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A probability distribution can be created in Mathematica (I am using 8.0.1) with e.g.

distribution1 = ProbabilityDistribution[(Sqrt[2])/Pi*(1/((1+x^4))), 

Random variates from this distribution can be created with RandomVariate easily:

Histogram[dataDistribution1](*Just an optical control*)

How can I create random variates from a 2-dimensional probability distribution? Let's say my 2-dimensional distribution is of the following form (very similar to the previous one):

distribution2=ProbabilityDistribution[((((Sqrt[2] π^(3/2) Gamma[5/4])/

I thought it would be logical to try


But that does not work. I get the following message then:

RandomVariate::noimp: Sampling from ProbabilityDistribution[Gamma[3/4]…}] is not implemented.

I tried a lot of variations of this approach to create random variates from such a distribution but without any success. Either I am doing a lot of things wrong (very likely in my experience ;-) or Mathematica cannot deliver random variates from 2-dimensional probability distributions. But in the help of RandomVariate (under “Simulate a multivariate continuous distribution”) one can see that this should be possible.

Perhaps somebody of you can tell me, how I can generate random variates from 2 dimensional probability distributions? I would be very happy about any help!

share|improve this question
Huh, from the docs I would expect the same thing. I'm looking forward to answers as well. –  tkott Mar 6 '12 at 17:40
Looks like you might need to go with a hand-made random number generator: reference.wolfram.com/mathematica/tutorial/… –  tkott Mar 6 '12 at 17:58
Hi! Thanks for commenting. Great to see that you share my expectation. I hope there will be more comments/answers too. –  partial81 Mar 6 '12 at 17:59
Doesn't seem to be implemented for ProbabilityDistribution? Interesting, let's wait for answers –  Rojo Mar 6 '12 at 18:27
The obvious way is to integrate out one of the variables to obtain its marginal distribution. Sample from the marginal, then sample the other variable from the conditional distribution. This process is quite slow, though, and requires the marginal to have a closed form that Mathematica can find. You're probably better off using Gibbs sampling. –  whuber Mar 6 '12 at 19:29
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1 Answer 1

up vote 24 down vote accepted

Mathematica v8 does not provide support for automated random number generation from multivariate distributions, specified in terms of its probability density functions, as you have already discovered it.

At the Wolfram Technology conference 2011, I gave a presentation "Create Your Own Distribution", where the issue of sampling from custom distribution is extensively discussed with many examples.

You can draw samples from the particular distribution at hand by several methods. Let

di = ProbabilityDistribution[
   Beta[3/4, 1/2]/(Sqrt[2] Pi^2) 1/(1 + x^4 + y^4), {x, -Infinity, 
    Infinity}, {y, -Infinity, Infinity}];

Conditional method

The idea here is to first generate the first component of the vector from a marginal, then a second one from a conditional distribution:

md1 = ProbabilityDistribution[
   PDF[MarginalDistribution[di, 1], x], {x, -Infinity, Infinity}];

cd2[a_] = 
   Simplify[PDF[di, {a, y}]/PDF[MarginalDistribution[di, 1], a], 
    a \[Element] Reals], {y, -Infinity, Infinity}, 
   Assumptions -> a \[Element] Reals];

Then the conditional method is easy to code:

diRNG[len_, prec_] := Module[{x1, x2},
  x1 = RandomVariate[md1, len, WorkingPrecision -> prec];
  x2 = Function[a, RandomVariate[cd2[a], WorkingPrecision -> prec]] /@
  Transpose[{x1, x2}]

You can not call it speedy:

In[196]:= AbsoluteTiming[sample1 = diRNG[10^3, MachinePrecision];]

