# Durrett, probability theory and examples exercise 2.7,5

The problem is as below.

Suppose $$EX_i=0$$ and $$E\exp(\theta X_i)=\infty$$ for all $$\theta>0$$. Let $$S_n=X_1+\dots+X_n$$.

Then $$\frac{1}{n}\log P(S_n \geq na) \rightarrow 0\quad\text{for all}\quad a>0.$$

The solution uses the fact that $$P(S_n\geq na)\;\geq\; P(S_{n-1}\geq -n\epsilon)\;P(X_n\geq n(a+\epsilon)).$$

This is different from the original large deviation proof, where we use $$P(S_n \geq na)\;\geq\; P(S_{n-1} \geq (a-a'+\epsilon)n)\;P(X_{n}\in[(a'-\epsilon)n,(a'+\epsilon)n]).$$

The solution says that $$Ee^{\theta X}=\infty$$ implies $$\limsup_{n\rightarrow\infty}\frac{1}{n}\log P(X_n >na)=0.$$

How do you prove this and how does the result follows from this fact?

The equality $$\tag{*} P(S_n\geq na)\;\geq\; P(S_{n-1}\geq -n\epsilon)\;P(X_n\geq n(a+\epsilon))$$ follows from the inclusion $$\left\{S_n\geq na\right\}\supset \{S_{n-1}\geq -n\epsilon\}\cap \{X_n\geq n(a+\epsilon)\}$$ and independence between $$S_{n-1}$$ and $$X_n$$. Then applying $$\log$$ and multiplying by $$-1/n$$ shows that $$0\leqslant -\frac 1n\log P(S_n\geq na)\leqslant -\frac{1}{n}\log P(S_{n-1}\geq -n\epsilon) -\frac 1n\log P(X_n\geq n(a+\epsilon)).$$ Since the random variables $$X_i$$ are centered and identically distributec, $$S_n/(n-1)\to 0$$ in probability hence $$P(S_{n-1}\geq -n\epsilon)\to 1$$. Moreover, $$X_n$$ has the same distribution as $$X_1$$ hence it suffices to check that $$\frac 1n\log P(X_1\geq n(a+\epsilon))\to 0.$$
For the first part, if the lim sup was $$<0$$, we could pick $$\beta>0$$ such that for $$n$$ large enough $$\frac{1}{n}\log P(X_n>na)\le-\beta$$ Then $$\begin{split} \int e^{\theta x}dF &\le P(X_i<0)+\sum_{n=0}^\infty\int_{na\le X_i\le (n+1)a}e^{\theta x}dF\\ &\le 1+\sum_{n=0}^{N-1}\int_{na\le X_i\le (n+1)a}e^{\theta x}dF+\sum_{n=N}^\infty\int_{na\le X_i\le (n+1)a}e^{\theta x}dF\\ &\le C+\sum_{n=N}^\infty e^{a(n+1)\theta}\cdot e^{-\beta n}\\ &\le C+\sum_{n=N}^\infty e^{-n(\beta-a\theta)+a\theta} \end{split}\tag{*}$$ Here $$C$$ and $$N$$ are constants. \
We can pick $$\theta>0$$ so small that $$\beta-a\theta>0$$ and thus the last term of $$(*)$$ would converge, which is a contradiction to $$Ee^{\theta X_i}=\infty$$.