Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

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

 Messages: [ Previous | Next ]
 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(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.
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) ]

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