# Consistent estimator for the variance of a normal distribution

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!

• Using relation between $\hat\sigma_n^2$ and $S^2$ (which you don't define) , find the mean and variance. Mar 9, 2019 at 14:06
• 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 $n\to \infty$ Mar 9, 2019 at 14:25
• @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 $\lim_{n\to\infty}Var\left(\hat{\sigma}_n^2\right)=0$ Mar 9, 2019 at 15:51
• 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 $Var(\hat{\sigma_n^2})=\frac{2\sigma^4}{n}$? Mar 9, 2019 at 16:10
$$\hat{\sigma}_n^2=\frac{1}{n}\cdot (\sum_{i=1}^n(X_i-\bar{X})^2)$$
$$\hat{\sigma}_n^2=\frac{\sigma^2}{n}\cdot \frac{(\sum_{i=1}^n(X_i-\bar{X})^2)}{\sigma^2}$$
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 $$n \to\infty$$