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Paul
Posts:
208
Registered:
2/23/10
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Re: Variance of the recursive union of events
Posted:
Feb 15, 2013 8:53 PM
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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.
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.
>
> For someone who does this a lot, I imagine that the logic above is
> elementary. Thanks for any confirmation on the above.
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