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Understanding Dirichlet Process (vs Dirichlet Distribution)
$begingroup$
I'm studying the Dirichlet process but I'm confused.
I think I can visualize it, through the Chinese Restaurant Example or The Pólya urn scheme.
But I cannot see the connection with Dirichlet Distribution.
Particularly on Wikipedia regarding it, it's stated that
The Dirichlet process can also be seen as the infinite-dimensional
generalization of the Dirichlet distribution.
How can it be? Can you provide me more insight regarding this sentence?
probability-distributions stochastic-processes
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
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add a comment |
$begingroup$
I'm studying the Dirichlet process but I'm confused.
I think I can visualize it, through the Chinese Restaurant Example or The Pólya urn scheme.
But I cannot see the connection with Dirichlet Distribution.
Particularly on Wikipedia regarding it, it's stated that
The Dirichlet process can also be seen as the infinite-dimensional
generalization of the Dirichlet distribution.
How can it be? Can you provide me more insight regarding this sentence?
probability-distributions stochastic-processes
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
$endgroup$
add a comment |
$begingroup$
I'm studying the Dirichlet process but I'm confused.
I think I can visualize it, through the Chinese Restaurant Example or The Pólya urn scheme.
But I cannot see the connection with Dirichlet Distribution.
Particularly on Wikipedia regarding it, it's stated that
The Dirichlet process can also be seen as the infinite-dimensional
generalization of the Dirichlet distribution.
How can it be? Can you provide me more insight regarding this sentence?
probability-distributions stochastic-processes
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
$endgroup$
I'm studying the Dirichlet process but I'm confused.
I think I can visualize it, through the Chinese Restaurant Example or The Pólya urn scheme.
But I cannot see the connection with Dirichlet Distribution.
Particularly on Wikipedia regarding it, it's stated that
The Dirichlet process can also be seen as the infinite-dimensional
generalization of the Dirichlet distribution.
How can it be? Can you provide me more insight regarding this sentence?
probability-distributions stochastic-processes
probability-distributions stochastic-processes
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
asked Jan 1 at 10:47
Tommaso BendinelliTommaso Bendinelli
1218
1218
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, dots, G(A_k)) sim Dir(MG_0(A_1), dots MG_0(A_k)$$
This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.
Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.
EDIT: (In simpler language)
Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.
answered Jan 1 at 10:59
twnlytwnly
534110
$endgroup$
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, dots, G(A_k)) sim Dir(MG_0(A_1), dots MG_0(A_k)$$
This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.
Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.
EDIT: (In simpler language)
Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.
answered Jan 1 at 10:59
twnlytwnly
534110
$endgroup$
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
add a comment |
$begingroup$
One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, dots, G(A_k)) sim Dir(MG_0(A_1), dots MG_0(A_k)$$
This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.
Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.
EDIT: (In simpler language)
Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.
answered Jan 1 at 10:59
twnlytwnly
534110
$endgroup$
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
add a comment |
$begingroup$
One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, dots, G(A_k)) sim Dir(MG_0(A_1), dots MG_0(A_k)$$
This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.
Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.
EDIT: (In simpler language)
Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.
answered Jan 1 at 10:59
twnlytwnly
534110
$endgroup$
One definition of the Dirichlet process is that given a probability space, $(X, A, G)$ and an arbitrary partition of $X$ given by $A_1, dots A_k$ a distribution $G$ is a Dirichlet process with probability measure $G_0$ and mass parameter $M$ if we have $$G(A_1, dots, G(A_k)) sim Dir(MG_0(A_1), dots MG_0(A_k)$$
This can be found from an article here. This explains the definition in terms of the Dirichlet distribution and where the Dirichlet name comes from.
Now then what you do is that you verify some conditions, and then apply the Kolmogorov Extension Theorem, which gives you the desired infinite-dimensional distribution from the earlier collection of finite-dimensional distributions.
EDIT: (In simpler language)
Take all subsets of the original probability space, the Dirichlet process is a distribution where any group of subsets follow the Dirichlet distribution. Now what you have is a collection of finite-dimensional distributions. To get an infinite dimensional distributions from this, you have to use the Kolmogorov extension theorem. So you check a few properties to check that the theorem holds, and what it states is that your definition from finite dimensional distributions determines an infinite-dimensional distribution uniquely.
answered Jan 1 at 10:59
twnlytwnly
534110
edited Jan 1 at 19:50
answered Jan 1 at 10:59
twnlytwnly
534110
answered Jan 1 at 10:59
twnlytwnly
534110
answered Jan 1 at 10:59
twnlytwnly
534110
534110
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
add a comment |
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Thank you for answering, but your explanation it's still too technical for my background
$endgroup$
– Tommaso Bendinelli
Jan 1 at 13:29
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Added a paragraph to describe what is going on. Does it help?
$endgroup$
– twnly
Jan 1 at 19:43
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
$begingroup$
Yes, thank you make more sense. Regarding the part " the Dirichlet process is a distribution where any group of subsets follows the Dirichlet distribution", can we rephrase the sentence by saying that if the random variable X defined in a given probability space follow a Dirichlet distribution, then a sequence of random variables (X1, X2, X3) forms a Dirichlet Process?
$endgroup$
– Tommaso Bendinelli
Jan 2 at 11:31
add a comment |
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