Suppose $X_1, X_2,\ldots$ are i.i.d. normal ($\mu=0, \sigma^2=1$) random variables and let $S_n$ denote the sum of first $n$ $X_i$'s. Show that $$\lim_{n\to \infty} \exp(2S_n – 2n)=0$$
I think I am supposed to use the Martingale Convergence Theorem here, where $M_n=\exp(2S_n-2n)$ is a martingale with respect to $\mathcal{\{F_n\}}$ (information in $X_1, \ldots, X_n$). I have shown that this martingale satisfies the properties of MCT so the limit exists, but I don't know how to show that it is $0$.
Best Answer
Let $X_1,...$ be i.i.d normal variables $\mathcal N(0,1)$. Let $S_n = \sum_{k=1}^n X_k$.
Let $M_n = \exp(2S_n - 2n) = \exp(2(S_n - n))$. You showed it's a martingale with respect to filtration $(\mathcal F_n)$. By martingale convergence theorem ($M_n$ is nonnegative) we have the existence of limit $M_\infty = \lim_{n \to \infty} M_n$ almost surely. Now, since $\exp$ is continuous, it means that $\lim_{n \to \infty} S_n - n = Z$ almost surely for some $Z$. Note that $S_n - n = n(\frac{S_n}{n} - 1)$. You should already know that $\frac{S_n}{n} \to 0$ almost surely by SLLN, so $S_n - n \to -\infty$ almost surely, which means that $\exp(2(S_n-n)) \to \exp(-\infty) = 0$ almost surely