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How are you, people? Before anything, let me tell you that my native language is spanish, and I never wrote about math in english, so please bear with me if I translate some terms incorrectly. I will try to keep it understandable.

Anyways, I need help with a certain problem. My teacher says that I did it wrong; I cannot really see why. She says that I cannot use the theorem I applied because the number of arrivals is unknown, which sounds really strange to me, but let me show you the statement of the problem:

Passengers arrive at a train platform following a Poisson process of intensity 30 per minute. The passengers start arriving at 4:00. Calculate the expected value of the total wait time of all the passengers if the train leaves at 4:15.

The statement was something like that; all the important info is there anyways; there were no other conditions. So, my solution is the following:

Let $T_i$ be the time of arrival of the $i$-th passenger, in minutes from 4:00. Let $N$ be the amount of arrivals between 4:00 and 4:15. Let $T$ be the total wait time, such that...

$$ T = \sum^N_{i = 1}\,(15 - T_i) $$

So far so good. Now what I did is the following. I condition by $N = n$ such that...

$$ E[\,T\,] = E[\,E[\,T\,|\,N = n\,]\,] = E[\,E[\,\sum^N_{i = 1}\,(15 - T_i)\,|\,N = n\,]\,]$$

Now there is a theorem that states that, given $n$ arrivals of a Poisson process in the time interval $[0, t]$, the arrival times are independent and identically distributed as uniform $[0, t]$ variables. This is where I supposedly messed up; my teacher says this is not valid since $n$ is unknown, and I had to use the Gamma. This sounds really strange to me; I mean, if the experiment was done, $N$ had to take a value, and the result I get doesn't really depend on the actual $n$, but the expected value of $N$, and I do not really see why the theorem will stop being valid because the amount of arrivals was not stated.

I proceed like this: by the previously stated theorem, this reduces to...

$$ E[\,T\,] = E[\,N(15 - E[T_i])\,] = E[\,N(15 - 7.5)\,] = 7.5\,E[\,N\,] $$

Now $N$ is the amount of arrivals of a Poisson process of intensity 30 per minute, with $t = 15$. So...

$$ E[\,T\,] = 7.5\,E[\,N\,] = 7.5 \times 30 \times 15 = 3375 $$

... and this is my final result, which makes sense to me. If, on average, 30 people arrive at the platform per minute, and on average, each one of them waits 7.5 minutes because all intervals are independent and no one is preferable over the others, then the average total wait time should be $7.5 \times 30 \times 15$.

Is my solution really wrong? If it is, please help me see why. Of course, any other details and comments are welcome.

Thank you very much for taking your time with this. Have a nice day.

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More precisely, given that $N=n$, the arrival times have the same distribution as the order statistics $U_{(1)}<U_{(2)}<\cdots<U_{(n)}$ of an iid sample $(U_1,U_2,\ldots,U_n)$ from the uniform(0,15) distribution. Because $\sum_{i=1}^n U_{(i)} =\sum_{i=1}^n U_i$, $$ \eqalign{ E\left(\sum_{i=1}^N(15-T_i)\Big|N=n\right)&=E\left(\sum_{i=1}^n(15-U_{(i)})\right)\cr &=E\left(\sum_{i=1}^n(15-U_{i})\right)=nE(15-U_1)\cr &=15n/2.\cr } $$ Your method is correct (if a little imprecisely expressed) and your answer is correct.

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  • $\begingroup$ Thank you very much! I need 15 reputation to up vote, however. $\endgroup$
    – learner
    Nov 6, 2014 at 18:27
  • $\begingroup$ "Your method is correct" Very far from it. $\endgroup$
    – Did
    Aug 9, 2017 at 8:53

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