Mixing
Symmetrization of Conditional Expectation without common density
$begingroup$
I know that the following rule for (factorizations of) conditional expectations is true:
Theorem
Let $X : Omega to mathcal{X}=mathbb{R}$ be an integrable, real valued random variable and let $Y:Omega to mathcal{Y}, Z : Omega to mathcal{Z}$ be two further random variables then
$$E[X|Y=y] = int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz$$
(remarks on the $dz$ below). I have a proof for that involving the common density $p(x,y,z)$.
Question: Does somebody know a direct proof for that which does neither involve nor assume the mere existence of this common density but only the existence of the common density $p(y,z)$ for the random variables we condition on?
More information:
Remarks on the dz: I only want to consider the case where many of the involved random variables have (common) densities. I.e. we may restrict ourselves to the case where $Y,Z$ are of the following type:
Either they are random variables mapping to a 'nice' subset of $mathbb{R}$ (an interval I guess) and they have a density w.r.t. the Lebesgue measure or they are discretely valued (either they map to a finite set or they map to $mathbb{N}_0$) and their density is to be regarded w.r.t. the counting measure or they are simple products of such variables.
proof with $p(x,y,z)$
It is basically proved in Formal definition of conditional probability that for any measurable function $g : mathbb{R} times mathcal{Y} to mathbb{R}$ such that $g(X,Y)$ is still integrable,
$$E[g(X,Y)|Y=y] = int_{mathcal{X}} g(x,y) p(x|y) dx$$
Consequently, applying this rule twice for $g(x,y) = x$ and the random variables $Y$, respectively $(Y,Z)$,
begin{align*}
E[X|Y=y] &= int_{mathcal{X}} x p(x,y)/p(y) dx \
&= int_{mathcal{X}} x int_{mathcal{Z}}p(x,y,z) dz/p(y) dx \
&= int_{mathcal{Z}} int_{mathcal{X}} p(x|y,z)p(y,z)/p(y) dx dz\
&= int_{mathcal{Z}} p(z|y) int_{mathcal{X}} x p(x|y,z) dx dz\
&= int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz
end{align*}
Origin of the question I want to understand the 'proof' of the so-called Bellman equation (see https://stats.stackexchange.com/questions/243384/deriving-bellmans-equation-in-reinforcement-learning/300360?noredirect=1#comment725912_300360). The 'answers' to that question I am referring to make use of the fact that only because one knows properties about every finite sum $G_{t}^{(K)} := sum_{k=0}^K gamma^k R_{t+1+k}$ one also knows properties about the variable $G_{t} = sum_{k=0}^infty gamma^k R_{t+1+k}$. This is clearly not proven in the answers (also not in the accepted answer) because one does not even know whether or not the random variable $G_{t}$ has a density or not (let alone proving the existence and some properties of its common density together with some other state variables).
probability-theory measure-theory conditional-expectation
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
$endgroup$
add a comment |
$begingroup$
I know that the following rule for (factorizations of) conditional expectations is true:
Theorem
Let $X : Omega to mathcal{X}=mathbb{R}$ be an integrable, real valued random variable and let $Y:Omega to mathcal{Y}, Z : Omega to mathcal{Z}$ be two further random variables then
$$E[X|Y=y] = int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz$$
(remarks on the $dz$ below). I have a proof for that involving the common density $p(x,y,z)$.
Question: Does somebody know a direct proof for that which does neither involve nor assume the mere existence of this common density but only the existence of the common density $p(y,z)$ for the random variables we condition on?
