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Forum index » Science and Technology » Math » Probability
A stochastic geometry problem
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yangyu05@gmail.com
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Joined: 28 Oct 2005
Posts: 7

PostPosted: Thu Feb 23, 2006 5:40 pm    Post subject: A stochastic geometry problem Reply with quote
Consider a circular field with k independently and uniformaly
distributed points. Suppose we randomly drop one more point, A, in the
field. Let the distance from A to the k points, d(A), be the shortest
distance between A and all k points. Can we get the expected value of
d(A)?

Thanks!
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One Usenet Poster
science forum beginner


Joined: 03 May 2005
Posts: 9

PostPosted: Thu Feb 23, 2006 9:09 pm    Post subject: Re: A stochastic geometry problem Reply with quote
yang wrote:
Quote:
Consider a circular field with k independently and uniformaly
distributed points. Suppose we randomly drop one more point, A, in the
field. Let the distance from A to the k points, d(A), be the shortest
distance between A and all k points. Can we get the expected value of
d(A)?

Thanks!


0.5 * sqrt( area / (k+1) )
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john2
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Joined: 01 Feb 2006
Posts: 15

PostPosted: Fri Feb 24, 2006 1:09 pm    Post subject: Re: A stochastic geometry problem Reply with quote
yang wrote:
Quote:
Consider a circular field with k independently and uniformaly
distributed points. Suppose we randomly drop one more point, A, in the
field. Let the distance from A to the k points, d(A), be the shortest
distance between A and all k points. Can we get the expected value of
d(A)?

Thanks!


If A is not too close to the edge of the field:

Consider a circle, centre A, radius r.
The mean number of points in the circle is
mu = pi.r^2.sigma
where sigma is the point density = field area/k.
Probability of no points inside r is given by the Poisson result
P_0(r) = exp(- mu)
This is the same probability as the nearest point being at distance r
(from rank statistics)
Next differentiate P_0(r) wrt r to find the pdf of the nearest point.
Find the expectation of this pdf.

When A is close to the edge of the field it will get a *lot* more
complicated

john2
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Mike
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Joined: 17 Sep 2005
Posts: 74

PostPosted: Sat Feb 25, 2006 11:41 pm    Post subject: Re: A stochastic geometry problem Reply with quote
"One Usenet Poster" <me@my.computer.org> wrote in message
news:11vs93h1vokrafd@corp.supernews.com...
Quote:
yang wrote:
Consider a circular field with k independently and uniformaly
distributed points. Suppose we randomly drop one more point, A, in the
field. Let the distance from A to the k points, d(A), be the shortest
distance between A and all k points. Can we get the expected value of
d(A)?

Thanks!


0.5 * sqrt( area / (k+1) )

Could you please explain how you got this result. Thanks.
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One Usenet Poster
science forum beginner


Joined: 03 May 2005
Posts: 9

PostPosted: Sun Feb 26, 2006 9:18 pm    Post subject: Re: A stochastic geometry problem Reply with quote
Mike wrote:
Quote:
"One Usenet Poster" <me@my.computer.org> wrote in message
news:11vs93h1vokrafd@corp.supernews.com...
yang wrote:
Consider a circular field with k independently and uniformaly
distributed points. Suppose we randomly drop one more point, A, in the
field. Let the distance from A to the k points, d(A), be the shortest
distance between A and all k points. Can we get the expected value of
d(A)?

Thanks!

0.5 * sqrt( area / (k+1) )

Could you please explain how you got this result. Thanks.

I didn't derive it; I just found this
www.css.cornell.edu/courses/620/lecture11.ppt via Google.

The general expression for the k_th order is:

d(k) = k * (2 * k)! / ( (2^k k!)^2 sqrt( N / area ) )

So for k = 1, we get the above result.

Mike
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Google
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