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Does Mathematica have an implementation of the Poisson Binomial Distribution?
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?RandomVariate from 2-dimensional probability distributionBayesian Inference with Continuous prior distributionGrain(Particle) Size Distribution (PSD) Analysis with MathematicaHow to get probabilities for multinomial & hypergeometric distribution ranges more quickly?Given an exact formula, how can Mathematica find a probability distribution whose PDF matches it?Probability of an eventLikelihood for BetaBinomialDistribution with variable number of trialsInverse Fourier Transform of Poisson Characteristic Function?How to use EventData correctly to model a trial sequenceComputing the Expectation
$begingroup$
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
edited 6 hours ago
Chris K
7,32722143
asked 6 hours ago
user120911user120911
81838
$endgroup$
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
edited 6 hours ago
Chris K
7,32722143
asked 6 hours ago
user120911user120911
81838
$endgroup$
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
edited 6 hours ago
Chris K
7,32722143
asked 6 hours ago
user120911user120911
81838
$endgroup$
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
probability-or-statistics
edited 6 hours ago
Chris K
7,32722143
asked 6 hours ago
user120911user120911
81838
edited 6 hours ago
Chris K
7,32722143
asked 6 hours ago
user120911user120911
81838
edited 6 hours ago
Chris K
7,32722143
edited 6 hours ago
Chris K
7,32722143
edited 6 hours ago
Chris K
7,32722143
7,32722143
asked 6 hours ago
user120911user120911
81838
asked 6 hours ago
user120911user120911
81838
asked 6 hours ago
user120911user120911
81838
81838
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 %
answered 5 hours ago
gwrgwr
8,74322862
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 %
answered 5 hours ago
gwrgwr
8,74322862
$endgroup$
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 %
answered 5 hours ago
gwrgwr
8,74322862
$endgroup$
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 %
answered 5 hours ago
gwrgwr
8,74322862
$endgroup$
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 %
answered 5 hours ago
gwrgwr
8,74322862
edited 5 hours ago
answered 5 hours ago
gwrgwr
8,74322862
answered 5 hours ago
gwrgwr
8,74322862
answered 5 hours ago
gwrgwr
8,74322862
8,74322862
add a comment |
add a comment |
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