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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.


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.

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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.

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  • $\begingroup$ Unfortunately that seems to be incorrect. My textbook is stating that the expected failure time was 8:43 AM on the last day of the week. I believe the issue is the interval given for $t$. Why do you say 14 instead of 7? $\endgroup$ – Nicklovn Apr 9 '18 at 14:35
  • $\begingroup$ I misread the problem as two weeks instead of one week. $\endgroup$ – Math1000 Apr 10 '18 at 18:14
  • $\begingroup$ This is correct! Thank you. The only part I don't understand is why do you use $P(\tau > t)$? $\endgroup$ – Nicklovn Apr 10 '18 at 20:21
  • $\begingroup$ Oh wait. I see there is a property about this! I may have heard of it a long time ago but I'd forgotten. For anyone else confused, this other question even proves it: stats.stackexchange.com/questions/10159/… Also: math.stackexchange.com/questions/64186/… $\endgroup$ – Nicklovn Apr 10 '18 at 21:16
  • $\begingroup$ Yes, that is known as the "Darth Vader rule." It's just another way of computing expectations. $\endgroup$ – Math1000 Apr 11 '18 at 1:06

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