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# Problem Set 10 Random Graph Estimation

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EECS 126: Probability and Random Processes
Problem Set 10

1. Random Graph Estimation
Consider a random graph on n vertices in which each edge appears independently with
probability p. Let D be the average degree of a vertex in the graph. Compute the maximum
likelihood estimator of p given D. You may approximate Binomial(n, p) = Poisson(np).
2. Introduction to Information Theory
Recall that the entropy of a discrete random variable X is defined as
H(X)
∆= −
X
x
p(x) log p(x) = −E[log p(X)],
where p(·) is the PMF of X. Here, the logarithm is taken with base 2, and entropy is measured
in bits.
(a) Prove that H(X) ≥ 0.
(b) Entropy is often described as the average information content of a random variable. If
H(X) = 0, then no new information is given by observing X. On the other hand, if
H(X) = m, then observing the value of X gives you m bits of information on average.
Let X be a Bernoulli random variable with P(X = 1) = p. Would you expect H(X) to
be greater when p = 1/2 or when p = 1/3? Calculate H(X) in both of these cases and
(c) We now consider a binary erasure channel (BEC).
Figure 1: The channel model for the BEC showing a mapping from channel input X to channel
output Y . The probability of erasure is pe.
The input X is a Bernoulli random variable with P(X = 0) = P(X = 1) = 1/2. Each
time that we use the channel the input X will either get get erased with probability pe,
or it will get transmitted correctly with probability 1 − pe. Using the character “?” to
denote erasures, the output Y of the channel can be written as
Y =
(
X, with probability 1 − pe
?, with probability pe.
Compute H(Y ).
1
(d) We defined the entropy of a single random variable as a measure of the uncertainty
inherent in the distribution of the random variable. We now extend this definition for a
pair of random variables (X, Y ), but there is nothing really new in this definition because
the pair (X, Y ) can be considered to be a single vector-valued random variable. Define
the joint entropy of a pair of discrete random variables (X, Y ) to be
H(X, Y )
∆= −E[log p(X, Y )],
where p(·, ·) is the joint PMF and the expectation is also taken over the joint distribution
of X and Y .
Compute H(X, Y ), for the BEC.
3. Info Theory Bounds
In this problem we explore some intuitive results which can be formalized using information
theory.
(a) (optional) Prove Jensen’s inequality: if f is a convex function and Z is random variable,
then f(E[Z]) ≤ E[f(Z)]. Hint: You can use fact that every convex function can be
represented by the pointwise supremum of affine functions that are bounded above by f,
i.e.
f(x) = sup{l(x) = ax + b : l(x) ≤ f(x) ∀x}.
(b) It turns out that there is actually a limit to how much “randomness” there is in a
random variable X which takes on |X | distinct values. Show that for any distribution
pX, H(X) ≤ log |X |. Use this to conclude that if a random variable X takes values in
[n] := {1, 2, . . . , n}, then the distribution which maximizes H(X) is X ∼ Uniform([n]).
(c) For two random variable X, Y we define the mutual information (this should have also
been covered in discussion) to be
I(X; Y ) = X
x
X
y
p(x, y) log p(x, y)
p(x)p(y)
,
where the sums are taken over all outcomes of X and Y . Show that I(X; Y ) ≥ 0. In
discussion, you have seen that I(X; Y ) = H(X) − H(X|Y ). Therefore the fact that
mutual information is nonnegative means intuitively that conditioning will only ever
reduce our uncertainty.
4. Relative Entropy and Stationary Distributions
We define the relative entropy, also known as Kullback-Leibler divergence, between two
distributions p and q as
D(p||q) = EX∼p

log 
p(X)
q(X)
 =
X
x
p(x) log p(x)
q(x)
(a) Show that D(p||q) ≥ 0, with equality if and only if p(x) = q(x) for all x. Thus, it is
useful to think about D(·||·) as a sort of distance function. Hint: For strictly concave
functions f, Jensen’s inequality states that f(E[Z]) ≥ E[f(Z)] with equality if and only
if Z is constant.
(b) Show that for any irreducible Markov chain with stationary distribution π, any other
stationary distribution µ must be equal to π. Hint: Consider D(π||µP).
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