How to find the bias of $\hat{\sigma}^2=\frac{\sum^n_{i=1}(X_i-\bar{X})^2}{n}$ for the population variance $\sigma^2$

Question

My question is

The bias of an estimator $\hat{\theta}$ for parameter $\theta$ is defined as $E(\hat{\theta})-\theta$

please find the bias of $\hat{\sigma}^2=\frac{\sum^n_{i=1}(X_i-\bar{X})^2}{n}$ for the population variance $\sigma^2$

I tried to do something below:

the bias of $\hat{\sigma}^2$

$=E(\frac{\sum^n_{i=1}(X_i-\bar{X})^2}{n})-\sigma^2$

$=\frac{1}{n}\sum^n_{i=1}E(X_i^2-nE((\bar{X})^2)-\sigma^2$

$=\frac{1}{n}[n\,\times\,(\sigma^2+\mu^2)-n\,\times\,(\frac{\sigma^2}{n}+\mu^2)]-\sigma^2$

$=\frac{-1}{n}\sigma^2$

I am not sure my answer is correct or not, can anyone help me? If my answer is incorrect, could you tell me how to do?

Answer 1

It is correct. You can also check a similar way of calculation from the wiki page. So, for the estimate to be unbiased, the formula you have should be multiplied by ${n \over n-1}$, which is called as Bessel's Correction.