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Topic:
Variance of the recursive union of events
Replies:
3
Last Post:
Feb 16, 2013 12:06 AM



Paul
Posts:
517
Registered:
2/23/10


Variance of the recursive union of events
Posted:
Feb 15, 2013 8:30 PM


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 with http://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(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.
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.
For someone who does this a lot, I imagine that the logic above is elementary. Thanks for any confirmation on the above.



