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?
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.)