Bound for sum of normal distributions I have encountered an exercise that was quite puzzling for me. Maybe someone can help me out here?
So let $(X_n)_n $ be $N(-a,1)$ distributed, independent random variables where $a>0$. I need to prove $$P(\sup_{n \in \mathbb N} S_n > x)\leq e^{-2ax}$$
where $S_n= \sum_{k=1}^n X_k$. I am tempted to use Markovs inequality but the supremum kind of irritates me.
Can I somehow use that $e^{h S_n}$ is a Martingale if $h=2a$ which I proved in the first part of the exercise?

Since $e^{2aS_n}$ is a martingale, by Doob's martingale inequality, for any fixed $N$ you have
$$P \left (\sup_{n \leq N} e^{2aS_n}>e^{2ax} \right ) \leq \frac{E[e^{2aS_N}]}{e^{2ax}}.$$
But $e^{2aS_n}$ is a martingale, so its expectation is fixed, equal to $E[e^{2aX_1}]$. This is the expectation of a lognormal r.v. where the underlying normal r.v. has mean $-2a^2$ and variance $4a^2$. Thus its expectation is $e^{-2a^2+(4a^2)/2}=1$, which gives the result up to $N$. Send $N \to \infty$ on both sides to get the desired result.
