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Write $b=\mathrm{esssup}{X}_1$. The hypothesis $\mathbb{P}(X_1>a)>0$ implies $a<b$ and rules out a constant random variable. In particular $I(a)<\infty$. The cumulant-generating function is strictly convex, since its second derivative is the positive variance under exponential tilting, and ${\psi }^{\prime }(0)=\mu$, ${\psi }^{\prime }(\theta )\to b$ as $\theta \to \infty$.

If $a>\mu$, there is a unique ${\theta }_a>0$ with ${\psi }^{\prime }({\theta }_a)=a$, and the Legendre transform gives $I(a)={\theta }_{a}a-\psi ({\theta }_a)$. ThusIf $a=\mu$, use differentiability at zero: $\psi (\theta )=\mu \theta +o(\theta )$, so for some sufficiently small $\theta >0$,In either case choose a parameter $\theta \ge 0$ with $J=\theta (a+\epsilon )-\psi (\theta )>I(a)$, and choose $0<\eta <J-I(a)$. The Cramér theorem gives, for all sufficiently large $n$,The Chernoff bound at the chosen parameter gives $\mathbb{P}(S_n/n\ge a+\epsilon )\le e^{-nJ}$. The conditioning event has positive probability for every $n$, since all summands can simultaneously exceed $a$. ThereforeTaking the complement within the conditioning event proves the conditional concentration at a large-deviation threshold:The farther tail has a strictly larger exponential cost. The argument also works when $a+\epsilon$ reaches or exceeds the upper support endpoint, without requiring an interior maximizer there.