# Thread: Another Poisson process question

1. ## Another Poisson process question

You and N of your friends are meeting for dinner, where N is a Poisson random variable with $\lambda=10$. All of you arrive independently and according to a uniform distribution over (0,1). Find the expected number of friends that arrive before you.

my book has an example, but they skip a lot of steps, so all I have is:

$\int_0^1 E[N](t) \cdot 1 dt$ and they conclude that it's $\frac{\lambda}{2}$ but I'm a little lost at picking my E[N](t)

2. Originally Posted by lllll
You and N of your friends are meeting for dinner, where N is a Poisson random variable with $\lambda=10$. All of you arrive independently and according to a uniform distribution over (0,1). Find the expected number of friends that arrive before you.

my book has an example, but they skip a lot of steps, so all I have is:

$\int_0^1 E[N](t) \cdot 1 dt$ and they conclude that it's $\frac{\lambda}{2}$ but I'm a little lost at picking my E[N](t)
The proibability that there are $N$ friends involved is $P(N,\lambda)$, if you arrive at $t \in (0,1)$, then probability than $n$ arrive before you is $B(n;N,t).$

So:

$
E(n)=\int_{t=0}^1 \left[ \sum_{N=0}^{\infty} P(N,\lambda) \left(\sum_{n=0}^N n B(n;N,t)\right) \right] \ dt=\int_{t=0}^1 \left[ \sum_{N=0}^{\infty} P(N,\lambda) (N t) \right] \ dt
$

CB

3. Originally Posted by lllll
You and N of your friends are meeting for dinner, where N is a Poisson random variable with $\lambda=10$. All of you arrive independently and according to a uniform distribution over (0,1). Find the expected number of friends that arrive before you.

my book has an example, but they skip a lot of steps, so all I have is:

$\int_0^1 E[N](t) \cdot 1 dt$ and they conclude that it's $\frac{\lambda}{2}$ but I'm a little lost at picking my E[N](t)
CaptainBlack's answer is a good one. What I would like to do is explain your book's answer, because it uses important properties of Poisson point processes (cf. the title of your post...).

The hypothesis about the number and the arrival times of the friends is equivalent (this is a theorem) to saying that the friends arrive according to a Poisson point process of intensity $\lambda$, stopped at time 1.

Then, the number $N_t$ of friends that arrive before time $t\in[0,1]$ equals the number of points of the point process before $t$. As a consequence, $N_t$ is a Poisson random variable of parameter $\lambda t$, and $E[N_t]=\lambda t$.

Finally, let us denote by $T$ your arrival time (uniformly distributed on $$[0,1]$$). Because of the independence between you and your friends, the expected number of friends that arrive before you is $E[N_T]=\int_0^1 E[N_t]dt=\int_0^1 \lambda t\,dt=\frac{\lambda}{2}$.