Use the distribution $R$ associated with the code in part b. Since $2^{L(x)}=(KR(x))^{-1}$,
$$
\mathbb E_P2^{\rho L(X)}
=K^{-\rho}\sum_xP(x)R(x)^{-\rho}
\geq\sum_xP(x)R(x)^{-\rho}.
$$
For $\alpha=1/(1+\rho)$, <Holder inequality>, equivalently the indicated <Jensen inequality>, gives
$$
\sum_xP(x)R(x)^{-\rho}
\geq\left(\sum_xP(x)^\alpha\right)^{1/\alpha}.
$$
Therefore
$$
\frac1\rho\log_2\mathbb E2^{\rho L(X_1^n)}
\geq\frac{1+\rho}{\rho}
\log_2\sum_xP_n(x)^{1/(1+\rho)}
=H_\alpha(X_1^n).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 224 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-224/2/c/solution