I have a follow-up question to this one:


[unbiased estimate of the variance of a weighted mean](https://mathoverflow.net/questions/11803/unbiased-estimate-of-the-variance-of-a-weighted-mean)


Specifically, how do I generalise the result given here (and on Wikipedia) for the unbiased sample estimate of the variance of a weighted population to the case where the weights are not normalised to 1? (or equivalently are not in the standard simplex, as in the previous question's answer derivation)


I'm not sure how much of the previous answer relied on the weights being in the unit simplex, but it's clear that the given answer contains denominator terms like $1 - \sum\_i w\_i^2$ which aren't going to be nice if $\sum\_i w\_i^2 > 1$! Maybe there's a simple ansatz for modification to unnormalized weights, but it's not obvious to me which to choose!


Thanks!


Andy



Unbiased estimate of the variance of an *unnormalised* weighted mean
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


