Fdtxjr

Python if-else code style for reduced code for rounding floats

How to make healing in an exploration game interesting

Official degrees of earth’s rotation per day

Is it good practice to use Linear Least-Squares with SMA?

Is a party consisting of only a bard, a cleric, and a warlock functional long-term?

What is a ^ b and (a & b) << 1?

Why do passenger jet manufacturers design their planes with stall prevention systems?

Why is the President allowed to veto a cancellation of emergency powers?

Does multi-classing into Fighter give you heavy armor?

Fastest way to pop N items from a large dict

How to explain that I do not want to visit a country due to personal safety concern?

Non-trivial topology where only open sets are closed

If I am holding an item before I cast Blink, will it move with me through the Ethereal Plane?

Employee lack of ownership

How could an airship be repaired midflight?

World War I as a war of liberals against authoritarians?

Bach's Toccata and Fugue in D minor breaks the "no parallel octaves" rule?

Recruiter wants very extensive technical details about all of my previous work

Is there a hypothetical scenario that would make Earth uninhabitable for humans, but not for (the majority of) other animals?

What does 高層ビルに何車線もの道路。mean?

How to deal with taxi scam when on vacation?

What is "focus distance lower/upper" and how is it different from depth of field?

What are substitutions for coconut in curry?

If I can solve Sudoku, can I solve the Travelling Salesman Problem (TSP)? If so, how?

What does it mean : the finite dimensional distribution determine the random process?

Does the distribution of a process on $mathbbR^[0,infty)$ uniquely define it?definition of stochastic process on countable spaceEqual distribution only for finite dimensional distributionsRecurrence of random walks in terms of the mean of each step.Finite dimensional distribution of a stochastic processDoes a Levy process have the property $mathbbE_x(X_t|mathcalG)=mathbbE(X_t+x|mathcalG)$?For a given sequence of random variables $R_n$, what does the stochastic order relations $o_p(R_n)$ and $O_p(R_n)$ mean intuitively?Pointwise convergence in limit of conditional probabilities implies almost sure convergence?What does the weak convergence of stochastic intensity tell us about the point process?Definition of a Markov process: What does $mathbb P_xX_uin Bmid mathcal F_t=p(u-t,X_t,Gamma)$ mean?

2

$begingroup$

In the book Random perturbation of dynamical system of Fredlin and Wantzell (2nd edition) page 17, it's written :

Let $(Omega ,mathcal F,mathbb P)$ a probability space and $(xi_t)_tin T$ a stochastic process with state space $(X,mathcal B )$. If $T$ is countable, then the finite dimensional distributions determine the random process to the degree of uniqueness usual in probability. If $T$ is an interval, we can have two process (one continuous and ond discontinuous) with the same finite dimensional distribution.

---

I'm not sure if I completly understand this... does it mean (in the countable case), that if for all $Bin mathcal B$, $$mathbb PX_nin B=mathbb PY_nin B$$
for all $ninmathbb N$, then $X_n=Y_n$ a.s. ?

probability

share|cite|improve this question

edited Mar 12 at 10:32

Pierre

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

2

$begingroup$

In the book Random perturbation of dynamical system of Fredlin and Wantzell (2nd edition) page 17, it's written :

Let $(Omega ,mathcal F,mathbb P)$ a probability space and $(xi_t)_tin T$ a stochastic process with state space $(X,mathcal B )$. If $T$ is countable, then the finite dimensional distributions determine the random process to the degree of uniqueness usual in probability. If $T$ is an interval, we can have two process (one continuous and ond discontinuous) with the same finite dimensional distribution.

---

I'm not sure if I completly understand this... does it mean (in the countable case), that if for all $Bin mathcal B$, $$mathbb PX_nin B=mathbb PY_nin B$$
for all $ninmathbb N$, then $X_n=Y_n$ a.s. ?

probability

share|cite|improve this question

edited Mar 12 at 10:32

Pierre

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

In the book Random perturbation of dynamical system of Fredlin and Wantzell (2nd edition) page 17, it's written :

Let $(Omega ,mathcal F,mathbb P)$ a probability space and $(xi_t)_tin T$ a stochastic process with state space $(X,mathcal B )$. If $T$ is countable, then the finite dimensional distributions determine the random process to the degree of uniqueness usual in probability. If $T$ is an interval, we can have two process (one continuous and ond discontinuous) with the same finite dimensional distribution.

