Re: is this integration possible with mathematica?
- To: mathgroup at smc.vnet.net
- Subject: [mg3093] Re: is this integration possible with mathematica?
- From: wagner at bullwinkle.cs.Colorado.EDU (Dave Wagner)
- Date: Wed, 31 Jan 1996 03:04:14 -0500
- Organization: University of Colorado, Boulder
- Sender: owner-wri-mathgroup 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 ==== [MESSAGE SEPARATOR] ====