1. ## 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

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

respectively, where $\displaystyle x$ and $\displaystyle y$ 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?

2. 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

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

respectively, where $\displaystyle x$ and $\displaystyle y$ 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, $\displaystyle d = \pi r^2 = \pi$ km

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

So

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

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

Then

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

Am I way off with this?

3. 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, $\displaystyle d = \pi r^2 = \pi$ km

$\displaystyle Pr(R>r) = Pr (N(D) = 0)$
I have quoted the correct part of your post. Next equality was wrong.

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

Let $\displaystyle N_g$ (resp. $\displaystyle N_i$) 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 )

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

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 $\displaystyle \lambda+\nu$ and have directly $\displaystyle P(N(D)=0)=e^{-\int_D (\lambda+\nu)}$.

I let you try that. Then the following questions rely on the same initial property: $\displaystyle N(D)$ 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 $\displaystyle A$ and $\displaystyle B$ are disjoint subsets, then the r.v. $\displaystyle N(A)$ and $\displaystyle N(B)$ are independent. This and the distribution of $\displaystyle N(A)$ characterize the fact that $\displaystyle N$ is a PPP.