Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?
Thanks in advance!
probability-theory conditional-expectation independence
asked Dec 7 at 8:07
bellcircle
1,327411
add a comment |
Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?
Thanks in advance!
probability-theory conditional-expectation independence
asked Dec 7 at 8:07
bellcircle
1,327411
1
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16
add a comment |
Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?
Thanks in advance!
probability-theory conditional-expectation independence
asked Dec 7 at 8:07
bellcircle
1,327411
Let $Z$ be a random variable independent of $(X,Y)$. Prove that $mathbf{E}(Y|sigma(X))=mathbf{E}(Y|sigma(X,Z))$
My attempt: It is obvious that $int_Amathbf{E}(Y|sigma(X,Z))dmathbf{P}=int_Amathbf{E}(Y|sigma(X))dmathbf{P}=int_AYdmathbf{P}$ for all $Ain sigma(X)$. However, I'm stuck in proving that $mathbf{E}(Y|sigma(X,Z))$ is $sigma(X)$-measurable. I think I should take advantage of independence, but I don't know how to do so. In statistics, dividing the case into discrete and continuous gives straightforward result since the joint pmf or pdf splits, but how about the general case?
Thanks in advance!
probability-theory conditional-expectation independence
probability-theory conditional-expectation independence
asked Dec 7 at 8:07
bellcircle
1,327411
asked Dec 7 at 8:07
bellcircle
1,327411
edited Dec 7 at 8:17
asked Dec 7 at 8:07
bellcircle
1,327411
asked Dec 7 at 8:07
bellcircle
1,327411
asked Dec 7 at 8:07
bellcircle
1,327411
1,327411
1
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16
add a comment |
1
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16
1
1
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16
add a comment |
1 Answer
1
active
oldest
votes
Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
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%2f3029631%2fprove-that-mathbfey-sigmax-mathbfey-sigmax-z%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
Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
add a comment |
Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
add a comment |
Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
Proving that RHS (or a version of it) is meaurable w.r.t. $sigma (X)$ is not easy. Since LHS is measurable w.r.t. $sigma (X,Z)$ it is enough to show that $int_{X^{-1}(A)cap Z^{-1}(B)} E(Y|X)dP=int_{X^{-1}(A)cap Z^{-1}(B)} YdP$ for all Borel sets $A,B$ in $mathbb R$. If you write both sides in terms of the joint distribution of $X,Y,Z$ and use the independence assumption the equation becomes $P{Z^{-1}(B)} int_{X^{-1}(A)} E(Y|X)dP=P{Z^{-1}(B)}int_{X^{-1}(A)} YdP$ which is true by definiton of $E(Y|X)$.
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
edited Dec 7 at 8:30
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
answered Dec 7 at 8:25
Kavi Rama Murthy
48.6k31854
48.6k31854
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3029631%2fprove-that-mathbfey-sigmax-mathbfey-sigmax-z%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
The correct hypothesis is that Z is independent of (X,Y).
– Did
Dec 7 at 8:16