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

Suppose $X\to Y\to Z$ is a Markov chain, so $I(X;Z\mid Y)=0$. The chain rule for mutual information givesand alsobecause conditional mutual information is nonnegative. Therefore $I(X;Z)\le I(X;Y)$. Similarly,while $I(X,Y;Z)=I(X;Z)+I(Y;Z\mid X)\ge I(X;Z)$, so $I(X;Z)\le I(Y;Z)$. These are the two data processing inequalities. In particular, applying any deterministic function or Markov kernel to either argument cannot increase mutual information.