Hi,
Rather long after your question, but it can be done directly in the same way Matus did it, or you can simply use the following:

Matus assumed weights $W_i$ which sum to $1$. Suppose you have weights Ui, and write $V_1 = \sum U_i$, and $V_2 = \sum U_i^2$, consistent with the Wikipedia entry for weighted sample variance.
Then we can put $\displaystyle W_i = \frac{U_i}{V_1}$. 

Now, look at the factor $\displaystyle \frac{1} {(1 - \sum W_i^2)}$, replace the $W_i$ with $\displaystyle\frac{U_i}{V_1}$, multiply top and bottom lines by $V_1^2$ and - voila! - you get $\displaystyle \frac{V_1^2}{ V_1^2 - V_2 }$  .

However, like Matus, I'm wondering when you would ever use such a "weighted sample variance" - see my question as a response to the original post.

I suspect there is much confusion over the different reasons for weighting.

Kathy