,
E(H1 ) =
1 µ 1
1
µ
+ · = + 2,
λ
λ λ
λ λ
i ≥ 1.
336
Chapter 12
Stochastic Processes
1
1
µ2
2µ 1
µ
µ
+
·
+ 2 =
+ 2 + 3,
2λ
2λ
λ λ
2λ λ
λ
1
1
µ2
µ2 µ3
3µ 1
µ
µ
E(H3 ) =
+
+ 2+ 3 =
+ 2 + 3 + 4,
3λ
3λ 2λ λ
λ
3λ 2λ
λ
λ
1
4µ 1
µ
µ
µ2 µ3
1
µ2
µ3 µ4
E(H4 ) =
+
+ 2+ 3 + 4 =
+ 2 + 3 + 4 + 5.
4λ
4λ 3λ 2λ
λ
λ
4λ 3λ
2λ
λ
λ
E(H2 ) =
Therefore, the answer is
4
E(Hi ) =
i=1
25λ4 + 34λ3 µ + 30λ2 µ2 + 24λµ3 + 12µ4
.
12λ5
18. Let X(t) be the population size at time t. Then X(t) : t ≥ 0 is a birth and death process
with rates λn = γ , n ≥ 0, and µn = nµ, n ≥ 1. To find πi ’s, we will first calculate the sum
in the relation (12.18):
∞
n=1
λ0 λ1 · · · λn−1
=
µ1 µ2 · · · µn
∞
n=1
∞
γn
1 γ
=
−1
+
n! µn
n! µ
n=0
n
= −1 + eγ /µ .
Thus, by (12.18), π0 = e−γ /µ and, by (12.17), for i ≥ 1,
πi =
γ n −γ /µ (γ /µ)n e−γ /µ
e
=
.
n! µn
n!
Hence the steady-state probability mass function of the population size is Poisson with parameter γ /µ.
19. By applying Theorem 12.9 to Y (t) : t ≥ 0 with t1 = 0, t2 = t, y1 = 0, y2 = y, and t = s,
we have
s
y−0
(s − 0) = y,
E Y (s) | Y (t) = y = 0 +
t −0
t
and
(t − s)(s − 0)
s
= σ 2 (t − s) .
Var Y (s) | Y (t) = y = σ 2 ·
t −0
t
20. First, suppose that s < t. By Example 10.23,
E X(s)X(t) | X(s) = X(s)E X(t) | X(s) .
Now, by Exercise 7, Section 12.5,
E X(t) | X(s) = X(s).
Chapter 12
Hence
Review Problems
337
E X(s)X(t) = E E X(s)X(t) | X(s)
= E X(s)E X(t) | X(s)
= E X(s)X(s) = E X(s)2
2
= Var X(s) + E X(s)
= σ 2 s + 0 = σ 2 s.
For t < s, by symmetry,
Therefore,
E X(s)X(t) = σ 2 t.
E X(s)X(t) = σ 2 min(s, t).
21. By Theorem 12.10,
2
P (U < x and T > y) = P no zeros in (x, y) = 1 − arccos
π
22. Let the current price of the stock, per share, be v0 . Noting that
√
(
x
.
y
27.04 = 5.2, we have
V (t) = v0 e3t+5.2W (t) ,
where W (t) : t ≥ 0 is a standard Brownian motion. Hence W (t) ∼ N (0, t). The desired
probability is calculated as follows:
P V (2) ≥ 2v0 = P v0 e6+5.2W (2) ≥ 2v0
= P 6 + 5.2W (2) ≥ ln 2 = P W (2) ≥ −1.02
W (2) − 0
≥ −0.72
=P
√
2
= P (Z ≥ −0.72) = 1 − P (Z < −0.72)
=1−
(−0.72) = 0.7642.
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