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Thread: The probability of two random points on a plane being closer than a certain distance?

  1. #1
    Oct 2011

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


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

    This point $\displaystyle x$ is used for Monte Carlo integration, and if I use the pdf $\displaystyle p(x) = p_1(x_1) p_2(x_2) \pi r^2$ 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, $\displaystyle \pi r^2$ 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
    MHF Contributor

    Aug 2008
    Paris, France

    Re: The probability of two random points on a plane being closer than a certain dista


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

    If so, first the law of $\displaystyle X$ should not symmetric with respect to $\displaystyle p_1$ and $\displaystyle p_2$ as they have different roles. And second there is no meaning to writing $\displaystyle p(x)$ as a function of $\displaystyle x_1,x_2$ which are not defined in terms of $\displaystyle x$? 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 $\displaystyle X$ 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):

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

    If $\displaystyle p_2$ is continuous at $\displaystyle x$, the integral at the numerator is close to $\displaystyle p_2(x)\pi r^2$ when $\displaystyle r$ is small. Then, for small $\displaystyle r$, one may say that the pdf $\displaystyle p$ of $\displaystyle X$ is close to $\displaystyle \frac1C p_1(x)p_2(x)$, where $\displaystyle C=\int p_1(y)p_2(y)dy$ (so that $\displaystyle p$ 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
    Oct 2011

    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 $\displaystyle x_1$ can be (e.g. the whole plane). Actually, I can already use $\displaystyle x \equiv x_1$ for the MC integration with PDF $\displaystyle p(x) = p_1(x_1)$. Just that I want to have another sampling strategy that takes $\displaystyle x_1$ only if a random point $\displaystyle x_2$ with PDF $\displaystyle p_2(x_2)$ lands within a distance $\displaystyle r$ of $\displaystyle x_1$.

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

    How does that sound?
    Last edited by ingenious; Oct 14th 2011 at 11:56 AM.
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