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Chapter 5: Q16E (page 345)

Suppose that five components are functioning simultaneously, that the lifetimes of the components are i.i.d., and that each life has the exponential distribution with parameter\(\beta \). Let\({T_1}\)denote the time from the beginning of the process until one of the components fails, and let\({T_5}\)denote the total time until all five components have failed. Evaluate\(Cov\left( {{T_{1,}}{T_5}} \right)\)

Short Answer

Expert verified

\(Cov\left( {{T_1},{T_5}} \right) = \frac{1}{{{5^2}}}\frac{1}{{{\beta ^2}}}\)

Step by step solution

01

Given information.

Five components are functioning simultaneously. The lifetime of the components are independent and identically distributed, following exponentially distributed with parameter\(\beta \).

\({T_1}\)be the time from the beginning of the process until one of the components fails;

\({T_5}\)be the total time until all five components have failed.

Hence\({T_5} = {T_1} + F\)

F is the time required after the component has failed pending the other four components failed.

02

 Step 2: Computing the covariance

\(\begin{aligned}{}Cov\left( {{T_1},{T_5}} \right) &= Cov\left( {{T_1},{T_1} + F} \right)\\ &= Cov\left( {{T_1},{T_1}} \right) + Cov\left( {{T_1},F} \right)\end{aligned}\)

We know that\(Cov\left( {{T_1},{T_1}} \right)\)is equal to the variance of\({T_1}\).

Also\(Cov\left( {{T_1},F} \right) = 0\)

Since there is no relation between the time from the beginning of the process until one of the components fails and the time required after one of the components has failed pending the other four components failed.

Therefore, we get

\(\begin{aligned}{}Cov\left( {{T_1},{T_5}} \right) &= Cov\left( {{T_1},{T_1}} \right) + Cov\left( {{T_1},F} \right)\\ &= Var\left( {{T_1}} \right) + 0\\ &= \frac{1}{{{5^2}Var\left( {{T_1}} \right)}}\end{aligned}\)

\(Cov\left( {{T_1},{T_5}} \right) = \frac{1}{{{5^2}}}\frac{1}{{{\beta ^2}}}\)

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