How to model the probability of a system failing if any of its subsystems fails?

Question

Suppose I have a system that consists of $m$ independent subsystems. The work expectancy for each of these subsystems is a random variable, let's say $X_i$, where $i$ denotes the $i$-th subsystem. Each of these variables has the same exponential distribution (with, let's say $\lambda$ as the parameter. Now, my system only works as long as all the subsystems work.

Now, I want to model a random variable $Z$ such that it denotes the work expectancy of the system of a whole and find its CDF. How can I achieve this?

Answer 1

You are asking for the distribution of $Z = \min(Y_i)$. The system "dies" when the first subsystem "dies."

If the reliability function of $Y_i$ is $$1 - F_i(t) = R_i(t) = P(X_i > t),$$ then can you write the formula for the reliability function for $t > 0$? Next, do you see how to express the reliability function $R_Z(t)$ of $Z$ in terms of the $R_i(t)?$

The 'parameter' of the exponential distribution can be either the rate or the mean. Your use of $\lambda$ suggests you are using the rate, but you need to clarify that. It makes a difference when you write $P(X_i > t).$

Maybe start with the case $m = 2.$ You can find that much on this site by searching for page 580279. Then see how easy it is to generalize to $m$ subsystems.

Once you have the reliability curve of $Z$, use it to find the CDF of $Z$, and differentiate to get the density function of $Z$ (easy calculus).

As another clue I am going to show you the results of a simulation that has $m = 3$ independent subsystems, each with parameter $\lambda = 1$ (the only case in which it doesn't matter whether the parameter is mean or rate). The histogram shows the results of a million simulated experiments. The density curve is consistent with $E(Z) = 1/3$ and should be a special case of the density curve of $Z$ as you derive it.

 m = 10^6;  y1 = rexp(m);  y2 = rexp(m);  y3 = rexp(m)
 z = pmin(y1, y2, y3)
 mean(z)
 ## 0.3337452 # simulation of E(Z)
 sd(z)
 ## 0.3334918 # simulation of SD(Z)

Answer 2

I would say

$P(X_i>t)=e^{-\lambda t}$, random variables $ x_i $ are independent, $\Rightarrow P(X>t)=(e^{-\lambda t})^m=e^{-m\lambda t}$

$\Rightarrow P(X\le t) = 1-e^{-m\lambda t}$

See For rate = 3