🚀 go-pugleaf

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
Kruislaan 413, 1098 SJ Amsterdam, Netherlands.	: fax +31 20 592 4199