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Re: A distribution problem using Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg111502] Re: A distribution problem using Mathematica
*From*: Daniel Lichtblau <danl at wolfram.com>
*Date*: Tue, 3 Aug 2010 06:40:48 -0400 (EDT)
S. B. Gray 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
You can move along edges from first to second, second to third, and
third to fourth vertices to get to a point. To do so with equal
probability, weight so that first move, for random 0<r1<1, goes a
distance that covers r1 of the volume. This means, I think going
r1^(1/3) of the distance along that edge. Then for 0<r2<1, go distance
that covers r2 of the area of the appropriate cross sectional triangle,
etc. Code below purports to do all this.
randomTetPoint[verts_] := Module[
{dirs = -Apply[Subtract, Partition[verts, 2, 1], 1],
vals = RandomReal[{0, 1}, 3]},
vals = Reverse[Flatten[MapIndexed[#^(1/#2) &, vals]]];
vals = Rest[FoldList[Times, 1, vals]];
First[verts] + vals.dirs
]
Quick visual sanity test:
vertices = {{0, 0, 0}, {1, 0, 0}, {0, 1, 0}, {0, 0, 1}};
rndpts = Table[randomTetPoint[vertices], {2000}];
ListPointPlot3D[rndpts, BoxRatios -> {1, 1, 1}]
Daniel Lichtblau
Wolfram Research
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