Let $(X_n)_{n \geq 0}$ be a sequence of independent random variables such that
$X_n = n $ with probability $\frac{1}{2n \log(n)}$
$X_n = -n$ with probability $\frac{1}{2n \log(n)}$
$X_n = 0$ with probability $1-\frac{1}{n \log(n)}$
Let S := $X_2 + … + X_{n+1}$.
I got the hint to consider the events $A_n = \{ |X_n| = n \}$ and to use the Borel-Cantelli lemma. I should consider what happens to $| \frac{S_n}{n}|$ under the event $A_n$.
Why, in this case, the law of strong numbers does NOT hold?
Thanks for any help.
Best Answer
The Borel–Cantelli lemma hint suggests considering something like $\sum\limits_{n=2}^\infty \mathbb P\left(\left| \frac{S_n}{n}\right| \ge 1\right)$
$P\left(\left| \frac{S_n}{n}\right| \ge 1\right) =\mathbb P\left(\left| {S_n}\right| \ge n\right) \ge \frac12 P\left(\left| {A_n}\right| \ge n\right) = \frac12 \frac{1}{n \log_e(n)}\ge \frac12 \int\limits_n^{n+1}\frac{1}{x \log_e(x)}\, dx$, since we might either add $n$ or subtract $n$ from whatever value $S_n-A_n$ takes and at least one of those $A_n=\pm n$ events will make $\left| {S_n}\right| \ge n$
So $\sum\limits_{n=2}^\infty \mathbb P\left(\left| \frac{S_n}{n}\right| \ge 1\right) \ge \frac12 \sum\limits_{n=2}^\infty P\left(\left| {A_n}\right| \ge n\right) \ge \frac12 \int\limits_2^{\infty}\frac{1}{x \log_e(x)}\, dx = \infty$.
Since the $A_n$ are independent we can use the converse of the Borel–Cantelli lemma to conclude that almost surely among the $A_n$ the necessary $\pm n$ occur infinitely often and so $\left| \frac{S_n}{n}\right| > 1$ infinitely often.
So the strong law of large numbers does not apply: $\frac{S_n}{n}$ does not converge to $0$ almost surely.