Assume we have a normal distribution with known mean $\mu$. How can we estimate the variance by sampling? The typical answer to this question is to use the unbiased sample variance estimator i.e. if the data points are indicated by $x_1, \ldots x_n$ the following: $$\frac{(x_1-\bar{x})^2+\ldots +(x_n-\bar{x})^2}{n-1}$$ Where $\bar{x}$ is the sample mean. Now can we use the actual mean in any meaningful way to get a better estimator of the variance? The first thing that comes to mind would be to replace $\bar{x}$ by $\mu$. This is also going to be unbiased, but is it a better estimator? why?

Assume we have a normal distribution with known mean $\mu$. How can we estimate the variance by sampling? The typical answer to this question is to use the unbiased sample variance estimator i.e. if the data points are indicated by $x_1, \ldots x_n$ the following: $$\frac{(x_1-\bar{x})^2+\ldots +(x_n-\bar{x})^2}{n-1}$$ Where $\bar{x}$ is the sample mean. Now can we use the actual mean in any meaningful way to get a better estimator of the variance? The first thing that comes to mind would be to replace $\bar{x}$ by $\mu$ and divide it by $n$ instead of $n-1$ (to keep it unbiased). Is this a better estimator? why?