Logarithm of martingale to the power of a Poisson random variable is a martingale

Question

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 $$\begin{align*} \mathbb{E}[(\log X_{n-1}+1)^{Z_n}|F_n]&=\mathbb{E}[\sum\limits_{k\in\mathbb{N}}\frac{1}{e\cdot k!}(\log X_{n-1}+1)^k|F_n]\\ &=e^{-1}\mathbb{E}[\exp(\log X_{n-1}+1)|F_n]\\ &=\mathbb{E}[X_{n-1}|F_{n-1}]\\ &=X_{n-1} \end{align*}$$

Answer 1

• Part (a):
  As @Davide pointed out, we can see $X_{n-1}\in F_{n-1} \implies X_n\in F_n$ easily. By induction, $(X_n,F_n)$ is an adapted process. Also note that $X_{n-1}\ge 1 \implies X_n\ge 1$, hence $X_n\ge 1$ a.s. We find that $\Bbb EX_1 =2<\infty$ as a base case and $$\begin{eqnarray} \Bbb E\left[X_{n+1}1_{\{Z_{n+1}\le N\}}|F_n\right] &=&\Bbb E\left[\left(\log X_n+1\right)^{Z_{n+1}}1_{\{Z_{n+1}\le N\}}|X_n\right]\\&= &\sum_{k=0}^N \frac{e^{-1}\left(\log X_n +1\right)^k}{k!}\\ &\stackrel{N\to\infty}\longrightarrow&e^{-1}e^{\log X_n +1}=X_n. \end{eqnarray}$$ By induction and monotone convergence theorem, it follows $\Bbb E[X_{n+1}|F_n]$ is integrable and $\Bbb E[X_{n+1}|F_n]= X_n$
  
• Part (b):
  

First note that since $(X_n, F_n)$ is a $L^1$-bounded martingale, there exists $$ X =\lim_{n\to\infty} X_n\in [1,\infty)\ \quad\text{a.s.} $$ by martingale convergence theorem. Let $A =\{X>1\}$. By taking logarithms, we find that $$ \log X_{n} =Z_n \log \left(\log X_{n-1}+1\right). $$ From this we obtain $$ Z_n \stackrel{n\to\infty}\longrightarrow \frac{\log X}{\log \left(\log X+1\right)} \ \ \quad\text{a.s. on }\ A. $$ On the other hand, by strong law of large numbers $$ \frac{Z_1+Z_2+\cdots +Z_n}{n}\stackrel{n\to\infty}\longrightarrow 1\ \ \quad\text{a.s.} $$ Since the Cesaro limit is equal to the limit if both exist, we have $$ X=\log X+1 \ \quad\text{a.s. on } \ A. $$ Since $\log t < t-1$ for all $t\ne 1$, this is impossible on $A=\{X> 1\}$. Hence $A\subset A^c$ a.s. and $\Bbb P(A)=0$ as desired.

• Part (c):
  

Note that if $(X_n,F_n)$ is $L^p$-bounded for $p>1$, martingale convergence theorem implies that $$ X_n \to X=1\quad\text{both a.s. and in} \ L^p. $$ But this implies that $$ X_1 =\Bbb E[X_n |F_1] \stackrel{n\to\infty}\longrightarrow \Bbb E [X|F_1] =1 \text{ in }\ L^p $$ leading to the contradiction.

Answer 2

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 $$\mathbb E\left\lvert X_n\right\rvert=\mathbb E\left[\sum_{k=0}^{+\infty}e^{-1}\frac{1}{k!}\left(1+\log X_{n-1}\right)^k\right]=e^{-1}\mathbb E\left[ \exp\left(1+\log X_{n-1}\right) \right]=\mathbb E\left\lvert X_{n-1}\right\rvert.$$