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CSC411 Fall 2018
Homework 2
Homework 2
Deadline: Wednesday, Oct. 3, at 11:59pm.
Submission: You need to submit one file through MarkUs1 :
• Your answers to Questions 1, 2, and 3 as a PDF file titled hw2_writeup.pdf. You can produce
the file however you like (e.g. LATEX, Microsoft Word, scanner), as long as it is readable.
Neatness Point: One of the 10 points will be given for neatness. You will receive this point as
long as we don’t have a hard time reading your solutions or understanding the structure of your code.
Late Submission: 10% of the marks will be deducted for each day late, up to a maximum of 3
days. After that, no submissions will be accepted.
Weekly homeworks are individual work. See the Course Information handout2 for detailed policies.
1. [4pts] Information Theory. The goal of this question is to help you become more familiar
with the basic equalities and inequalities of information theory. They appear in many contexts
in machine learning and elsewhere, so having some experience with them is quite helpful. We
review some concepts from information theory, and ask you a few questions.
Recall the definition of the entropy
a discrete random variable X with probability mass
of
P
1
function p: H(X) = x p(x) log2 p(x) . Here the summation is over all possible values of
x ∈ X , which (for simplicity) we assume is finite. For example, X might be {1, 2, . . . , N }.
(a) [1pt] Prove that the entropy H(X) is non-negative.
An important concept in information theory is the relative entropy or the KL-divergence of
two distributions p and q. It is defined as
X
p(x)
KL(p||q) =
p(x) log2
.
q(x)
x
The KL-divergence is one of the most commonly used measure of difference (or divergence)
between two distributions, and it regularly appears in information theory, machine learning,
and statistics. For this question, you may assume p(x) > 0 and q(x) > 0 for all x.
If two distributions are close to each other, their KL divergence is small. If they are exactly
the same, their KL divergence is zero. KL divergence is not a true distance metric (since it
isn’t symmetric and doesn’t satisfy the triangle inequality), but we often use it as a measure
of dissimilarity between two probability distributions.
(b) [2pt] Prove that KL(p||q) is non-negative. Hint: you may want to use Jensen’s Inequality, which is described in the Appendix.
(c) [1pt] The Information Gain or Mutual Information between X and Y is I(Y ; X) =
H(Y ) − H(Y |X). Show that
I(Y ; X) = KL(p(x, y)||p(x)p(y)),
where p(x) =
1
2
P
y
p(x, y) is the marginal distribution of X.
https://markus.teach.cs.toronto.edu/csc411-2018-09
http://www.cs.toronto.edu/~rgrosse/courses/csc411_f18/syllabus.pdf
1
CSC411 Fall 2018
Homework 2
2. [2pts] Benefit of Averaging. Consider m estimators h1 , . . . , hm , each of which accepts an
input x and produces an output y, i.e., yi = hi (x). These estimators might be generated
through a Bagging procedure, but that is not necessary to the result that we want to prove.
Consider the squared error loss function L(y, t) = 12 (y − t)2 . Show that the loss of the average
estimator
m
1 X
h̄(x) =
hi (x),
m
i=1
is smaller than the average loss of the estimators. That is, for any x and t, we have
m
1 X
L(hi (x), t).
m
L(h̄(x), t) ≤
i=1
Hint: you may want to use Jensen’s Inequality, which is described in the Appendix.
3. [3pts] AdaBoost. The goal of this question is to show that the AdaBoost algorithm changes
the weights in order to force the weak learner to focus on difficult data points. Here we consider
the case that the target labels are from the set {−1, +1} and the weak learner also returns a
classifier whose outputs belongs to {−1, +1} (instead of {0, 1}). Consider the t-th iteration
of AdaBoost, where the weak learner is
ht ← argmin
h∈H
N
X
wi I{h(x(i) ) 6= t(i) },
i=1
the w-weighted classification error is
PN
errt =
(i)
i=1 wi I{ht (x )
PN
i=1 wi
6= t(i) }
,
t
and the classifier coefficient is αt = 12 log 1−err
errt . (Here, log denotes the natural logarithm.)
AdaBoost changes the weights of each sample depending on whether the weak learner ht
classifies it correctly or incorrectly. The updated weights for sample i is denoted by wi0 and is
wi0 ← wi exp −αt t(i) ht (x(i) ) .
0 ) is exactly 1 . That is, show that
Show that the error w.r.t. (w10 , . . . , wN
2
PN
w0 I{ht (x(i) ) 6= t(i) }
1
err0t = i=1 i PN
= .
0
2
i=1 wi
Note that here we use the weak learner of iteration t and evaluate it according to the new
weights, which will be used to learn the t + 1-st weak learner. What is the interpretation of
this result?
Tips:
• Start from err0t and divide the summation to two sets of E = {i : ht (x(i) ) 6= t(i) } and its
complement E c = {i : ht (x(i) ) = t(i) }.
• Note that
P
i∈E
PN
wi
i=1 wi
2
= errt .
CSC411 Fall 2018
Homework 2
Appendix: Convexity and Jensen’s Inequality. Here, we give some background on convexity which you may find useful for some of the questions in this assignment. You may assume
anything given here.
Convexity is an important concept in mathematics with many uses in machine learning. We
briefly define convex set and function and some of their properties here. Using these properties are
useful in solving some of the questions in the rest of this homework. If you are interested to know
more about convexity, refer to Boyd and Vandenberghe, Convex Optimization, 2004.
A set C is convex if the line segment between any two points in C lies within C, i.e., if for any
x1 , x2 ∈ C and for any 0 ≤ λ ≤ 1, we have
λx1 + (1 − λ)x2 ∈ C.
For example, a cube or sphere in Rd are convex sets, but a cross (a shape like X) is not.
A function f : Rd → R is convex if its domain is a convex set and if for all x1 , x2 in its domain,
and for any 0 ≤ λ ≤ 1, we have
f (λx1 + (1 − λ)x2 ) ≤ λf (x1 ) + (1 − λ)f (x2 ).
This inequality means that the line segment between (x1 , f (x1 )) and (x2 , f (x2 )) lies above the
graph of f . A convex function looks like `. We say that f is concave if −f is convex. A concave
function looks like a.
Some examples of convex and concave functions are (you do not need to use most of them in
your homework, but knowing them is useful):
• Powers: xp is convex on the set of positive real numbers when p ≥ 1 or p ≤ 0. It is concave
for 0 ≤ p ≤ 1.
• Exponential: eax is convex on R, for any a ∈ R.
• Logarithm: log(x) is concave on the set of positive real numbers.
• Norms: Every norm on Rd is convex.
• Max function: f (x) = max{x1 , x2 , . . . , xd } is convex on Rd .
• Log-sum-exp: The function f (x) = log(ex1 + . . . + exd ) is convex on Rd .
An important property of convex and concave functions, which you may need to use in your
homework, is Jensen’s inequality. Jensen’s inequality states that if φ(x) is a convex function of x,
we have
φ(E [X]) ≤ E [φ(X)] .
In words, if we apply a convex function to the expectation of a random variable, it is less than or
equal to the expected value of that convex function when its argument is the random variable. If
the function is concave, the direction of the inequality is reversed.
Jensen’s inequality has a physical interpretation: Consider a set X = {x1 , . . . , xN } of points
on R. Corresponding to each point, we have a probability p(xi ). If we interpret the probability as
mass, and we put an object with mass p(xi ) at location (xi , φ(xi )), then the centre of gravity of
these objects, which is in R2 , is located at the point (E [X] , E [φ(X)]). If φ is convex `, the centre
of gravity lies above the curve x 7→ φ(x), and vice versa for a concave function a.
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