Clarification on a probability theory statement
$begingroup$
In a text I'm reading, it says the following:
Suppose that $(Omega, mathcal{F})$ is a measureable space and that $X_1, dots, X_n: Omega to mathbf{R}$ are $n$ (real-valued) random variables. If $f: mathbf{R}^n to mathbf{R}$ is a measurable function, then $Z = f(X_1, dots, X_n)$ is measurable with respect to the $sigma$-algebra $sigma(X_1, dots, X_n)$.
I guess my first question is what does it mean for the random variable $Z$ to be measurable with respect to $sigma(X_1, dots, X_n)$? Does it mean that we are considering $Z(omega) = f(X_1(omega), dots, X_n(omega))$ as a map $Z: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}))$?
If so, in this case I think we can do it in two steps: first, we show that $X: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}^n))$ is measurable, and then we apply the result that $Z = f circ X$, the composition of measurable maps, hence measurable.
That $X$ as defined above (the vector random variable) is measurable is clear. Let $B = B_1 times cdots B_n$, where $B_i in mathcal{B}(mathbf{R})$ for all $i$. Note that
$$
X^{-1}(B) = cap_{i=1}^n X_i^{-1}(B_i) in sigma(X_1, dots, X_n),
$$
where the membership follows since for each $i$, $X_i^{-1}(B_i) in sigma(X_i) subset sigma(X_1, dots X_n)$, and latter is closed under intersection. But then now use that the Borels on $mathbf{R}^n$ are generated by these `rectangular' sets, so hence $X$ is measurable. Is this roughly the idea?
probability-theory measure-theory
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
$endgroup$
add a comment |
$begingroup$
In a text I'm reading, it says the following:
Suppose that $(Omega, mathcal{F})$ is a measureable space and that $X_1, dots, X_n: Omega to mathbf{R}$ are $n$ (real-valued) random variables. If $f: mathbf{R}^n to mathbf{R}$ is a measurable function, then $Z = f(X_1, dots, X_n)$ is measurable with respect to the $sigma$-algebra $sigma(X_1, dots, X_n)$.
I guess my first question is what does it mean for the random variable $Z$ to be measurable with respect to $sigma(X_1, dots, X_n)$? Does it mean that we are considering $Z(omega) = f(X_1(omega), dots, X_n(omega))$ as a map $Z: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}))$?
If so, in this case I think we can do it in two steps: first, we show that $X: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}^n))$ is measurable, and then we apply the result that $Z = f circ X$, the composition of measurable maps, hence measurable.
That $X$ as defined above (the vector random variable) is measurable is clear. Let $B = B_1 times cdots B_n$, where $B_i in mathcal{B}(mathbf{R})$ for all $i$. Note that
$$
X^{-1}(B) = cap_{i=1}^n X_i^{-1}(B_i) in sigma(X_1, dots, X_n),
$$
where the membership follows since for each $i$, $X_i^{-1}(B_i) in sigma(X_i) subset sigma(X_1, dots X_n)$, and latter is closed under intersection. But then now use that the Borels on $mathbf{R}^n$ are generated by these `rectangular' sets, so hence $X$ is measurable. Is this roughly the idea?
probability-theory measure-theory
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
$endgroup$
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27
add a comment |
$begingroup$
In a text I'm reading, it says the following:
Suppose that $(Omega, mathcal{F})$ is a measureable space and that $X_1, dots, X_n: Omega to mathbf{R}$ are $n$ (real-valued) random variables. If $f: mathbf{R}^n to mathbf{R}$ is a measurable function, then $Z = f(X_1, dots, X_n)$ is measurable with respect to the $sigma$-algebra $sigma(X_1, dots, X_n)$.
I guess my first question is what does it mean for the random variable $Z$ to be measurable with respect to $sigma(X_1, dots, X_n)$? Does it mean that we are considering $Z(omega) = f(X_1(omega), dots, X_n(omega))$ as a map $Z: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}))$?
If so, in this case I think we can do it in two steps: first, we show that $X: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}^n))$ is measurable, and then we apply the result that $Z = f circ X$, the composition of measurable maps, hence measurable.
That $X$ as defined above (the vector random variable) is measurable is clear. Let $B = B_1 times cdots B_n$, where $B_i in mathcal{B}(mathbf{R})$ for all $i$. Note that
$$
X^{-1}(B) = cap_{i=1}^n X_i^{-1}(B_i) in sigma(X_1, dots, X_n),
$$
where the membership follows since for each $i$, $X_i^{-1}(B_i) in sigma(X_i) subset sigma(X_1, dots X_n)$, and latter is closed under intersection. But then now use that the Borels on $mathbf{R}^n$ are generated by these `rectangular' sets, so hence $X$ is measurable. Is this roughly the idea?
probability-theory measure-theory
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
$endgroup$
In a text I'm reading, it says the following:
Suppose that $(Omega, mathcal{F})$ is a measureable space and that $X_1, dots, X_n: Omega to mathbf{R}$ are $n$ (real-valued) random variables. If $f: mathbf{R}^n to mathbf{R}$ is a measurable function, then $Z = f(X_1, dots, X_n)$ is measurable with respect to the $sigma$-algebra $sigma(X_1, dots, X_n)$.
I guess my first question is what does it mean for the random variable $Z$ to be measurable with respect to $sigma(X_1, dots, X_n)$? Does it mean that we are considering $Z(omega) = f(X_1(omega), dots, X_n(omega))$ as a map $Z: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}))$?
If so, in this case I think we can do it in two steps: first, we show that $X: (Omega, sigma(X_1, dots, X_n)) to (mathbf{R}, mathcal{B}(mathbf{R}^n))$ is measurable, and then we apply the result that $Z = f circ X$, the composition of measurable maps, hence measurable.
That $X$ as defined above (the vector random variable) is measurable is clear. Let $B = B_1 times cdots B_n$, where $B_i in mathcal{B}(mathbf{R})$ for all $i$. Note that
$$
X^{-1}(B) = cap_{i=1}^n X_i^{-1}(B_i) in sigma(X_1, dots, X_n),
$$
where the membership follows since for each $i$, $X_i^{-1}(B_i) in sigma(X_i) subset sigma(X_1, dots X_n)$, and latter is closed under intersection. But then now use that the Borels on $mathbf{R}^n$ are generated by these `rectangular' sets, so hence $X$ is measurable. Is this roughly the idea?
probability-theory measure-theory
probability-theory measure-theory
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
asked Dec 8 '18 at 1:25
Drew BradyDrew Brady
682315
682315
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27
add a comment |
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27
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%2f3030594%2fclarification-on-a-probability-theory-statement%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%2f3030594%2fclarification-on-a-probability-theory-statement%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
$begingroup$
Yes. Your argument is correct.
$endgroup$
– Kavi Rama Murthy
Dec 8 '18 at 5:27