More information:
Remarks on the dz: I only want to consider the case where many of the involved random variables have (common) densities. I.e. we may restrict ourselves to the case where $Y,Z$ are of the following type:
Either they are random variables mapping to a 'nice' subset of $mathbb{R}$ (an interval I guess) and they have a density w.r.t. the Lebesgue measure or they are discretely valued (either they map to a finite set or they map to $mathbb{N}_0$) and their density is to be regarded w.r.t. the counting measure or they are simple products of such variables.
proof with $p(x,y,z)$
It is basically proved in Formal definition of conditional probability that for any measurable function $g : mathbb{R} times mathcal{Y} to mathbb{R}$ such that $g(X,Y)$ is still integrable,
$$E[g(X,Y)|Y=y] = int_{mathcal{X}} g(x,y) p(x|y) dx$$
Consequently, applying this rule twice for $g(x,y) = x$ and the random variables $Y$, respectively $(Y,Z)$,
begin{align*}
E[X|Y=y] &= int_{mathcal{X}} x p(x,y)/p(y) dx \
&= int_{mathcal{X}} x int_{mathcal{Z}}p(x,y,z) dz/p(y) dx \
&= int_{mathcal{Z}} int_{mathcal{X}} p(x|y,z)p(y,z)/p(y) dx dz\
&= int_{mathcal{Z}} p(z|y) int_{mathcal{X}} x p(x|y,z) dx dz\
&= int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz
end{align*}
Origin of the question I want to understand the 'proof' of the so-called Bellman equation (see https://stats.stackexchange.com/questions/243384/deriving-bellmans-equation-in-reinforcement-learning/300360?noredirect=1#comment725912_300360). The 'answers' to that question I am referring to make use of the fact that only because one knows properties about every finite sum $G_{t}^{(K)} := sum_{k=0}^K gamma^k R_{t+1+k}$ one also knows properties about the variable $G_{t} = sum_{k=0}^infty gamma^k R_{t+1+k}$. This is clearly not proven in the answers (also not in the accepted answer) because one does not even know whether or not the random variable $G_{t}$ has a density or not (let alone proving the existence and some properties of its common density together with some other state variables).
probability-theory measure-theory conditional-expectation
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
$endgroup$
add a comment |
$begingroup$
I know that the following rule for (factorizations of) conditional expectations is true:
Theorem
Let $X : Omega to mathcal{X}=mathbb{R}$ be an integrable, real valued random variable and let $Y:Omega to mathcal{Y}, Z : Omega to mathcal{Z}$ be two further random variables then
$$E[X|Y=y] = int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz$$
(remarks on the $dz$ below). I have a proof for that involving the common density $p(x,y,z)$.
Question: Does somebody know a direct proof for that which does neither involve nor assume the mere existence of this common density but only the existence of the common density $p(y,z)$ for the random variables we condition on?
More information:
Remarks on the dz: I only want to consider the case where many of the involved random variables have (common) densities. I.e. we may restrict ourselves to the case where $Y,Z$ are of the following type:
Either they are random variables mapping to a 'nice' subset of $mathbb{R}$ (an interval I guess) and they have a density w.r.t. the Lebesgue measure or they are discretely valued (either they map to a finite set or they map to $mathbb{N}_0$) and their density is to be regarded w.r.t. the counting measure or they are simple products of such variables.
proof with $p(x,y,z)$
It is basically proved in Formal definition of conditional probability that for any measurable function $g : mathbb{R} times mathcal{Y} to mathbb{R}$ such that $g(X,Y)$ is still integrable,
$$E[g(X,Y)|Y=y] = int_{mathcal{X}} g(x,y) p(x|y) dx$$
Consequently, applying this rule twice for $g(x,y) = x$ and the random variables $Y$, respectively $(Y,Z)$,
begin{align*}
E[X|Y=y] &= int_{mathcal{X}} x p(x,y)/p(y) dx \
&= int_{mathcal{X}} x int_{mathcal{Z}}p(x,y,z) dz/p(y) dx \
&= int_{mathcal{Z}} int_{mathcal{X}} p(x|y,z)p(y,z)/p(y) dx dz\
&= int_{mathcal{Z}} p(z|y) int_{mathcal{X}} x p(x|y,z) dx dz\
&= int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz
end{align*}
Origin of the question I want to understand the 'proof' of the so-called Bellman equation (see https://stats.stackexchange.com/questions/243384/deriving-bellmans-equation-in-reinforcement-learning/300360?noredirect=1#comment725912_300360). The 'answers' to that question I am referring to make use of the fact that only because one knows properties about every finite sum $G_{t}^{(K)} := sum_{k=0}^K gamma^k R_{t+1+k}$ one also knows properties about the variable $G_{t} = sum_{k=0}^infty gamma^k R_{t+1+k}$. This is clearly not proven in the answers (also not in the accepted answer) because one does not even know whether or not the random variable $G_{t}$ has a density or not (let alone proving the existence and some properties of its common density together with some other state variables).
