Calculating the variance of sample, knowing the mean of population Suppose that I somehow know the mean of the population. And I want to calculate the variance of a sample.
Should I subtract population mean or sample mean?
Is there any situation in which I should use population mean?
 A: No one would like to use sample estimator, when population parameter is known. So, if population mean is given, then use population mean, but then use $n$ in denominators instead of $n-1$ because no degree of freedom will be lost in this case. 
So, sample variance (when population mean is not known)
$$s^2 = \frac{1}{n-1}\sum_{i = 1}^{n}\left(X_i - \bar X \right)^2$$
So, sample variance (when population mean is known)
$$s^2 = \frac{1}{n}\sum_{i = 1}^{n}\left(X_i - \mu \right)^2$$
where, $\mu$ is population mean. 
A: The true answer is vague: it depends on your prior model of the data. If you assumed the data to be normally distributed with a known mean, you could use the gamma distribution as the conjugate prior for Bayesian inference. Maybe use moment-matching estimates rather than using optimizing procedures for simplicity.
The Bayesian approach might sound like overkill in such a simple context, but frequentist statistics depend a lot more on best-practices-like procedures that don't extend in a simple fashion to nonstandard problems like the one you're posing.
