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

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 Paul Posts: 493 Registered: 2/23/10
Variance of the recursive union of events
Posted: Feb 15, 2013 8:30 PM
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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(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.

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.

Date Subject Author
2/15/13 Paul
2/15/13 Paul
2/15/13 quasi
2/16/13 Paul

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