Poisson Distribution from first arrival?

Question

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?

Answer 1

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 $$\frac{1}{\lambda}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right).$$

Answer 2

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.

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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 $\lambda$.