```Date: Feb 15, 2013 10:50 PM
Author: quasi
Subject: Re: Variance of the recursive union of events

Paul wrote:>Paul wrote:>>>> Using a simplification of the notation in the paper, >> consider variance of the recursive relationship:>>>> 0) c(n) = u(n) + c(n-1) - c(n-1) u(n)>>>> for n=1,2,... and c(0)=0.  All c(n) and u(n) values represent>> probabilities i.e. lie with [0,1].  Furthermore, in the above>> expression (0), u(n) and c(n-1) are independent.>>Actually, consider:>>   U(n) = event to which probability u(n) is assigned>   C(n-1) = event to which probability c(n-1) is assigned>   C(n) = union[ U(n) , C(n-1) ]>>It is the *events* U(n) and C(n-1) that are independent, not >the probabilities u(n) and c(n-1).>>> In evaluating the variance of (0), the indices are rather>> meaningless, as we are completely focused on the right hand >> side of the equation.  I only include them in case a reader >> has access to the paper.  The variance of (0) is presented as:>>>> 1) var c(n) = var u(n) + var C(n-1) + var[ c(n-1) u(n) ]>>                          - 2 cov[ u(n) , c(n-1) ]>>                          - 2 cov[ c(n-1) , c(n-1) u(n) ]>>>> According to the above wikipedia page, however, it should be:>>>> 2) var c(n) = var u(n) + var C(n-1) + var[ c(n-1) u(n) ]>>                          - 2 cov[ u(n) , c(n-1) ]>>                          - 2 cov[ u(n) , c(n-1) u(n) ]>>                          - 2 cov[ c(n-1) , c(n-1) u(n) ]>>>> Since u(n) and c(n-1) are independent, their covariance disappears,>> so (2) becomes:>>>> 3) var c(n) = var u(n) + var C(n-1) + var[ c(n-1) u(n) ]>>                          - 2 cov[ u(n) , c(n-1) u(n) ]>>                          - 2 cov[ c(n-1) , c(n-1) u(n) ]>>>> This still differs from (1).  It is plausible that (1) is a >> typo,though not all that likely.Actually, it's very likely.I think you should _assume_ it's a typo, correct it, and see if the corrected version is consistent with the rest of the paper.quasi
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