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Can anyone give a concrete example to illustrate what is an uniform prior?

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2

$begingroup$

I googled "uniform prior" and got a link to Prior probability, which uses the term without an explanation or a definition.

anther link is to quora, which does not give an concrete example.

Can anyone give a concrete example, such as coin flipping or dice tossing, to illustrate the uniform prior?

probability bayesian mathematical-statistics

share|cite|improve this question

asked 9 hours ago

czlswsczlsws

16966 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly useful link.
  $endgroup$
  – BruceET
  8 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

I googled "uniform prior" and got a link to Prior probability, which uses the term without an explanation or a definition.

anther link is to quora, which does not give an concrete example.

Can anyone give a concrete example, such as coin flipping or dice tossing, to illustrate the uniform prior?

probability bayesian mathematical-statistics

share|cite|improve this question

asked 9 hours ago

czlswsczlsws

16966 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possibly useful link.
  $endgroup$
  – BruceET
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I googled "uniform prior" and got a link to Prior probability, which uses the term without an explanation or a definition.

anther link is to quora, which does not give an concrete example.

Can anyone give a concrete example, such as coin flipping or dice tossing, to illustrate the uniform prior?

probability bayesian mathematical-statistics

share|cite|improve this question

asked 9 hours ago

czlswsczlsws

16966 bronze badges

$endgroup$

share|cite|improve this question

asked 9 hours ago

czlswsczlsws

16966 bronze badges

share|cite|improve this question

asked 9 hours ago

czlswsczlsws

16966 bronze badges

asked 9 hours ago

czlswsczlsws

16966 bronze badges

asked 9 hours ago

czlswsczlsws

16966 bronze badges

3 Answers
3

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oldest

votes

3

$begingroup$

Let's say you don't know the probability of head, $p$, of a coin. You decide to conduct an experiment to estimate what it is, via Bayesian analysis. It requires you to choose a prior, and in general you're free to choose one of the feasible ones. If you don't know or don't want to assume anything about this $p$, you can say that it is uniformly distributed in $[0,1]$, in which $f_P(p)=1, 0leq pleq 1$, $0$ otherwise. This is quite similar to saying that any $p$ value in $[0,1]$ is equally likely. This prior distribution is a uniform prior.

You can also choose other priors, that focus on different regions in $[0,1]$, for example, if you choose a prior like $f_P(p)=frac32(1-(1-2p)^2), 0leq pleq 1$, you'll assume that $p$ is more likely to be around $0.5$ compared to edge cases, such as $p=0,p=1$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Does the function $f_P(p)$ represent the random variables, Bernoulli random variable in this case?
  $endgroup$
  – czlsws
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The notion of uniform prior as a prior with a constant density is not well-defined (or even meaningful) as it depends on both

1. the dominating measure that determines the density function of the prior


2. the parameterisation of the model for which the prior is constructed.

If either entry is modified, the density of the prior changes as well and stops being constant.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Another way to express this, perhaps a little more forcibly, is that all (absolutely continuous) priors are uniform.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

When the prior distribution $pi$, of the parameter $theta$ to be estimated is the Uniform distribution, i.e. $pi(theta)sim U(a,b)$, we refer to prior $pi$ as a uniform or uninformative prior. I'm not sure what's not to understand here except the basics of Bayesian inference and the Uniform distribution.

The best way to understand the uniform distribution is via a Monte Carlo sample of fair die rolls. The probability of every outcome $P(X=x_i),~ i=1,...,6$ is equal to $1/6$ and the histogram of the empirical distribution should be approximating a horizontal line (i.e. is uniform).

share|cite|improve this answer

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

$endgroup$

add a comment | 

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3

$begingroup$

Let's say you don't know the probability of head, $p$, of a coin. You decide to conduct an experiment to estimate what it is, via Bayesian analysis. It requires you to choose a prior, and in general you're free to choose one of the feasible ones. If you don't know or don't want to assume anything about this $p$, you can say that it is uniformly distributed in $[0,1]$, in which $f_P(p)=1, 0leq pleq 1$, $0$ otherwise. This is quite similar to saying that any $p$ value in $[0,1]$ is equally likely. This prior distribution is a uniform prior.

You can also choose other priors, that focus on different regions in $[0,1]$, for example, if you choose a prior like $f_P(p)=frac32(1-(1-2p)^2), 0leq pleq 1$, you'll assume that $p$ is more likely to be around $0.5$ compared to edge cases, such as $p=0,p=1$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Does the function $f_P(p)$ represent the random variables, Bernoulli random variable in this case?
  $endgroup$
  – czlsws
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Let's say you don't know the probability of head, $p$, of a coin. You decide to conduct an experiment to estimate what it is, via Bayesian analysis. It requires you to choose a prior, and in general you're free to choose one of the feasible ones. If you don't know or don't want to assume anything about this $p$, you can say that it is uniformly distributed in $[0,1]$, in which $f_P(p)=1, 0leq pleq 1$, $0$ otherwise. This is quite similar to saying that any $p$ value in $[0,1]$ is equally likely. This prior distribution is a uniform prior.

