What mean $int mathcal D(n(x)) delta left(n(x)-frac{1}{N}sum_{k=1}^N delta (x-x_i)right)$?
$begingroup$
In formula (4.5) page 27 of this document, what mean
$$int mathcal D(n(x)) delta left(n(x)-frac{1}{N}sum_{k=1}^N delta (x-x_i)right),$$
where the integral run over all possible normalized, non negative smooth function $n(x)$, and $delta $ is the $delta $ distribution. Could someone explain what is this integral, how it work and why this integral is $1$ ?
probability-theory mathematical-physics
edited Dec 14 '18 at 12:19
Did
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
$endgroup$
add a comment |
$begingroup$
In formula (4.5) page 27 of this document, what mean
$$int mathcal D(n(x)) delta left(n(x)-frac{1}{N}sum_{k=1}^N delta (x-x_i)right),$$
where the integral run over all possible normalized, non negative smooth function $n(x)$, and $delta $ is the $delta $ distribution. Could someone explain what is this integral, how it work and why this integral is $1$ ?
probability-theory mathematical-physics
edited Dec 14 '18 at 12:19
Did
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
$endgroup$
2
$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20
add a comment |
$begingroup$
In formula (4.5) page 27 of this document, what mean
$$int mathcal D(n(x)) delta left(n(x)-frac{1}{N}sum_{k=1}^N delta (x-x_i)right),$$
where the integral run over all possible normalized, non negative smooth function $n(x)$, and $delta $ is the $delta $ distribution. Could someone explain what is this integral, how it work and why this integral is $1$ ?
probability-theory mathematical-physics
edited Dec 14 '18 at 12:19
Did
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
$endgroup$
In formula (4.5) page 27 of this document, what mean
$$int mathcal D(n(x)) delta left(n(x)-frac{1}{N}sum_{k=1}^N delta (x-x_i)right),$$
where the integral run over all possible normalized, non negative smooth function $n(x)$, and $delta $ is the $delta $ distribution. Could someone explain what is this integral, how it work and why this integral is $1$ ?
probability-theory mathematical-physics
probability-theory mathematical-physics
edited Dec 14 '18 at 12:19
Did
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
edited Dec 14 '18 at 12:19
Did
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
edited Dec 14 '18 at 12:19
Did
246k23222458
edited Dec 14 '18 at 12:19
Did
246k23222458
edited Dec 14 '18 at 12:19
Did
246k23222458
246k23222458
asked Dec 14 '18 at 11:53
user623855user623855
927
asked Dec 14 '18 at 11:53
user623855user623855
927
asked Dec 14 '18 at 11:53
user623855user623855
927
927
2
$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20
add a comment |
2
$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20
2
2
$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20
$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20
add a comment |
1 Answer
1
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Looks like a path integral. Actually the result should be $frac{1}{N}sum_{k=1}^N delta (x-x_i)$. Then when you integrate over $x$ one more time it is $=1$.
Look into distributions to get a explanation for the $delta()$. The one appearing in your integral is a sort of infinite dimensional delta distribution.
answered Dec 14 '18 at 12:13
tonydotonydo
1477
$endgroup$
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Looks like a path integral. Actually the result should be $frac{1}{N}sum_{k=1}^N delta (x-x_i)$. Then when you integrate over $x$ one more time it is $=1$.
Look into distributions to get a explanation for the $delta()$. The one appearing in your integral is a sort of infinite dimensional delta distribution.
answered Dec 14 '18 at 12:13
tonydotonydo
1477
$endgroup$
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
add a comment |
$begingroup$
Looks like a path integral. Actually the result should be $frac{1}{N}sum_{k=1}^N delta (x-x_i)$. Then when you integrate over $x$ one more time it is $=1$.
Look into distributions to get a explanation for the $delta()$. The one appearing in your integral is a sort of infinite dimensional delta distribution.
answered Dec 14 '18 at 12:13
tonydotonydo
1477
$endgroup$
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
add a comment |
$begingroup$
Looks like a path integral. Actually the result should be $frac{1}{N}sum_{k=1}^N delta (x-x_i)$. Then when you integrate over $x$ one more time it is $=1$.
Look into distributions to get a explanation for the $delta()$. The one appearing in your integral is a sort of infinite dimensional delta distribution.
answered Dec 14 '18 at 12:13
tonydotonydo
1477
$endgroup$
Looks like a path integral. Actually the result should be $frac{1}{N}sum_{k=1}^N delta (x-x_i)$. Then when you integrate over $x$ one more time it is $=1$.
Look into distributions to get a explanation for the $delta()$. The one appearing in your integral is a sort of infinite dimensional delta distribution.
answered Dec 14 '18 at 12:13
tonydotonydo
1477
answered Dec 14 '18 at 12:13
tonydotonydo
1477
answered Dec 14 '18 at 12:13
tonydotonydo
1477
answered Dec 14 '18 at 12:13
tonydotonydo
1477
1477
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
add a comment |
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
I in fact have more difficultie to understand $mathcal D(n(x))$ that the $delta $.
$endgroup$
– user623855
Dec 14 '18 at 12:54
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
you can approximate it by discretizing n(x) into K pieces with K very large. Then $int mathcal D(n(x))$ is a K dimensional integral.
$endgroup$
– tonydo
Dec 14 '18 at 13:02
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
$begingroup$
It's not so easy to make it rigorous. Its an infinite dimensional integration.
$endgroup$
– tonydo
Dec 14 '18 at 13:05
add a comment |
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$begingroup$
Please do not force people to download the whole PDF, simply to be able to know the title and authors and abstract of the document.
$endgroup$
– Did
Dec 14 '18 at 12:20