Why Kolmogorov extension theorem is so important. I don't really see the implications behind.
$begingroup$
In many books/articles, the Kolmogorov extension theorem is mentioned, but I don't really understand in what this theorem is so important. The statement is the following :
Let $T$ denote some interval (thought of as time), and let $ninmathbb N$. For each $kinmathbb N$ and finite sequence of distinct times $t_1,...,t_kin T$, let $nu_{t_1,...,t_k}$ be a probability measure on $(mathbb R^n)^k$. Suppose that these measure satisfy two consistency conditions :
1) For all permutation $pi$ of ${1,...,k}$ and measurable set $F_isubset mathbb R^n$,$$nu_{pi(1),...,pi(k)}(F_{pi(1)}times ...times F_{pi(k)}),$$
2) For all measurable sets $F_isubset mathbb R^n$,
$minmathbb N$ $$nu_{t_1,...,t_k}(F_1times ...times F_k)=nu_{t_1,...,t_k,t_{k+1},...,t_{k+m}}(F_1times ...times F_ktimes mathbb R^ntimes ...times mathbb R^n).$$
Then, there is a probability space $(Omega ,mathcal F,mathbb P)$ and a stochastic process $X:Ttimes Omegato mathbb R^n$ s.t. $$nu_{t_1...t_k}(F_1times ...times F_k)=mathbb P(X_{t_1}in F_1,...,X_{t_{k}}in F_k),$$ for all $t_iin T$, $kinmathbb N$ and measurable sets $F_isubset mathbb R^n$, i.e. $X$ has $nu_{t_1,...,t_k}$ as its finite dimensional distribution relative to times $t_1,...,t_k$.
First, the two consistency conditions looks quite natural, no ? How can such condition doesn't hold . I don't have any example, could someone give one ? After, I don't really understand the conclusion. If I have a stochastic process, then of course $$mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k,X_{t_{k+1}}in mathbb R^n}= mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k}$$
probability measure-theory
asked Jan 2 at 17:42
idmidm
8,57821445
$endgroup$
|
show 9 more comments
$begingroup$
In many books/articles, the Kolmogorov extension theorem is mentioned, but I don't really understand in what this theorem is so important. The statement is the following :
Let $T$ denote some interval (thought of as time), and let $ninmathbb N$. For each $kinmathbb N$ and finite sequence of distinct times $t_1,...,t_kin T$, let $nu_{t_1,...,t_k}$ be a probability measure on $(mathbb R^n)^k$. Suppose that these measure satisfy two consistency conditions :
1) For all permutation $pi$ of ${1,...,k}$ and measurable set $F_isubset mathbb R^n$,$$nu_{pi(1),...,pi(k)}(F_{pi(1)}times ...times F_{pi(k)}),$$
2) For all measurable sets $F_isubset mathbb R^n$,
$minmathbb N$ $$nu_{t_1,...,t_k}(F_1times ...times F_k)=nu_{t_1,...,t_k,t_{k+1},...,t_{k+m}}(F_1times ...times F_ktimes mathbb R^ntimes ...times mathbb R^n).$$
Then, there is a probability space $(Omega ,mathcal F,mathbb P)$ and a stochastic process $X:Ttimes Omegato mathbb R^n$ s.t. $$nu_{t_1...t_k}(F_1times ...times F_k)=mathbb P(X_{t_1}in F_1,...,X_{t_{k}}in F_k),$$ for all $t_iin T$, $kinmathbb N$ and measurable sets $F_isubset mathbb R^n$, i.e. $X$ has $nu_{t_1,...,t_k}$ as its finite dimensional distribution relative to times $t_1,...,t_k$.
First, the two consistency conditions looks quite natural, no ? How can such condition doesn't hold . I don't have any example, could someone give one ? After, I don't really understand the conclusion. If I have a stochastic process, then of course $$mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k,X_{t_{k+1}}in mathbb R^n}= mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k}$$
probability measure-theory
asked Jan 2 at 17:42
idmidm
8,57821445
$endgroup$
1
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
1
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01
|
show 9 more comments
$begingroup$
In many books/articles, the Kolmogorov extension theorem is mentioned, but I don't really understand in what this theorem is so important. The statement is the following :
Let $T$ denote some interval (thought of as time), and let $ninmathbb N$. For each $kinmathbb N$ and finite sequence of distinct times $t_1,...,t_kin T$, let $nu_{t_1,...,t_k}$ be a probability measure on $(mathbb R^n)^k$. Suppose that these measure satisfy two consistency conditions :
1) For all permutation $pi$ of ${1,...,k}$ and measurable set $F_isubset mathbb R^n$,$$nu_{pi(1),...,pi(k)}(F_{pi(1)}times ...times F_{pi(k)}),$$
2) For all measurable sets $F_isubset mathbb R^n$,
$minmathbb N$ $$nu_{t_1,...,t_k}(F_1times ...times F_k)=nu_{t_1,...,t_k,t_{k+1},...,t_{k+m}}(F_1times ...times F_ktimes mathbb R^ntimes ...times mathbb R^n).$$
Then, there is a probability space $(Omega ,mathcal F,mathbb P)$ and a stochastic process $X:Ttimes Omegato mathbb R^n$ s.t. $$nu_{t_1...t_k}(F_1times ...times F_k)=mathbb P(X_{t_1}in F_1,...,X_{t_{k}}in F_k),$$ for all $t_iin T$, $kinmathbb N$ and measurable sets $F_isubset mathbb R^n$, i.e. $X$ has $nu_{t_1,...,t_k}$ as its finite dimensional distribution relative to times $t_1,...,t_k$.
