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Why does the likelihood function of a binomial distribution not include the combinatorics term? [duplicate]

What does “likelihood is only defined up to a multiplicative constant of proportionality” mean in practice?When using the beta distribution as a prior distribution for binomial, why won't the distribution results match with the calculated probability?Using the binomial distribution to identify chance-level responsesMaximum likelihood estimation of a Poisson binomial distributionWhy is not the definition of probability distribution consistent with the definition of Binomial Distributions?understanding the binomial distributionBinomial distribution as likelihood in Bayesian modeling. When (not) to use it?Deriving likelihood function of binomial distribution, confusion over exponents

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7

$begingroup$

This question already has an answer here:

• 
  What does “likelihood is only defined up to a multiplicative constant of proportionality” mean in practice?
  
  5 answers
  
  

So the likelihood function for a binomial distribution is:

Why is the likelihood function above not multiplied by a combinatorics term: n! / (x! * (n - x)!)

If the likelihood function is interpreted as the probability of an outcome occurring x times out of n trials as the parameter, θ, varies, shouldn't there be a combinatorics term as that would provide the actual probability?

Thanks!

bayesian binomial loss-functions combinatorics

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edited Oct 14 at 16:54

confused

asked Oct 14 at 1:25

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$begingroup$

This question already has an answer here:

• 
  What does “likelihood is only defined up to a multiplicative constant of proportionality” mean in practice?
  
  5 answers
  
  

So the likelihood function for a binomial distribution is:

Why is the likelihood function above not multiplied by a combinatorics term: n! / (x! * (n - x)!)

If the likelihood function is interpreted as the probability of an outcome occurring x times out of n trials as the parameter, θ, varies, shouldn't there be a combinatorics term as that would provide the actual probability?

Thanks!

bayesian binomial loss-functions combinatorics

share|cite|improve this question

edited Oct 14 at 16:54

confused

asked Oct 14 at 1:25

confusedconfused

31288 bronze badges

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Oct 15 at 4:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$begingroup$

This question already has an answer here:

• 
  What does “likelihood is only defined up to a multiplicative constant of proportionality” mean in practice?
  
  5 answers
  
  

So the likelihood function for a binomial distribution is:

Why is the likelihood function above not multiplied by a combinatorics term: n! / (x! * (n - x)!)

If the likelihood function is interpreted as the probability of an outcome occurring x times out of n trials as the parameter, θ, varies, shouldn't there be a combinatorics term as that would provide the actual probability?

Thanks!

bayesian binomial loss-functions combinatorics

share|cite|improve this question

edited Oct 14 at 16:54

confused

asked Oct 14 at 1:25

confusedconfused

31288 bronze badges

$endgroup$

share|cite|improve this question

edited Oct 14 at 16:54

confused

asked Oct 14 at 1:25

confusedconfused

31288 bronze badges

share|cite|improve this question

edited Oct 14 at 16:54

confused

asked Oct 14 at 1:25

confusedconfused

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asked Oct 14 at 1:25

confusedconfused

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asked Oct 14 at 1:25

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1 Answer
1

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oldest

votes

10

$begingroup$

We often don’t care about the likelihood, just the value for which the likelihood is maximized.

When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maximum whether you include constants out front or not. Sure, that maximum value will be different, but that is not our concern.

So let’s make it convenient for ourselves and drop constants out in front, especially bulky combinatorics terms!

While we’re at it, we usually take the logarithm of the likelihood function since its derivative is easier to calculate, and log doesn’t change the point at which the maximum occurs.

Edit

This is in my comment but ought to be in the main post. We typically care about the argmax of a likelihood function, not the max itself.

share|cite|improve this answer

edited Oct 14 at 15:57

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I'm guessing there isn't a case where we actually need the likelihood itself? I know when we do Bayes Analysis the constant would cancel out when dividing the joint by the marginal.
  $endgroup$
  – confused
  Oct 14 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t want to go as far as saying that we’d never care about the likelihood, but we often can get away with dropping coefficients because they cancel or because we care about the argmax instead of the max itself.
  $endgroup$
  – Dave
  Oct 14 at 2:25
  

  
  
  
  

add a comment
 | 

10

$begingroup$

We often don’t care about the likelihood, just the value for which the likelihood is maximized.

When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maximum whether you include constants out front or not. Sure, that maximum value will be different, but that is not our concern.

So let’s make it convenient for ourselves and drop constants out in front, especially bulky combinatorics terms!

While we’re at it, we usually take the logarithm of the likelihood function since its derivative is easier to calculate, and log doesn’t change the point at which the maximum occurs.

Edit

This is in my comment but ought to be in the main post. We typically care about the argmax of a likelihood function, not the max itself.

share|cite|improve this answer

edited Oct 14 at 15:57

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I'm guessing there isn't a case where we actually need the likelihood itself? I know when we do Bayes Analysis the constant would cancel out when dividing the joint by the marginal.
  $endgroup$
  – confused
  Oct 14 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t want to go as far as saying that we’d never care about the likelihood, but we often can get away with dropping coefficients because they cancel or because we care about the argmax instead of the max itself.
  $endgroup$
  – Dave
  Oct 14 at 2:25
  

  
  
  
  

add a comment
 | 

10

$begingroup$

We often don’t care about the likelihood, just the value for which the likelihood is maximized.

When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maximum whether you include constants out front or not. Sure, that maximum value will be different, but that is not our concern.

So let’s make it convenient for ourselves and drop constants out in front, especially bulky combinatorics terms!

While we’re at it, we usually take the logarithm of the likelihood function since its derivative is easier to calculate, and log doesn’t change the point at which the maximum occurs.

Edit

This is in my comment but ought to be in the main post. We typically care about the argmax of a likelihood function, not the max itself.

share|cite|improve this answer

edited Oct 14 at 15:57

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! I'm guessing there isn't a case where we actually need the likelihood itself? I know when we do Bayes Analysis the constant would cancel out when dividing the joint by the marginal.
  $endgroup$
  – confused
  Oct 14 at 2:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I don’t want to go as far as saying that we’d never care about the likelihood, but we often can get away with dropping coefficients because they cancel or because we care about the argmax instead of the max itself.
  $endgroup$
  – Dave
  Oct 14 at 2:25
  

  
  
  
  

add a comment
 | 

$begingroup$

We often don’t care about the likelihood, just the value for which the likelihood is maximized.

When you use the likelihood function to find a maximum likelihood estimator, you get the same point giving the maximum whether you include constants out front or not. Sure, that maximum value will be different, but that is not our concern.

So let’s make it convenient for ourselves and drop constants out in front, especially bulky combinatorics terms!

While we’re at it, we usually take the logarithm of the likelihood function since its derivative is easier to calculate, and log doesn’t change the point at which the maximum occurs.

Edit

This is in my comment but ought to be in the main post. We typically care about the argmax of a likelihood function, not the max itself.

share|cite|improve this answer

edited Oct 14 at 15:57

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

$endgroup$

share|cite|improve this answer

edited Oct 14 at 15:57

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges

answered Oct 14 at 1:47

DaveDave

2,38822 silver badges1919 bronze badges