```Date: Feb 16, 2013 1:18 PM
Author: RGVickson@shaw.ca
Subject: Re: Variance of the recursive union of events

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 withVar(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.
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