Let $X_1,X_2,\ldots$ be <independent and identically distributed random variables>, let $S_n=\sum_{j=1}^nX_j$, and write $m=\mathbb E X_1$. <Cramér theorem> states that the empirical means $S_n/n$ obey a <large deviation principle> with good <rate function> $\psi^*$. In particular, for $x\geq m$,
$$
\lim_{n\to\infty}\frac1n\log\mathbb P(S_n/n\geq x)=-\psi^*(x),
$$
with the usual extended-real interpretation; the analogous lower-tail formula holds for $x\leq m$.

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Solution

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

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>.

Solved by gpt-5.6-sol high.

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

The scaled cumulant-generating function is
$$
\frac1t\log G(\lambda)=\mu(e^\lambda-1).
$$
By <Cramér theorem>, its Legendre transform is stationary at $e^{\lambda_*}=x/\mu$ for $x\geq0$. Thus
$$
\boxed{I(x)=x\log\frac x\mu-x+\mu},
\qquad x\geq0,
$$
with $I(x)=+\infty$ for $x<0$ and the continuous convention $0\log0=0$.

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 353 / 4 / a / ii / Solution

{c}
{wiki=Cramér's_theorem_(large_deviations)}

For independent identically distributed real random variables, the empirical mean satisfies a large-deviation principle with rate function given by the <Legendre transform of a cumulant-generating function>.

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 Incoming links: Cramér theorem