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MathGroup Archive 1996

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Re: is this integration possible with mathematica?

  • Subject: [mg3093] Re: is this integration possible with mathematica?
  • From: wagner at bullwinkle.cs.Colorado.EDU (Dave Wagner)
  • Date: 31 Jan 1996 04:21:07 -0600
  • Approved: usenet@wri.com
  • Distribution: local
  • Newsgroups: wri.mathgroup
  • Organization: University of Colorado, Boulder
  • Sender: daemon at wri.com

In article <4ekf0l$6gj at dragonfly.wri.com>,
Mickael Salabasis <md88-msa at nada.kth.se> wrote:
>i have encountered a problem in statistics that i cannot seem
>to solve without help.
>
>i have a line separating the xy-plane. the line is given by the
>following equation:
>
>f(x,y) : f1(x,y)/f2(x,y) = 1
>
>where 
>f1(x,y) = exp[-k* ((x-m11)^2 + (y-m12)^2)] + exp[-k*((x-m21)^2 + (y-m22)^2)]
>f2(x,y) = exp[-k* ((x-m31)^2 + (y-m32)^2)] + exp[-k*((x-m41)^2 + (y-m42)^2)]
>the m's and k are known constants
>
>now i want to calculate the (double) integral of f1 in the area ABOVE the
>line defined by f and respectively the integral of f2 for the area BELOWE 
>the same line. but i have problems with one of the integrals limits as the
>function f cannot be manipulated to give an expression of the type y=h(x)...

This problem exceeds my analytical skills, but since you're a statistician,
why not try to get a statistical approximation to the integral?
The integral is a volume, so first, define a cuboid large enough to contain
that volume.  Next, choose a random triple {x, y, z} in the
cuboid, see if f1(x,y) > f2(x,y) and z < f1(x,y); repeat umpteen times.
The approximate value of the integral is the fraction of points that meet
the criteria times the volume of the cuboid.

		Dave Wagner
		Principia Consulting
		(303) 786-8371
		dbwagner at princon.com
		http://www.princon.com/princon


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