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Does Mathematica have an implementation of the Poisson Binomial Distribution?



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I need to work out the probability of having $k$ successful trials out of a total of $n$ when success probabilities are heterogeneous. This calculation relates to the Poisson Binomial Distribution. Does Mathematica, or perhaps the Mathstatica add-on, have an implementation for that?










share|improve this question











$endgroup$

















    7












    $begingroup$


    I need to work out the probability of having $k$ successful trials out of a total of $n$ when success probabilities are heterogeneous. This calculation relates to the Poisson Binomial Distribution. Does Mathematica, or perhaps the Mathstatica add-on, have an implementation for that?










    share|improve this question











    $endgroup$















      7












      7








      7


      2



      $begingroup$


      I need to work out the probability of having $k$ successful trials out of a total of $n$ when success probabilities are heterogeneous. This calculation relates to the Poisson Binomial Distribution. Does Mathematica, or perhaps the Mathstatica add-on, have an implementation for that?










      share|improve this question











      $endgroup$




      I need to work out the probability of having $k$ successful trials out of a total of $n$ when success probabilities are heterogeneous. This calculation relates to the Poisson Binomial Distribution. Does Mathematica, or perhaps the Mathstatica add-on, have an implementation for that?







      probability-or-statistics






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 3 hours ago









      Chris K

      7,32722143




      7,32722143










      asked 3 hours ago









      user120911user120911

      81338




      81338






















          1 Answer
          1






          active

          oldest

          votes


















          9












          $begingroup$

          Mathematica does not know about the PoissonBinomialDistribution, but you can use the formula given for the PDF on Wikipedia:



          PoissonBinomialDistribution[ plist : { __Real } ] := With[
          {
          n = Length @ plist,
          c = Exp[(2 I [Pi])/(Length@plist + 1)]
          }
          ,
          ProbabilityDistribution[
          1/(n + 1) Sum[c^(-l k) Product[1 + (c^l - 1) plist[[m]], {m, 1, n }], {l, 0, n}]
          ,
          {k, 0, n, 1}
          ]
          ]


          Now we may model a quality control where fault type 1 has a prob of 4% and fault types 2 and 3 have a prob of 7%:



          dist = PoissonBinomialDistribution[ {0.04, 0.07, 0.07} ];


          With this we find the probability for 3 faults:



          Probability[ k == 3, k [Distributed] dist ]// PercentForm



          0.0196 %







          share|improve this answer











          $endgroup$














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            1 Answer
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            active

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            9












            $begingroup$

            Mathematica does not know about the PoissonBinomialDistribution, but you can use the formula given for the PDF on Wikipedia:



            PoissonBinomialDistribution[ plist : { __Real } ] := With[
            {
            n = Length @ plist,
            c = Exp[(2 I [Pi])/(Length@plist + 1)]
            }
            ,
            ProbabilityDistribution[
            1/(n + 1) Sum[c^(-l k) Product[1 + (c^l - 1) plist[[m]], {m, 1, n }], {l, 0, n}]
            ,
            {k, 0, n, 1}
            ]
            ]


            Now we may model a quality control where fault type 1 has a prob of 4% and fault types 2 and 3 have a prob of 7%:



            dist = PoissonBinomialDistribution[ {0.04, 0.07, 0.07} ];


            With this we find the probability for 3 faults:



            Probability[ k == 3, k [Distributed] dist ]// PercentForm



            0.0196 %







            share|improve this answer











            $endgroup$


















              9












              $begingroup$

              Mathematica does not know about the PoissonBinomialDistribution, but you can use the formula given for the PDF on Wikipedia:



              PoissonBinomialDistribution[ plist : { __Real } ] := With[
              {
              n = Length @ plist,
              c = Exp[(2 I [Pi])/(Length@plist + 1)]
              }
              ,
              ProbabilityDistribution[
              1/(n + 1) Sum[c^(-l k) Product[1 + (c^l - 1) plist[[m]], {m, 1, n }], {l, 0, n}]
              ,
              {k, 0, n, 1}
              ]
              ]


              Now we may model a quality control where fault type 1 has a prob of 4% and fault types 2 and 3 have a prob of 7%:



              dist = PoissonBinomialDistribution[ {0.04, 0.07, 0.07} ];


              With this we find the probability for 3 faults:



              Probability[ k == 3, k [Distributed] dist ]// PercentForm



              0.0196 %







              share|improve this answer











              $endgroup$
















                9












                9








                9





                $begingroup$

                Mathematica does not know about the PoissonBinomialDistribution, but you can use the formula given for the PDF on Wikipedia:



                PoissonBinomialDistribution[ plist : { __Real } ] := With[
                {
                n = Length @ plist,
                c = Exp[(2 I [Pi])/(Length@plist + 1)]
                }
                ,
                ProbabilityDistribution[
                1/(n + 1) Sum[c^(-l k) Product[1 + (c^l - 1) plist[[m]], {m, 1, n }], {l, 0, n}]
                ,
                {k, 0, n, 1}
                ]
                ]


                Now we may model a quality control where fault type 1 has a prob of 4% and fault types 2 and 3 have a prob of 7%:



                dist = PoissonBinomialDistribution[ {0.04, 0.07, 0.07} ];


                With this we find the probability for 3 faults:



                Probability[ k == 3, k [Distributed] dist ]// PercentForm



                0.0196 %







                share|improve this answer











                $endgroup$



                Mathematica does not know about the PoissonBinomialDistribution, but you can use the formula given for the PDF on Wikipedia:



                PoissonBinomialDistribution[ plist : { __Real } ] := With[
                {
                n = Length @ plist,
                c = Exp[(2 I [Pi])/(Length@plist + 1)]
                }
                ,
                ProbabilityDistribution[
                1/(n + 1) Sum[c^(-l k) Product[1 + (c^l - 1) plist[[m]], {m, 1, n }], {l, 0, n}]
                ,
                {k, 0, n, 1}
                ]
                ]


                Now we may model a quality control where fault type 1 has a prob of 4% and fault types 2 and 3 have a prob of 7%:



                dist = PoissonBinomialDistribution[ {0.04, 0.07, 0.07} ];


                With this we find the probability for 3 faults:



                Probability[ k == 3, k [Distributed] dist ]// PercentForm



                0.0196 %








                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                gwrgwr

                8,73322861




                8,73322861






























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