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Let $(X_n)_{n\geq1}$ be a sequence of i.i.d. random variables such that the moment generating function $M_{X_1}(t)<\infty$ for all $t$. Let $S_n=\sum_{i=1}^n X_i$ and

$\displaystyle{M_n=\frac{e^{tS_n}}{M_{X_1}(t)^n}, n = 1,2,\dots}$

Show that $(M_n)$ is a martingale w.r.t. $(F_n=\sigma\{X_m:m\leq n\})$.

How to show $E[M_{n+1}|F_n]=M_n$?

If I want to show $M_n$ is integrable, then I have to show $E[M_n]<\infty$. It is easy to show the numerator of $M_n$ is integrable, but how to show $M_n$ is integrable?

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Note that $M_{X_1}(t)^n$ is a constant (i.e. not random); thus, since $e^{t S_n}$ is in $L^1$, so is $M_n$. Once you have that, simply compute: \begin{align*} E[M_{n+1} |F_n] &= E\left[\frac{e^{tS_n}e^{tX_{n+1}}}{M_{X_1}(t)^n M_{X_1}(t)} \bigg| F_n\right] \\ &=\frac{e^{tS_n}}{M_{X_1}(t)^n}E\left[\frac{e^{tX_{n+1}}}{M_{X_1}(t)} \bigg| F_n\right] \\ &= \frac{e^{tS_n}}{M_{X_1}(t)^n} \cdot 1 = M_n. \end{align*}

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  • $\begingroup$ Why $M_{X_1}(t)^n$ is a constant? As a function, shouldn't $M_{X_1}(t)$ change as $t$ changes? $\endgroup$ – user430647 May 10 '17 at 15:26
  • $\begingroup$ Yes, but for each $t$, $M_{X_1}(t)$ is constant (as in not a random variable) so you can pull it out of the expectation. In this problem, $M_n$ depends on $t$, so technically what you're trying to prove is that for each fixed $t$ we have $(M_n)$ is a martingale. $\endgroup$ – Marcus M May 10 '17 at 15:28
  • $\begingroup$ Thanks. I previously understood the problem as "t not fixed". That is $M_n(t)=\frac{e^{tS}}{M_{X_1}(t)}$. Now I think it is clear for me to do this problem. $\endgroup$ – user430647 May 10 '17 at 15:37

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