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