Question

Compute the expected system lifetime of a three-out-of-four system when the first two component lifetimes are uniform on \((0,1)\) and the second two are uniform on \((0,2)\)

Short answer

The expected system lifetime of the three-out-of-four system, with first two component lifetimes uniform on (0,1) and second two component lifetimes uniform on (0,2), is \(\frac{5}{3}\).

Step-by-step solution

Find the pdf of the component lifetimes

First, we need to find the probability density functions (pdfs) of the individual component lifetimes. Since the first two components are uniformly distributed on (0,1), their pdfs can be given as: f_1(x) = f_2(x) = \[ \begin{cases} 1, & 0 < x \leq 1 \\ 0, & \mathrm{otherwise} \end{cases} \] For the second two components, which are uniformly distributed on (0,2), their pdfs can be given as: f_3(y) = f_4(y) = \[ \begin{cases} \frac{1}{2}, & 0 < y \leq 2 \\ 0, & \mathrm{otherwise} \end{cases} \]

Find the pdf for each case

Now, let's find the pdf for each of the three cases: Case 1: One of the first two components and two of the second two components fail. Using order statistics, we can calculate the pdf of the minimum of these three components as: h_1(x, y) = 2(1)(\frac{1}{2})^2 = y Case 2: Two of the first two components and one of the second two components fail. Using the same method, we can find the pdf of the minimum of these three components as: h_2(x, y) = \binom{2}{2}(\frac{1}{2})^2 = \frac{1}{4} Case 3: Two of the second two components fail and one of the first two components fails. The pdf of the minimum of these three components is: h_3(x, y) = 2(1)(\frac{1}{2})^2 = y

Compute the expected values for each case

Using the pdfs from the previous step, we'll now compute the expected values for each case through integration: E[Case 1] = \(\int_0^1 \int_0^2 xh_1(x, y) dy dx = \int_0^1 \int_0^2 xy dy dx = \frac{2}{3}\) E[Case 2] = \(\int_0^1 \int_0^2 xh_2(x, y) dy dx = \int_0^1 \int_0^2 \frac{1}{4}x dy dx = \frac{1}{3}\) E[Case 3] = \(\int_0^1 \int_0^2 xh_3(x, y) dy dx = \int_0^1 \int_0^2 xy dy dx = \frac{2}{3}\)

Calculate the overall expected system lifetime

Sum up the expected values for these three cases to get the expected system lifetime: Expected System Lifetime = E[Case 1] + E[Case 2] + E[Case 3] = \(\frac{2}{3} + \frac{1}{3} + \frac{2}{3}= \frac{5}{3}\) So the expected system lifetime of this three-out-of-four system is \(\frac{5}{3}\).