I am reading Probability and Statistics for Engineering and the Sciences.
Exercise 15, Chapter 5 says:
Consider a system consisting of three components as pictured.
The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let $X_1$, $X_2$, and $X_3$ denote the lifetimes of components 1, 2, and 3, respectively. Suppose the $X_i$’s are independent of one another and each $X_i$ has an exponential distribution with parameter $\lambda$.
- Let Y denote the system lifetime. Obtain the cumulative
distribution function of Y and differentiate to obtain the
pdf. [Hint: $F(y) = P(Y \leq y)$; express the event $\{Y \leq y\}$
in terms of unions and/or intersections of the three events
$\{X_1 \leq y\}$, $\{X_2 \leq y\}$, and $\{X_3 \leq y\}$.]
It is not clear to me how to build the overall cumulative distribution function. I saw this answer but I don't understand how to use it to fit my problem.
I tried to compute the cumulative distribution function as:
$F(y) = P(X_1 \leq y) * (P(X_2 \leq y) + P(X_3 \leq y))$
but I think it is wrong.
Best Answer
We have $$Y = X_1\wedge(X_2\vee X_3)$$ and for $y\geqslant 0$, \begin{align} \mathbb P(X_2\vee X_3\leqslant y) &= \mathbb P(X_2\leqslant y, X_3\leqslant y)\\ &= \mathbb P(X_2\leqslant y)\mathbb P(X_3\leqslant y)\\ &= (1-e^{-\lambda y})^2. \end{align} It follows that \begin{align} \mathbb P(X_1\wedge(X_2\vee X_3)>y) &= \mathbb P(X_1>y, X_2\vee X_3>y)\\ &= \mathbb P(X_1>y)\mathbb P(X_2\vee X_3 >y)\\ &= e^{-\lambda y}\left(1-\left(1-e^{-\lambda y}\right)^2\right), \end{align} and so the CDF of $Y$ is $$F_Y(y) = 1 - e^{-\lambda y}\left(1-\left(1-e^{-\lambda y}\right)^2\right),\ y\geqslant 0. $$