Independence makes
$$
X\longrightarrow X+Y\longrightarrow X+Y+Z
$$
a <Markov chain>. The <data processing inequality for mutual information> therefore gives $I(X;X+Y+Z)\leq I(X;X+Y)$. Translation by the known value of $X$ is a <bijection>, so <conditional entropy> and independence give
$$
I(X;X+Y+Z)=H(X+Y+Z)-H(Y+Z),
$$
and $I(X;X+Y)=H(X+Y)-H(Y)$. Rearranging proves the stated <entropy submodularity for three independent sums>:
$$
H(X+Y+Z)+H(Y)\leq H(X+Y)+H(Y+Z).
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

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