# Poisson process on R^2

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• December 9th 2009, 05:05 PM
kingwinner
Poisson process on R^2
Consider a Poisson process of rate λ on R^2. Let N(r) be the number of points in a circle of radius r centered at the origin and Y2 be the distance from the origin to the 2nd closest point.
Calculate E(Y2) and E[Y2 |N(1)=10].

I have no problem with computing E(Y2), but I am stuck with calcuating E[Y2 |N(1)=10].

So my first step is to try to compute the tail distribution function P(Y2>y |N(1)=10), and then differentiate to get the density function and then take the conditional expectation using the density.

By definition,
P(Y2>y |N(1)=10)
= P(Y2>y and N(1)=10) / P(N(1)=10)

Now I am having trouble computing P(Y2>y and N(1)=10). How can we express it in terms of N(y) (which is the number of points in a circle of radius y about the origin)?

Can someone please go through the idea of how to compute this?
Any help is greatly appreciated! :)

[note: also under discussion in talk stats forum]
• December 10th 2009, 01:06 AM
Laurent
Quote:

Originally Posted by kingwinner
Consider a Poisson process of rate λ on R^2. Let N(r) be the number of points in a circle of radius r centered at the origin and Y2 be the distance from the origin to the 2nd closest point.
Calculate E(Y2) and E[Y2 |N(1)=10].

You should probably use the fact that, conditionally to $N(1)=10$, the points of the Poisson process inside the circle of radius 1 have same joint distribution as 10 independent random variables uniformly distributed in the unit disk. (In particular, $Y_2$ depends only on these points) So you can forget about the Poisson process and only consider a situation involving uniform distribution in the unit disk.
• December 10th 2009, 03:42 AM
kingwinner
Quote:

Originally Posted by Laurent
You should probably use the fact that, conditionally to $N(1)=10$, the points of the Poisson process inside the circle of radius 1 have same joint distribution as 10 independent random variables uniformly distributed in the unit disk. (In particular, $Y_2$ depends only on these points) So you can forget about the Poisson process and only consider a situation involving uniform distribution in the unit disk.

hmm...I haven't encountered this theorem in my course, so I think I would have to do the computations by direct method.

I know that N(y)~Poisson(λ pi y^2), and I know that for Poisson process, things happening in non-overlapping sets are independent. I think the use of these would allow me to compute directly.
But I am not sure how to express the event {Y2>y and N(1)=10} entirely in terms of N(y). Could you please give me some help on this?
[I don't really bother with all the computations, but I know, at least in theory, that if we can express the event {Y2>y and N(1)=10} entirely in terms of N(y), then the problem is basically solved.]

Thank you very much!
• December 10th 2009, 09:23 AM
Laurent
Quote:

Originally Posted by kingwinner
hmm...I haven't encountered this theorem in my course, so I think I would have to do the computations by direct method.

(in fact, this theorem is the exact analog to the theorem about ordered uniform random variables in the 1-dimensional case)

You could find the distribution of $Y_2$ given $N(1)=10$. To that aim, for any $0<r<1$, you have:

$P(Y_2> r, N(1)=10)= P(N(r)\leq 1, N(1)=10)=$ $P(N(r)=0, N(1)=10) + P(N(r)=1, N(1)=10)$

and all these probabilities can be computed easily using annuli. For instance, $P(N(r)=1,N(1)=10)$ is the probability that there is 1 point in $D(0,r)$ and 9 points in $D(0,1)\setminus D(0,r)$, and since these sets are disjoint, the numbers of points therein are independent (and Poisson distributed with parameter proportional to the area...).

Finally, you can use $E[Y_2|N(1)=10]=\int_0^1 P(Y_2>t|N(1)=10)dt$ to compute the expectation.

I let you try that.
• December 10th 2009, 02:19 PM
kingwinner
Quote:

Originally Posted by Laurent
(in fact, this theorem is the exact analog to the theorem about ordered uniform random variables in the 1-dimensional case)

You could find the distribution of $Y_2$ given $N(1)=10$. To that aim, for any $0<r<1$, you have:

$P(Y_2> r, N(1)=10)= P(N(r)\leq 1, N(1)=10)=$ $P(N(r)=0, N(1)=10) + P(N(r)=1, N(1)=10)$

and all these probabilities can be computed easily using annuli. For instance, $P(N(r)=1,N(1)=10)$ is the probability that there is 1 point in $D(0,r)$ and 9 points in $D(0,1)\setminus D(0,r)$, and since these sets are disjoint, the numbers of points therein are independent (and Poisson distributed with parameter proportional to the area...).

Finally, you can use $E[Y_2|N(1)=10]=\int_0^1 P(Y_2>t|N(1)=10)dt$ to compute the expectation.

I let you try that.

Thanks! :)

If I use
E(Y2|N(1)=10) =
1
∫ y f(y) dy
0
where f(y) is the conditional density function of Y2 given that N(1)=10,
will I get the same answer?
• December 10th 2009, 02:27 PM
Laurent
Quote:

Originally Posted by kingwinner
If I use
E(Y2|N(1)=10) =
1
∫ y f(y) dy
0
where f(y) is the conditional density function of Y2 given that N(1)=10,
will I get the same answer?

Sure, even if it's less direct. It's like applying an integration by part from the formula I gave.
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