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Topic:
Variance of the recursive union of events
Replies:
3
Last Post:
Feb 16, 2013 12:06 AM



quasi
Posts:
12,067
Registered:
7/15/05


Re: Variance of the recursive union of events
Posted:
Feb 15, 2013 10:50 PM


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(n1)  c(n1) 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(n1) are independent. > >Actually, consider: > > U(n) = event to which probability u(n) is assigned > C(n1) = event to which probability c(n1) is assigned > C(n) = union[ U(n) , C(n1) ] > >It is the *events* U(n) and C(n1) that are independent, not >the probabilities u(n) and c(n1). > >> 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(n1) + var[ c(n1) u(n) ] >>  2 cov[ u(n) , c(n1) ] >>  2 cov[ c(n1) , c(n1) u(n) ] >> >> According to the above wikipedia page, however, it should be: >> >> 2) var c(n) = var u(n) + var C(n1) + var[ c(n1) u(n) ] >>  2 cov[ u(n) , c(n1) ] >>  2 cov[ u(n) , c(n1) u(n) ] >>  2 cov[ c(n1) , c(n1) u(n) ] >> >> Since u(n) and c(n1) are independent, their covariance disappears, >> so (2) becomes: >> >> 3) var c(n) = var u(n) + var C(n1) + var[ c(n1) u(n) ] >>  2 cov[ u(n) , c(n1) u(n) ] >>  2 cov[ c(n1) , c(n1) 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



