Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 1 / 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 yieldsThe 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 iswhereas the right side is $H(X-Z)-H(Z)$. Hencewhere the final step is subadditivity of information entropy. Substituting this inequality into the definition of $d_R$ gives the Entropic Ruzsa triangle inequality