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Topic: Variance of the recursive union of events
Replies: 4   Last Post: Feb 16, 2013 1:18 PM

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 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(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) ]
>>
>>
>> 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

Date Subject Author
2/15/13 Paul
2/15/13 Paul
2/15/13 quasi
2/16/13 Paul
2/16/13 RGVickson@shaw.ca