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Let $X_1,X_2,...$ be independent random variables with $P[X_n=0]=1-1/n$, $P[X_n=1]=1/2n$, $P[X_n=-1]=1/2n$ Does $X_n$ converge almost surely? , Does $X_n$ converge in probability?

I just started to study the types of convergence and I´m not very sure how to proceed: for example to check convergence in probability by definition I need to have that $\lim_{n\to \infty}P[\omega\in \Omega: |X_n(\omega)-X(\omega)|>\epsilon]=0$ for all $\epsilon$ but to use this I think I need to have the limit random variable $X$, but how can I get it?

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$X_n$ converges in probability to $0$ since

$$\mathbb{P}(|X_n|>\epsilon) = \mathbb{P}(|X_n|=1) = \frac{1}{n} \stackrel{n \to \infty}{\to} 0$$

for any $\epsilon>0$. However, $X_n$ does (in general) not converge almost surely:

Since the sets $\{|X_n|>\frac{1}{2}\}$, $n \in \mathbb{N}$, are independent, we have by the Borel Cantelli lemma that

$$\mathbb{P}(|X_n|>\frac{1}{2} \, \text{happens infinitely often}) = 1 \Leftrightarrow \sum_{n=1}^{\infty} \mathbb{P}(|X_n|>\frac{1}{2}) = \infty.$$

As $\sum_n \mathbb{P}(|X_n|>\frac{1}{2}) = \sum_n \frac{1}{n} = \infty$, this means that $|X_n(\omega)|> \frac{1}{2}$ happens infinitely often for almost all $\omega \in \Omega$. Consequently, $X_n$ does not converge almost surely to $0$. (Note: Since convergence almost surely implies convergence in probability and limits in probability are unique, $(X_n)_n$ cannot converge almost surely to any other random variable than $X=0$.)

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