The problem:
A system has 2 components: A and B. These components have independent lifetimes that are
exponentially distributed with parameters 2 and 3 respectively. (Recall an exponential
prob. density function with parameter $λ$ is $λe^{−λt}$. The system fails as soon as one component fails.
Part C asks the following:
(c) What is the mean time to failure for the system?
Hint: Define $Z = min(X, Y )$. Then the continuous distribution function of Z is
$F_Z(z) = P(Z ≤ z) = P(X ≤ z, Y ≥ z) + P(X ≥ z, Y ≤ Z) = F_X(z)(1 − F_Y (z)) + F_Y (z)(1 − F_X(z))$
The answer is given as $1/5$, which I cannot seem to be able to get. Can someone explain why I am setting $Z=min(X,Y)$? Where do i go once I've done what the hint says?
Best Answer
The event $Z \ge z$ occurs if and only if $X \ge z$ and $Y \ge z.$ So $$P[Z \ge z]=P[X \ge z]P[Y \ge z]$$
$$P[Z \ge z]=e^{- \lambda_1z} e^{- \lambda_2 z}=e^{-(\lambda_1 + \lambda_2)z}$$
Now it can be seen that $Z$ is itself an exponential random variable with parameter $\lambda_1 + \lambda_2.$ So it has mean $$E[Z]={1 \over {\lambda_1 + \lambda_2}}= {1 \over 5}.$$