# If $E(X_n) = 1/n$ and $\mathrm{Var}(X_n) = 1/n^2$ then $X_n\to0$ almost surely

I would like to show that

If $E(X_n) = 1/n$ and $\mathrm{Var}(X_n) = 1/n^2$ then $X_n\to0$ almost surely.

I can not find some relation between the almost sure convergence and $E(X_n)$ and $Var(X_n)$.

• To complete the proof above, you might want to check that $$\liminf\{|X_n-1/n|<\epsilon\}\subseteq\{\limsup|X_n|\leqslant\epsilon\}$$ – Did Jun 7 '17 at 7:52

Chebyshev gives $P(|X_n-1/n| \ge \epsilon )\le \frac{1}{\epsilon^2n^2}.$ Then apply Borel-Cantelli. The key relationship here is that often "in probability" + "Borel-Cantelli" = "almost sure"
• I am trying to prove $\left[X_n \nrightarrow 0 \right] \subset \hbox{lim sup} \left[|X_n - \frac{1}{n}| > \epsilon \right]$ – orrillo Jun 7 '17 at 6:26
• @orrillo Intuition here is that for any $\epsilon$ borel cantelli shows that $|X_n-1/n|< \epsilon$ for all sufficiently large $n$ with probability $1$. Claim this means that for any $\epsilon,$ $|X_n|< \epsilon$ for all sufficiently large $n$ With prob 1. Claim this means $X_n \to 0$ w p 1. Now prove and fill in gaps. This is really just definition chasing in real analysis, not a lot of probability after the application of borel cantelli. – spaceisdarkgreen Jun 7 '17 at 15:52
• @orrillo to put it more simply $|X_n-1/n|\to 0$ implies $X_n\to 0.$ so if you can show borel cantelli implies the first is true a.s. then you're done. – spaceisdarkgreen Jun 7 '17 at 15:58
• Let $Y_n = (X_n - \frac{1}{n})$. By Borel Cantelli, we have that $P(\hbox{lim sup}\left[|Y_n|> \epsilon \right])= 0$ for all $\epsilon > 0$. We have that $\left[Y_n \nrightarrow 0 \right] = \bigcup_{\epsilon>0} \hbox{lim sup}\left[|Y_n|> \epsilon \right] = \bigcup_{n \in \mathbb{N}} \hbox{lim sup}\left[|Y_n|> 1/n \right]$. So, $P(\left[Y_n \nrightarrow 0 \right] ) = P(\bigcup_{n \in \mathbb{N}} \hbox{lim sup}\left[|Y_n|> 1/n \right]) \leq \Sigma_{n \in \mathbb{N}} P(\hbox{lim sup}\left[|Y_n|> 1/n \right])= 0$. It is OK? – orrillo Jun 7 '17 at 17:59
• @orrillo looks fine to me other than a bit of a notation confusion using $n$ in two different ways. – spaceisdarkgreen Jun 7 '17 at 18:45