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$X_n$ s are a sequence off iid random variables with E($X_n$) = $\mu$, V($X_n$)= $\sigma$$^2$ and $\bar X = \sum$ $\frac{X_i}{n}$. Then show that
$\frac1n$ $\sum (X_i - \bar X )^2\to\sigma^2$ in probability.

$\mathbb P(|S_n - \sigma^2|>\epsilon) < V(S_n)/\epsilon^2$ by Chebyshev inequality. How to proceed?

Davide Giraudo
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kris91
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1 Answers1

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$$ \frac 1n \sum (X_i - \bar{X})^2 = \frac 1n \sum (X_i^2 - 2 X_i \bar{X}) + \bar{X}^2 $$ Now apply the large law of large numbers: $$ \frac 1n \sum X_i^2 \to EX_i^2 = \sigma^2 + \mu^2\\ \bar{X}^2\to \mu^2\\ \frac 1n \sum2 X_i \bar{X} \to 2\times \mu\times \mu $$because operations are compatible with convergence almost sure.

Eventually, sum everything and you will find $\sigma^2$.

Thus, $\frac1n\sum (X_i - \bar X )^2\to\sigma^2$ almost surely. Since almost sure convergence implies convergence in probability, this proves that $\frac1n\sum (X_i - \bar X )^2\to\sigma^2$ in probability, as desired.

Did
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mookid
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  • To get the convergence in probability using Chebyshev, one should evaluate the variance of $\sum(X_i-\bar X)^2$, not the variance of $S_n=n\bar X$. This is complicated (and assumes that the $X_i$s are in $L^4$) hence one prefers very much the detour by the almost sure convergence (under $L^2$ condition) proved in the first part of your answer. – Did Mar 30 '14 at 07:28
  • You are right, as often. – mookid Mar 30 '14 at 09:21
  • +1. Added a more explicit paragraph at the end. Just delete it if you oppose it. – Did Mar 30 '14 at 09:52