Let $S_n = X_1 + ... + X_n$, where $X_1, X_2, ...$ are iid with $X_1 \sim \operatorname{Exp}(1)$. I'd like to verify that $$ M_n = \frac{n!}{(1+S_n)^{n+1}} \cdot e^{S_n} $$ is a martingale, that is $\mathbb{E}(M_{n+1} \mid \mathcal{F}_n) = M_n$, where $\mathcal{F}_n$ is the natural filtration.

So far I've tried to 'separate' $X_{n+1}$ from $S_{n+1}$: $$ \mathbb{E}(M_{n+1} \mid \mathcal{F}_n) = \mathbb{E} \Big( \frac{(n+1)!}{(1+S_n+X_{n+1})^{n+2}} \cdot e^{S_n} e^{X_{n+1}} \mid \mathcal{F}_n\Big) $$ But I don't see what to do with the denominator there. Is this a correct way to start (and if yes, what should I do next?) or is there some other way to show that $M_n$ is a martingale?

| cite | improve this question | | | | |

Fix $n \in \mathbb{N}$. By the pull-out property, we have

$$\mathbb{E}(M_{n+1} \mid \mathcal{F}_n) = e^{S_n} (n+1)! \mathbb{E} \left( \frac{1}{(1+S_n+X_{n+1})^{n+2}} e^{X_{n+1}} \mid \mathcal{F}_n \right). \tag{1}$$

Consequently, it remains to calculate

$$\mathbb{E} \left( \frac{1}{(1+S_n+X_{n+1})^{n+2}} e^{X_{n+1}} \mid \mathcal{F}_n \right).$$

Since the random variables $(X_k)_{k \in \mathbb{N}}$ are independent, we know that $X_{n+1}$ and $\mathcal{F}_n$ are independent. On the other hand, $S_n$ is $\mathcal{F}_n$-measurable and so

$$\mathbb{E} \left( \frac{1}{(1+S_n+X_{n+1})^{n+2}} e^{X_{n+1}} \mid \mathcal{F}_n \right) = f(S_n) \tag{2}$$


$$f(s) := \mathbb{E} \left( \frac{1}{(1+s+X_{n+1})^{n+2}} e^{X_{n+1}} \right), \qquad s \geq 0.$$

Using that $X_{n+1}$ is exponentially distributed, we find

$$\begin{align*} f(s) &= \int_0^{\infty} \frac{1}{(1+s+x)^{n+2}} e^x \, \mathbb{P}_{X_{n+1}}(dx) \\ &= \int_0^{\infty} \frac{1}{(1+s+x)^{n+2}} \, dx. \end{align*}$$

This integral can be easily calculated. Plugging the result into $(2)$ and using $(1)$, this proves that $(M_n)_{n \in \mathbb{N}}$ is a martingale.

| cite | improve this answer | | | | |
  • $\begingroup$ Clever argument...is there any particular name by which this type of martingale is known? $\endgroup$ – Math1000 Sep 26 '16 at 13:04
  • $\begingroup$ @Math1000 Not as far as I know. $\endgroup$ – saz Sep 26 '16 at 13:47
  • $\begingroup$ Thanks for the answer, it was clear this way. I'll try no to forget this in the future. :)) $\endgroup$ – Faragó Dávid Sep 26 '16 at 14:30
  • $\begingroup$ @FaragóDávid You are welcome. $\endgroup$ – saz Sep 26 '16 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.