Please see clarification below regarding independence of the input terms.
On Feb 15, 8:30 pm, Paul <paul.domas...@gmail.com> wrote: > I am studying a reliability paper "Analytical propagation of > uncertainties through fault trees" (Hauptmanns 2002). > Unfortunately, I cannot find an online copy to link to. > > The paper expresses the variance of the union of two events in a way > that doesn't seem to be consistent > withhttp://en.wikipedia.org/wiki/Variance#Weighted_sum_of_variables, > at least to my (rather novice) eyes. > > 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.
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. > > For someone who does this a lot, I imagine that the logic above is > elementary. Thanks for any confirmation on the above.