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 Wi which sum to 1. Suppose you have weights Ui, and write V1 = sum of the Ui, and V2 = sum of the Ui^2, consistent with the Wikipedia entry for weighted sample variance. Then we can put Wi = Ui/V1.

Now, look at the factor 1 / (1 - sum(Wi^2)), replace the Wi with Ui/V1, multiply top and bottom lines by V1^2 and - voila! - you get V1^2 / { V1^2 - V2 } .

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

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