Variance of exponential distribution hooked up in parallel
- To: mathgroup at smc.vnet.net
- Subject: [mg36098] Variance of exponential distribution hooked up in parallel
- From: b90401114 at ms90.ntu.edu.tw (Sam Yang)
- Date: Wed, 21 Aug 2002 05:52:06 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
I do not use English as mother tongue. If my phrases is strange,
please correct them. I will be grateful to you.
I have a problem:
n component are arranged in parallel, and each of them has a failure
rule given by an exponential distribution with parameter lamba and
fuctions idependently. That is, the whole system's lifetime, denoted
by T, have the cumulative distribution function (1-e^(-lamba*t))^n,
prove that the variance of T,
Variance[T]=(1/lamda^2)*Sum[(1/i^2),{i,1,n}]
It's not difficult to calculate E[T] and E[T^2], but I can't figure a
way to trasform E[T^2]-(E[t])^2 into the simple form as above.
Thanks very much.