Measure on $C([0,1]^n)$
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We can define a measure on $C[0,1]$ by viewing a continuous function as a path of 1-dimensional Brownian motion. This is called classical Wiener measure. I am wondering if there is a generalization to function space of higher dimension?
I knew from Wikipedia that there is something called abstract Wiener space, which uses a canonical Gaussian cylinder set measure. However I do not know what is the "quotient inner product" mentioned in the construction. Could anyone explain for me what is a quotient inner product, and how does it coincide with the classical case?
probability functional-analysis measure-theory stochastic-processes
asked Nov 19 at 10:25
1830rbc03
39046
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We can define a measure on $C[0,1]$ by viewing a continuous function as a path of 1-dimensional Brownian motion. This is called classical Wiener measure. I am wondering if there is a generalization to function space of higher dimension?
I knew from Wikipedia that there is something called abstract Wiener space, which uses a canonical Gaussian cylinder set measure. However I do not know what is the "quotient inner product" mentioned in the construction. Could anyone explain for me what is a quotient inner product, and how does it coincide with the classical case?
probability functional-analysis measure-theory stochastic-processes
asked Nov 19 at 10:25
1830rbc03
39046
To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
Oh, you are right!
– p4sch
Nov 19 at 17:21
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
We can define a measure on $C[0,1]$ by viewing a continuous function as a path of 1-dimensional Brownian motion. This is called classical Wiener measure. I am wondering if there is a generalization to function space of higher dimension?
I knew from Wikipedia that there is something called abstract Wiener space, which uses a canonical Gaussian cylinder set measure. However I do not know what is the "quotient inner product" mentioned in the construction. Could anyone explain for me what is a quotient inner product, and how does it coincide with the classical case?
probability functional-analysis measure-theory stochastic-processes
asked Nov 19 at 10:25
1830rbc03
39046
We can define a measure on $C[0,1]$ by viewing a continuous function as a path of 1-dimensional Brownian motion. This is called classical Wiener measure. I am wondering if there is a generalization to function space of higher dimension?
I knew from Wikipedia that there is something called abstract Wiener space, which uses a canonical Gaussian cylinder set measure. However I do not know what is the "quotient inner product" mentioned in the construction. Could anyone explain for me what is a quotient inner product, and how does it coincide with the classical case?
probability functional-analysis measure-theory stochastic-processes
probability functional-analysis measure-theory stochastic-processes
asked Nov 19 at 10:25
1830rbc03
39046
asked Nov 19 at 10:25
1830rbc03
39046
edited Nov 20 at 19:07
asked Nov 19 at 10:25
1830rbc03
39046
asked Nov 19 at 10:25
1830rbc03
39046
asked Nov 19 at 10:25
1830rbc03
39046
39046
To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
Oh, you are right!
– p4sch
Nov 19 at 17:21
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08
|
show 1 more comment
To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
Oh, you are right!
– p4sch
Nov 19 at 17:21
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08
To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
Oh, you are right!
– p4sch
Nov 19 at 17:21
Oh, you are right!
– p4sch
Nov 19 at 17:21
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08
|
show 1 more comment
1 Answer
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0
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Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
add a comment |
up vote
0
down vote
Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
add a comment |
up vote
0
down vote
up vote
0
down vote
Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
Your question is ill-posed as is. There exists probability measures on $C([0,1]^n)$, for example the unit Dirac mass at the constant function equal to zero. It would help if you said which properties you would like your measure to satisfy. Note that the natural generalization of Brownian motion to higher dimensional domains is the so-called Gaussian free field but it is a measure supported on generalized functions/Schwartz distributions instead of continuous functions.
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
answered Nov 20 at 16:01
Abdelmalek Abdesselam
396110
396110
add a comment |
add a comment |
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To 1: Of course, there is a simple generalization to the $d$-dimensional case. Just take the $d$-dimensional browinan motion. The $d$-dimensional BM can be defined by setting $B_t = (B_t^{(1)}, ldots B_t^{(d)})$, where $B_t^{(1)},ldots B_t^{(d)}$ are continuous independent Brownian Motions.
– p4sch
Nov 19 at 14:50
@p4sch But isn't that a measure on $R$ to $R^{n}$ functions instead of $R^{n}$ to $R$ functions?
– 1830rbc03
Nov 19 at 17:15
Oh, you are right!
– p4sch
Nov 19 at 17:21
I find interesting the idea of using a continuous function of ${C[0,1]}^n$ to $C([0,1]^n)$ to induce a distribution in $C([0,1]^n)$.
– Daniel Camarena Perez
Nov 20 at 5:18
@Daniel Camarena Perez I just realized the "measure" I construct in 1 is wrong because not every function $h(x,y)$ can be represented as $f(x)+g(y)$ so please ignore it.
– 1830rbc03
Nov 20 at 19:08