# 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:

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

?

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 $$\sigma^2.$$

• 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)
– tvbc
Commented Jul 29, 2020 at 1:38
• 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 $\mu.$ Commented Jul 29, 2020 at 1:44