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?
Best Answer
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 $\text{Var}(T_{(5)})$.