Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 224 / 1 / b

Table of contents
- Solution b

Solution

Write

p = (1 + μ)^{- 1}

. Since

X

and the geometric random variable

Z

have the same expected value

μ

,

D (P_{X} ∥ P_{Z}) = k \geq 0 \sum P_{X} (k) lo g_{2} \frac{P _{X} ( k )}{p ( 1 - p ) ^{k}} = - H (X) - lo g_{2} p - μ lo g_{2} (1 - p) = H (Z) - H (X) .

This is also the entropy deficit relative to the maximum entropy distribution on the nonnegative integers.

The random variables

X

and

- Z

are independent, so part a gives

H (X - Z) \geq H (- Z) = H (Z)

. Consequently

2 d_{R} (X, Z) - D (P_{X} ∥ P_{Z}) = 2 H (X - Z) - H (X) - H (Z) - {H (Z) - H (X)} = 2 {H (X - Z) - H (Z)} \geq 0,

which is the required bound in terms of the Entropic Ruzsa distance.

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