10
$\begingroup$

I have an EmpiricalDistribution

  d = EmpiricalDistribution[{1, 1, 1, 2, 2, 2, 3, 3, 3, 3}]

now, I want to sample twice without replacement $x_1,x_2$.

afterward, compute probability distribution of $|x_1-x_2|$ to get:

  X=|x_1-x_2|   0        1       2
      P        4/15    7/15     4/15

What command should be used?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ d = {1,1,1,2,2,2,3,3,3,3}; samples=Subsets[d, {2}]; Map[{First[#], Length[#]/Length[samples]} &, Split[Sort[Map[Abs[First[#]-Last[#]]&, samples]]]] which gives you {{0, 4/15}, {1, 7/15}, {2, 4/15}} $\endgroup$
    – Bill
    Nov 30 '16 at 1:00
  • $\begingroup$ This is a brute force algorithm. Is there a little more simple method? $\endgroup$
    – tiankonghewo
    Nov 30 '16 at 1:13
  • $\begingroup$ be aware if you want to work with EmpiricalDistribution the second draw comes from a different distribution. Likely more complicated than Bills method $\endgroup$
    – george2079
    Nov 30 '16 at 4:01
  • $\begingroup$ Hi tiankonghewo If your problem has more than 3 types of numbers or you need more than 3 sums then the solution that Bill gave is going to be a lot more attractive to you than the one I posted below. $\endgroup$
    – bobbym
    Nov 30 '16 at 4:45
  • $\begingroup$ Hi tiankonghewo; Always wait about 24 hours or so before accepting an answer. The first answer is not always the best and never is when it is mine. It is just to get the ball rolling and if you wait without accepting you will encourage others to answer. $\endgroup$
    – bobbym
    Nov 30 '16 at 6:15
11
$\begingroup$

You can do it this way but it is a bit clumsy:

For $ \left | x2-x1 \right |=0$

Probability[x == 2 || y == 2 || z == 2, {x, y, z} \[Distributed]
MultivariateHypergeometricDistribution[2, {3, 3, 4}]]

For $ \left | x2-x1 \right |=1$

Probability[(x == 1 && y == 1) || (y == 1 && z == 1), {x, y, z}
\[Distributed]MultivariateHypergeometricDistribution[2, {3, 3, 4}]]

For $ \left | x2-x1 \right |=2$

Probability[(x == 1 && z == 1), {x, y, z}
\[Distributed]MultivariateHypergeometricDistribution[2, {3, 3, 4}]]
Improve this answer
$\endgroup$
4
  • $\begingroup$ Great! this is just what I need. I have learned something new. Thank you very much. $\endgroup$
    – tiankonghewo
    Nov 30 '16 at 5:19
  • 1
    $\begingroup$ You can write this as p[x_] = Piecewise[{Probability[#[[1]], {x, y, z} \[Distributed] MultivariateHypergeometricDistribution[2, {3, 3, 4}]], x == #[[2]]} & /@ {{x == 2 || y == 2 || z == 2, 0}, {(x == 1 && y == 1) || (y == 1 && z == 1), 1}, {x == 1 && z == 1, 2}}] $\endgroup$
    – Bob Hanlon
    Nov 30 '16 at 6:07
  • $\begingroup$ Neater and cleaner, thanks +1 $\endgroup$
    – bobbym
    Nov 30 '16 at 6:16
  • $\begingroup$ OK, Piecewise is a new thing to me. I learned something new. $\endgroup$
    – tiankonghewo
    Nov 30 '16 at 7:18
11
$\begingroup$

I am a bit late but another way to use MultivariateHypergeometricDistribution:

md = MultivariateHypergeometricDistribution[2, {3, 3, 4}];
f[{___, 2, ___}] := 0
f[{1, 0, 1}] := 2
f[{___, 1, 1, ___}] := 1
td = TransformedDistribution[
   f[{x, y, z}], {x, y, z} \[Distributed] md];
res = Probability[x == #, x \[Distributed] td] & /@ Range[0, 2]
Histogram[RandomVariate[td, 10000], Automatic, "PDF", 
 Epilog -> {Red, PointSize[0.02], 
   Point[MapIndexed[{#2[[1]] - 1/2, #1} &, res]]}]

enter image description here

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ That is nice, I played with the Histogram command but could not make it work.+1 $\endgroup$
    – bobbym
    Nov 30 '16 at 6:21
  • $\begingroup$ @bobbym thank you, your answer got to the guts of the matter (reciprocal +1). I just wanted to show another way to use MultivariateHypergeometricDistribution :) $\endgroup$
    – ubpdqn
    Nov 30 '16 at 6:23
  • $\begingroup$ This method is more simple and convenient. $\endgroup$
    – tiankonghewo
    Nov 30 '16 at 7:30
  • $\begingroup$ @tiankonghewo please do not feel compelled to accept my answer. I think MultivariateHypergeometricDistribution was the key insight and bobbym and all our subsequent answers. I do thank you for the accept. $\endgroup$
    – ubpdqn
    Nov 30 '16 at 7:34
6
$\begingroup$

Amplifying on answer by @bobbym

p[x_] = Piecewise[{Probability[#[[1]], {x, y, z} \[Distributed] 
       MultivariateHypergeometricDistribution[2, {3, 3, 4}]], 
     x == #[[2]]} & /@
   {{x == 2 || y == 2 || z == 2, 0},
    {(x == 1 && y == 1) || (y == 1 && z == 1), 1},
    {x == 1 && z == 1, 2}}]

enter image description here

The distribution for |x1-x2| is then

dist = ProbabilityDistribution[p[x], {x, 0, 2, 1}];

This distribution can be used like any other distribution

PDF[dist, x] // Simplify

enter image description here

CDF[dist, x]

enter image description here

Mean[dist]

(*  1  *)

Variance[dist]

(*  8/15  *)

SeedRandom[1]

RandomVariate[dist, 10]

(*  {2, 0, 2, 0, 0, 0, 1, 0, 1, 1}  *)
Improve this answer
$\endgroup$
1
  • $\begingroup$ instructive amplification +1 (TransformedDistribution also works nicely in this case) $\endgroup$
    – ubpdqn
    Nov 30 '16 at 6:45
4
$\begingroup$

in case you wanted to work with EmpericalDistribution :

list = {1, 1, 1, 2, 2, 2, 3, 3, 3, 3};
outcomes = Tuples[ConstantArray[Union[list], 2]]

{{1, 1}, {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}

{Abs[Subtract @@ #], PDF[EmpiricalDistribution[list], #[[1]]]  
     PDF[EmpiricalDistribution[
       Drop[list, First@Position[list, #[[1]]]]], #[[2]]] } & /@ 
  outcomes;
{#[[1, 1]], Total[#[[All, 2]]]} & /@ GatherBy[%, First]

{{0, 4/15}, {1, 7/15}, {2, 4/15}}

of course , PDF[EmpiricalDistribution[list], i] just gives you Count[list,i]/Length@list

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.