Question about biased estimator $\hat{\sigma}=\dfrac{1}{n}\sum^n_{i=1} (x_i-\mu)^2$

Question

I was trying to solve the problem asked here why is my solution wrong? $$E[\hat{\sigma}]=\dfrac{1}{n}\sum^n_{i=1} E[(x_i-\mu)^2]$$ $$E[\hat{\sigma}]=\dfrac{1}{n}\sum^n_{i=1} E[(\dfrac{(x_i-\mu)}{\sigma}\sigma)^2]$$ $$E[\hat{\sigma}]=\dfrac{1}{n}\sum^n_{i=1} E[(z\sigma)^2]$$ $$E[\hat{\sigma}]=\dfrac{1}{n}\sum^n_{i=1} \sigma^2E[z^2]$$ $$E[\hat{\sigma}]=\dfrac{1}{n}\sigma^2E[\sum^n_{i=1}z^2]$$ $$E[\hat{\sigma}]=\dfrac{1}{n}\sigma^2E[v]$$ $v\sim\chi^2(n) \implies$ $E[\hat{\sigma}]=\dfrac{1}{n}\sigma^2n=\sigma^2$

Answer 1

What you've written is correct. The random variable $$ \frac 1 n \sum_{i=1}^n (X_i-\mu)^2 $$ is an unbiased estimator of $\sigma^2$ provided $\mu$ is (somehow) known. If $\mu$ is not known, then this cannot be used as an estimator at all.

A frequently seen biased estimator of $\sigma^2$ is $$ \frac 1 n \sum_{i=1}^n (X_i - \bar X)^2, \text{ where } \bar X = \frac 1 n\sum_{i=1}^n X_i. $$ This uses $\bar X$ rather than $\mu$. The expected value of that is $\dfrac{n-1} n \sigma^2.$

Here's a slightly simpler argument than what you wrote: \begin{align} & \operatorname{E} \left( \frac 1 n \sum_{i=1}^n (X_i-\mu)^2 \right) = \frac 1 n \operatorname{E} \sum_{i=1}^n (X_i - \mu)^2 \\[10pt] = {} & \frac 1 n \sum_{i=1}^n \operatorname{E}((X_i-\mu)^2) = \frac 1 n \sum_{i=1}^n \sigma^2 = \frac 1 n n\sigma^2 = \sigma^2. \end{align}