First, the two consistency conditions looks quite natural, no ? How can such condition doesn't hold . I don't have any example, could someone give one ? After, I don't really understand the conclusion. If I have a stochastic process, then of course $$mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k,X_{t_{k+1}}in mathbb R^n}= mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k}$$
probability measure-theory
asked Jan 2 at 17:42
idmidm
8,57821445
$endgroup$
In many books/articles, the Kolmogorov extension theorem is mentioned, but I don't really understand in what this theorem is so important. The statement is the following :
Let $T$ denote some interval (thought of as time), and let $ninmathbb N$. For each $kinmathbb N$ and finite sequence of distinct times $t_1,...,t_kin T$, let $nu_{t_1,...,t_k}$ be a probability measure on $(mathbb R^n)^k$. Suppose that these measure satisfy two consistency conditions :
1) For all permutation $pi$ of ${1,...,k}$ and measurable set $F_isubset mathbb R^n$,$$nu_{pi(1),...,pi(k)}(F_{pi(1)}times ...times F_{pi(k)}),$$
2) For all measurable sets $F_isubset mathbb R^n$,
$minmathbb N$ $$nu_{t_1,...,t_k}(F_1times ...times F_k)=nu_{t_1,...,t_k,t_{k+1},...,t_{k+m}}(F_1times ...times F_ktimes mathbb R^ntimes ...times mathbb R^n).$$
Then, there is a probability space $(Omega ,mathcal F,mathbb P)$ and a stochastic process $X:Ttimes Omegato mathbb R^n$ s.t. $$nu_{t_1...t_k}(F_1times ...times F_k)=mathbb P(X_{t_1}in F_1,...,X_{t_{k}}in F_k),$$ for all $t_iin T$, $kinmathbb N$ and measurable sets $F_isubset mathbb R^n$, i.e. $X$ has $nu_{t_1,...,t_k}$ as its finite dimensional distribution relative to times $t_1,...,t_k$.
First, the two consistency conditions looks quite natural, no ? How can such condition doesn't hold . I don't have any example, could someone give one ? After, I don't really understand the conclusion. If I have a stochastic process, then of course $$mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k,X_{t_{k+1}}in mathbb R^n}= mathbb P{X_{t_1}in F_1,...,X_{t_k}in F_k}$$
probability measure-theory
probability measure-theory
asked Jan 2 at 17:42
idmidm
8,57821445
asked Jan 2 at 17:42
idmidm
8,57821445
edited Jan 2 at 21:27
idm
asked Jan 2 at 17:42
idmidm
8,57821445
asked Jan 2 at 17:42
idmidm
8,57821445
asked Jan 2 at 17:42
idmidm
8,57821445
8,57821445
1
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
1
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01
|
show 9 more comments
1
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
1
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01
1
1
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
1
1
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01
|
show 9 more comments
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059741%2fwhy-kolmogorov-extension-theorem-is-so-important-i-dont-really-see-the-implica%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059741%2fwhy-kolmogorov-extension-theorem-is-so-important-i-dont-really-see-the-implica%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Kolmogorov's existence theorem is saying that for any family of probability measures satisfying the consistency condition, we can construct an associated process. What you are saying is that the finite-dimensional distributions of a stochastic process satisfy the consistency conditions... that's also true, but it's not the statement of the theorem. You are talking about necessary conditions for the existence of a stochastic process (with given fdds) whereas Kolmogorov's existence theorem is about sufficient conditions for the existence of a process (with given fdd's).
$endgroup$
– saz
Jan 2 at 18:23
$begingroup$
@saz : I think I see better thank you. But could you please give me an example of situation where it's necessary ? Let $(Omega ,mathcal F,mathbb P)$ a probability space. Let $(X_t)$ a Brownian motion (or any other stochastic process) on this probability space. Where do I have to use Kolmogorov theorem ? If it's to general, consider $([0,1], mathscr B(mathbb R),m)$ where $mathscr B(mathbb R)$ is the Borel $sigma -$algebra and $m$ Lebesgue measure.
$endgroup$
– idm
Jan 2 at 21:31
$begingroup$
Kolmogorov's result allows us to prove the existence of a Brownian motion (and many other stochastic processes).
$endgroup$
– saz
Jan 3 at 7:32
$begingroup$
@saz : Ok, I think I get it. Just a precision, let $(X_i)_{iinmathbb N}$ a sequence of random variable on $(Omega ,mathcal F,mathbb P)$. Does Kolmogorov extension tel us that there is a probability space $(tilde Omega , mathcal G,mathbb Q)$ s.t. $(X_i)_{iinmathbb N}$ is measurable and such that $mathbb P_k(X_{t_1}in A_1,..., X_{t_k}in A_k)=mathbb Q(X_{t_1}in A_1,..., X_{t_k}in A_k)$ where $mathbb P_k=mathbb P_{t_1}otimes...otimes mathbb P_{t_k}$,
$endgroup$
– idm
Jan 3 at 13:52
1
$begingroup$
@idm You are getting closer, but it seems to me that you are still thinking the wrong way round. Note that the family of random variables $(X_i)_i$ is not given. You are given the family of measures $nu_{t_1,ldots,t_n}$ and you have to construct $(X_i)_{i in T}$ such that the distribution of $(X_{t_1},ldots,X_{t_n})$ equals $nu_{t_1,ldots,t_n}$ for any choice of $t_i$.
$endgroup$
– saz
Jan 4 at 13:01