Let $X,Y$ have the same distribution on common prob space, do they generate same $sigma$-algebra?
$begingroup$
So let $X,Y$ be real random variables on common probability space $(Omega, mathcal{F}, P)$, the measures on Borel $(mathbb{R},mathcal{B}_{mathbb{R}})$ induced by $X$ and $Y$ are equal, that is for all $A in mathcal{B}_{mathbb{R}}$.
$$ P_{X}(A) = P_{Y}(A)$$
where
$$P_{X}(A) := P(X^{-1}(A))$$ and
$$P_{Y}(A) := P(Y^{-1}(A))$$
Do $X,Y$ generate the same $sigma-$algebra? I feel that it might not be necessarily the case but was not able to construct a counter example.
Any help would be appreciated
probability measure-theory lebesgue-measure
asked Apr 10 '16 at 16:42
themthem
296214
$endgroup$
add a comment |
$begingroup$
So let $X,Y$ be real random variables on common probability space $(Omega, mathcal{F}, P)$, the measures on Borel $(mathbb{R},mathcal{B}_{mathbb{R}})$ induced by $X$ and $Y$ are equal, that is for all $A in mathcal{B}_{mathbb{R}}$.
$$ P_{X}(A) = P_{Y}(A)$$
where
$$P_{X}(A) := P(X^{-1}(A))$$ and
$$P_{Y}(A) := P(Y^{-1}(A))$$
Do $X,Y$ generate the same $sigma-$algebra? I feel that it might not be necessarily the case but was not able to construct a counter example.
Any help would be appreciated
probability measure-theory lebesgue-measure
asked Apr 10 '16 at 16:42
themthem
296214
$endgroup$
8
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02
add a comment |
$begingroup$
So let $X,Y$ be real random variables on common probability space $(Omega, mathcal{F}, P)$, the measures on Borel $(mathbb{R},mathcal{B}_{mathbb{R}})$ induced by $X$ and $Y$ are equal, that is for all $A in mathcal{B}_{mathbb{R}}$.
$$ P_{X}(A) = P_{Y}(A)$$
where
$$P_{X}(A) := P(X^{-1}(A))$$ and
$$P_{Y}(A) := P(Y^{-1}(A))$$
Do $X,Y$ generate the same $sigma-$algebra? I feel that it might not be necessarily the case but was not able to construct a counter example.
Any help would be appreciated
probability measure-theory lebesgue-measure
asked Apr 10 '16 at 16:42
themthem
296214
$endgroup$
So let $X,Y$ be real random variables on common probability space $(Omega, mathcal{F}, P)$, the measures on Borel $(mathbb{R},mathcal{B}_{mathbb{R}})$ induced by $X$ and $Y$ are equal, that is for all $A in mathcal{B}_{mathbb{R}}$.
$$ P_{X}(A) = P_{Y}(A)$$
where
$$P_{X}(A) := P(X^{-1}(A))$$ and
$$P_{Y}(A) := P(Y^{-1}(A))$$
Do $X,Y$ generate the same $sigma-$algebra? I feel that it might not be necessarily the case but was not able to construct a counter example.
Any help would be appreciated
probability measure-theory lebesgue-measure
probability measure-theory lebesgue-measure
asked Apr 10 '16 at 16:42
themthem
296214
asked Apr 10 '16 at 16:42
themthem
296214
edited Apr 10 '16 at 16:48
them
asked Apr 10 '16 at 16:42
themthem
296214
asked Apr 10 '16 at 16:42
themthem
296214
asked Apr 10 '16 at 16:42
themthem
296214
296214
8
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02
add a comment |
8
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02
8
8
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Take a probability space $left(Omega,mathcal F, mathbb Pright)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.
I rewrite the example given by Did: let $Omega=[0,1]^2$ with the Borel $sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $sigma$-algebra.
$endgroup$
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
});
}
});
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%2f1736253%2flet-x-y-have-the-same-distribution-on-common-prob-space-do-they-generate-same%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take a probability space $left(Omega,mathcal F, mathbb Pright)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.
I rewrite the example given by Did: let $Omega=[0,1]^2$ with the Borel $sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $sigma$-algebra.
$endgroup$
add a comment |
$begingroup$
Take a probability space $left(Omega,mathcal F, mathbb Pright)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.
I rewrite the example given by Did: let $Omega=[0,1]^2$ with the Borel $sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $sigma$-algebra.
$endgroup$
add a comment |
$begingroup$
Take a probability space $left(Omega,mathcal F, mathbb Pright)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.
I rewrite the example given by Did: let $Omega=[0,1]^2$ with the Borel $sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $sigma$-algebra.
$endgroup$
Take a probability space $left(Omega,mathcal F, mathbb Pright)$ with two independent identically distributed random variables $X$ and $Y$ which are not almost surely constant. Then the law are equal but the the generated $sigma$-algebras can not the same, since they would be independent and would contain an event whose probability is neither $0$ nor $1$.
I rewrite the example given by Did: let $Omega=[0,1]^2$ with the Borel $sigma$-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
Overall, we could not define an independent identically distributed sequence of non-constant random variable if having the same distribution would imply having the same generated $sigma$-algebra.
answered Dec 2 '18 at 15:55
community wiki
Davide Giraudo
add a comment |
add a comment |
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%2f1736253%2flet-x-y-have-the-same-distribution-on-common-prob-space-do-they-generate-same%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
8
$begingroup$
Your feeling is right, try $Omega=[0,1]^2$ with the Borel sigma-algebra and the Lebesgue measure and the random variables $X$ and $Y$ defined by $X(x,y)=x$ and $Y(x,y)=y$.
$endgroup$
– Did
Apr 10 '16 at 16:57
$begingroup$
@Did, Thanks, that answers my question.
$endgroup$
– them
Apr 10 '16 at 17:02