---

I'm not sure if I completly understand this... does it mean (in the countable case), that if for all $Bin mathcal B$, $$mathbb PX_nin B=mathbb PY_nin B$$
for all $ninmathbb N$, then $X_n=Y_n$ a.s. ?

probability

share|cite|improve this question

edited Mar 12 at 10:32

Pierre

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 12 at 10:32

Pierre

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 12 at 10:32

Pierre

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 12 at 10:04

PierrePierre

5610

New contributor

Pierre is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 12 at 10:04

PierrePierre

5610

1 Answer
1

active

oldest

votes

1

$begingroup$

No.

It means that if two stochastic processes $xi^1$ and $xi^2$ are such that
$$
forall t_1,cdots,t_nin T,forall A_1,cdots,A_ninmathcal B,mathbb P(xi^1_t_1in A_1,cdots,xi^1_t_nin A_n)=mathbb P(xi^2_t_1in A_1,cdots,xi^2_t_nin A_n),
$$
then $xi^1$ and $xi^2$ have the same distribution, that is to say
$$
forall Ainmathcal B^otimes T,mathbb P((xi^1_t)_tinmathbb Rin A)=mathbb P((xi^2_t)_tinmathbb Rin A).
$$

The short way of saying this is that if two processes $xi^1$ and $xi^2$ have the same finite dimensional distributions, then they have the same distribution. In the general case, you cannot deduce the almost sure equality between the two processes.

Let us now illustrate the last sentence. Let $T=[0,1]$, $U$ be a random variable uniformly distributed on $[0,1]$ and $xi^1$ and $xi^2$ be defined for all $tin T$ by $xi^1_t=0$ and $xi^2_t=1_U=t$ (that is $1$ if $U=t$ and $0$ otherwise). Then for all $tin [0,T]$, $xi^1_t=xi^2_t$ a.s. (since $mathbb P(U=t)=0$) so we easily deduce that $xi^1$ and $xi^2$ have equal finite dimensional distributions and therefore equal distributions. However, $xi^1$ is clearly continuous and $xi^2$ discontinuous.

share|cite|improve this answer

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, but my question was rather for $T=mathbb N$ (or $1,...,n$). The statement looks stronger. (I know that for the case $T=[0,1]$ is not true that the process are a.s. equal when they have same distribution)
  $endgroup$
  – Pierre
  Mar 12 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Everything I said (except the last paragraph of course) is valid for $T$ countable. If $T=mathbb N$ or $1,cdots,n$ you can deduce the equality of the distributions but not the equality a.s.
  $endgroup$
  – Will
  Mar 12 at 12:08
  

  
  
  
  

add a comment | 

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
);



);

Pierre is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3144894%2fwhat-does-it-mean-the-finite-dimensional-distribution-determine-the-random-pro%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

No.

It means that if two stochastic processes $xi^1$ and $xi^2$ are such that
$$
forall t_1,cdots,t_nin T,forall A_1,cdots,A_ninmathcal B,mathbb P(xi^1_t_1in A_1,cdots,xi^1_t_nin A_n)=mathbb P(xi^2_t_1in A_1,cdots,xi^2_t_nin A_n),
$$
then $xi^1$ and $xi^2$ have the same distribution, that is to say
$$
forall Ainmathcal B^otimes T,mathbb P((xi^1_t)_tinmathbb Rin A)=mathbb P((xi^2_t)_tinmathbb Rin A).
$$

The short way of saying this is that if two processes $xi^1$ and $xi^2$ have the same finite dimensional distributions, then they have the same distribution. In the general case, you cannot deduce the almost sure equality between the two processes.

