Why is $E[f(X,Y) mid X]=g(X)$?
up vote
0
down vote
favorite
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 at 12:32
user3503589
1,1961721
add a comment |
up vote
0
down vote
favorite
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 at 12:32
user3503589
1,1961721
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
asked Dec 5 at 12:32
user3503589
1,1961721
Let $X,Y$ be independent and let $f$ be Borel such that $f(X,Y) in L^1(Omega,mathcal{A},P)$. Moreover let
$$
g(x)=E[f(x,Y)] text{ if } vert E[f(x,Y)]vert <infty text{ and } 0 text{ otherwise}
$$
Then show that $g$ is Borel on $mathbb{R}$ and that
$$
E[f(X,Y)mid X]=g(X)
$$
My attempt
1. I am not sure how to go about showing the Borel measurability of $g$ so a hint here would be highly appreciated
- In order to show that $E[f(X,Y)mid X]=g(X)$ we need to show that g(X) is $sigma(X)$ measurable which is the case once part 1)(Borel measurability of $g$) is proved.
And we must also show that for all $A in sigma(X)$ we have
begin{equation} label{1}
E[f(X,Y) 1_A]=E[g(X) 1_A]
end{equation}
Now I can show that if $A$ is of the form $A={X=x}$ then the equation above holds since
$$
E[f(X,Y) 1_A]=E[f(X,Y) 1_{{X=x}}]=E[f(x,Y) 1_{{X=x}}]=E[f(x,Y)] E[1_{{X=x}}]=g(x)E[1_{{X=x}}]=E[g(x)1_{{X=x}}]=E[g(X)1_{{X=x}}]
$$
where I used the independence of $X$ and $Y$ and the properties of expectation. How can I extend this to all sets in $sigma(X)$ if $X$ we discrete taking countably many values this would work but how can I make this argument work in the general case. Even hints would be highly appreciated or a nudge in the right direction
probability-theory conditional-expectation
probability-theory conditional-expectation
asked Dec 5 at 12:32
user3503589
1,1961721
asked Dec 5 at 12:32
user3503589
1,1961721
asked Dec 5 at 12:32
user3503589
1,1961721
asked Dec 5 at 12:32
user3503589
1,1961721
asked Dec 5 at 12:32
user3503589
1,1961721
1,1961721
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 at 13:36
drhab
96.2k543126
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',
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%2f3027017%2fwhy-is-efx-y-mid-x-gx%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
up vote
1
down vote
accepted
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 at 13:36
drhab
96.2k543126
add a comment |
up vote
1
down vote
accepted
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 at 13:36
drhab
96.2k543126
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 at 13:36
drhab
96.2k543126
Second part something like:$$mathbb Ef(X,Y)mathbf1_{{Xin B}}=intint f(x,y)mathbf1_B(x);dF_Y(y)dF_X(x)=intmathbf1_B(x)int f(x,y);dF_Y(y)dF_X(x)=$$$$intmathbf1_B(x) g(x)dF_X(x)=mathbb Eg(X)mathbf1_{{Xin B}}$$where the first equality is based on independence.
What we use here too is that $sigma(X)={{Xin B}mid Binmathcal B(mathbb R)}$
answered Dec 5 at 13:36
drhab
96.2k543126
answered Dec 5 at 13:36
drhab
96.2k543126
answered Dec 5 at 13:36
drhab
96.2k543126
answered Dec 5 at 13:36
drhab
96.2k543126
96.2k543126
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%2f3027017%2fwhy-is-efx-y-mid-x-gx%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
