Regard the alphabet as the cyclic group $\mathbb Z/m\mathbb Z$ and put $E=X-Y$. For each $y$, subtraction by $y$ is a bijection, so
$$
H(X\mid Y=y)=H(E\mid Y=y)\leq\phi(d_y),
\qquad
d_y=\mathbb P(X\ne Y\mid Y=y).
$$
By concavity of $\phi$, its monotonicity in the distortion allowance, and $\mathbb E d_Y=\mathbb P(X\ne Y)\leq d$,
$$
H(X\mid Y)
\leq\mathbb E\phi(d_Y)
\leq\phi(\mathbb E d_Y)
\leq\phi(d).
$$
Therefore
$$
I(X;Y)=H(X)-H(X\mid Y)\geq H(X)-\phi(d).
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 224 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-224/3/b/solution