Mixing
Conditioning on an union of sigma algebras
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X(w)=
begin{cases}
1 quad w in [0,1/2] \
0 quad w notin [0,1/2]
end{cases}$
Let $Y(w)=
begin{cases}
1 quad w in [0,3/4] \
0 quad w notin [0,3/4]
end{cases}$
Let $Z(w)=
begin{cases}
1 quad w in [1/4,3/4] \
0 quad w notin [1/4,3/4]
end{cases}$
I succeeded in proving that X is independent from Z and E(X|Y=1)=2/3 conditioning on the sub-sigma algebras generated by Z and Y, respectively.
Now my aim is discovering the value of E{X|(Y,Z)} but I don't know how to condition X on an union of sigma algebras.
measure-theory lebesgue-measure conditional-expectation
asked Jan 18 at 11:46
ank13ank13
255
$endgroup$
add a comment |
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X(w)=
begin{cases}
1 quad w in [0,1/2] \
0 quad w notin [0,1/2]
end{cases}$
Let $Y(w)=
begin{cases}
1 quad w in [0,3/4] \
0 quad w notin [0,3/4]
end{cases}$
Let $Z(w)=
begin{cases}
1 quad w in [1/4,3/4] \
0 quad w notin [1/4,3/4]
end{cases}$
I succeeded in proving that X is independent from Z and E(X|Y=1)=2/3 conditioning on the sub-sigma algebras generated by Z and Y, respectively.
Now my aim is discovering the value of E{X|(Y,Z)} but I don't know how to condition X on an union of sigma algebras.
measure-theory lebesgue-measure conditional-expectation
asked Jan 18 at 11:46
ank13ank13
255
$endgroup$
add a comment |
$begingroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X(w)=
begin{cases}
1 quad w in [0,1/2] \
0 quad w notin [0,1/2]
end{cases}$
Let $Y(w)=
begin{cases}
1 quad w in [0,3/4] \
0 quad w notin [0,3/4]
end{cases}$
Let $Z(w)=
begin{cases}
1 quad w in [1/4,3/4] \
0 quad w notin [1/4,3/4]
end{cases}$
I succeeded in proving that X is independent from Z and E(X|Y=1)=2/3 conditioning on the sub-sigma algebras generated by Z and Y, respectively.
Now my aim is discovering the value of E{X|(Y,Z)} but I don't know how to condition X on an union of sigma algebras.
measure-theory lebesgue-measure conditional-expectation
asked Jan 18 at 11:46
ank13ank13
255
$endgroup$
Let $Omega = [0,1]$, $mathcal{F} = mathcal{B}(0,1)$, P=Lebesgue measure.
Let $X(w)=
begin{cases}
1 quad w in [0,1/2] \
0 quad w notin [0,1/2]
end{cases}$
Let $Y(w)=
begin{cases}
1 quad w in [0,3/4] \
0 quad w notin [0,3/4]
end{cases}$
Let $Z(w)=
begin{cases}
1 quad w in [1/4,3/4] \
0 quad w notin [1/4,3/4]
end{cases}$
I succeeded in proving that X is independent from Z and E(X|Y=1)=2/3 conditioning on the sub-sigma algebras generated by Z and Y, respectively.
Now my aim is discovering the value of E{X|(Y,Z)} but I don't know how to condition X on an union of sigma algebras.
measure-theory lebesgue-measure conditional-expectation
measure-theory lebesgue-measure conditional-expectation
asked Jan 18 at 11:46
ank13ank13
255
asked Jan 18 at 11:46
ank13ank13
255
edited Jan 18 at 11:58
ank13
asked Jan 18 at 11:46
ank13ank13
255
asked Jan 18 at 11:46
ank13ank13
255
asked Jan 18 at 11:46
ank13ank13
255
255
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
- $mathbb E[Xmid Y=1,Z=1]=P(win[0,frac12]mid win[frac14,frac34])=P(win[frac14,frac12]/P(win[frac14,frac34])=frac12$
- $mathbb E[Xmid Y=1,Z=0]=P(win[0,frac12]mid win[0,frac14))=P(win[0,frac14)/P(win[0,frac14))=1$
- $mathbb E[Xmid Y=0,Z=0]=P(win[0,frac12]mid win(frac34,1])=P(winvarnothing)/P(win(frac34,1])=0$
Observe that ${Y=0,Z=1}=varnothing$ so conditioning on that event can be left out.
