Consider estimating the variance of a RV $X$, we start with the sample variance:
$$ \begin{array}{ll} V_1 & = \frac{1}{N-1} \sum_{i=1}^N (X_i - \bar{X})^2\\ & = \frac{1}{N-1} \left(\sum_{i=1}^N X_i^2 - 2\bar{X}\sum_{i=1}^N X_i + N\bar{X}^2 \right)\\ &= \frac{1}{N-1} \left(\sum_{i=1}^N X_i^2 - N\bar{X}^2\right) \end{array} $$
Where $\bar{X} = N^{-1}\sum_{i=1}^N X_i$.
My question is, imagine we had a better estimate of the population mean: $\hat{X}$ that was guaranteed to have lower variance than the sample mean $\bar{X}$. Could it be used to obtain an improved estimate of the population variance? Also, notice how the above derivation no longer holds:
** Correction: previous question used $\frac{1}{N-1}$ which is only need if the sample mean is used. ** $$ \begin{array}{ll} V_2 & = \frac{1}{N} \sum_{i=1}^N (X_i - \hat{X})^2\\ & = \frac{1}{N} \left(\sum_{i=1}^N X_i^2 - 2\hat{X}\sum_{i=1}^N X_i + N\hat{X}^2 \right)\\ &\ne \frac{1}{N} \left(\sum_{i=1}^N X_i^2 - N\hat{X}^2\right) \end{array} $$
Unless $\bar{X} = \hat{X}$. Also, is there a way to measure the difference between $var(V_1)$ and $var(V_2)$ as a function of $var(\bar{X})$ and $var(\hat{X})$?
Edit: Let me further motivate why this might be useful. Imagine we have $K$ different estimators of the same mean, for simplicity imagine each of these are importance sampling estimators each with a different proposal distribution. So each of these estimators has a very different variance, and I want to know which estimator is the best one (or likely to be).
We could take $N$ samples from each estimator and estimate the sample variance for each one using the estimator $V_1$. Alternatively, we could use the samples from all of them together to get an improved estimate of the population mean and use $V_2$ instead. I would expect this would be advantageous if $K$ is very large (like $K = N^2$).
Consider a simple illustration, where one estimator has extremely high variance and returns $0$ in almost every single sample except with very low probability returns a huge value (enough to make it an unbiased estimate still). The sample variance ($V_1$) for this estimator is likely to be zero for any reasonable $N$, where $V_2$ would indicate that the estimator is actually quite poor.