How to prove $X_n =e^{aS_n-bn} \to 0$ a.s.(where $S_n$ is the sum of standard normal random variables)?

Question

Imagine we have iid. standard normal random variables $z_1, z_2,\cdots$. Let $S_n = \sum_{k=1}^{n}z_k$, and $X_n = e^{aS_n-bn}$ with $a, b\in \mathbb{R}$ . How to prove that if $b>0$, then $X_n \to 0$ almost surely?

I have made some attemptes but can't think of any approaches at the moment, thank you very much in advance for any hints and helps!

Answer 1

You need to prove that your exponent converges a.s. to $-\infty$

to do that, observe that your exponent can be rewritten in

$$aS_n-bn=n\left(a\underbrace{\overline{X}_n}_{\xrightarrow{a.s.}0}-b \right)$$

Applying Strong Law of Large number, the sample mean converges almost surely to the population's mean, zero, and thus, if $b>0$ your exponent converges a.s. to $-\infty$ and thus, applying again the Continuous Mapping theorem, your original sequence

$$X_n\xrightarrow{a.s.}0$$