# verification of convergence of random variable

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 $$L^2$$?

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 $$L^1$$.
• Thanks. Just one more thing. Does your last statement implies convergence contradiction $X = 0$ in $L^2$ or in probability? Aug 31, 2019 at 4:02
• @BhavitSharma You are welcome. $L^2$ of course, since we just have proven that $X_n$ can't converge to $0$ in$L^2$.