How do I show that $Z_n$ is a martingale?

Question

Let me assume $(X_n)$ is an i.i.d. sequence of random variables and define $S_n=X_1+...+X_n$. Assume further that $X_1\sim N(0;1)$. I want to show that $Z_n=\left(e^{\frac{\lambda S_n-\lambda^2 n}{2}}\right)_{n\geq 1}$ is a martingale with respect to $F_n=\sigma(X_1,...,X_n)$

My Idea was the following, $$\Bbb{E}(Z_{n+1}|F_n)=\Bbb{E}\left(e^{\frac{\lambda S_{n+1}-\lambda^2 (n+1)}{2}}\big|F_n\right)=\Bbb{E}\left(e^{\frac{\lambda S_n-\lambda^2 n}{2}}e^{\frac{\lambda X_{n+1}-\lambda^2 }{2}}\big| F_n\right)=Z_n\Bbb{E}\left(e^{\frac{\lambda X_{n+1}-\lambda^2 }{2}}\big| F_n\right)$$ now using at all $X_i$ are independent we have that $$\Bbb{E}\left(e^{\frac{\lambda X_{n+1}-\lambda^2 }{2}}\big| F_n\right)=\Bbb{E}\left(e^{\frac{\lambda X_{n+1}-\lambda^2 }{2}}\right)=\Bbb{E}\left(e^{\frac{\lambda X_{1}-\lambda^2 }{2}}\right)$$ Then I wanted to compute this by $$\Bbb{E}\left(e^{\frac{\lambda X_{1}-\lambda^2 }{2}}\right)=\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}} e^{\frac{\lambda x-\lambda^2 }{2}} e^{-\frac{x^2}{2}}dx$$

But this somehow does not work and I don't see where the error is.

Could maybe someone help me?

Answer 1

There is an error in the question. You have to replace $\frac {\lambda S_n-n\lambda^{2}} 2$ in the exponent defining $Z_n$ by $\lambda S_n-\frac {n\lambda^{2}} 2$. As it stands, $(Z_n)$ is not a martingale.

$\Bbb{E}\left(e^{\lambda X_{1}-\lambda^2/2 }\right)=e^{-\lambda^{2}/2} \Bbb{E} \left(e^{\lambda X_{1}}\right)$ and $\Bbb{E}\left(e^{\lambda X_{1}}\right)=e^{\lambda^{2}/2}$.

For the last equaion refer to MGF of normal distribution in Wikipedia.