This shows that: $$mathbb E[Xmid (Y,Z)]=Y-frac12Z$$
answered Jan 18 at 12:22
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
Hint: Note that $$
Y^{-1}({1})=[0,frac{3}{4}],
$$
$$
Z^{-1}({1})=[frac{1}{4},frac{3}{4}].
$$ It says that the $sigma$-algebra generated by $(Y,Z)$ is also generated by
$
[0,frac{1}{4}]=Y^{-1}({1})setminus Z^{-1}({1})
$ and $[frac{1}{4},frac{3}{4}]$.
answered Jan 18 at 12:24
SongSong
15.6k1736
$endgroup$
add a comment |
$begingroup$
For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $sigma(Y,Z)$ as the sigma algebra generated by a partition ${A_1,A_2,cdots ,A_n}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=sum c_iI_{A_i}$ where $c_i=frac {int_{A_i} XdP} {P(A_i)}$.
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
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%2f3078140%2fconditioning-on-an-union-of-sigma-algebras%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
- $mathbb E[Xmid Y=1,Z=1]=P(win[0,frac12]mid win[frac14,frac34])=P(win[frac14,frac12]/P(win[frac14,frac34])=frac12$
- $mathbb E[Xmid Y=1,Z=0]=P(win[0,frac12]mid win[0,frac14))=P(win[0,frac14)/P(win[0,frac14))=1$
- $mathbb E[Xmid Y=0,Z=0]=P(win[0,frac12]mid win(frac34,1])=P(winvarnothing)/P(win(frac34,1])=0$
Observe that ${Y=0,Z=1}=varnothing$ so conditioning on that event can be left out.
This shows that: $$mathbb E[Xmid (Y,Z)]=Y-frac12Z$$
answered Jan 18 at 12:22
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
- $mathbb E[Xmid Y=1,Z=1]=P(win[0,frac12]mid win[frac14,frac34])=P(win[frac14,frac12]/P(win[frac14,frac34])=frac12$
- $mathbb E[Xmid Y=1,Z=0]=P(win[0,frac12]mid win[0,frac14))=P(win[0,frac14)/P(win[0,frac14))=1$
- $mathbb E[Xmid Y=0,Z=0]=P(win[0,frac12]mid win(frac34,1])=P(winvarnothing)/P(win(frac34,1])=0$
Observe that ${Y=0,Z=1}=varnothing$ so conditioning on that event can be left out.
This shows that: $$mathbb E[Xmid (Y,Z)]=Y-frac12Z$$
answered Jan 18 at 12:22
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
- $mathbb E[Xmid Y=1,Z=1]=P(win[0,frac12]mid win[frac14,frac34])=P(win[frac14,frac12]/P(win[frac14,frac34])=frac12$
- $mathbb E[Xmid Y=1,Z=0]=P(win[0,frac12]mid win[0,frac14))=P(win[0,frac14)/P(win[0,frac14))=1$
- $mathbb E[Xmid Y=0,Z=0]=P(win[0,frac12]mid win(frac34,1])=P(winvarnothing)/P(win(frac34,1])=0$
Observe that ${Y=0,Z=1}=varnothing$ so conditioning on that event can be left out.
This shows that: $$mathbb E[Xmid (Y,Z)]=Y-frac12Z$$
answered Jan 18 at 12:22
drhabdrhab
102k545136
$endgroup$
- $mathbb E[Xmid Y=1,Z=1]=P(win[0,frac12]mid win[frac14,frac34])=P(win[frac14,frac12]/P(win[frac14,frac34])=frac12$
- $mathbb E[Xmid Y=1,Z=0]=P(win[0,frac12]mid win[0,frac14))=P(win[0,frac14)/P(win[0,frac14))=1$
- $mathbb E[Xmid Y=0,Z=0]=P(win[0,frac12]mid win(frac34,1])=P(winvarnothing)/P(win(frac34,1])=0$
Observe that ${Y=0,Z=1}=varnothing$ so conditioning on that event can be left out.
