Unbiased estimate of the variance of an *unnormalised* weighted mean - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:58:48Z http://mathoverflow.net/feeds/question/22203 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22203/unbiased-estimate-of-the-variance-of-an-unnormalised-weighted-mean Unbiased estimate of the variance of an *unnormalised* weighted mean Andy Buckley 2010-04-22T15:39:45Z 2011-04-21T10:22:13Z <p>I have a follow-up question to this one:</p> <p><a href="http://mathoverflow.net/questions/11803/unbiased-estimate-of-the-variance-of-a-weighted-mean" rel="nofollow">http://mathoverflow.net/questions/11803/unbiased-estimate-of-the-variance-of-a-weighted-mean</a></p> <p>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)</p> <p>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!</p> <p>Thanks!</p> <p>Andy</p> http://mathoverflow.net/questions/22203/unbiased-estimate-of-the-variance-of-an-unnormalised-weighted-mean/59403#59403 Answer by Kathy for Unbiased estimate of the variance of an *unnormalised* weighted mean Kathy 2011-03-24T08:28:54Z 2011-03-24T08:28:54Z <p>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:</p> <p>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. </p> <p>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 } .</p> <p>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.</p> <p>I suspect there is much confusion over the different reasons for weighting.</p> <p>Kathy</p>