For every $\theta\geq0$, the <exponential Markov bound> and <independent random variables>[independence] give
$$
\mathbb P(S_n/n\geq x)
=\mathbb P(e^{\theta S_n}\geq e^{n\theta x})
\leq e^{-n\theta x}\mathbb E e^{\theta S_n}
=\exp\{-n(\theta x-\psi(\theta))\}.
$$
Taking the <infimum> over $\theta\geq0$ yields
$$
\limsup_{n\to\infty}\frac1n\log\mathbb P(S_n/n\geq x)
\leq-\sup_{\theta\geq0}(\theta x-\psi(\theta)).
$$
For $x\geq m$, <convex function>[convexity] makes this supremum equal to $\psi^*(x)$, proving the upper bound in the stated tail form of <Cramér theorem>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 201 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-201/2/c/solution