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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?

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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

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

where

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

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.

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  • $\begingroup$ 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

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