Discrete Exponential Martingale - Properties

Question

This question is about the discrete exponential martingale.
Let $(Y_n)_n$ be a sequence of independent and identically distributed random variables with $m_{Y}(t) :=\mathbb{E}\left[e^{t Y_{1}}\right]<\infty, t \in \mathbb{R}$. I want to show that
\begin{align*} M_{n} :=\frac{1}{m_{Y}(t)} \exp \left(t \sum_{i=1}^{n} Y_{i}\right) \end{align*} is a martingale. But I didn't know how to proof that \begin{align*} \begin{aligned} \mathbb{E}\left[M_{n+1} | \mathcal{F}_{n}\right] &=\frac{1}{m_{Y}(t)} \mathbb{E}\left[\exp \left(t \sum_{i=1}^{n+1} Y_{i}\right) | \mathcal{F}_{n}\right] \\ &=\frac{1}{m_{Y}(t)}\left\{\mathbb{E}\left[\exp \left(t Y_{n+1}\right) | \mathcal{F}_{n}\right]+\sum_{i=1}^{n} \mathbb{E}\left[\exp \left(t Y_{i}\right) | \mathcal{F}_{n}\right]\right\} = ... = M_n\end{aligned} \end{align*}

Answer 1

The candidate martingale has perhaps been miscopied. As per the comment by @Sesame, if instead one defines $M_n$ by $$M_n = \frac{\exp(t\sum_{i=1}^n Y_i)}{m_Y(t)^n},$$ then the usual computations carry through to show this is a martingale.

Summarizing the steps presented in the comment, we compute $$\mathbb{E}(M_{n+1}|\mathscr{F}_n)=\frac{1}{m_Y(t)^{n+1}}\exp(t \sum_{i=1}^n Y_i)\mathbb{E}(\exp(tY_{n+1})|\mathscr{F}_n)$$ $$=\frac{M_n}{m_Y(t)}\mathbb{E}(\exp(tY_{n+1}))=M_n$$ where we have, first, used the "pulling out what is known" property since $\exp(t \sum_{i=1}^n Y_i) \in \mathscr{F}_n$ and, second, used that $\exp(t Y_{n+1})$ is independent of $\mathscr{F}_n$. Thus $\mathbb{E}(M_{n+1}|\mathscr{F}_n)=M_n$, so $M_n$ is a martingale as desired.

Answer 2

Yeah, it sounds good. I did not realize that $\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}}>0$ so $\mathbb{E}|\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}}|=\mathbb{E}(\frac{\exp^{(t\sum_{i=1}^{n}Y_{i})}}{m_Y(t)^{n}})$. Next $\frac{1}{m_Y(t)^{n}}\in\mathbb{R}$ and all $\mathbb{E}e^{tY_{i}}<\infty$.