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- Path: sparky!uunet!math.fu-berlin.de!news.netmbx.de!Germany.EU.net!mcsun!sun4nl!cwi.nl!adrianb
- From: adrianb@cwi.nl (Adrian Baddeley)
- Newsgroups: sci.math.stat
- Subject: Re: Spatial Statistics Problem.
- Message-ID: <7184@charon.cwi.nl>
- Date: 30 Aug 92 14:42:37 GMT
- References: <cbc.714687996@milton>
- Sender: news@cwi.nl
- Lines: 99
-
- cbc@milton.u.washington.edu (Charles Cook) writes:
-
- >I'm interested in knowing the distance between random points on a plane.
- ...
- >.. to reach every point, we are going to start down our 'swath' in the y
- >direction, touching every point as we reach it in y. (traveling from
- >the last point directly to the next.)
-
- If I understand it, you have a point process in the
- 'swath' S = [0,X] x [-infty,+infty], and you are considering
- the distance between a given point in the process, (x*,y*),
- and the next lowest point, defined as that
- point of the process with the largest y coordinate subject
- to y < y*.
-
- >Now, ideally, the points are
- >uniformily distributed over the whole plane, but for this simplification
- >I've assumed a uniform distribution in x and a poisson distribution in
- >y.
-
- So you assume a Poisson process on S with
- homogeneous intensity Td.
-
- Given that there is a point at (x*,y*), the rest of the process is just
- a Poisson process with intensity Td again. (Slivnyak's theorem).
-
- So the successive y coordinates form a 1-dimensional Poisson process
- with intensity Td*X, and the corresponding x coordinates are
- i.i.d. uniform on [0,X].
-
- So the distance from (x*,y*) to the next lowest point is
-
- D = sqrt( (x - x*)^2 + e^2) (1)
-
- where e = y*-y has an exponential (Td*X) distribution,
- and x is uniform on [0,X] independently of e.
-
-
- > The problem that I've run into is I can't solve the resulting integral:
-
- > / infinity / X-xo
- >E(D) = 1/(X^2*Td) | | exp(-y/Td*X) * Sqrt(x^2 +y^2) dx dy
- > / 0 / -xo
-
- >Where E(D) is the expectation value of the distance from the point you're
- >at (xo,0) to the next point.
-
- OK; this is the expectation of (1) above.
-
- But now you're going to step from point to point in this fashion
- effectively running a Markov chain where you incur
- a random cost D for each step. The x coordinates of the successive
- points are i.i.d. uniform so the average cost **per step** is
- the expectation of (1) when both x and x* are uniform [0,X].
- [Warning: The average cost *per unit height of strip* is this quantity
- divided by the mean height of each step, viz 1/(Td*X).]
-
- It seems to me that you want to evaluate
-
- E*(D) = E sqrt((u-v)^2 + e^2) (2)
-
- where u,v are i.i.d. uniform [0,X] and e is exponential (Td * X).
-
- The second moment is easy :-)
-
- The density of (u-v) is f(w) = 8/(X^2) min(x, X-x) so we get
-
- /infty /(X/2)
- E*(D) = 2(8/X^2) (1/Td*X) | | exp(-y/Td*X).w.sqrt(w^2+y^2) dw dy
- /0 /0
-
-
- >My eventual goal is to have a closed form solution for this expectation
- >value dependent on X and Td. From this I can find the optimal value for X
- >to minimize the distance to touch every point.
-
- A remark: by scaling properties of the Poisson process,
-
- r(Td/a , a* X) = a * r(Td, X)
-
- where r(Td,X) denotes the value of (2) for parameter values Td, X.
- So
- r(Td, X) = (1/Td) * r(1, X * Td)
-
- So you only need to consider the case of a unit rate Poisson process.
-
-
- That's all I can think of right now.
-
- Related literature: R. Coleman (1972) Sampling procedures for the
- lengths of random straight lines. Biometrika 59 (1972) 415-426.
-
-
-
- --
- =========== "Honk if you use the Axiom of Choice" ===============
- Adrian Baddeley : adrianb@cwi.nl
- Centre for Mathematics & Computer Science : tel +31 20 592 4050
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