Bayesian statistics and convergence to “the true distribution”

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Suppose we have a random variable $mathbb X : Omega to E$ and let's define random variables $mathbb Y_1, mathbb Y_2, mathbb Y_3, ldots, $ all i.i.d to $mathbb X$. Now we choose an arbitrary distribution $mathbb Z_0$ with the same range as $mathbb X$, called our prior.

Now we define our posterior random variable $mathbb Z_1$ via Bayesian adjustment according to Bayes' law observing $mathbb Y_1$. Consider $mathbb Z_1$ to be our prior. Repeat this and define a sequence $leftmathbb Z_n right_n=1^infty$.

Does it make sense to talk about a limit whereas $mathbb Z_n to mathbb X$ in probability, meaning that for all $varepsilon >0$ we have $displaystylelim_ntoinfty mathbb P(mathbb| Z_n - mathbb X| < varepsilon) = 1$? Does a stochastic random variable $mathbb Z_0$ always exist?

probability probability-theory measure-theory bayesian bayes-theorem

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edited Jul 25 at 0:34

asked Jul 24 at 23:42

Markus Klyver

350314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it makes sense.
  – David G. Stork
  Jul 24 at 23:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
  – Michael Hardy
  Jul 24 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
  – Markus Klyver
  Jul 25 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
  – David G. Stork
  Jul 25 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
  – Markus Klyver
  Jul 25 at 0:37
  
  

  
  
  
  

 | 
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up vote
2
down vote

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Suppose we have a random variable $mathbb X : Omega to E$ and let's define random variables $mathbb Y_1, mathbb Y_2, mathbb Y_3, ldots, $ all i.i.d to $mathbb X$. Now we choose an arbitrary distribution $mathbb Z_0$ with the same range as $mathbb X$, called our prior.

Now we define our posterior random variable $mathbb Z_1$ via Bayesian adjustment according to Bayes' law observing $mathbb Y_1$. Consider $mathbb Z_1$ to be our prior. Repeat this and define a sequence $leftmathbb Z_n right_n=1^infty$.

Does it make sense to talk about a limit whereas $mathbb Z_n to mathbb X$ in probability, meaning that for all $varepsilon >0$ we have $displaystylelim_ntoinfty mathbb P(mathbb| Z_n - mathbb X| < varepsilon) = 1$? Does a stochastic random variable $mathbb Z_0$ always exist?

probability probability-theory measure-theory bayesian bayes-theorem

share|cite|improve this question

edited Jul 25 at 0:34

asked Jul 24 at 23:42

Markus Klyver

350314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it makes sense.
  – David G. Stork
  Jul 24 at 23:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
  – Michael Hardy
  Jul 24 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
  – Markus Klyver
  Jul 25 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
  – David G. Stork
  Jul 25 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
  – Markus Klyver
  Jul 25 at 0:37
  
  

  
  
  
  

 | 
show 1 more comment

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Suppose we have a random variable $mathbb X : Omega to E$ and let's define random variables $mathbb Y_1, mathbb Y_2, mathbb Y_3, ldots, $ all i.i.d to $mathbb X$. Now we choose an arbitrary distribution $mathbb Z_0$ with the same range as $mathbb X$, called our prior.

Now we define our posterior random variable $mathbb Z_1$ via Bayesian adjustment according to Bayes' law observing $mathbb Y_1$. Consider $mathbb Z_1$ to be our prior. Repeat this and define a sequence $leftmathbb Z_n right_n=1^infty$.

Does it make sense to talk about a limit whereas $mathbb Z_n to mathbb X$ in probability, meaning that for all $varepsilon >0$ we have $displaystylelim_ntoinfty mathbb P(mathbb| Z_n - mathbb X| < varepsilon) = 1$? Does a stochastic random variable $mathbb Z_0$ always exist?

probability probability-theory measure-theory bayesian bayes-theorem

share|cite|improve this question

edited Jul 25 at 0:34

asked Jul 24 at 23:42

Markus Klyver

350314

Suppose we have a random variable $mathbb X : Omega to E$ and let's define random variables $mathbb Y_1, mathbb Y_2, mathbb Y_3, ldots, $ all i.i.d to $mathbb X$. Now we choose an arbitrary distribution $mathbb Z_0$ with the same range as $mathbb X$, called our prior.

Now we define our posterior random variable $mathbb Z_1$ via Bayesian adjustment according to Bayes' law observing $mathbb Y_1$. Consider $mathbb Z_1$ to be our prior. Repeat this and define a sequence $leftmathbb Z_n right_n=1^infty$.

Does it make sense to talk about a limit whereas $mathbb Z_n to mathbb X$ in probability, meaning that for all $varepsilon >0$ we have $displaystylelim_ntoinfty mathbb P(mathbb| Z_n - mathbb X| < varepsilon) = 1$? Does a stochastic random variable $mathbb Z_0$ always exist?

probability probability-theory measure-theory bayesian bayes-theorem

share|cite|improve this question

edited Jul 25 at 0:34

asked Jul 24 at 23:42

Markus Klyver

350314

share|cite|improve this question

share|cite|improve this question

edited Jul 25 at 0:34

edited Jul 25 at 0:34

edited Jul 25 at 0:34

asked Jul 24 at 23:42

Markus Klyver

350314

asked Jul 24 at 23:42

Markus Klyver

350314

asked Jul 24 at 23:42

Markus Klyver

350314

350314

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it makes sense.
  – David G. Stork
  Jul 24 at 23:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
  – Michael Hardy
  Jul 24 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
  – Markus Klyver
  Jul 25 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
  – David G. Stork
  Jul 25 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
  – Markus Klyver
  Jul 25 at 0:37
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Yes, it makes sense.
  – David G. Stork
  Jul 24 at 23:46
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
  – Michael Hardy
  Jul 24 at 23:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
  – Markus Klyver
  Jul 25 at 0:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
  – David G. Stork
  Jul 25 at 0:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
  – Markus Klyver
  Jul 25 at 0:37
  
  

  
  
  
  

Yes, it makes sense.
– David G. Stork
Jul 24 at 23:46

Yes, it makes sense.
– David G. Stork
Jul 24 at 23:46

3

3

I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
– Michael Hardy
Jul 24 at 23:47

I prefer to call it the "population distribution" rather than the "true distribution". Other probability distributions involved in the problem are not less "true". $qquad$
– Michael Hardy
Jul 24 at 23:47

@DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
– Markus Klyver
Jul 25 at 0:18

@DavidG.Stork In what way does it make sense? Will a such initial $mathbb Z_0$ always exist?
– Markus Klyver
Jul 25 at 0:18

Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
– David G. Stork
Jul 25 at 0:22

Frankly, my answer was somewhat ironic, because your question was so vague, so unfocused and un-answerable. You asked "does it make sense...?" to which the answer is "yes." If you want to ask a better question, you'll more likely get a better answer.
– David G. Stork
Jul 25 at 0:22

@DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
– Markus Klyver
Jul 25 at 0:37

@DavidG.Stork I think it's quite clear. We want to "find" $mathbb X$ by defining a sequence of $mathbb Z_k$s and each iteration is defined by using Bayes' law to update the distribution in the light of a new observation $mathbb Y_k$. The question is about a limiting process as $ktoinfty.$
– Markus Klyver
Jul 25 at 0:37

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