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Math Help - Poisson process on R^2

  1. #1
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    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]
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  2. #2
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    Quote Originally Posted by kingwinner View Post
    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.
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  3. #3
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    Quote Originally Posted by Laurent View Post
    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!
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  4. #4
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    Quote Originally Posted by kingwinner View Post
    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.
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  5. #5
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    Quote Originally Posted by Laurent View Post
    (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?
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  6. #6
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    Quote Originally Posted by kingwinner View Post
    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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