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Given $X_1, X_2,\dots, X_n$ independent random variables with the same distribution, if we define $S^2_N = \displaystyle\frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar X_n)^2$ show that $S_N^2$ converges almost surely to $\sigma^2$ -variance-

It seems that i have to prove the following: $P\{(\lim {S_N^2} = \sigma^2) \} = 1$, but how could i do it?.

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up vote 3 down vote accepted

First recall this bit of algebra: $$ \frac{1}{n-1}\sum_{i=1}^n \left(X_i-\overline X_n\right)^2 = \left(\frac{1}{n-1} \sum_{i=1}^n(X_i-\mu)^2\right) + \frac{n}{n-1}\left(\overline X_n-\mu\right)^2. $$

In the first of the two terms, the $n$ terms in the sum are now independent.

Then suppose $\mu=\mathbb E X_i$ and see if you can prove something about almost sure limits of the two terms separately.

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Hint: Take a look at the variance of $S_N^2$ itself! As $N\rightarrow\infty$, what happens to $var(S_N^2)$?

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