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Sorry if the question is too obvious or strange, I'm learning about Poisson distributions by myself.

Say I have some independent process that follows a Poisson distribution of unknown rate (10 particles in 1D that have to get to a specific position where the process finishes) and that I know the mean time at which the first particle gets to that position (I run simulations and I can identify the first time a particle reaches that position, for example). How can I estimate the mean time at which the 10 particles have arrived to that position?

Thanks a lot for any help! I thought it was just 10*time of the first one but that doesn't seem right, does it?

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  • $\begingroup$ Not clear what you mean. A poisson process is a counting process - what is being counted? Time implies a continuous variable - is this time an exponential process of some sort? $\endgroup$
    – Paul
    Commented Apr 7, 2016 at 15:47
  • $\begingroup$ Thanks for your comments! I'm counting how many particles get to the finish line. Time is the usual time, I'm measuring how much time passes until the different particles get to the finish line. I know that the first particle gets there in a mean time of, say 10 seconds. I want to know how long I have to wait until every particle got to the finish line $\endgroup$
    – Sebasen
    Commented Apr 7, 2016 at 15:52
  • $\begingroup$ Let the parameter be $\lambda$. The minimum time has exponential distribution mean $1/(n\lambda)$, so if we know that we know $\lambda$. Then in principle we can find the mean of the maximum. It is a moderately standard result that the maximum has mean $\frac{1}{\lambda}\left(1+\frac{1}{2}+\cdots +\frac{1}{n}\right)$. If you can run simulations for min then you can run simulations for max. Your results should be pretty close to the theoretical mean for the max given above. $\endgroup$ Commented Apr 7, 2016 at 15:53
  • $\begingroup$ Thanks @AndréNicolas! I'm guessing n is the number of elements, right? So, if I have 10 elements and I know that the first element that gets to the finish line does so in 10 seconds, \lambda would be 1/100 and the mean of the maximum would be 100(1+1/2+...+1/10)? Are you familiar with any book that may teach me about this? $\endgroup$
    – Sebasen
    Commented Apr 7, 2016 at 15:56
  • $\begingroup$ Yes, the $n$ is $10$, and the mean of the max is what you wrote. $\endgroup$ Commented Apr 7, 2016 at 16:00

2 Answers 2

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Let the parameter of the Poisson be $\lambda$. Then the distribution of the minimum arrival time is exponential with mean $\frac{1}{10\lambda}$. If we know that then we know $\lambda$.

To finish, we use the moderately standard result that the mean of the maximum of $n$ independent exponentially distributed random variables with parameter $\lambda$ is $$\frac{1}{\lambda}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right).$$

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  • $\begingroup$ @JKnecht: Not an expert, computed the mean of the maximum once upon a time and remembered doing it. $\endgroup$ Commented Apr 7, 2016 at 16:21
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Let $T_n$ be the time of the $n$-th arrival of the poisson process $X(t)$ with rate $\lambda$. Where $\lambda$ is $1/(\text{simulated mean time of first particle})$.

$T_n = Z_1 + Z_2 + \cdots + Z_n$

where $Z_n,\: n=1,2...$ are the interarrival times.

It is a straight forward exercise to show that $Z_n$ are iid exponential random variables with parameter $\lambda$. Note that an exponential random variable with parameter $\lambda$ is a gamma random variable with parameters $(1,\lambda)$.

It is another straight forward excercise to show, by induction, that the sum $T_n$ is a gamma random variable with parameter $(n, \lambda)$.

Hence $T_n$ has a pdf given by

$$ f_{T_n}(t) = \begin{cases} \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!}, & t > 0 \\ 0, & t < 0 \end{cases} $$

Use the gamma pdf $f_{T_{10}}(t)$ to calculate the mean time at which the 10 particles have arrived at that position.


You can show that the $Z_n's$ are iid exponentials like this:

$$P(Z_1 > t) = P\{X(t) = 0\} = e^{-\lambda t}$$

$$F_{Z_1}(t) = P(Z_1 \leq t) = 1-e^{-\lambda t}$$

Hence, $Z_1$ is an exponential r.v.

Let $f_1(t)$ be the pdf of $Z_1$.

Then we have that

$$ \begin{align} P(Z_2 > t) & = \int P(Z_2 > t | Z_1 = \tau)f_1(\tau) d \tau \\ & \\ & = \int P[X(t + \tau) - X(\tau)=0]f_1(\tau) d \tau \\ & \\ & = e^{-\lambda t}\int f_1(\tau) d \tau \\ & \\ & = e^{-\lambda t} \end{align} $$

which indicates that $Z_2$ is also an exponential r.v. with parameter $\lambda$ and is independent of $Z_1$. Repeating the same argument, you can conclude that $Z_1, Z_2, ....$ are iid exponential r.v.'s with parameter $\lambda$.

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