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




Re: Variance of the recursive union of events
Posted:
Feb 16, 2013 1:18 PM


On Friday, February 15, 2013 9:06:38 PM UTC8, 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(n1)  c(n1) 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(n1) are independent. > > >> > > >> Actually, consider: > > >> > > >> U(n) = event to which probability u(n) is assigned > > >> C(n1) = event to which probability c(n1) is assigned > > >> C(n) = union[ U(n) , C(n1) ] > > >> > > >> It is the *events* U(n) and C(n1) that are independent, not > > >> the probabilities u(n) and c(n1). > > >> > > >>> 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(n1) + var[ c(n1) u(n) ] > > >>>  2 cov[ u(n) , c(n1) ] > > >>>  2 cov[ c(n1) , c(n1) u(n) ] > > >>> > > >>> According to the above wikipedia page, however, it should be: > > >>> > > >>> 2) var c(n) = var u(n) + var C(n1) + var[ c(n1) u(n) ] > > >>>  2 cov[ u(n) , c(n1) ] > > >>>  2 cov[ u(n) , c(n1) u(n) ] > > >>>  2 cov[ c(n1) , c(n1) u(n) ] > > >>> > > >>> Since u(n) and c(n1) are independent, their covariance > > >>> disappears, so (2) becomes: > > >>> > > >>> 3) var c(n) = var u(n) + var C(n1) + var[ c(n1) u(n) ] > > >>>  2 cov[ u(n) , c(n1) u(n) ] > > >>>  2 cov[ c(n1) , c(n1) 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 driveby 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 surefire 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(n1)).
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.



