For $\mu>0$, let $Q$ be the <geometric distribution>
$$
Q(k)=\frac1{1+\mu}\left(\frac\mu{1+\mu}\right)^k,
\qquad k\geq0.
$$
If $P$ is the law of $X$, <Gibbs inequality> gives
$$
0\leq D(P\Vert Q)
=-H(X)+\log(1+\mu)+\mu\log\frac{1+\mu}{\mu}.
$$
Therefore
$$
H(X)\leq\log(1+\mu)+\mu\log\frac{1+\mu}{\mu}
=(1+\mu)h\left(\frac1{1+\mu}\right).
$$
For $\mu=0$, the nonnegative random variable $X$ is zero almost surely and both sides vanish. This proves that the <maximum entropy distribution on the nonnegative integers> with fixed <expected value> is geometric.

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Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 4 / a / Solution

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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-224/4/a/solution