Arsthbt

What is /etc/mtab in Linux?

How do I check if a string is entirely made of the same substring?

Are these square matrices always diagonalisable?

What to do with someone that cheated their way through university and a PhD program?

Split coins into combinations of different denominations

Suing a Police Officer Instead of the Police Department

The art of proof summarizing. Are there known rules, or is it a purely common sense matter?

Co-worker works way more than he should

Is Diceware more secure than a long passphrase?

Can I criticise the more senior developers around me for not writing clean code?

Married in secret, can marital status in passport be changed at a later date?

Map material from china not allowed to leave the country

How to count in linear time worst-case?

Israeli soda type drink

What is the term for a person whose job is to place products on shelves in stores?

Password Generator in batch

Why did Israel vote against lifting the American embargo on Cuba?

Retract an already submitted recommendation letter (written for an undergrad student)

What is "leading note" and what does it mean to "raise a note"?

Does the set of sets which are elements of every set exist?

How would this chord from "Rocket Man" be analyzed?

Align column where each cell has two decimals with siunitx

Raising a bilingual kid. When should we introduce the majority language?

Implementing 3DES algorithm in Java: is my code secure?

Does Mathematica have an implementation of the Poisson Binomial Distribution?

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?

probability-or-statistics

share|improve this question

edited 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

$endgroup$

add a comment | 

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?

probability-or-statistics

share|improve this question

edited 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

$endgroup$

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 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

$endgroup$

share|improve this question

edited 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

share|improve this question

edited 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

edited 2 hours ago

Chris K

7,32722143

edited 2 hours ago

Chris K

7,32722143

asked 3 hours ago

user120911user120911

81338

asked 3 hours ago

user120911user120911

81338

1 Answer
1

active

oldest

votes

8

$begingroup$

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 1 hour ago

answered 2 hours ago

gwrgwr

8,72322861

$endgroup$

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "387"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f196962%2fdoes-mathematica-have-an-implementation-of-the-poisson-binomial-distribution%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

8

$begingroup$

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 1 hour ago

answered 2 hours ago

gwrgwr

8,72322861

$endgroup$

add a comment | 

8

$begingroup$

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 1 hour ago

answered 2 hours ago

gwrgwr

8,72322861

$endgroup$

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 1 hour ago

answered 2 hours ago

gwrgwr

8,72322861

$endgroup$

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 1 hour ago

answered 2 hours ago

gwrgwr

8,72322861

answered 2 hours ago

gwrgwr

8,72322861

answered 2 hours ago

gwrgwr

8,72322861