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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:

$$\hat \sigma^2=\frac{\sum_ix_i^2 + n(\mu)^2}{n-1}=\frac{\sum_i(x_i-\mu)^2}{n-1}$$

?

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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 $\sigma^2.$

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    $\begingroup$ But then this means that the use of the true parameter $$\mu$$ has led to a minor reduction in estimator variance (practically no difference if n is very large) $\endgroup$
    – tvbc
    Jul 29, 2020 at 1:38
  • $\begingroup$ Knowing $\mu$ is useful information, and additional information often reduces variance. Somewhat similarly, if $\sigma$ is known and $\mu$ is not, then a z confidence interval is used to estimate $\mu.$ However, if $\sigma^2$ is also unknown and estimated by $S^2,$ then the (wider) t confidence interval is used for $\mu.$ $\endgroup$
    – BruceET
    Jul 29, 2020 at 1:44

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