Discrete Exponential Martingale - Properties 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*}

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 $\mathbb{E}e^{tY_{i}}<\infty$.
