# Show that $E(X_n)=0$ but a.s. The sum of $\frac{1}{n}(X_1+X_2+…X_n)\rightarrow -1$ as $n\rightarrow\infty$

If $$X_1,X_2,…$$ is a sequence of independent random variables such that, $$\begin{cases} n^2-1 & \text{with probability } \frac{1}{n^2} \\ -1 & \text{with probability } 1-\frac{1}{n^2} \end{cases}$$ How do I show that for every $$n$$ we have $$E(X_n)=0$$, but almost surely we have $$\frac{X_1+X_2+\dots+X_n}{n}\longrightarrow-1$$ as $$n\rightarrow\infty$$.

This is what I've managed to do.

First, let's calculate the expectation of $$X_n$$. \begin{align*} E(X_n) = (n^2-1)\frac{1}{n^2}+(-1)(1-\frac{1}{n^2}) &= 1-\frac{1}{n^2}-1+\frac{1}{n^2} \\ &= 0 \end{align*} Given the $$P(X_n\in\{n^2-1,-1\})=1$$ and $$P(X_n=n^2-1)=\frac{1}{n^2}$$ and $$P(X_n=-1)=1-\frac{1}{n^2}$$ \begin{align*} &\sum^\infty_{n=1}P(X_n=n^2-1) = \sum^\infty_{n=1}\frac{1}{n^2}\rightarrow\frac{\pi^2}{6} \\ &\sum^\infty_{n=1}P(X_n=-1) = \sum^\infty_{n=1}1-\frac{1}{n^2}\rightarrow\infty \end{align*} Now, $$P(X_n=-1)\longrightarrow1$$ as $$n\rightarrow\infty$$ by Borel Cantelli, since $$P(\limsup(X_n=n^2-1))\rightarrow0$$. Following this, \begin{align} \lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_k)=\lim_{n\rightarrow\infty}\big(\sum^{n^*}_{k=1}\frac{X_k}{n}+\sum^{n}_{k=n^*}\frac{-1}{n} \big) \longrightarrow -1 \end{align} where, $$n^*$$ is such that $$X_n=-1$$ for $$n>n^*$$ and $$n^*>0$$. \ \ As $$n\rightarrow\infty, \sum^{n^*}_{k=1}\frac{X_k}{n}\rightarrow0$$ and $$\sum^{n}_{k=n^*}\frac{-1}{n}=\frac{n^*}{n}-1$$. Therefore, \begin{align*} \lim_{n\rightarrow\infty}\frac{1}{n}(X_1+X_2+\dots+X_n)\longrightarrow -1 \end{align*}

• You need to explain what you have done and where this problem comes from, otherwise the moderators are going to close your question.
– Momo
Commented Apr 24, 2022 at 2:10
• yep just updated. apologies Commented Apr 24, 2022 at 11:43
• In your updated work, only one of those sums is relevant to proving $X_n=-1$ finitely often (with prob 1). I found your work after those sums to be confusing, since you never directly concluded that $X_n=-1$ happens finitely often, and you had a statement "$P[X_n=-1]\rightarrow 1$ by Borel Cantelli" which is not relevant to the problem at hand and it does not require Borel Cantelli to prove that statement. Commented Apr 24, 2022 at 16:07

Use the Borel-Cantelli lemma to conclude that $$X_n=-1$$ for all but finitely many $$n$$.