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

 Messages: [ Previous | Next ]
 Paul Posts: 517 Registered: 2/23/10
Re: Variance of the recursive union of events
Posted: Feb 16, 2013 12:06 AM

On Feb 15, 10:50 pm, quasi <qu...@null.set> wrote:
>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.

The problem is that this is only a drive-by perusal of the paper,
since it is plumbing far deeper than I can rationalize for the results
that I'll be using (which occur in a paper that is 2 citations removed
from this one). I'd love to be able to take the time and become
familiar with the weird and wonderful PDFs and to program the Monte
Carlo simulation that follows the above steps, but it's just not in
the cards right now. I was hoping that key derivation steps would
make sense, and I could mentally give it the green label of "Yup,
seems quite reasonable, onward ho". Even that's going to take some
time considering the new territory (for me) that it wanders into, so I
was hoping to get a confirmation on the above logic thus far.

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