Let $(Z_n)_n$ be independently Poisson(1) distributed random variables.
Let $X_1=Z_1+1$ and
$$X_n=(\log X_{n-1}+1)^{Z_n}$$
for $n>1$ and define $F_n=\sigma(Z_1,…,Z_n)$.
How do I show that
a) $X_n$ is a $F_n$-martingale
b) its limit is $1$ almost surely
c) it is not bounded in $L^p$ for any $p>1$?
For a) we need to show that $\mathbb{E}|X_n|<\infty$, but how do I write $\mathbb{E}|(\log X_{n-1}+1)^{Z_n}|$?
Further we want $\mathbb{E}[X_n|f_{n-1}]=X_{n-1}$, but again I don't know what to do with the power of $Z_n$?
EDIT:
Based on Davide Giraudo's reply, I write for the martingale condition
$$\begin{align*}
\mathbb{E}[(\log X_{n-1}+1)^{Z_n}|F_n]&=\mathbb{E}[\sum\limits_{k\in\mathbb{N}}\frac{1}{e\cdot k!}(\log X_{n-1}+1)^k|F_n]\\
&=e^{-1}\mathbb{E}[\exp(\log X_{n-1}+1)|F_n]\\
&=\mathbb{E}[X_{n-1}|F_{n-1}]\\
&=X_{n-1}
\end{align*}$$
Best Answer
Observe that we can prove by induction that $X_n$ is $\mathcal F_n$-measurable and that $X_n\geqslant 1$ almost surely. Since $Z_n$ is independent of $X_{n-1}$, it follows that $$\mathbb E\left\lvert X_n\right\rvert=\mathbb E\left[\sum_{k=0}^{+\infty}e^{-1}\frac{1}{k!}\left(1+\log X_{n-1}\right)^k\right]=e^{-1}\mathbb E\left[ \exp\left(1+\log X_{n-1}\right) \right]=\mathbb E\left\lvert X_{n-1}\right\rvert.$$