Out[196]= {20.450045, Null}

But it works:

enter image description here

Transformed distribution method

This is somewhat a craft, but if such an approach pans out, it typically yields the best performing random number generation method. We start with a mathematical identity $$ \frac{1}{1+x^4+y^4} = \int_0^\infty \mathrm{e}^{-t(1+x^4+y^4)} \mathrm{d} t = \mathbb{E}_Z( \exp(-Z (x^4+y^4))) $$ where $Z \sim \mathcal{E}(1)$, i.e. $Z$ is exponential random variable with unit mean. Thus, for a random vector $(X,Y)$ with the distribution in question we have $$ \mathbb{E}_{X,Y}(f(X,Y)) = \mathbb{E}_{X,Y,Z}\left( f(X,Y) \exp(-Z X^4) \exp(-Z Y^4) \right) $$ This suggests to introduce $U = X Z^{1/4}$ and $V = Y Z^{1/4}$. It is easy to see, that the probability density function for $(Z, U, V)$ factors: $$ f_{Z,U,V}(t, u, v) = \frac{\operatorname{\mathrm{Beta}}(3/4,1/2)}{\sqrt{2} \pi^2} \cdot \frac{1}{\sqrt{t}} \mathrm{e}^{-t} \cdot \mathrm{e}^{-u^4} \cdot \mathrm{e}^{-v^4} $$ It is easy to generate $(W, U, V)$, since they are independent. Then $(X,Y) = (U, V) W^{-1/4}$, where $f_W(t) = \frac{1}{\sqrt{\pi}} \frac{1}{\sqrt{t}} \mathrm{e}^{-t}$, i.e. $W$ is $\Gamma(1/2)$ random variable.

This gives much more efficient algorithm:

  prec_] := (RandomVariate[NormalDistribution[], len, 
       WorkingPrecision -> prec]^2/2)^(-1/4) RandomVariate[
    1/(2 Gamma[5/4]) Exp[-x^4], {x, -Infinity, Infinity}], {len, 2}, 
   WorkingPrecision -> prec]

Noticing that $|W|$ is in fact a power of gamma random variable we can take it much further:

In[40]:= diRNG3[len_, prec_] := 
 Power[RandomVariate[GammaDistribution[1/4, 1], {len, 2}, 
     WorkingPrecision -> 
      prec]/(RandomVariate[NormalDistribution[], len, 
        WorkingPrecision -> prec]^2/2), 1/4] RandomChoice[
   Range[-1, 1, 2], {len, 2}]

In[42]:= AbsoluteTiming[sample3 = diRNG3[10^6, MachinePrecision];]

Out[42]= {0.7230723, Null}

Rejection method

Here the idea is to sample from a relatively simple to draw from hat distribution. It is again a craft to choose a good one. Once the one is chosen, we exercise the rejection sampling algorithm:

enter image description here

In the case at hand, a good hat is the bivariate T-distribution with 2 degrees of freedom, as it is easy to draw from, and it allows for easy computation of the scaling constant:

In[49]:= Maximize[(1/(1 + x^4 + y^4))/
  PDF[MultivariateTDistribution[{{1, 0}, {0, 1}}, 2], {x, y}], {x, y}]

Out[49]= {3 Pi, {x -> -(1/Sqrt[2]), y -> -(1/Sqrt[2])}}

This gives another algorithm:

diRNG4[len_, prec_] := Module[{dim = 0, bvs, u, res},
  res = Reap[While[dim < len,
      bvs = 
       RandomVariate[MultivariateTDistribution[{{1, 0}, {0, 1}}, 2], 
        len - dim, WorkingPrecision -> prec];
      u = RandomReal[3/2, len - dim, WorkingPrecision -> prec];
      bvs = 
        Sign[(Total[bvs^2, {2}]/2 + 1)^2 - u (1 + Total[bvs^4, {2}])],
      dim += Length[Sow[bvs]];
      ]][[2, 1]];
  Apply[Join, res]

This one proves to be quite efficient as well:

In[77]:= AbsoluteTiming[sample4 = diRNG4[10^6, MachinePrecision];]

Out[77]= {0.6910000, Null}
share|improve this answer
+1, love the 3D plot. –  rcollyer Mar 6 '12 at 20:20
+1 Beautifully done! This will serve as a working reference for many people. –  whuber Mar 6 '12 at 20:24
Hi @Sasha! Sorry for my late reply. It took me a while to adapt your solutions to my real problem. But it works! Thanks for this detailed answer! As whuber said, it will surely help a lot of people with similar problems! –  partial81 Mar 12 '12 at 23:21
it is so helpful, thanks so much. –  user6196 Mar 4 '13 at 22:15
Thanks for the detailed answer. I was just wondering if it is possible to extend the Conditional Method to construct a 3 dimensional distribution? –  user29165 2 days ago
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