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Does Mathematica have an implementation of the Poisson Binomial Distribution?

5

1

$begingroup$

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?

probability-or-statistics

share|improve this question

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

$endgroup$

add a comment | 

5

1

$begingroup$

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?

probability-or-statistics

share|improve this question

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

$endgroup$

add a comment | 

$begingroup$

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?

probability-or-statistics

share|improve this question

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

$endgroup$

share|improve this question

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

share|improve this question

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

edited 1 hour ago

Chris K

7,32722143

edited 1 hour ago

Chris K

7,32722143

asked 2 hours ago

user120911user120911

80338

asked 2 hours ago

user120911user120911

80338

1 Answer
1

active

oldest

votes

4

$begingroup$

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

edited 58 mins ago

answered 1 hour ago

gwrgwr

8,68322861

$endgroup$

add a comment | 

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4

$begingroup$

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

edited 58 mins ago

answered 1 hour ago

gwrgwr

8,68322861

$endgroup$

add a comment | 

4

$begingroup$

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

edited 58 mins ago

answered 1 hour ago

gwrgwr

8,68322861

$endgroup$

add a comment | 

$begingroup$

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

edited 58 mins ago

answered 1 hour ago

gwrgwr

8,68322861

$endgroup$

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
 ]
 ]

dist = PoissonBinomialDistribution[ 0.04, 0.07, 0.07 ];

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

share|improve this answer

edited 58 mins ago

answered 1 hour ago

gwrgwr

8,68322861

answered 1 hour ago

gwrgwr

8,68322861

answered 1 hour ago

gwrgwr

8,68322861