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

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?

• In the definition of $Z_n$ you should not divide $\lambda S_n$ by $2$. As it stands, $(Z_n)$ is not a martingale. Check your definition again. Oct 16, 2022 at 10:10
• @geetha290krm oh right I'm so sorry this was my mistake! Oct 16, 2022 at 10:11
• Then it should work! Oct 16, 2022 at 10:11
• At least, since $\mu := \mathbb{E}\left(e^{\frac{\lambda X_1 - \lambda^2}{2}}\right)$ is a non-zero constant, you can always "salvage" your to-be martinagle by looking at $\left(\frac{Z_n}{\mu^n}\right)_n$, which becomes a martingale. Could have been worse I suppose. Oct 16, 2022 at 10:20

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 $$\Bbb{E}\left(e^{\lambda X_{1}}\right)=e^{\lambda^{2}/2}$$.
• Sorry can I ask you something, is it true that $\Bbb{E}(|Z_n|)=\Bbb{E}(Z_n)$? Oct 16, 2022 at 11:14
• Because I asked myself if one could compute $\lim_{n\rightarrow \infty} Z_n$ Oct 16, 2022 at 11:33
• $|Z_n|=Z_n$ because $Z_n \geq 0$. $Z_n$ converges a.s to some r.v. , but I don't think you can fins the limit explicitly. @Haveaniceday Oct 16, 2022 at 11:39
• Hmm but this seems really strange to me. Because somehow I can should be able to show that $Z_n\rightarrow 0$ a.s. but on the same hand I know that $\Bbb{E}(|Z_n|)=0$ so $Z_n\rightarrow 0$ in $L^1$. But in the exercise the convergence in $L^1$ should not hold Oct 16, 2022 at 11:41
• Here $EZ_n=1$ not $0$. @Haveaniceday Oct 16, 2022 at 11:48