Skip to main content
deleted 8 characters in body
Source Link
Benoit Sanchez
  • 8.8k
  • 28
  • 51

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is aThis prior that generalizes the uniform prior for translation invariant problems. Jeffrey's prior somehow adapts to the (information theoretic) Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you answer 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approaches simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (over $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to Bayes estimators. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. It might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happen in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is a prior that generalizes the uniform prior for translation invariant problems. Jeffrey's prior somehow adapts to the (information theoretic) Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you answer 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approaches simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (over $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to Bayes estimators. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. It might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happen in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". This prior generalizes the uniform prior for translation invariant problems. Jeffrey's prior somehow adapts to the (information theoretic) Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you answer 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approaches simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (over $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to Bayes estimators. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. It might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happen in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

edited body
Source Link
Benoit Sanchez
  • 8.8k
  • 28
  • 51

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is a prior that generalizes the uniform prior for translation invariant problems. The JeffreyJeffrey's prior somehow adapts to the (information theoretic) Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you anwseranswer 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approachapproaches simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (aboutover $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to the Bayes criterionestimators. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. IIt might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happenshappen in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is a prior that generalizes the uniform prior for translation invariant problems. The Jeffrey prior somehow adapts to the Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you anwser 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approach simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (about $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to the Bayes criterion. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. I might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happens in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is a prior that generalizes the uniform prior for translation invariant problems. Jeffrey's prior somehow adapts to the (information theoretic) Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you answer 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approaches simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (over $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to Bayes estimators. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. It might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happen in more complicated problems in my opinion. What is important here is to understand what a certain prior says.

Source Link
Benoit Sanchez
  • 8.8k
  • 28
  • 51

Q1: Is the absence of a prior equivalent (in the strict theoretical sense) to having an uninformative prior?

No.

First, there is no mathematical definition for an "uninformative prior". This word is only used informally to describe some priors.

For example, Jeffrey's prior is often called "uninformative". It is a prior that generalizes the uniform prior for translation invariant problems. The Jeffrey prior somehow adapts to the Riemannian geometry of the model and thus is independent of parametrization, only dependent on the geometry of the manifold (in the space of distributions) that is the model. It might be perceived as canonical, but it's only a choice. It's just the uniform prior according to Riemannian structure. It's not absurd to define "uninformative = uniform" as a simplification of the question. This applies to many cases and helps to ask a clear and simple question.

Doing Bayesian inference without a prior is like "how can I guess $E(X)$ without any assumption about the distribution of $X$ only knowing that $X$ has values in $[0;1]$?" This question obviously makes no sense. If you anwser 0.5, you probably have a distribution in mind.

The Bayesian and frequentist approach simply answer different questions. For example, about estimators which is maybe the simplest:

  • Frequentist (for example): "How can I estimate $\theta$ such that my answer has the smallest error (only averaged over $x$) in the worst case (about $\theta$)?". This leads to minimax estimators.

  • Bayesian: "How can I estimate $\theta$ such that my answer has the smallest error in average (over $\theta$) ?". This leads to the Bayes criterion. But the question is incomplete and must specify "average in what sense?". Thus the question is only complete when it contains a prior.

Somehow, frequentist aims at worst case control and does not need a prior. Bayesian aims at average control and requires a prior to say "average in what sense?".

Q2 : If the answer to Q1 is "No", does this mean that, in cases where there are no priors, the Bayesian approach is not applicable from the beginning, and we have to first form a prior by some non-Bayesian way, so that we can subsequently apply the Bayesian approach?

Yes.

But beware of canonical prior construction. I might sound mathematically appealing but is not automatically realistic from a Bayesian point of view. It is possible a mathematically nice prior actually corresponds to a dumb belief system. For example if you study $X\sim N(\mu,1)$, Jeffrey's prior on $\mu$ is uniform and if about people's average size, this might not be a very realistic system. However with only a few observations, the problem actually disappears quite fast. The choice is not very important.

True problems with prior specification happens in more complicated problems in my opinion. What is important here is to understand what a certain prior says.