1. ## probability - variance

Hi,

I'm pretty stuck with the following question, any help would be appreciated

A system is composed of 5 subsystems, each of them has a random independent lifetime, exponentially distributed with mean = 1.
The system works only if all the subsystems are working.
What is the variance of the whole system's lifetime?

Thanks

2. Originally Posted by maria clara
The system works only if all the subsystems are working.
This means that the lifetime of the whole system is the minimum of the lifetimes of the subsystems.
Do you know the distribution of the minimum of two independent exponentially distributed random variables?

3. well... I calculated it now.
if X~exp(1) and Y~exp(1) and Z = min(X,Y)
then the probability density function of Z is 2exp(-2z) for z>0 and 0 otherwise.
So it's also exponential with parameter=2.
does that mean that the minimum of the five would be also exponential with a parameter of 5? and then the answer should be 1/25..

but as far as I recall, the sum of two exponentials is an exponential with a parameter that equals the sum of parameters, is it true? does it mean that minimum and sum of exponentially distributed random variables are equivalent?...

thanks again!

4. Your computation is right. More generally, if $\displaystyle X$ and $\displaystyle Y$ are exponential r.v. of parameters $\displaystyle \lambda$ and $\displaystyle \mu$, then $\displaystyle \min(X,Y)$ is an exponential r.v. with parameter $\displaystyle \lambda+\mu$. If you apply this 4 times (first, the min of the lifetimes of the first two subsystems, then the minimum of the first two and the third,...), you get exactly what you guessed: the lifetime is an exponential r.v. of parameter 5.

As for the sum, be careful, it is called a gamma distribution, and not exponential.

5. thanks a lot!

and regarding the sum - I think I confused the exponential distribution with poisson, where the rule does hold...