Finding an expectation involving the increment of a symmetric random walk

Question

I am given:

• $M_n$ is a symmetric random walk with $X_n=M_n-M_{n-1}$
• Since $X_n$ is the increment of a symmetric random walk, we know that $E[X_n]=0$

The question has two parts:

First, I want to get the following expectation: $g(\alpha)=E[e^{\alpha X_n}]$

I have tried using the definition of the expectation:

$g(\alpha)=e^\alpha*0.5 + e^{-\alpha}*0.5$

I was wondering if there is a way to simplify $g$ further by using $E[X_n]=0$?

Second part of the question asks us to show that $Y_n=exp\{\alpha M_n -n \log (g(\alpha))\}$ is a martingale. I think I will have clearer idea how to tackle this one once I get the answer for the first part. But I was thinking of using the definition of martingale:

$E[Y_m-Y_n|F_n]=0$, where $m>n$.

Any help would be appreciated!

Answer 1

You could write $g(\alpha)=\cosh\alpha$. There’s no further simplification using $E[X_n]=0$.