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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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 RGVickson@shaw.ca Posts: 1,677 Registered: 12/1/07
Re: Variance of the recursive union of events
Posted: Feb 16, 2013 1:18 PM

On Friday, February 15, 2013 9:06:38 PM UTC-8, Paul wrote:
> 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) ]
>
> >>>
>
> >>> 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.
>
>
>
> 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.

The sure-fire way to compute variance of a random variable Y is as VarY = E(Y^2) - (EY)^2. In your case, look at Y = X + U - U*X (using Y instead of c(n), U instead of u(n) and X instead of u(n-1)).

You can compute E(Y^2) - (EY)^2 using the following (where we use VU, MU, VX, MX as Var(U), EU, Var(X) and EX, respectively) and apply the following:
E(X^2) = VX + MX^2, E(U^2) = VU + MU^2, E(U^2*X^2) = E(U^2)* E(X^2),
E(X*U) = MX*MU, E(X*U^2) = MX*E(U^2), E(X*U^2) = MX*E(U^2). Finally, you will end up with
Var(Y) = VX + VU + VX*VU + VX*MU^2 + MX^2 * VU - 2*MX*VU - 2*VX*MU.

One can also apply the results that Cov(A,B) = E(A*B) - EA * EB to the paper's expression and to yours, and compare the results with Var(Y) above. Yours works and the paper's fails.

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