Let $(Z_i)_{i\in \mathbb{N}}$ be a sequence of independent and bounded r.v. with $\mathbb{E}(Z_i)=m_i$ and $\operatorname{Var}(Z_i)=\sigma_i$. Let $\mathcal{F}_n=\sigma({Z_1,\dots,Z_n})$. Find a $(b_n)$ such that for a $t\in\mathbb{R}$ $$b_n\exp\left(\sum\limits_{i=1}^ntZ_i\right)$$ becomes a martingale.
I get stucked in this exercise because of the exponential function. What shall I do: $$b_{n+1}\exp\left(\sum\limits_{i=1}^ntZ_i\right)\mathbb{E}\left(\exp\left(tZ_{n+1}\right)\mid\mathcal{F}_{n}\right)$$ I can't compute $$\mathbb{E}\left(\exp\left(tZ_{n+1}\right)\mid\mathcal{F}_{n}\right)=\mathbb{E}\left(\exp\left(tZ_{n+1}\right)\right)$$ Because it's a composition and I only know the expectation of $Z_i$, not of the composition. Do I have to make somehow a substitution?
Since $Z_{n+1}$ is independent from $\mathcal F_n$ you have that $$\mathbb E[\exp(tZ_{n+1})\mid \mathcal F_n]=\mathbb E[\exp(tZ_{n+1})]$$ and the last term is the moment generating function (MGF) of $Z_{n+1}$, which you may denote with $M_{n+1}(t)$. Hence $$b_{n+1}\mathbb E[\exp(tZ_{n+1})]=1\implies b_{n+1}=\frac1{M_{n+1}(t)}$$ so the martingale becomes $$W_n:=\frac{\exp(tS_n)}{M_n(t)}$$ which is known as the Wald's martingale.
The most common case is when $Z_{n+1}\sim \mathcal N(m_{n+1},σ^2_{n+1})$, (Brownian motion) and the MGF of $Z_{n+1}$ is given by $$M_{n+1}(t)=\mathbb E[\exp(tZ_{n+1})]=\exp\left(tm_{n+1}+\frac12σ^2_{n+1}t^2\right)$$ Hence $$b_{n+1}\mathbb E[\exp(tZ_{n+1})]=1\implies b_{n+1}=\exp\left(-tm_{n+1}-\frac12σ^2_{n+1}t^2\right)$$ and for $m_{n+1}=0, σ^2_{n+1}=σ^2$ this becomes $$W_n=\exp\left(tS_n-\frac12σ^2t^2\right)$$ (also known with continuous time, $t\ge0$, instead of discrete $n\ge0$).