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?


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*}

  • $\begingroup$ 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

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.