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Math Help - Average distance between 2 random points

  1. #1
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    Average distance between 2 random points

    Here's a problem I've been struggling with:

    If I select two numbers at random between 0 and 100 (inclusive and they can be the same number) what is the average difference between these numbers? Assume difference is always positive (i.e. if i select 2 and 40 or 41 and 3 the difference is 38) So if I select 1000000 pairs, what will the average difference be? I'm pretty sure it's less than 50 but I'd like a proof or integral etc...

    Ideally I would get a generic solution in the form of "N" where the range is between 0 and N. ( I really want the average between 0 and 1)


    Thanks.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by jaxon4 View Post
    Here's a problem I've been struggling with:

    If I select two numbers at random between 0 and 100 (inclusive and they can be the same number) what is the average difference between these numbers? Assume difference is always positive (i.e. if i select 2 and 40 or 41 and 3 the difference is 38) So if I select 1000000 pairs, what will the average difference be? I'm pretty sure it's less than 50 but I'd like a proof or integral etc...

    Ideally I would get a generic solution in the form of "N" where the range is between 0 and N. ( I really want the average between 0 and 1)


    Thanks.
    Let's assume, for simplicity, that the two points are uniformly and continuously distributed on the entire intervall [0;N].
    Suppose you now pick a first point x from [0;N] at random, then this point splits [0;N] into two segments, [0;x] and [x;N].
    If you now pick the second point, at random, it will be from the lower intervall with probability x/N and from the upper intervall with probability (N-x)/N.
    In the first case, the avarage distance of the second point from the first will be \tfrac{x}{2}, while in the second case, the average distance will be \tfrac{N-x}{2}.

    So, given the first point x from [0;N] the average distance of the second point is \tfrac{x}{2}\cdot \tfrac{x}{N}+\tfrac{N-x}{2}\cdot\tfrac{N-x}{N}=\tfrac{x^2+(N-x)^2}{2N}.
    You now only have to integrate this, multiplied with the appropriate density function \tfrac{1}{N}, with respect to x to get the average distance of two points randomly chosen from the interval [0;N].
    Last edited by Failure; April 10th 2010 at 02:07 AM.
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  3. #3
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    Should be \frac1{N^2}\int_0^N\int_0^N|x-y|\,\mathrm{d}x\,\mathrm{d}y for random real points in [0,N] shouldn't it? If so, then it should come out to N/3, which is less than half, as you thought.
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  4. #4
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    Quote Originally Posted by jaxon4 View Post
    Here's a problem I've been struggling with:

    If I select two numbers at random between 0 and 100 (inclusive and they can be the same number) what is the average difference between these numbers? Assume difference is always positive (i.e. if i select 2 and 40 or 41 and 3 the difference is 38) So if I select 1000000 pairs, what will the average difference be? I'm pretty sure it's less than 50 but I'd like a proof or integral etc...

    Ideally I would get a generic solution in the form of "N" where the range is between 0 and N. ( I really want the average between 0 and 1)


    Thanks.
    You are being vague are these reals numbers of integers?

    The real number problem is equivalent to choosing two numbers x,y\sim U(0,1) and asking for the mean of |x-y|. Well without loss we may assume that x\sim U(0,1) and y\sim U(0,x), then we have:

    E(|x-y|)=2\int_{x=0}^1 \int_{y=0}^x (x-y) \;dydx=1/3

    (the 2 is needed to normalise the probability over the lower half unit square)

    CB
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