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For $n \in \mathbb{N}$, let $X_n$ be a random variable such that $\mathbb{P} [X_n = \frac{1}{n}] = 1 − \frac{1}{n^2}$ and $\mathbb{P}[X_n = n] = \frac{1}{n^2}$. Does $X_n$ converge in probability? In $L^2$?

My attempt: To converge in probability we must have $$ P(|X_n - X| > \epsilon) = 0 $$ Since we can see that as $n$ gets larger, $X_n = \frac{1}{n}$ because its probability tends to 1. So I did the following

$$ P(|X_n - \frac{1}{n}| > \epsilon) = P(X_n = n) = \frac{1}{n^2} \\ \lim_{n \to \infty}\frac{1}{n^2} = 0 $$ Is this approach correct? Or do I have to use some version of Chebyshev's inequality to do this? Also I have little idea how to prove it for $L^2$?

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Not really. Here $X$ is chosen to let $X_n\to X$ in probability as $n\to\infty$, so we can not let "$n$" appear in $X$. Actually, we can choose $X=0$.Let $\epsilon>0$, for $n>\frac1\epsilon$, we have $$P(|X_n-0|>\epsilon)=P(X_n=n)=\frac1{n^2}\to0,$$ so $X_n\to 0$ in probability.

For the $L^2$ convergence, since $E|X_n-0|^2=\frac1{n^2}-\frac1{n^4}+1\to 1\neq 0$, $X_n$ is not convergent to $0$ in $L^2$, so $X_n$ is not convergent in $L^2$. In fact, if $X_n\to X$ in $L^2$ then $X_n\to X$ in probability so $X=0$, a contradiction.

Addendum: There is an alternative way to show the convergence in probability: Just note that $X_n\to 0$ in $L^1$.

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  • $\begingroup$ Thanks. Just one more thing. Does your last statement implies convergence contradiction $X = 0$ in $L^2$ or in probability? $\endgroup$ Aug 31, 2019 at 4:02
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    $\begingroup$ @BhavitSharma You are welcome. $L^2$ of course, since we just have proven that $X_n$ can't converge to $0$ in$L^2$. $\endgroup$
    – Feng
    Aug 31, 2019 at 4:11

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