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

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.

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Hint: To prove almost sure convergence, you generally want to use Borel Cantelli. Can you guess at what the answer should be in this case? One way to approach this would be to couple the random variables you are working with with something you can do this problem for that looks essentially the same, then to show that the probability that they differ infinitely often is zero. – Chris Janjigian Nov 26 '12 at 6:23

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.

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@martini's answer is correct. But he(or She) fails to mention something which may confuse you. I'm going to give more explanations on his(or her) answer.

What we have by Borel Cantelli lemma is

$$P(\cap_{n=1}\cup_{k\geq n}\{X_n\neq Y_n\}) = 0$$

i.e.

$$P(\cup_{n=1}\cap_{k\geq n}\{X_n = Y_n\}) = 1$$

By using the VERY important fact that

$$\big\{\omega:\cup_{n=1}\cap_{k\geq n}\{X_n(\omega) = Y_n(\omega)\}\big\} \subseteq\big\{\omega: \lim X_n(\omega) =\lim Y_n(\omega) \big\}$$

the above relation is very important, but most of the text books fail to mention. You can prove it yourself and it's not hard. Thus we have

$$P( \lim X_n =\lim Y_n ) = 1,\quad \text{i.e.}\quad \lim X_n =\lim Y_n\quad \text{a.s.}$$

By using another lemma that, for real numbers $a_n\in R$, if

$$\lim a_n = a, \text{ then }\lim_{n\to\infty} \frac{\sum_{j=1}^na_j}{n} = a$$

we thus have

$$\big\{\omega: \lim X_n(\omega) =\lim Y_n(\omega) \big\}\subseteq\big\{\omega: \lim \frac{\sum_{k=1}^nX_j(\omega)}{n} =\lim \frac{\sum_{k=1}^nY_j(\omega)}{n} \big\}$$

Finnaly

$$P\big(\big\{\omega: \lim \frac{\sum_{k=1}^nX_j(\omega)}{n} =\lim \frac{\sum_{k=1}^nY_j(\omega)}{n} \big\}\big)=1$$

Then we can say it's a complete prrof.

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