Convergence almost surely of the sample mean

almost-everywhereconvergence-divergenceprobabilityrandom variablesstatistics

In a probability textbook I have been working through, I came across the following exercise involving almost sure convergence for the sample mean of a given sequence of random variables and was unsure how to proceed with the problem.

For an independent sequence of random variables, for each value of $i\geq 1$, $\mathbb{P}(X_i=i^2-1)=i^{-2}$ and $\mathbb{P}(X_i=-1)=1-i^{-2}$. The sequence of random variables $\displaystyle \frac{1}{n}\sum \limits _{i=1}^nX_i$ converges almost surely to a constant. Find this constant.

The progress that I have made on the problem is that we know that$$\sum \limits _{i=1}^\infty \mathbb{P}(X_i\neq -1)=\sum \limits _{i=1}^\infty i^{-2}<\infty .$$And so by the first Borel-Cantelli Lemma, the probability of this event occurring infinitely often is equal to $0$.

For almost sure convergence, we need to show that$$\mathbb{P}\left (\left \{\left |\overline X_n-\overline X\right |\geq \varepsilon \right \}_{\text{io}}\right )=0$$

(where $\text{io}$ refers to an event that occurs “infinitely often”)

However, it is unclear to me how the progress I have made helps us in this particular instance and would be grateful for any additional guidance.

Best Answer

We have $P(X_k\neq -1\textrm{ i.o.})=0$ so for a.a. $\omega\in \Omega$, $\exists k(\omega)$ such that for all $j\geq k(\omega)$, $|X_j(\omega)+1|=0$. But then for all $n\geq k(\omega)$ $$\sum_{\ell\leq n}|X_\ell(\omega)+1|=\sum_{\ell \leq k(\omega)}|X_\ell(\omega)+1|$$ If we divide by $n$, we have that $\frac{1}{n}\sum_{\ell\leq n}|X_\ell+1|\to 0$ $P$-a.s. Therefore we conclude: $$\bigg|\frac{1}{n}\sum_{\ell \leq n}X_\ell -(-1)\bigg|\leq \frac{1}{n}\sum_{\ell\leq n}|X_\ell+1|\to 0,\quad P\textrm{-a.s.}$$

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