Sequence of martingales 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.
 A: 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:


*

*$(Y_n)_{n\geq 1}$ 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?

*$Y_n$ 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.

*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
  $$
E[X\mid \mathcal{G}]=E[X]\quad\text{a.s.}
$$

-

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

-

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.

A: 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 $\mathcal{F}_n=\sigma(X_1,\dots,X_n)$
