Let $\left(Z_{n}\right)_{n \geq 1}$ be a sequence of i.i.d r.v's such that : $$ P\left(Z_{n}=1\right)=P\left(Z_{n}=-1\right)=\frac{1}{2 n}, \quad P\left(Z_{n}=0\right)=1-\frac{1}{n}$$ Let $\mathcal{F}_{n}=\sigma\left(Z_{1}, Z_{2}, \ldots, Z_{n}\right), X_{0}=0$ and : $$ X_{n}=Z_{n} 1_{\left\{X_{n-1}=0\right\}}+n\left|Z_{n}\right| X_{n-1} 1_{\left\{X_{n-1} \neq 0\right\},} n \geq 1 $$
- I have shown that for $n \in \mathbb{N}, X_{n} \in \mathbb{Z}$ and $\left|X_{n}\right| \leq n !$
- I have shown that $\left(X_{n}\right)_{n \geq 1}$ is a martingale.
- I have shown that $P(X_n=0) = P(Z_n=0) = 1 - \frac1n$
I need to discuss the convergence of $(X_n)_{n \geq 1}$ in probability, $L^1$ and almost surely.
for the convergence in probability, since $P(X_n = 0) \to 1$ then $P(X_n \neq 0) = P(|X_n|>0) \to 0$
let $\varepsilon > 0$ then $P(|X_n| \geq \epsilon) \leq P(|X_n|>0) \to 0 $, thus $X_n$ converges to $0$ in probability.
for the convergence in $L^1$ :
$E|X_n| = \sum_{n \geq 0}P(|X_n|\geq n) \geq \sum_{n \geq 0} P(X_n = 0) = +\infty$
Thus it doesn't converge in $L^1$.
How to derive the convergence or non-convergence almost surely though, and is my work is correct for the two convergence modes above ?
EDIT : reasoning in $L^1$ convergence is wrong, it should be $E|X_n| = \sum_{n \geq 0}P(|X_n| > n)$