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Customers arrive at a service facility according to a Poisson process of rate $\lambda = 5$ customers/hour. Let N(t) be the number of customers that have arrived up to time t hours. Let $W_1,W_2,W_3,...$ be the successive arrival times of the customers.

a) Find the expected arrival time of the 7th customer $E[W_7]$

b) Given N(1) = 4, determine the expected arrival time of the 7th customer, $E[W_7|N(1)=4]$

so far I have calculated that each one customer arrives at a rate of $\frac{1}{5}$ and used that to calculate $E[W_7] = \sum_{1}^7\frac{1}{5} = \frac{7}{5}$

I am currently having trouble figuring out how to approach b). Any help would be appreciated

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1 Answer 1

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Hint: $$\require{cancel} \xcancel{\operatorname{E}[W_7 \mid N(1) = 4] = 4 + \operatorname{E}[W_6].}$$

Edit. I apologize--I misread the question. The notation $N(1) = 4$ means that we observed exactly four customers arriving within $1$ hour. Then this means we are waiting for another $3$ customers to arrive, so the correct relationship is $$\operatorname{E}[W_7 \mid N(1) = 4] = \color{red}{1} + \operatorname{E}[W_{\color{red}{3}}].$$ This is because the expected additional waiting time after one hour corresponds to having to wait to see $3$ more customers, which by the memorylessness property, is just $\operatorname{E}[W_3]$. Then we have to add back the $1$ hour that has already elapsed.

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  • $\begingroup$ Is it because we have to take into account the amount of time that has already passed, 4 people have arrived in an hour and due to the memoryless property. $E[W_7]$ depends on $E[W_6]$ and so $E[W_7|N(1)=4] = 4 + \frac{6}{5}$? $\endgroup$
    – waterr
    Commented Feb 8 at 18:15
  • $\begingroup$ @waterr The memoryless property is the key. Once you have observed $N(1) = 4$, then starting from time $t = 4$, the Poisson process doesn't "remember" what happened before, so the additional waiting time distribution for the next six events is the same as if we were looking at a "new" process. Then we have to remember to add back in the time we already waited to see the first event. $\endgroup$
    – heropup
    Commented Feb 8 at 18:56
  • $\begingroup$ I think I see what the logic behind it is, and so since we waited for 4 people to arrive, that would mean that $E[W_6] = \frac{2}{5}$ since in this case we are only waiting for 2 more to arrive because $W_i$ represents the arrival time of a customer and since 4 already arrived, we only need $E[W_5] = \frac{1}{5}$ and $E[W_6] = E[W_5} + \frac{1}{5}$? $\endgroup$
    – waterr
    Commented Feb 8 at 19:34
  • $\begingroup$ @waterr I am so very sorry. I just realized I misread the problem. $N(1) = 4$ means four people arrived in the first hour. Then the correct equation should be $$\operatorname{E}[W_7 \mid N(1) = 4] = 1 + \operatorname{E}[W_3].$$ $\endgroup$
    – heropup
    Commented Feb 8 at 20:49
  • $\begingroup$ It's alright, so by this equation, the answer would be $E[W_7|N(1)=4] = 1 + E[W_3] = 1 + \frac{3}{5} = \frac{8}{5}$ where we are waiting on 3 more customers and 4 have already arrived? $\endgroup$
    – waterr
    Commented Feb 8 at 21:21

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