Let us now illustrate the last sentence. Let $T=[0,1]$, $U$ be a random variable uniformly distributed on $[0,1]$ and $xi^1$ and $xi^2$ be defined for all $tin T$ by $xi^1_t=0$ and $xi^2_t=1_U=t$ (that is $1$ if $U=t$ and $0$ otherwise). Then for all $tin [0,T]$, $xi^1_t=xi^2_t$ a.s. (since $mathbb P(U=t)=0$) so we easily deduce that $xi^1$ and $xi^2$ have equal finite dimensional distributions and therefore equal distributions. However, $xi^1$ is clearly continuous and $xi^2$ discontinuous.

share|cite|improve this answer

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, but my question was rather for $T=mathbb N$ (or $1,...,n$). The statement looks stronger. (I know that for the case $T=[0,1]$ is not true that the process are a.s. equal when they have same distribution)
  $endgroup$
  – Pierre
  Mar 12 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Everything I said (except the last paragraph of course) is valid for $T$ countable. If $T=mathbb N$ or $1,cdots,n$ you can deduce the equality of the distributions but not the equality a.s.
  $endgroup$
  – Will
  Mar 12 at 12:08
  

  
  
  
  

add a comment | 

1

$begingroup$

No.

It means that if two stochastic processes $xi^1$ and $xi^2$ are such that
$$
forall t_1,cdots,t_nin T,forall A_1,cdots,A_ninmathcal B,mathbb P(xi^1_t_1in A_1,cdots,xi^1_t_nin A_n)=mathbb P(xi^2_t_1in A_1,cdots,xi^2_t_nin A_n),
$$
then $xi^1$ and $xi^2$ have the same distribution, that is to say
$$
forall Ainmathcal B^otimes T,mathbb P((xi^1_t)_tinmathbb Rin A)=mathbb P((xi^2_t)_tinmathbb Rin A).
$$

The short way of saying this is that if two processes $xi^1$ and $xi^2$ have the same finite dimensional distributions, then they have the same distribution. In the general case, you cannot deduce the almost sure equality between the two processes.

Let us now illustrate the last sentence. Let $T=[0,1]$, $U$ be a random variable uniformly distributed on $[0,1]$ and $xi^1$ and $xi^2$ be defined for all $tin T$ by $xi^1_t=0$ and $xi^2_t=1_U=t$ (that is $1$ if $U=t$ and $0$ otherwise). Then for all $tin [0,T]$, $xi^1_t=xi^2_t$ a.s. (since $mathbb P(U=t)=0$) so we easily deduce that $xi^1$ and $xi^2$ have equal finite dimensional distributions and therefore equal distributions. However, $xi^1$ is clearly continuous and $xi^2$ discontinuous.

share|cite|improve this answer

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for your answer, but my question was rather for $T=mathbb N$ (or $1,...,n$). The statement looks stronger. (I know that for the case $T=[0,1]$ is not true that the process are a.s. equal when they have same distribution)
  $endgroup$
  – Pierre
  Mar 12 at 11:33
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Everything I said (except the last paragraph of course) is valid for $T$ countable. If $T=mathbb N$ or $1,cdots,n$ you can deduce the equality of the distributions but not the equality a.s.
  $endgroup$
  – Will
  Mar 12 at 12:08
  

  
  
  
  

add a comment | 

$begingroup$

No.

It means that if two stochastic processes $xi^1$ and $xi^2$ are such that
$$
forall t_1,cdots,t_nin T,forall A_1,cdots,A_ninmathcal B,mathbb P(xi^1_t_1in A_1,cdots,xi^1_t_nin A_n)=mathbb P(xi^2_t_1in A_1,cdots,xi^2_t_nin A_n),
$$
then $xi^1$ and $xi^2$ have the same distribution, that is to say
$$
forall Ainmathcal B^otimes T,mathbb P((xi^1_t)_tinmathbb Rin A)=mathbb P((xi^2_t)_tinmathbb Rin A).
$$

The short way of saying this is that if two processes $xi^1$ and $xi^2$ have the same finite dimensional distributions, then they have the same distribution. In the general case, you cannot deduce the almost sure equality between the two processes.

Let us now illustrate the last sentence. Let $T=[0,1]$, $U$ be a random variable uniformly distributed on $[0,1]$ and $xi^1$ and $xi^2$ be defined for all $tin T$ by $xi^1_t=0$ and $xi^2_t=1_U=t$ (that is $1$ if $U=t$ and $0$ otherwise). Then for all $tin [0,T]$, $xi^1_t=xi^2_t$ a.s. (since $mathbb P(U=t)=0$) so we easily deduce that $xi^1$ and $xi^2$ have equal finite dimensional distributions and therefore equal distributions. However, $xi^1$ is clearly continuous and $xi^2$ discontinuous.

share|cite|improve this answer

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 11:10

WillWill

1663

New contributor

Will is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 12 at 11:10

WillWill

1663