Durrett, probability theory and examples exercise 2.7,5

Question

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?

Answer 1

Here is an answer to the second part of your question; I will try to find time tomorrow to address the first part.

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 $$ \frac 1n\log P(X_1\geq n(a+\epsilon))\to 0. $$

Answer 2

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 $Ee^{\theta X_i}=\infty$.