For unweighted variance $$\text{Var}(X):=\frac{1}{n}\sum_i(x_i - \mu)^2$$ there exists the bias corrected sample variance, when the mean was estimated from the same data: $$\text{Var}(X):=\frac{1}{n-1}\sum_i(x_i - E[X])^2$$

I'm looking into weighted mean and variance, and wondering what the appropriate bias correction for the weighted variance is. Using: $$\text{mean}(X):=\frac{1}{\sum_i \omega_i}\sum_i \omega_i x_i$$

The "naive", non-corrected variance I'm using is this: $$\text{Var}(X):=\frac{1}{\sum_i \omega_i}\sum_i\omega_i(x_i - \text{mean}(X))^2$$

So I'm wondering whether the correct way of correcting bias is

A) $$\text{Var}(X):=\frac{1}{\sum_i \omega_i - 1}\sum_i\omega_i(x_i - \text{mean}(X))^2$$

or B) $$\text{Var}(X):=\frac{n}{n-1}\frac{1}{\sum_i \omega_i}\sum_i\omega_i(x_i - \text{mean}(X))^2$$

or C) $$\text{Var}(X):=\frac{\sum_i \omega_i}{(\sum_i \omega_i)^2-\sum_i \omega_i^ 2}\sum_i\omega_i(x_i - \text{mean}(X))^2$$

A) does not make sense to me when the weights are small. The normalization value could be 0 or even negative. But how about B) ($n$ is the number of observations) - is this the correct approach? Do you have some reference that shows this? I belive "Updating mean and variance estimates: an improved method", D.H.D. West, 1979 uses this. The third, C) is my interpretation of the answer to this question: http://mathoverflow.net/questions/22203/unbiased-estimate-of-the-variance-of-an-unnormalised-weighted-mean

All three of them seem to "survive" the sanity check of setting all $\omega_i=1$. So which one should I used, under which premises?