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?

  • $\begingroup$ 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. $\endgroup$ Oct 16, 2022 at 10:10
  • $\begingroup$ @geetha290krm oh right I'm so sorry this was my mistake! $\endgroup$ Oct 16, 2022 at 10:11
  • $\begingroup$ Then it should work! $\endgroup$ Oct 16, 2022 at 10:11
  • $\begingroup$ 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. $\endgroup$
    – Bruno B
    Oct 16, 2022 at 10:20

1 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.

  • $\begingroup$ Sorry can I ask you something, is it true that $\Bbb{E}(|Z_n|)=\Bbb{E}(Z_n)$? $\endgroup$ Oct 16, 2022 at 11:14
  • $\begingroup$ Because I asked myself if one could compute $\lim_{n\rightarrow \infty} Z_n$ $\endgroup$ Oct 16, 2022 at 11:33
  • $\begingroup$ $|Z_n|=Z_n$ 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 $\endgroup$ Oct 16, 2022 at 11:39
  • $\begingroup$ 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 $\endgroup$ Oct 16, 2022 at 11:41
  • $\begingroup$ Here $EZ_n=1$ not $0$. @Haveaniceday $\endgroup$ Oct 16, 2022 at 11:48

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