I came across a description of the problem at: http://geomblog.blogspot.com/2005/10/sampling-from-simplex.html
Specifically, I would like a uniform sample from the set
X = { (x1, x2, ..., xD) | 0 <= xi <= 1, x1 + x2 + ... + xD = 1}.
D is the dimension of the simplex.
So, the coordinates of any point on the simplex would sum to 1 and I need to sample points on the simplex.
geomblog's solution suggested:
Generating IID random samples from an exponential distribution by sampling X from [0,1] uniformly, and returning -log(X)).
Take n samples, then normalize.
This should result in a list of numbers which is a uniform sample from the simplex.
I've searched extensively for a Mathematica implementation of something like this, to no avail.
I keep trying different things but haven't made much headway.
Any suggestions for how to develop this (or an equivelant) in Mathematica much appreciated
A
sampleSimplex[sampleSize_, dimensions_] :=
Module[{sample, sampleOutput},
sample = RandomReal[1, {sampleSize, dimensions}];
sampleOutput =
Table[1 / Total[sample[[i, All]]] *sample[[i, All]], {i, 1,
sampleSize}];
sampleOutput
]
example:
sampleSimplex[4, 3]
{{0.163972, 0.445817, 0.390211}, {0.470828, 0.291766,
0.237406}, {0.56663, 0.3813, 0.0520696}, {0.182715, 0.600876, 0.216409}}
This doesn't draw from an exponential distribution, but would this work?
Also, if it does work can I do it a better (faster, more elegant) way?
Hi. Try this (using Version 6 or 7):
ranSimp[d_] := (#/Plus @@ #) &[-Log[RandomReal[1, d]]]
ranSimp[d_, n_] := (#/Plus @@ #) & /@ (-Log[RandomReal[1, {n, d}]])
d is the "dimension" of the simplex and n is the number of draws.
--Mark
The following is code from Mathematica tutorials for sampling from the
Dirichlet distribution.
DirichletDistribution /: Random`DistributionVector[
DirichletDistribution[alpha_?(VectorQ[#, Positive] &)], n_Integer,
prec_?Positive] :=
Block[{gammas},
gammas =
Map[RandomReal[GammaDistribution[#, 1], n,
WorkingPrecision -> prec] &, alpha];
Transpose[gammas]/Total[gammas]]
Asim
The following example in R^3 should be close to what you are looking
for, though I am not a statistician and I may have failed to fully grasp
the suggested solution.
In[1]:= Module[{x},
Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {5}]]
Total[Transpose[%]]
Out[1]= {{0.545974, 0.204439, 0.249587}, {0.36947, 0.0545329,
0.575997}, {0.523704, 0.319784, 0.156512}, {0.490651, 0.398176,
0.111173}, {0.0332044, 0.406806, 0.55999}}
Out[2]= {1., 1., 1., 1., 1.}
In[3]:= Graphics3D[
Point /@ Module[{x},
Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {1000}]]]
Hope this helps,
-- Jean-Marc
randomSimplex[n_] := (#[[2]] - #[[1]]) & /@ (Partition[Union[{0, 1},
RandomReal[{0, 1}, n - 1]], 2, 1])
You can also use this method to sample points from a convex polytope
with a little extra work. See http://cg.scs.carleton.ca/~luc/rnbookindex.html
, page 568 (theorem 2.1).
Cheers,
Jason
--
Dr J. McKenzie Alexander
Department of Philosophy, Logic and Scientific Method
London School of Economics and Political Science
Houghton Street, London WC2A 2AE
On 14 Dec 2008, at 12:38, Jean-Marc Gulliet wrote:
> The following example in R^3 should be close to what you are looking
> for, though I am not a statistician and I may have failed to fully
> grasp
> the suggested solution.
>
> In[1]:= Module[{x},
> Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {5}]]
> Total[Transpose[%]]
>
> Out[1]= {{0.545974, 0.204439, 0.249587}, {0.36947, 0.0545329,
> 0.575997}, {0.523704, 0.319784, 0.156512}, {0.490651, 0.398176,
> 0.111173}, {0.0332044, 0.406806, 0.55999}}
>
> Out[2]= {1., 1., 1., 1., 1.}
>
> In[3]:= Graphics3D[
> Point /@ Module[{x},
> Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {1000}]]]
>
> Hope this helps,
> -- Jean-Marc
>
>
Please access the attached hyperlink for an important electronic communications disclaimer: http://www.lse.ac.uk/collections/secretariat/legal/disclaimer.htm
randomSimplex[n_] :=
Rest@# - Most@# &[Union[{0, 1}, RandomReal[{0, 1}, n - 1]]]
randomSimplex[10]
{0.00288906, 0.00240762, 0.0739569, 0.252289, 0.188723, 0.00997999, \
0.10469, 0.172935, 0.114573, 0.0775569}
Bobby
On Mon, 15 Dec 2008 06:43:14 -0600, J. McKenzie Alexander
<ja...@lse.ac.uk> wrote:
> Yet another solution would be to use a "stick-breaking" algorithm.
> Generate N-1 random reals in the interval [0,1], then use the length
> of the N subintervals as the values.
>
> randomSimplex[n_] := (#[[2]] - #[[1]]) & /@ (Partition[Union[{0, 1},
> RandomReal[{0, 1}, n - 1]], 2, 1])
>
> You can also use this method to sample points from a convex polytope
> with a little extra work. See
> http://cg.scs.carleton.ca/~luc/rnbookindex.html
> , page 568 (theorem 2.1).
>
> Cheers,
>
> Jason
>
> --
> Dr J. McKenzie Alexander
> Department of Philosophy, Logic and Scientific Method
> London School of Economics and Political Science
> Houghton Street, London WC2A 2AE
>
>
> On 14 Dec 2008, at 12:38, Jean-Marc Gulliet wrote:
>
>> Andreas wrote:
>>
>> The following example in R^3 should be close to what you are looking
>> for, though I am not a statistician and I may have failed to fully
>> grasp
>> the suggested solution.
>>
>> In[1]:= Module[{x},
>> Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {5}]]
>> Total[Transpose[%]]
>>
>> Out[1]= {{0.545974, 0.204439, 0.249587}, {0.36947, 0.0545329,
>> 0.575997}, {0.523704, 0.319784, 0.156512}, {0.490651, 0.398176,
>> 0.111173}, {0.0332044, 0.406806, 0.55999}}
>>
>> Out[2]= {1., 1., 1., 1., 1.}
>>
>> In[3]:= Graphics3D[
>> Point /@ Module[{x},
>> Table[x = -Log[RandomReal[1, {3}]]; x/Total[x], {1000}]]]
>>
>> Hope this helps,
>> -- Jean-Marc
>>
>>
>
>
>
>
>
> Please access the attached hyperlink for an important electronic
> communications disclaimer:
> http://www.lse.ac.uk/collections/secretariat/legal/disclaimer.htm
>