# Prove that the following is a martingale

Let $M_1=X_1+1$ and let $M_n=(\log(M_{n-1})+1)^{X_{n}}$ where $(X_n)$ is an i.i.d. sequence of Poisson(1) distributed random variables. Also, let $(\mathbb{F}_n)$ be the filtration defined by $\mathbb{F_n}=\sigma(X_1,...,X_n)$.

I can see that $M_n$ is indeed $\mathbb{F}_n$ measurable, but how to prove the martingale property $\mathbb{E}(M_n|\mathbb{F_{n-1}})=M_{n-1}$?

Let $X\sim\operatorname{Poi}(\lambda)$ be a Poisson rv with expectation $\lambda$, let $s\in\mathbb R$. Then $$Es^X = \sum_{k\ge0} s^k \frac{\lambda^k }{k!} e^{-\lambda} = \sum_{k\ge0} \frac{(s\lambda)^k }{k!} e^{-\lambda} = e^{s\lambda}e^{-\lambda} = e^{\lambda(s-1)}.$$ Use this with $X=X_n$, $\lambda=1$ and $s=\log M_{n-1}+1$ to get the desired martingale expectation formula $E(M_n|\mathbb F_{n-1}) = M_{n-1}$.
• Why do we calculate $\mathbb{E}(s^{X{_n}})$ with respect to the law of $X_n$ only instead of the joint law of $X_1,...,X_n$? – Joogs Jan 16 '18 at 12:02
• $X_n$ is independent of the previous $X_i$. By conditioning on $\mathbb F_{n-1}$ we have "taken into account" the previous $X_i$, whose effect is manifested in the value of $M_{n-1}$. On each atom of $\mathbb F_{n-1}$ the quantities $M_{n-1}$ and the previous $X_i$ are constants and $X_n$ is an independent Poisson. Atom by atom the expectation of $M_n$ works out to the "constant" $M_{n-1}$, which of course would typically be different on different atoms. – kimchi lover Jan 16 '18 at 12:23