# Poisson Process: Finding the sum of interarrivial time

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 $E[T]$.

My attempt: Let $T_i$ be the interarrivial time between the $(i-1)$th failure and $i$th failure. Then we have $$T= \sum_{i=1}^5 T_i$$ which is quickly followed by $$E[T]= \sum_{i=1}^5 E[T_i].$$

I guess there is no problem above. I now claim that interarrivial time is independent exponential r.v.. Thus, I have $E[T_i] = \frac{1}{200}$, which differs from the book answer: $E[T_i] = \frac{200}{101-i}$. Can anyone point out my mistake?

The minimum of $n=100$ i.i.d exponential random variables with parameter $λ=\frac{1}{200}$ is again exponentially distributed with parameter $$n\cdotλ=\frac{100}{200}=\frac{1}{2}$$ Thus the expected time $E[T_1]$ until the first failure is equal to $$E[T_1]=\frac{1}{\frac{1}{2}}=2$$ Now after the first failure there remain $n_1=99$ i.i.d. exponential random variables and their minimum is again exponentially distributed with parameter $$n_1\cdotλ=\frac{100-1}{200}$$ Therefore the expected time $E[T_2]$ between the first and the second failure is equal to $$E[T_2]=\frac{1}{\frac{100-1}{200}}=\frac{200}{100-1}$$ Proceeding like this and due to linearity of expectation you obtain that the expected time $E[T]$ until the $5$th failure is equal to $$E[T]=\sum_{i=1}^{5}E[T_i]=\sum_{i=1}^{5}\frac{200}{100-(i-1)}=\sum_{i=1}^{5}\frac{200}{101-i}$$ as your book suggests. Now using calculator you can find that this sum is equal to $$E[T]=\frac{200}{101-1}+\ldots+\frac{200}{101-5}=10.2062$$
The key you are missing is that this is not a Poisson process because the items are simultaneously tested, not sequentially. So $T = X_{(5)}$, the fifth order statistic of $X_1, \ldots, X_{100}$ IID exponential random variables.
To find this, recall the formula for the PDF of the $k^{\rm th}$ order statistic of a set of $n$ IID random variables: $$f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} F_X(x)^{k-1} (1-F_X(x))^{n-k} f_X(x),$$ where $f_X(x) = \frac{1}{\mu} e^{-x/\mu}$ and $F_X(x) = 1-e^{-x/\mu}$ where $\mu = 200$ is the mean lifetime of an individual item. Then with $k = 5$, $n = 100$, we get $$\operatorname{E}[T] = \int_{x=0}^\infty x \cdot 1882188 e^{-x/2} \left(e^{x/200}-1\right)^4 \, dx = \frac{19210001}{1882188} \approx 10.2062.$$
Let $T_i$ denote the time until the $i^{\mathrm{th}}$ failure, $i=1,2,3,4,5$. Then $T_i$ is the minimum of $100-(i-1)$ exponential random variables with rate $\frac1{200}$. The distribution of $\min\{X,Y\}$ where $X\sim \mathrm{Exp}(\lambda)$ and $Y\sim \mathrm{Exp}(\mu)$ is $\mathrm{Exp}(\lambda + \mu)$, so the distribution of $T_i$ is Exp$\left(\frac{100-(i-1)}{200}\right)$. As $T=T_1 + T_2 + \cdots + T_5$ we have
$$\mathbb E[T] = \sum_{i=1}^5\mathbb E[T_i] = \sum_{i=1}^5 \frac{200}{100-(i-1)} = \frac{19210001}{1882188}.$$