Consider iid $X_1,X_2,...N(\mu,\sigma^2)$ and let $M_n=e^{tS_n-\lambda n}$, $S_n=\sum_{i=1}^{n} X_i$. For a given $\mu,\sigma^2,t$, find the value of $\lambda$ for which $M_n$ is a martingale, submartingale, and supermartingale.
Martingale:
$E(M_{n+1}|F_n)=E(e^{tS_{n+1}-\lambda(n+1)}|F_n)=e^{-\lambda(n+1)}*e^{tS_n}*E(e^{tX_{n+1}})=e^{-\lambda(n+1)}*e^{tS_n}*e^{\frac{\sigma^2t^2}{2}+t\mu}=M_ne^{-\lambda}e^{\frac{\sigma^2t^2}{2}+t\mu}$. Setting this equal to $M_n$ we get $M_ne^{-\lambda}e^{\frac{\sigma^2t^2}{2}+t\mu}=M_n\implies e^{\frac{\sigma^2t^2}{2}+t\mu-\lambda}=1\implies \dfrac{\sigma^2t^2}{2}+t\mu-\lambda=0\implies \lambda=\dfrac{\sigma^2t^2}{2}+t\mu$. For the submartingale case I get $\lambda\leq \dfrac{\sigma^2t^2}{2}+t\mu $. And for the supermartingale case I get $\lambda\geq \dfrac{\sigma^2t^2}{2}+t\mu $. The thing I am not sure about is if I have the martingale case correct. Thanks
Is $F_n = \sigma(X_1, X_2, ..., X_n)$? I think you should be more explicit:
Let $X_1, X_2, ...$ be a iid random variables in a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$, where $\mathscr{F}_n = \mathscr{F}_n^X = \sigma(X_1, X_2, ..., X_n)$ and $X_i$ ~ $N(\mu, \sigma^2)$.
We want to find $\lambda$ that makes $M = (M_n)_{n \in \mathbb{N}}$ a $(\{\mathscr{F}_n\}_{n \in \mathbb{N}}, \mathbb{P})$-martingale where $M_n$ is given as above.
You said:
This particular equality relies on:
taking out $e^{tS_n}$ because $S_n$ is $\mathscr{F}_n$-measurable because $\mathscr{F}_n = \mathscr{F}_n^X$
removing $|\mathscr{F}_n$ because $X_{n+1}$ is independent of $\mathscr{F}_n$ because $\mathscr{F}_n = \mathscr{F}_n^X$.
Btw, I think you might be interested in the 3 martingales of Brownian motion (see Theorem 1.4.1 and Theorem 1.4.2).