Re: A distribution problem using Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg111494] Re: A distribution problem using Mathematica*From*: Ray Koopman <koopman at sfu.ca>*Date*: Tue, 3 Aug 2010 06:38:52 -0400 (EDT)*References*: <i368np$r7k$1@smc.vnet.net>

On Aug 2, 4:04 am, "S. B. Gray" <stev... at ROADRUNNER.COM> wrote: > I want 3 random variables v1,v2,v3 with uniform distribution from 0 to > 1, but modified or normalized so that their sum is a uniform > distribution from 0 to 1. > These variables are for placing a point at a random place inside a > tetrahedron defined by 4 vertex vectors p1,p2,p3,p4. The internal point > is given by p=v1(p1-p4)+v2(p2-p4)+v3(p3-p4) (barycentric coordinates). > The 0 to 1 constraints assure that the point p will be inside. > The closest I have come is this function giving the triplet > frs={v1,v2,v3}: > > mul = RandomReal[{0, 1}, {3}]; > mul = mul/Total[mul]; > frs = Sqrt[RandomReal[{0, 1}, {3}]]*Power[mul, (3)^-1]; > > which I tested with the histogram nhist: (The +1 avoids trying to access > the 0th element of the list nhist.) > > nhist = Table[0, {1000}]; > Do [ mul = RandomReal[{0, 1}, {3}]; > mul = mul/Total[mul]; > frs = Sqrt[RandomReal[{0, 1}, {3}]]*Power[mul, (3)^-1]; > ip = IntegerPart[1000 frs][[2]]; > nhist[[ip+1]]++, {10000} > ]; > Print[frs]; > ListPlot[nhist] > > This ad-hoc method gives a distribution that covers the range 0-1 but is > too heavy in the region 0.3 to 0.7. This would put too many points near > the middle of the tetrahedron. Something tells me there must be a better > and more elegant solution. Any ideas? > > Steve Gray This will give n points uniformly distributed inside an m-dimensional polytope whose vertices are given by the rows of a table p whose dimensions are {m+1,m}: Table[ #/Tr@# & @ RandomReal[ExponentialDistribution[1], m+1], {n}].p