Because $X$ and $Y$ are <independent random variables>, conditioning on $Y$ merely translates $X$. The <conditional entropy under a deterministic change of variables> and the fact that <conditioning reduces entropy> therefore give
$$
H(X+Y)\geq H(X+Y\mid Y)=H(X\mid Y)=H(X).
$$
Interchanging $X$ and $Y$ similarly gives $H(X+Y)\geq H(Y)$, and hence
$$
H(X+Y)\geq\max\{H(X),H(Y)\}.
$$
The proof applies to countable alphabets whenever the displayed <information entropies> are finite.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 224 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-224/1/a/solution