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Let $(X_n)$ be a sequence of independent, identically distributed random variables with finite moment-generating function $M(t) = \mathbb{E}\left[\exp(tX_1\right)] < \infty$ for $t \in \mathbb{R}$. Let $S_n = \sum_{i=1}^n X_i$ and $Y_n = \frac{1}{(M(t))^n}\exp(tS_n)$ for $n \geq 0$ and $t \in \mathbb{R}$. I would like to show that $(Y_n)_{n\geq 0}$ are martingals with respect to the filterings $\left(\sigma\left(X_1,X_2,\ldots,X_n \right)\right)_{n \geq 0}$.

Could you please help me with some ideas or suggestions.

Thanks.

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I guess you mean $M(t) = E[\exp{(tX_i)}]$. Not you forgot the $\exp(x)$. –  user8 Jan 12 '13 at 16:53
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Is there a typo in your question? Shouldn't $\mathbb{E}[ tX_1] < \infty$ be $\mathbb{E}[\exp(tX_1)] < \infty$? –  Harald Hanche-Olsen Jan 12 '13 at 16:54
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@eugen1806 What exactly is stopping you? –  Did Jan 12 '13 at 18:00
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@eugen1806 I've just realized that you are looking for a suggestion / ideas and not a complete solution. Sorry for overlooking it. If you want, I will delete my answer and add some hints instead –  user8 Jan 12 '13 at 18:03
    
thanks, indeed there was a type in the problem description. I have corrected it. –  eugen1806 Jan 14 '13 at 17:09
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2 Answers 2

up vote 0 down vote accepted

Just plugging in:

$$E[Y_{n+1}|\mathcal{F}_n]=\frac{1}{(M(t))^{n+1}}E[\exp{(tS_n)}\exp{(tX_{n+1})}|\mathcal{F}_n]=\frac{1}{(M(t))^{n+1}}\exp{(tS_n)}E[\exp{(tX_{n+1})}]=\frac{1}{(M(t))^{n+1}}M(t)\exp{(tS_n)}=Y_n$$

where I've used, that $X_i$ are iid, and $Y_n$ is $\mathcal{F}_n$ measurable, where $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$

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The term martingale refers to the whole sequence $(Y_n)_{n\geq 1}$, so a "sequence of martingales" is not the correct term to use in this case. Instead $(Y_n)_{n\geq 1}$ is a sequence that forms a martingale.

Let $(\Omega,\mathcal F,(\mathcal F_n)_{n\geq 1},P)$ be a filtered probability space with $\mathcal{F_n}=\sigma(X_1,\ldots,X_n)$. To show that any sequence $(Y_n)_{n\geq 1}$ is a martingale with respect to $(\mathcal{F}_n)_{n\geq 1}$ you must show the following three items:

  1. $(Y_n)_{n\geq 1}$ is adapted to the filtration $(\mathcal{F}_n)_{n\geq 1}$, i.e. for every $n\geq 1$ we have that $Y_n$ is $\mathcal{F}_n$-measurable. Why is this true?

  2. $Y_n$ is integrable for every $n\geq 1$, i.e. $E[|Y_n|]<\infty$. This item is often something that is assumed. You have probably made a typo as Harald Hanche-Olsen points out in the comment.

  3. The defining equation: $$ E[Y_{n+1}\mid\mathcal{F}_n]=Y_n \quad\text{a.s.} $$ for all $n\geq 1$. To show this defining equality you simply fix an $n\geq 1$ and start with the left-hand side. Plug in $Y_{n+1}$ and use everything you know about conditional expectations.

In situations where a sequence $Y_n$ is formed from another sequence $X_n$, there are some properties about the conditional expectations that is used alot:

If $X$ is an integrable variable that is independent of a sub $\sigma$-field $\mathcal{G}$, then $$ E[X\mid \mathcal{G}]=E[X]\quad\text{a.s.} $$

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If $X$ is an integrable, $\mathcal{G}$-measurable variable, then $$ E[X\mid\mathcal{G}]=X\quad\text{a.s.} $$

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If $X$ is integrable, and $U$ is another random variable such that $UX$ is integrable, then $$ E[UX\mid\mathcal{G}]=UE[X\mid\mathcal{G}]\quad\text{a.s.} $$ if $U$ is $\mathcal{G}$-measurable.

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