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!

  • $\begingroup$ Using relation between $\hat\sigma_n^2$ and $S^2$ (which you don't define) , find the mean and variance. $\endgroup$ Mar 9, 2019 at 14:06
  • $\begingroup$ See here. Just replace $\frac1{n-1}$ by $\frac1n$. You will get $E\left(\hat \sigma_n^2\right)=\frac{n}{n-1}\cdot \sigma^2$ Then let $n\to \infty$ $\endgroup$ Mar 9, 2019 at 14:25
  • $\begingroup$ @callculus, I have already shown that $\lim_{n\to\infty}E\left[\hat{\sigma}_n^2\right]=\sigma^2$. I'm having some trouble with showing $\lim_{n\to\infty}Var\left(\hat{\sigma}_n^2\right)=0$ $\endgroup$
    – statman
    Mar 9, 2019 at 15:51
  • $\begingroup$ Ahh so since $\frac{(n-1)S^2}{\sigma^2}$ is $\chi^2$ distributed with n-1 degrees of freedom and $2(n-1)$ variance then $2(n-1)=Var(\frac{(n-1)S^2}{\sigma^2}=\frac{(n-1)^2}{\sigma^4}Var(S^2) \Rightarrow Var(S^2)=\frac{(2(n-1)\sigma^4}{(n-1)^2}=\frac{2\sigma^4}{n-1}$. So when $\hat{\sigma}_n^2=\frac{1}{n}\cdot (\sum_{i=1}^n(X_i-\bar{X})^2)$ then $Var(\hat{\sigma_n^2})=\frac{2\sigma^4}{n}$? $\endgroup$
    – statman
    Mar 9, 2019 at 16:10
  • $\begingroup$ Ask your queries/show your work in the body of the question, not in comments only. $\endgroup$ Mar 9, 2019 at 19:25

1 Answer 1


$\hat{\sigma}_n^2=\frac{1}{n}\cdot (\sum_{i=1}^n(X_i-\bar{X})^2)$

$\hat{\sigma}_n^2=\frac{\sigma^2}{n}\cdot \frac{(\sum_{i=1}^n(X_i-\bar{X})^2)}{\sigma^2}$

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 $n \to\infty$


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