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Math Help - The probability of two random points on a plane being closer than a certain distance?

  1. #1
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    The probability of two random points on a plane being closer than a certain distance?

    Hi,

    I've been struggling with the following problem for some time now, and it seems that I need some pro help:

    I have a process that generates a point x on a plane, and I need to find the pdf p(x) of that point. The process goes like this: First, points x_1 and x_2 are independently generated on the plane with pdfs p_1(x_1) and p_2(x_2). Then, only if the two points are closer than a certain distance r, i.e. ||x_1 - x_2|| < r, I take as my point x \equiv x_1.

    This point x is used for Monte Carlo integration, and if I use the pdf p(x) = p_1(x_1) p_2(x_2) \pi r^2 it all works nicely. However, I'm having a hard time justifying this particular pdf. I see two possible ways: (1) this pdf is a decent approximation of the real pdf; (2) this pdf is an unbiased estimate of the real pdf. More specifically, \pi r^2 is the result of a one-sample Monte Carlo estimation of the integral that appears due to the distance probability.

    I'm having a hard time mathematically formulating this pdf. Any ideas? Thanks very much in advance!
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  2. #2
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    Re: The probability of two random points on a plane being closer than a certain dista

    Hi,

    As far as I understand, you define X to be X_1 conditioned to be at distance smaller than r to X_2, right? In other words, you sample X_1 and X_2 repeatedly until they are closer than r from each other and then you set X=X_1. Is that indeed it?

    If so, first the law of X should not symmetric with respect to p_1 and p_2 as they have different roles. And second there is no meaning to writing p(x) as a function of x_1,x_2 which are not defined in terms of x? So something has to be wrong; it may be my understanding, but let's carry on.

    Given my rewriting of the problem, the pdf of X is derived as follows (I use a slightly sloppy notation "dx" that stands for an "infinitesimal subset around x"; you can turn it to fully rigorous by writing a Borel set A instead of dx, but I feel this writing with dx gives more feeling of what happens):

    P(X\in dx)=P(X_1\in dx |\, \|X_2-X_1\|<r)=\frac{P(X_1\in dx, X_2\in D(x,r))}{P(\|X_2-X_1\|<r)} =\frac{P(X_1\in dx)P(X_2\in D(x,r))}{P(\|X_1-X_2\|<r)}
    hence
    P(X\in dx)=\frac{p_1(x) \int_{D(x,r)}p_2(y)dy}{\int_{\{\|z-y\|<r\}}p_1(y)p_2(z)dy dz}\,dx
    so the pdf is the function on the right hand side.

    If p_2 is continuous at x, the integral at the numerator is close to p_2(x)\pi r^2 when r is small. Then, for small r, one may say that the pdf p of X is close to \frac1C p_1(x)p_2(x), where C=\int p_1(y)p_2(y)dy (so that p integrates to 1).

    I hope this answers your question. Feel free to ask for further details. Or to make your question more specific, if I got it wrong.
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  3. #3
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    Re: The probability of two random points on a plane being closer than a certain dista

    Thanks for the reply, Laurent!

    I kind of understand what you mean. I'm not sure this is exactly what I want, but it's close, I feel.

    In the following, please forgive any incorrect terminology/notations used. I'm really a computer scientist with a shallow background in probability theory.

    The generated point will be used for Monte Carlo integration of a function that is defined everywhere where x_1 can be (e.g. the whole plane). Actually, I can already use x \equiv x_1 for the MC integration with PDF p(x) = p_1(x_1). Just that I want to have another sampling strategy that takes x_1 only if a random point x_2 with PDF p_2(x_2) lands within a distance r of x_1.

    Therefore, I'm not sure about whether this normalization is needed. Otherwise the nominator looks OK: If I assume that p_2 is constant within the r-neighborhood around x_1, then I can take p_2(y) out of the integral as p_2(x_2) and then I get my result p_1(x_1) p_2(x_2) \pi r^2.

    How does that sound?
    Last edited by ingenious; October 14th 2011 at 12:56 PM.
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