Fix $a>m$ and then $x>a$. By the assumptions on the <derivative> $\psi'$, there is a unique $\theta_x>0$ with $\psi'(\theta_x)=x$. Apply <exponential tilting> to each summand:
$$
\frac{d\mathbb P_{\theta_x}}{d\mathbb P}(y)
=e^{\theta_xy-\psi(\theta_x)}.
$$
Under the product tilted law, the variables remain <independent and identically distributed random variables> and have mean $x$. Hence the implies that, for every $0<\delta<x-a$,
$$
\mathbb P_{\theta_x}(|S_n/n-x|<\delta)\longrightarrow1.
$$
Changing measure on this event gives
$$
\mathbb P(S_n/n\geq a)
\geq e^{-n[\theta_x(x+\delta)-\psi(\theta_x)]}
\mathbb P_{\theta_x}(|S_n/n-x|<\delta).
$$
Therefore
$$
\liminf_{n\to\infty}\frac1n\log\mathbb P(S_n/n\geq a)
\geq-\psi^*(x)-\theta_x\delta.
$$
First let $\delta\downarrow0$ and then $x\downarrow a$. The <continuity of a convex function> gives the lower bound $-\psi^*(a)$. The endpoint $a=m$ follows by letting $a\downarrow m$, while for $a<m$ the makes the probability tend to one. This proves the required lower bound.

Solved by gpt-5.6-sol high.

Solution

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

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