Let $X_{k}$ be i.i.d random variables with $E[X_{k}]=\mu$ and $Var(X_{k})=\sigma^{2}<\infty$ and let $\bar{X}=\frac{1}{n}\sum_{k=1}^{n}{X_{k}}$. Let $$S^{2}=\frac{1}{n-1}\sum_{k=1}^{n}{(X_{k}-\bar{X})^{2}}(n\geq 2)$$ be the sample variance. Does $S^{2}$ converges almost surely and what is (if it exists) its limit?
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Take a look at this question: https://math.stackexchange.com/q/243348/36150 – saz Jun 10 '18 at 11:46
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Your $S_n^2$ has the same distribution as $\frac{\sigma^2}{n-1}\sum_{i=1}^{n-1}U_i^2$ where the $U_i$ are iid with standard normal distribution. – drhab Jun 10 '18 at 11:50
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As shown in the question linked to in the comments,
$$ \frac n{n-1}S^2 $$
converges to $\sigma^2$ almost surely. Since $n/(n-1)$ converges to $1$, the sequence is almost surely the product of two convergent sequences, in which case its limit is the product of the limits, i.e. $\sigma^2$.
joriki
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