Suppose $X\to Y\to Z$ is a <Markov chain>, so $I(X;Z\mid Y)=0$. The <chain rule for mutual information> gives
$$
I(X;Y,Z)=I(X;Y)+I(X;Z\mid Y)=I(X;Y)
$$
and also
$$
I(X;Y,Z)=I(X;Z)+I(X;Y\mid Z)\geq I(X;Z),
$$
because <conditional mutual information> is nonnegative. Therefore $I(X;Z)\leq I(X;Y)$. Similarly,
$$
I(X,Y;Z)=I(Y;Z)+I(X;Z\mid Y)=I(Y;Z)
$$
while $I(X,Y;Z)=I(X;Z)+I(Y;Z\mid X)\geq I(X;Z)$, so $I(X;Z)\leq I(Y;Z)$. These are the two <data processing inequality for mutual information>[data processing inequalities]. In particular, applying any deterministic function or <Markov kernel> to either argument cannot increase <mutual information>.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-224/1/a/solution