Results 1 to 3 of 3

Math Help - Spatial Poisson Problem

  1. #1
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member BrooketheChook's Avatar
    Joined
    Sep 2009
    From
    Gold Coast
    Posts
    27
    Quote Originally Posted by BrooketheChook View Post
    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?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by BrooketheChook View Post
    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 [tex]P(N(D)=k)=e^{-\lambda_D}\frac{(\lambda_D)^k}{k!}[/Math].

    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 )

    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....

    --
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Spatial Autocorrelation.
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: October 4th 2010, 06:39 PM
  2. Spatial geometry
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 9th 2009, 06:57 AM
  3. Spatial Correlation Matrix
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: May 7th 2009, 11:10 PM
  4. Volume of a Spatial Figure
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 6th 2009, 03:03 PM
  5. spatial geometry
    Posted in the Geometry Forum
    Replies: 3
    Last Post: December 19th 2008, 09:19 PM

Search Tags


/mathhelpforum @mathhelpforum