Since $\mathbb Ef(X)=0$ and the moment-generating function is finite near zero, $\log F(u)=O(u^2)$. Consequently
$$
2^m\log F(\lambda/2^m)=O(2^{-m})\to0,
$$
which proves $F(\lambda/2^m)^{2^m}\to1$.

For $\lambda_*=C_P(X)^{-1/2}$, part (a) and the elementary bound $-\log(1-x)\leq4x/3$ for $0\leq x\leq1/4$ give
$$
\log F(\lambda_*)
\leq\sum_{k\geq0}2^k[-\log(1-4^{-(k+1)})]
\leq\frac23<\log3.
$$
Thus $F(\lambda_*)\leq3$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 208 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-208/2/b/solution