# Proving $\frac{1}{n}\sum X_k\to -1$ almost surely for given $X_n$

I'm doing an exercise that shows what can go wrong if we do without some of the assumptions made in the law of large numbers.

I have a sequence of events $E_n$ with $\mathbb{P}(E_n)=1/n^2$, and random variables $X_n=n^2\mathbf{1}_{E_n}-1$, and we set $M_m=\frac{1}{m}\sum_{n=1}^m X_n$. I have shown that $\mathbb{E}[M_n]=0$, and I want to show that $M_m\to -1$ almost surely.

I'm stuck because I don't know how to relate the $\mathbb{P}(E_n)$ to the $X_n$. I've noticed that: $$\sum_{k=1}^{\infty} \mathbb{P}(E_n)<\infty$$ So by the Borel Canteli lemma, $\mathbb{P}(\limsup E_n)=0$. But how do I use this to even prove what I want? I think $X_n\to 0$ almost surely might be enough.

## 1 Answer

Since $\mathbb{P}\{X_n>-1\}=n^{-2}$ we have $$\sum_{n\ge 1}\mathbb{P}\{X_n>-1\}<\infty.$$

Hence, $\mathbb{P}\{X_{n}>-1 \text{ i.o.}\}=0$ so that $\mathbb{P}\{X_{n}=-1 \text{ e.v.}\}=1$. The latter implies that for almost all $\omega\in \Omega$, $n^{-1}\sum_{k\ge 1}X_k(\omega)\to-1$ (for any such $\omega$, $X_k=-1$ eventually).