supremum of normal distributed random variables 
Let $X_i$ iid normal distributed random variables on $(\Omega,F,\mathbb P)$ with mean $\mu <0$ and variance $\sigma^2 >0$. Let $S_o:=0,\quad S_n:=X_1+\dots X_n \, (n=1,2,\dots), F_n:= \sigma(S_0,\dots S_n)\, n=(0,1,\dots)$ and define $S_\infty^\star :=\sup_{n \ge 0}S_n$.
  I want to show that there exists  $\lambda_0 >0$, such that $(e^{\lambda_0S_n})$ is a $F_n$ martingale. (1)

$$E(e^{\lambda_0S_{n+1}}\mid F_n)=E(e^{\lambda_0 S_{n}}\mid F_n)E(e^{\lambda_0X_{n+1}}\mid F_n)= e^{\lambda_0 S_{n}}\cdot e^{\lambda_0E(X_{n+1})} $$ I used independence and $F_n$ measerubility 
$\Rightarrow$ since $\mu < 0 $ it follows that $ \lambda_0=0$
What am I doing wrong? 

Furthermore I want to show that $S_\infty^\star$ is a.s. finite random variable. (2)

I thought about using the law of large numbers, but could not really conclude anything,i.e. $$\lim_{n\to \infty}S_n \to \mu n$$$\mu n$ is decreasing that means $P(\sup_{n \ge 0}S_n > -\infty)=1,$ but$\dots$

Can someone show me what I am doing wrong in (1) and how to show (2) correctly? 

 A: Let $M_n=e^{\lambda_0S_n}$. Clearly $\{M_n\}$ is adapted to $\mathcal F_n$ as the map $t\to e^{\lambda_0 t}$ is measurable. For each $n$ we have
\begin{align}
\mathbb E[|M_n|] &= \mathbb E\left[e^{\lambda_0 S_n}\right]\\
&=\mathbb E\left[e^{\lambda_0\sum_{i=1}^n X_i} \right]\\
&=\mathbb E\left[\prod_{i=1}^n e^{\lambda_0X_i} \right]\\
&= \prod_{i=1}^n \mathbb E\left[e^{\lambda_0 X_i}\right]\\
&= e^{n\left(\mu\lambda_0 + \frac12\sigma^2\lambda_0^2\right)}\\
&<\infty.
\end{align}
Moreover,
\begin{align}
\mathbb E[M_{n+1}\mid\mathcal F_n] &= \mathbb E\left[e^{\lambda_0(S_n+X_{n+1})}\mid\mathcal F_n \right]\\
&= E\left[e^{\lambda_0S_n}e^{\lambda_0X_{n+1}}\mid\mathcal F_n \right]\\
&= e^{\lambda_0S_n}\mathbb E\left[e^{\lambda_0X_{n+1}}\right]\\
&= M_ne^{\mu\lambda_0 + \frac12\sigma^2\lambda_0^2}.
\end{align}
It follows that $\{M_n\}$ is a martingale with respect to $\mathcal \{F_n\}$ if and only if $e^{\mu\lambda_0 + \frac12\sigma^2\lambda_0^2}=1$, which is true when $\lambda_0=0$ and when $\lambda_0 = \frac{-2\mu}{\sigma^2}$. Since $\mu<0$,  $\frac{-2\mu}{\sigma^2}>0$.
(Part 2 to come later.)