probability-theory measure-theory conditional-expectation
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
$endgroup$
I know that the following rule for (factorizations of) conditional expectations is true:
Theorem
Let $X : Omega to mathcal{X}=mathbb{R}$ be an integrable, real valued random variable and let $Y:Omega to mathcal{Y}, Z : Omega to mathcal{Z}$ be two further random variables then
$$E[X|Y=y] = int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz$$
(remarks on the $dz$ below). I have a proof for that involving the common density $p(x,y,z)$.
Question: Does somebody know a direct proof for that which does neither involve nor assume the mere existence of this common density but only the existence of the common density $p(y,z)$ for the random variables we condition on?
More information:
Remarks on the dz: I only want to consider the case where many of the involved random variables have (common) densities. I.e. we may restrict ourselves to the case where $Y,Z$ are of the following type:
Either they are random variables mapping to a 'nice' subset of $mathbb{R}$ (an interval I guess) and they have a density w.r.t. the Lebesgue measure or they are discretely valued (either they map to a finite set or they map to $mathbb{N}_0$) and their density is to be regarded w.r.t. the counting measure or they are simple products of such variables.
proof with $p(x,y,z)$
It is basically proved in Formal definition of conditional probability that for any measurable function $g : mathbb{R} times mathcal{Y} to mathbb{R}$ such that $g(X,Y)$ is still integrable,
$$E[g(X,Y)|Y=y] = int_{mathcal{X}} g(x,y) p(x|y) dx$$
Consequently, applying this rule twice for $g(x,y) = x$ and the random variables $Y$, respectively $(Y,Z)$,
begin{align*}
E[X|Y=y] &= int_{mathcal{X}} x p(x,y)/p(y) dx \
&= int_{mathcal{X}} x int_{mathcal{Z}}p(x,y,z) dz/p(y) dx \
&= int_{mathcal{Z}} int_{mathcal{X}} p(x|y,z)p(y,z)/p(y) dx dz\
&= int_{mathcal{Z}} p(z|y) int_{mathcal{X}} x p(x|y,z) dx dz\
&= int_{mathcal{Z}} p(z|y) E[X|Y=y,Z=z] dz
end{align*}
Origin of the question I want to understand the 'proof' of the so-called Bellman equation (see https://stats.stackexchange.com/questions/243384/deriving-bellmans-equation-in-reinforcement-learning/300360?noredirect=1#comment725912_300360). The 'answers' to that question I am referring to make use of the fact that only because one knows properties about every finite sum $G_{t}^{(K)} := sum_{k=0}^K gamma^k R_{t+1+k}$ one also knows properties about the variable $G_{t} = sum_{k=0}^infty gamma^k R_{t+1+k}$. This is clearly not proven in the answers (also not in the accepted answer) because one does not even know whether or not the random variable $G_{t}$ has a density or not (let alone proving the existence and some properties of its common density together with some other state variables).
probability-theory measure-theory conditional-expectation
probability-theory measure-theory conditional-expectation
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
asked Jan 22 at 17:49
Fabian WernerFabian Werner
499212
499212
add a comment |
add a comment |
0
active
oldest
votes
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%2f3083461%2fsymmetrization-of-conditional-expectation-without-common-density%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3083461%2fsymmetrization-of-conditional-expectation-without-common-density%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
