Write $p_k=\mathbb P(X=k)$. Monotonicity gives
$$
1\geq\sum_{j=1}^kp_j\geq kp_k,
$$
so $p_k\leq1/k$ and therefore $\log_2k\leq\log_2(1/p_k)$ whenever $p_k>0$. Hence
$$
\mathbb E[\log_2X]
=\sum_{k\geq1}p_k\log_2k
\leq\sum_{k\geq1}p_k\log_2\frac1{p_k}
=H(X)<\infty.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 224 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-224/4/b/solution