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One hundred items are simultaneously put on a life test. Suppose the lifetimes of the individual items are independent exponential random variables with mean $200$ hours. The test will end when there have been a total of $5$ failures. If T is the time at which the test ends, find $E[T]$ and $Var(T)$.

I'm stuck in this exercise.

If $T$ is the time at which test ends, then $T$ is the time of fifth failure, suppose $T_i$ for $i=1,2,3,4,5$ are the times that the five failures occurred, how I can find the distribution of each $T_i$?

In a previous exercise I saw that the failure rate is $$r(t)=\frac{f(t)}{1-F(t)}$$ this means that failure time is a exponential random variable with parameter $r(t)$?

EDIT: The answer of @Did in this post rate parameter seems to have a relationship with what I'm asking, the failure time has no distribution?

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Roland
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1 Answers1

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For the first moment we can write \begin{align} \text{E}(T_{(5)}) =& \text{E}(T_{(1)}) + \\ & \text{E}(T_{(2)} - T_{(1)}) + \\ & \text{E}(T_{(3)} - T_{(2)}) + \\ & \text{E}(T_{(4)} - T_{(3)}) + \\ & \text{E}(T_{(5)} - T_{(4)}) . \end{align} The next observation is that $T_{(1)}$ is a minimum of $100$ exponential$(\lambda = 1/200)$ random variables and therefore is itself exponential with rate $100 / 200$. Whatever the value of $T_{(1)}$ happens to be, by the memoryless property $T_{(2)} - T_{(1)}$ has the same distribution as the minimum of $99$ exponential$(\lambda = 1/200)$ random variables and so has rate $99/200$. We can reason in the same way for the other terms to get \begin{align} \text{E}(T_{(5)}) &= \frac{200}{100} + \frac{200}{99} + \frac{200}{98} + \frac{200}{97} + \frac{200}{96} \\ &\cong 10.2 . \end{align}

You should be able to use this same strategy to get $\text{Var}(T_{(5)})$.

dsaxton
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  • I think I understand almost everything, why do you use the difference between times $T_{(i)}-T_{(i-1)}$ and not just $T_{(i)}$ the order statistics? – Roland Jul 20 '15 at 20:31
  • It just allows us to use some known properties of the exponential distribution and avoid explicitly finding the distribution of $T_{(i)}$ for $i > 1$. – dsaxton Jul 20 '15 at 20:36
  • I did not understand well this property you used in these differences – Roland Jul 20 '15 at 20:39
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    Exponential random variables are "memoryless," which means if you condition on them being larger than some value, the amount by which they exceed that value has the same distribution as the original random variable before conditioning. And if you take the minimum of $n$ exponentials with rates $\lambda_1, \ldots , \lambda_n$, that minimum is exponential with rate $\sum_{i=1}^{n} \lambda_i$. – dsaxton Jul 20 '15 at 20:42
  • @JohnSheridan BTW, when you compute the variance it's key to notice that we've written $T_{(5)}$ as a sum of independent increments. – dsaxton Jul 21 '15 at 00:20