Variance Estimator Change if we know Population Mean? (Normal dist. example) For a normal distribution $N(\mu, \sigma^2)$ a commonly used unbiased and consistent estimator of variance is
$$\hat \sigma^2=\frac{\sum_ix_i^2 + n(\bar x)^2}{n-1}=\frac{\sum_i(x_i-\bar x)^2}{n-1}$$
However, suppose we know the population (true) mean, should the estimator be adjusted to reflect that, meaning should my estimator instead be:
$$\hat \sigma^2=\frac{\sum_ix_i^2 + n(\mu)^2}{n-1}=\frac{\sum_i(x_i-\mu)^2}{n-1}$$
?
 A: Your estimated variance in the case where the population mean $\mu$ is known, has
the wrong denominator (if it is to be unbiased). The correct denominator of the unbiased estimate $\hat{\sigma^2}$ of $\sigma^2$ is especially easy to verify in the case of normal data.
To begin, $Z_i = \frac{X_i -\mu}{\sigma} \sim \mathsf{Norm}(0,1).$
So, $Q_i = Z_i^2 = \frac{(X_i - \mu)^2}{\sigma^2} \sim \mathsf{Chisq}(\nu=1)$
and $Q = \sum_{i=1}^n Z_i^2 = \sum_{i=1}^n \frac{(X_i - \mu)^2}{\sigma^2}
\sim \mathsf{Chisq}(\nu = n),$ which has $E(Q) = n.$
Thus (upon multiplying by $\sigma^2/n)$ we have $V =\hat{\sigma^2} = \frac{1}{n}\sum_{i=1}^n (X_i - \mu)^2$ has
$E\left(\hat{\sigma^2}\right) = \sigma^2.$
In the (most common) case where data are normal and $\mu$ is unknown (estimated by $\bar X),$ the relationship
$\frac{(n-1)S^2}{\sigma^2} \sim \mathsf{Chisq}(n-1),$
where $S^2 = \frac{1}{n-1}\sum_{i=1}^n(X_i-\bar X)^2,$ is important
for making confidence intervals for $\sigma^2$ and testing hypotheses about $\sigma^2.$
Similarly, when data are normal and $\mu$ is known , the relationship
$\frac{nV}{\sigma^2} \sim \mathsf{Chisq}(n)$
is important
for making confidence intervals for $\sigma^2$ and testing hypotheses about $\sigma^2.$
