TITLE: Sample variance converge almost surely
QUESTION [13 upvotes]: Suppose $X_{1},X_{2},\ldots$ be i.i.d. random variables such that
$E\left[X_{i}\right]=\mu$ and $Var(X_{i})=\sigma^{2}<\infty$. Let
$\bar{X}=\left(X_{1}+\cdots+X_{n}\right)/n$. Show that $\frac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(X_{i}-\bar{X}\right)^{2}\rightarrow\sigma^{2}$
a.s.
Solution: Let $S_{n}=\frac{1}{n}\overset{n}{\underset{i=1}{\sum}}\left(X_{i}-\bar{X}\right)^{2}$.
By Chebyshev's Inequality, $\sum_{n=1}^{\infty}P\left(\left|S_{n}-E\left[S_{n}\right]\right|>\varepsilon\right)\leq\sum_{n=1}^{\infty}\frac{Var(S_{n})}{\varepsilon^{2}}$.
And we know that $E\left[S_{n}\right]\rightarrow\sigma^{2}$, so I
would like to finish this proof by using first Borel-Cantelli Lemma
if I can show the right hand side is summable. However, $Var(S_{n})=\frac{2\sigma^{4}}{n-1}$,
so the sum is infinity, which means it doesn't converge almost surely.
Where am I wrong?

REPLY [13 votes]: This is much after 2012, but since I chanced upon this, I'd like to just offer an alternative solution (but similar in spirit):
$$\frac{1}{n}  \sum_{i=1}^n (X_i-\bar{X})^2 = \frac{1}{n}\sum_{i=1}^n (X_i^2-2X_i\bar{X}+(\bar{X})^2)=\frac{\sum_{i=1}^n X_i^2}{n}-\left(\frac{\sum_{i=1}^n X_i}{n}\right)^2$$
Since $X_i^2$ are i.i.d. with finite mean $\mu^2 + \sigma^2$, by strong law of large numbers (SLLN), $\frac{\sum_{i=1}^n X_i^2}{n} \to \mu^2 + \sigma^2$ almost surely. 
In addition, SLLN says $\frac{\sum_{i=1}^n X_i}{n} \to \mu$ almost surely. By Continuous Mapping Theorem (or direct reasoning), $\left(\frac{\sum_{i=1}^n X_i}{n}\right)^2 \to \mu^2$.
Hence, $S_n \to \sigma^2$ almost surely.