A problem about almost sure converges of a random variable with infinite variance.

Question

Let $X_1,X_2,...$ be independent random variables such that for all positive integers $k$, we have $P\left( X_{k}=k^{2}\right) =\dfrac {1} {k^{2}}$, $P\left( X_{k}=2\right)=\dfrac{1}{2}$, and $P\left( X_{k}=0\right) =\dfrac{1}{2}-\dfrac {1} {k^{2}}$.

Define $S_n=X_1+X_2+...+X_n$. Show that there is a real number $c$ such that $\dfrac{S_n}{n}\rightarrow c\space a.s.$ and find the value of c.

Answer 1

Following Chris's advice, define $Y_k$ by $$ Y_k = \begin{cases} X_k & X_k \in\{0,2\}\\ 0 & \text{otherwise}\end{cases} $$ Then the $(Y_k)$ are independent and identically distributed, therefore $\frac 1n\sum_{k=1}^n Y_k \to \mathbb E(Y_1) = 1$ almost surely. We have that \begin{align*} \sum_{k=1}^\infty \mathbb P(X_k \ne Y_k) &= \sum_{k=1}^\infty \mathbb P(X_k = 2^k)\\ &= \sum_{k=1}^\infty \frac 1{2^k}\\ &< \infty. \end{align*} So, by Borel-Cantelli, we have that $X_k \ne Y_k$ hold almost surely only finitely many times. And on the set $\{X_k = Y_k \text{ finally}\}$ we have $\lim_n \frac 1n\sum_{k=1}^n X_k = \lim_n \frac 1n \sum_{k=1}^n Y_k = 1$. So $\frac 1n \sum_{k=1}^n X_k \to 1$ almost surely.