So I have to show that $\hat{\sigma}_n^2=\frac{1}{n}\cdot (\sum_{i=1}^n(X_i-\bar{X})^2)$ is a consistent estimator for the variance $\sigma^2$ when $X_1,X_2,…,X$ are i.i.d. from a normal distribution with mean $\mu$ and variance $\sigma^2$.
I have already shown that $\lim_{n\to\infty}E\left[\hat{\sigma}_n^2\right]=\sigma^2$ but I still need to show that $\lim_{n\to\infty}Var\left(\hat{\sigma}_n^2\right)=0$.
I have a theorem which states that if $X_1,…,X_n$ are from a normal distribution with mean $\mu$ and variance $\sigma^2$ then
1) $\bar{X}$ and $S^2$ are independent random variables,
2) $\frac{(n-1)\cdot S^2}{\sigma^2}$ has a $\chi^2$ distribution with $n-1$ degrees of freedom.
So we have that $2(n-1)=Var(\frac{(n-1)\cdot S^2}{\sigma^2})=\frac{(n-1)^2}{\sigma^4}\cdot Var(S^2)\Rightarrow Var(S^2)=\frac{2(n-1)\sigma^4}{(n-1)^2}=\frac{2\sigma^4}{n-1}$.
Since $S^2=\frac{1}{n-1}(\sum_{i=1}^n(X_i-\bar{X})^2)$ and $\hat{\sigma_n^2}=\frac{1}{n}(\sum_{i=1}^n (X_i-\bar{X})^2)$ does this mean that $Var(\hat{\sigma_n^2})=\frac{2\sigma^4}{n}$?
However, I'm not sure how to go on from here. Any help would be appreciated!
Best Answer
$\hat{\sigma}_n^2=\frac{1}{n}\cdot (\sum_{i=1}^n(X_i-\bar{X})^2)$
$\hat{\sigma}_n^2=\frac{\sigma^2}{n}\cdot \frac{(\sum_{i=1}^n(X_i-\bar{X})^2)}{\sigma^2}$
Now you know $\frac{(\sum_{i=1}^n(X_i-\bar{X})^2)}{\sigma^2}\sim \chi^2_{(n-1)}$
$V(\hat{\sigma}_n^2)=\frac{\sigma^4(2n-2)}{n^2} \to 0 $ as $n \to\infty$