# Spatial Poisson Problem

• Oct 6th 2009, 08:12 AM
BrooketheChook
Spatial Poisson Problem
A bee needs to collect pollen for his hive. The positions of flowering flowering gum trees and flowering iron bark trees can be described by a pair of spatial Poisson processes with rates

$\lambda (x,y) = \frac{1 - (\cos(x^2+y^2))^2}{1+\sqrt{(x^2+y^2)}} \$ and $\ \nu (x,y) = \frac{1 - (\sin(x^2+y^2))^2}{1+\sqrt{(x^2+y^2)}}$

respectively, where $x$ and $y$ are measured in kilometers from the beehive.
Find

(a) the probability that the bee will have to fly more than 1km to reach a flowering tree.

(b) the probability the there are exactly 5 flowering trees within 1 km?

(c) the expected number of flowering trees within 9km?
• Oct 7th 2009, 08:48 AM
BrooketheChook
Quote:

Originally Posted by BrooketheChook
A bee needs to collect pollen for his hive. The positions of flowering flowering gum trees and flowering iron bark trees can be described by a pair of spatial Poisson processes with rates

$\lambda (x,y) = \frac{1 - (\cos(x^2+y^2))^2}{1+\sqrt{(x^2+y^2)}} \$ and $\ \nu (x,y) = \frac{1 - (\sin(x^2+y^2))^2}{1+\sqrt{(x^2+y^2)}}$

respectively, where $x$ and $y$ are measured in kilometers from the beehive.
Find

(a) the probability that the bee will have to fly more than 1km to reach a flowering tree.

I've had a go, but im really not confident about it being correct......

Let R be the distance from the hive to the nearest flowering tree.

R > 1 it there aren't any flowering trees in a circle D with a radius of 1 (with the hive being at the centre (0,0).

The area of D, $d = \pi r^2 = \pi$ km

$Pr(R>r) = Pr (N(D) = 0) = e^{-\lambda d} \ \frac{(\lambda d)^0}{0!}= e^{-\lambda \pi r^2}$ where r is the radius of the circle.

So

$\lambda (1,1) = \frac{1-(\cos(1+1))^2}{1+\sqrt{(1^2+1^2)}} \ \approx \ 0.3424$

Similarly $\nu (1,1) \approx \ 0.0717$

Then

$Pr(R>1) = e^{-0.4141 \pi} \approx \ 0.27$

Am I way off with this?
• Oct 7th 2009, 12:39 PM
Laurent
Quote:

Originally Posted by BrooketheChook
I've had a go, but im really not confident about it being correct......

Let R be the distance from the hive to the nearest flowering tree.

R > 1 it there aren't any flowering trees in a circle D with a radius of 1 (with the hive being at the centre (0,0).

The area of D, $d = \pi r^2 = \pi$ km

$Pr(R>r) = Pr (N(D) = 0)$

I have quoted the correct part of your post. Next equality was wrong.

You need to remember the following: if $N$ is a Poisson point process with rate $\lambda$ on $\mathbb{R}^2$ and $D\subset\mathbb{R}^2$, then $N(D)$ is a Poisson random variable of parameter $\lambda_D=\int_D\lambda(x,y)dx dy$. In particular, $P(N(D)=0)=e^{-\lambda_D}$ and in general $$P(N(D)=k)=e^{-\lambda_D}\frac{(\lambda_D)^k}{k!}$$.

Let $N_g$ (resp. $N_i$) be associated to the process of gum trees (resp. iron bark trees). (No idea what these trees are... I should have a look at a dictionary (Thinking))

Then $P(N(D)=0)=P(N_g(D)=0\mbox{ and }N_i(D)=0)=P(N_g(D)=0)P(N_i(D)=0)$ (using independence between these processes (even though the text doesn't specify it)). Using the above, we have $P(N(D)=0)=e^{-\int_D \lambda(x,y)dx dy}e^{-\int_D\nu(x,y)dx dy}$ $=e^{-\int_D (\lambda(x,y)+\nu(x,y))dx dy}$. You can compute this integral and conclude (By the way, one probably wouldn't be able to compute $\int_D \lambda(x,y) dx dy$ but there happens to be a nice simplification when you add $\nu(x,y)$).

Perhaps you know that the union of two independent Poisson point processes is a Poisson point process whose rate function is the sum of the two initial rate functions. Then you could say that the positions of flowering trees are described by a Ppp of rate function $\lambda+\nu$ and have directly $P(N(D)=0)=e^{-\int_D (\lambda+\nu)}$.

I let you try that. Then the following questions rely on the same initial property: $N(D)$ is a Poisson random variable with parameter etc....

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You don't need this here, but let me remind you the second characteristic property of PPP: if $A$ and $B$ are disjoint subsets, then the r.v. $N(A)$ and $N(B)$ are independent. This and the distribution of $N(A)$ characterize the fact that $N$ is a PPP.