This shows that: $$mathbb E[Xmid (Y,Z)]=Y-frac12Z$$
answered Jan 18 at 12:22
drhabdrhab
102k545136
answered Jan 18 at 12:22
drhabdrhab
102k545136
answered Jan 18 at 12:22
drhabdrhab
102k545136
answered Jan 18 at 12:22
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
Hint: Note that $$
Y^{-1}({1})=[0,frac{3}{4}],
$$
$$
Z^{-1}({1})=[frac{1}{4},frac{3}{4}].
$$ It says that the $sigma$-algebra generated by $(Y,Z)$ is also generated by
$
[0,frac{1}{4}]=Y^{-1}({1})setminus Z^{-1}({1})
$ and $[frac{1}{4},frac{3}{4}]$.
answered Jan 18 at 12:24
SongSong
15.6k1736
$endgroup$
add a comment |
$begingroup$
Hint: Note that $$
Y^{-1}({1})=[0,frac{3}{4}],
$$
$$
Z^{-1}({1})=[frac{1}{4},frac{3}{4}].
$$ It says that the $sigma$-algebra generated by $(Y,Z)$ is also generated by
$
[0,frac{1}{4}]=Y^{-1}({1})setminus Z^{-1}({1})
$ and $[frac{1}{4},frac{3}{4}]$.
answered Jan 18 at 12:24
SongSong
15.6k1736
$endgroup$
add a comment |
$begingroup$
Hint: Note that $$
Y^{-1}({1})=[0,frac{3}{4}],
$$
$$
Z^{-1}({1})=[frac{1}{4},frac{3}{4}].
$$ It says that the $sigma$-algebra generated by $(Y,Z)$ is also generated by
$
[0,frac{1}{4}]=Y^{-1}({1})setminus Z^{-1}({1})
$ and $[frac{1}{4},frac{3}{4}]$.
answered Jan 18 at 12:24
SongSong
15.6k1736
$endgroup$
Hint: Note that $$
Y^{-1}({1})=[0,frac{3}{4}],
$$
$$
Z^{-1}({1})=[frac{1}{4},frac{3}{4}].
$$ It says that the $sigma$-algebra generated by $(Y,Z)$ is also generated by
$
[0,frac{1}{4}]=Y^{-1}({1})setminus Z^{-1}({1})
$ and $[frac{1}{4},frac{3}{4}]$.
answered Jan 18 at 12:24
SongSong
15.6k1736
answered Jan 18 at 12:24
SongSong
15.6k1736
answered Jan 18 at 12:24
SongSong
15.6k1736
answered Jan 18 at 12:24
SongSong
15.6k1736
15.6k1736
add a comment |
add a comment |
$begingroup$
For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $sigma(Y,Z)$ as the sigma algebra generated by a partition ${A_1,A_2,cdots ,A_n}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=sum c_iI_{A_i}$ where $c_i=frac {int_{A_i} XdP} {P(A_i)}$.
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
add a comment |
$begingroup$
For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $sigma(Y,Z)$ as the sigma algebra generated by a partition ${A_1,A_2,cdots ,A_n}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=sum c_iI_{A_i}$ where $c_i=frac {int_{A_i} XdP} {P(A_i)}$.
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
add a comment |
$begingroup$
For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $sigma(Y,Z)$ as the sigma algebra generated by a partition ${A_1,A_2,cdots ,A_n}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=sum c_iI_{A_i}$ where $c_i=frac {int_{A_i} XdP} {P(A_i)}$.
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
$endgroup$
For whatever it is worth, let me describe a general procedure for answering such questions. It is easy to write $sigma(Y,Z)$ as the sigma algebra generated by a partition ${A_1,A_2,cdots ,A_n}$. Once you identify the sets $A_i$ you can compute $E(X|Y,Z)$ by the formula $E(X|Y,Z)=sum c_iI_{A_i}$ where $c_i=frac {int_{A_i} XdP} {P(A_i)}$.
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
answered Jan 18 at 12:30
Kavi Rama MurthyKavi Rama Murthy
63.2k42362
63.2k42362
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%2f3078140%2fconditioning-on-an-union-of-sigma-algebras%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
