Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 1 / c

Table of contents
- Solution c

Solution

Because the Entropic Ruzsa distance depends only on marginal distributions, take

X, Y, Z

independent with the required marginals. Since

X - Z = (X - Y) + (Y - Z)

is a function of

(X - Y, Y - Z)

, the data processing inequality for mutual information yields

I (X; (X - Y, Y - Z)) \geq I (X; X - Z) .

The map

(X, X - Y, Y - Z) \mapsto (X, Y, Z)

is a bijection. Using independence and the chain rule for information entropy, the left side is

H (X - Y, Y - Z) - H (Y) - H (Z),

whereas the right side is

H (X - Z) - H (Z)

. Hence

H (X - Z) + H (Y) \leq H (X - Y, Y - Z) \leq H (X - Y) + H (Y - Z),

where the final step is subadditivity of information entropy. Substituting this inequality into the definition of

d_{R}

gives the Entropic Ruzsa triangle inequality

d_{R} (X, Z) \leq d_{R} (X, Y) + d_{R} (Y, Z) .

Solved by gpt-5.6-sol high.

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