Poisson Process: Computers in a lab fail, on average, twice a day, according to a Poisson process

Question

Computers in a lab fail, on average, twice a day, according to a Poisson process. Last week, 10 computers failed. Find the expected time of the last failure, and give an approximate time of day when the last afilure occured.

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I'm seriously struggling with this and I feel that it may be straight forward but I'm not sure that I'm conceptualizing it correctly. Below is what I tried but didn't derive the correct answer. Please forgive me if this is totally off.

I let $S_{10}$ be the time of the 10th failure,

$\lambda = 2$ be the rate of failure,

$N_7 = 10$ be the event that there were 10 failures by time 7.

I then write,

$E(S_{10} | N_7 = 10) = \frac{1}{P(N_7 = 10)} \int_0^{10} t f_{S_{10}}(t) dt$

where

$f_{S_{10}}(t)$ is the Gamma distribution since arrival time follows a Gamma distribution for Poisson processes.

EDIT: This gives me 15.0121. But everything is in days so I know the 10th arrival can't be on day 15.

Answer 1

Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda=2$. Let $T_0=0$ and define the jump times by $$T_{n+1}=\inf\{t>T_n: N(t)>N(T_n)\}.$$ Conditioned on $\{N(7)=10\}$, the joint distribution of the order statistics $\left(T_{(1)}, T_{(2)},\ldots, T_{(10)}\right)$ is the same as that of $\left(U_{(1)},U_{(2)},\ldots,U_{(10)}\right)$ where $U_1,\ldots, U_{10}\stackrel{\mathsf{i.i.d.}}\sim \mathsf{Unif}(0,7)$. The distribution $\tau$ of the $10^{\mathsf{th}}$ failure is given by \begin{align} \mathbb P(\tau \leqslant t) &= \mathbb P\left(\bigcap_{j=1}^{10} \{U_j\leqslant t\}\right)\\ &= \prod_{j=1}^{10}\mathbb P(U_j\leqslant t)\\ &= \mathbb P(U_1\leqslant t)^{10}\\ &= \left(\frac t7\right)^{10},\quad 0\leqslant t\leqslant 7. \end{align} The expected time of the last failure is then $$ \int_0^7\mathbb P(\tau>t)\ \mathsf dt = \int_0^7 \left(1-\left(\frac t7\right)^{10}\right)\ \mathsf dt = \frac{70}{11}\approx 6.36364. \\ $$ It follows that the last failure occurred approximately in the first half of the $7^{\mathsf{th}}$ day.