Let $Y_1,Y_2,...$ be i.i.d $\mathcal(N)(0,1)$ random variables and $\mathcal{F}_n=\sigma(Y_1,Y_2,..,Y_n)$ the corresponding natural filtration. Let further $X_n=Y_1+Y_2+...+Y_n$. Show that for all $u\in\mathbb{R}$ $$Z_n=\exp\left(uX_n-\frac{1}{2}nu^2\sigma^2\right)$$ is a martingale.
According the suggestion of my lecturer, I am trying to compute the expectation of it:
$$E(Z_n|\mathcal{F_n})=E\left(\exp\left(uX_n-\frac{1}{2}nu^2\sigma^2\right)|\mathcal{F}_n\right)$$
However, I don´t know how to continue to proof that is a martingale.
I think that I canthe next argument, since $X_n=Y_1+...+Y_n\sim N(0,n)$ then $$uX_n-\frac{1}{2}nu^2\sigma^2\sim N\left(-\frac{1}{2}nu^2\sigma^2,u^2n\right).$$ Now I can use the moment generating function for a normal distribution, but How to conclude it?
The $\sigma$-algebra generated by $$Z_n=\exp\left(uX_n-\frac{1}{2}nu^2\sigma^2\right)=\exp\left(u\sum_{ i=1}^n Y_i-\frac{1}{2}nu^2\sigma^2\right)$$
is contained in $\mathcal{F}_n=\sigma(Y_1,Y_2,..,Y_n)$ . In other words: $Z_n$ is completely determined if $Y_1,Y_2,\cdots, Y_n$ are given. As a result $$E\left[\exp\left(uX_n-\frac{1}{2}nu^2\sigma^2\right)\big |\mathcal{F}_n\right]=\exp\left(uX_n-\frac{1}{2}nu^2\sigma^2\right)=Z_n.$$
That is, $Z_n$ is a Martingale. (The expectations are finite.)