How to show Sub-independent Random Variables are uncorrelated.
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
$endgroup$
add a comment |
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
$endgroup$
add a comment |
$begingroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
$endgroup$
I want to prove the following:
If two RVs $X, Y$ are sub-independent, i.e., $phi_{X+Y}(t) = phi_X(t)phi_Y(t), tinmathbb{R}$ then $X, Y$are uncorrelated. Keep $Cov(X,Y) = E(XY)-E(X)E(Y) = 0$ in mind, I'm considering the way to connect $E(X)$ with characteristic function $phi_X(t)$. In fact, $phi_{X}'(0) = iE(X), phi_{Y}'(0) = iE(Y).$ Finding $E(XY)$ will end the proof, but I could not proceed further.
Any idea or tips would be really helpful.
random-variables independence correlation characteristic-functions
random-variables independence correlation characteristic-functions
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
edited Dec 5 '18 at 0:44
Euduardo
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
asked Dec 3 '18 at 14:12
EuduardoEuduardo
1288
1288
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
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%2f3024101%2fhow-to-show-sub-independent-random-variables-are-uncorrelated%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
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
$endgroup$
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
$endgroup$
The trick is to introduce auxiliary variables $X'$ and $Y'$ with known joint distribution. Define $X'$ and $Y'$ to be independent but with the same marginal distributions as $X$ and $Y$ respectively. Then
$$
phi_{X'+Y'}(t)stackrel{(a)}=phi_{X'}(t)phi_{Y'}(t) stackrel{(b)}=phi_X(t)phi_Y(t)stackrel{(c)}=phi_{X+Y}(t);
$$
in (a) we use independence of $X'$ and $Y'$, in (b) we use equality of marginal distributions; and (c) is the definition of sub-independence. Therefore, by uniqueness of characteristic functions, $X'+Y'$ has the same distribution as $X+Y$. In particular,
$$
E(X'+Y')^2=E(X+Y)^2
$$
(assuming second moments exist), which in turn implies $$E(XY)=E(X'Y')stackrel{(a)}=E(X')E(Y')stackrel{(b)}=E(X)E(Y).$$
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
answered Dec 5 '18 at 1:39
grand_chatgrand_chat
20.2k11226
20.2k11226
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
add a comment |
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
$begingroup$
Thanks a lot! It really helped me a lot and gave me insights.
$endgroup$
– Euduardo
Dec 5 '18 at 3:46
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%2f3024101%2fhow-to-show-sub-independent-random-variables-are-uncorrelated%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