You can also choose other priors, that focus on different regions in $[0,1]$, for example, if you choose a prior like $f_P(p)=frac32(1-(1-2p)^2), 0leq pleq 1$, you'll assume that $p$ is more likely to be around $0.5$ compared to edge cases, such as $p=0,p=1$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you so much! Does the function $f_P(p)$ represent the random variables, Bernoulli random variable in this case?
  $endgroup$
  – czlsws
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Let's say you don't know the probability of head, $p$, of a coin. You decide to conduct an experiment to estimate what it is, via Bayesian analysis. It requires you to choose a prior, and in general you're free to choose one of the feasible ones. If you don't know or don't want to assume anything about this $p$, you can say that it is uniformly distributed in $[0,1]$, in which $f_P(p)=1, 0leq pleq 1$, $0$ otherwise. This is quite similar to saying that any $p$ value in $[0,1]$ is equally likely. This prior distribution is a uniform prior.

You can also choose other priors, that focus on different regions in $[0,1]$, for example, if you choose a prior like $f_P(p)=frac32(1-(1-2p)^2), 0leq pleq 1$, you'll assume that $p$ is more likely to be around $0.5$ compared to edge cases, such as $p=0,p=1$.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

answered 8 hours ago

gunesgunes

12.2k11 gold badge55 silver badges2222 bronze badges

1

$begingroup$

The notion of uniform prior as a prior with a constant density is not well-defined (or even meaningful) as it depends on both

1. the dominating measure that determines the density function of the prior


2. the parameterisation of the model for which the prior is constructed.

If either entry is modified, the density of the prior changes as well and stops being constant.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Another way to express this, perhaps a little more forcibly, is that all (absolutely continuous) priors are uniform.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The notion of uniform prior as a prior with a constant density is not well-defined (or even meaningful) as it depends on both

1. the dominating measure that determines the density function of the prior


2. the parameterisation of the model for which the prior is constructed.

If either entry is modified, the density of the prior changes as well and stops being constant.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. Another way to express this, perhaps a little more forcibly, is that all (absolutely continuous) priors are uniform.
  $endgroup$
  – whuber♦
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

The notion of uniform prior as a prior with a constant density is not well-defined (or even meaningful) as it depends on both

1. the dominating measure that determines the density function of the prior


2. the parameterisation of the model for which the prior is constructed.

If either entry is modified, the density of the prior changes as well and stops being constant.

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

$endgroup$

share|cite|improve this answer

edited 6 hours ago

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

answered 8 hours ago

Xi'anXi'an

61.8k88 gold badges9999 silver badges377377 bronze badges

0

$begingroup$

When the prior distribution $pi$, of the parameter $theta$ to be estimated is the Uniform distribution, i.e. $pi(theta)sim U(a,b)$, we refer to prior $pi$ as a uniform or uninformative prior. I'm not sure what's not to understand here except the basics of Bayesian inference and the Uniform distribution.

The best way to understand the uniform distribution is via a Monte Carlo sample of fair die rolls. The probability of every outcome $P(X=x_i),~ i=1,...,6$ is equal to $1/6$ and the histogram of the empirical distribution should be approximating a horizontal line (i.e. is uniform).

share|cite|improve this answer

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

When the prior distribution $pi$, of the parameter $theta$ to be estimated is the Uniform distribution, i.e. $pi(theta)sim U(a,b)$, we refer to prior $pi$ as a uniform or uninformative prior. I'm not sure what's not to understand here except the basics of Bayesian inference and the Uniform distribution.

The best way to understand the uniform distribution is via a Monte Carlo sample of fair die rolls. The probability of every outcome $P(X=x_i),~ i=1,...,6$ is equal to $1/6$ and the histogram of the empirical distribution should be approximating a horizontal line (i.e. is uniform).

share|cite|improve this answer

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

$endgroup$

add a comment | 

$begingroup$

When the prior distribution $pi$, of the parameter $theta$ to be estimated is the Uniform distribution, i.e. $pi(theta)sim U(a,b)$, we refer to prior $pi$ as a uniform or uninformative prior. I'm not sure what's not to understand here except the basics of Bayesian inference and the Uniform distribution.

The best way to understand the uniform distribution is via a Monte Carlo sample of fair die rolls. The probability of every outcome $P(X=x_i),~ i=1,...,6$ is equal to $1/6$ and the histogram of the empirical distribution should be approximating a horizontal line (i.e. is uniform).

share|cite|improve this answer

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

$endgroup$

share|cite|improve this answer

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges

answered 5 hours ago

DigioDigio

1,87388 silver badges1717 bronze badges