I recently found the AEP algorithm. Curious how Mathematica would handle the situation, I made an experiment:

  x + y + z + a + b < 1, {x, y, z, a, b} \[Distributed] 
   UniformDistribution[{{0, 1}, {0, 1}, {0, 1}, {0, 1}, {0, 1}}]]


{0.435173, 1/120}

Which copula did Mathematica internally assume? (I suppose that the problem does not have a unique solution if no copula is given, like here. In this case, the interdependence between the single random variables could be arbitrary.) And how is the value calculated? (with Monte Carlo simulation?)

  • 1
    $\begingroup$ It is an exact result. Same as Integrate[Boole[x+y+z+a+b<1],{x,0,1},{y,0,1},{z,0,1},{a,0,1},{b,0,1}], which also gives 1/120. No Monte Carlo. Why would the result not be unique? $\endgroup$
    – user293787
    Nov 16, 2022 at 11:44
  • $\begingroup$ There really is no reason why this probability isn't unique. Or what do you expect to happen when you use MC to calculate it to high accuracy? $\endgroup$ Nov 16, 2022 at 12:07
  • $\begingroup$ Probability[x < 1, x \[Distributed] UniformSumDistribution[5]] // AbsoluteTiming $\endgroup$
    – Bob Hanlon
    Nov 16, 2022 at 15:22
  • $\begingroup$ I was a blockhead: Clearly $f: (X_1, \dots, X_d) \mapsto X_1 + \dots + X_d$ is a measurable function of $(X_1, \dots, X_d)$. Should I delete the question or would one of you like to write his comment as an answer which I will accept? $\endgroup$ Nov 17, 2022 at 11:55


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