Mixing
Finite variance of a linear combination
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$Let be $X$ and $Y$ two random variables, such that $E[X^2]<infty$ and $E[Y^2]<infty$, can we conclude that $Var[aX+bY]$, with $a,bin{mathbb{R}}$ is finite?
My attemp:
begin{align}
& left|Var[aX+bY]right|=|a^2Var[X]+b^2Var[Y]+abCov(X,Y)| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b| left|Cov(X,Y)right| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b|sqrt{(Var[X])}sqrt{(Var[Y]})<+infty
end{align}
So,in particular, we can conclude that $E[(aX+bY)^2]<+infty$
statistics statistical-inference
edited Oct 26 '17 at 19:57
Michael Hardy
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
$endgroup$
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$Let be $X$ and $Y$ two random variables, such that $E[X^2]<infty$ and $E[Y^2]<infty$, can we conclude that $Var[aX+bY]$, with $a,bin{mathbb{R}}$ is finite?
My attemp:
begin{align}
& left|Var[aX+bY]right|=|a^2Var[X]+b^2Var[Y]+abCov(X,Y)| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b| left|Cov(X,Y)right| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b|sqrt{(Var[X])}sqrt{(Var[Y]})<+infty
end{align}
So,in particular, we can conclude that $E[(aX+bY)^2]<+infty$
statistics statistical-inference
edited Oct 26 '17 at 19:57
Michael Hardy
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
$endgroup$
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$Let be $X$ and $Y$ two random variables, such that $E[X^2]<infty$ and $E[Y^2]<infty$, can we conclude that $Var[aX+bY]$, with $a,bin{mathbb{R}}$ is finite?
My attemp:
begin{align}
& left|Var[aX+bY]right|=|a^2Var[X]+b^2Var[Y]+abCov(X,Y)| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b| left|Cov(X,Y)right| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b|sqrt{(Var[X])}sqrt{(Var[Y]})<+infty
end{align}
So,in particular, we can conclude that $E[(aX+bY)^2]<+infty$
statistics statistical-inference
edited Oct 26 '17 at 19:57
Michael Hardy
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
$endgroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$Let be $X$ and $Y$ two random variables, such that $E[X^2]<infty$ and $E[Y^2]<infty$, can we conclude that $Var[aX+bY]$, with $a,bin{mathbb{R}}$ is finite?
My attemp:
begin{align}
& left|Var[aX+bY]right|=|a^2Var[X]+b^2Var[Y]+abCov(X,Y)| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b| left|Cov(X,Y)right| \[10pt]
leq {} & a^2Var[X]+b^2Var[Y]+|a||b|sqrt{(Var[X])}sqrt{(Var[Y]})<+infty
end{align}
So,in particular, we can conclude that $E[(aX+bY)^2]<+infty$
statistics statistical-inference
statistics statistical-inference
edited Oct 26 '17 at 19:57
Michael Hardy
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
edited Oct 26 '17 at 19:57
Michael Hardy
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
edited Oct 26 '17 at 19:57
Michael Hardy
1
edited Oct 26 '17 at 19:57
Michael Hardy
1
edited Oct 26 '17 at 19:57
Michael Hardy
1
1
asked Oct 26 '17 at 13:23
mathlifemathlife
679
asked Oct 26 '17 at 13:23
mathlifemathlife
679
asked Oct 26 '17 at 13:23
mathlifemathlife
679
679
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57
add a comment |
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$You have $$Var[X] = E[X^2] - E[X]^2 le E[X^2] < infty$$ and $$Var[a X + b Y] = a^2 Var[X] + b^2 Var[Y] + 2ab Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$Cov[X, Y]^2 le Var[X] Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$E[(aX+bY)^2] = E[a^2 X^2 + b^2 Y^2 + 2ab , X , Y] \= a^2 E[X^2] + b^2 E[Y^2] + 2ab E[XY] \le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , |E[XY]|$$
From the triangle inequality again you have $$|E[XY]| le |Cov[X,Y]| + |E[X]|,|E[Y]|$$ with the inequalities above and $E[X]^2 le E[X^2]$ that becomes $$|E[XY]| le 2 sqrt{E[X^2]} sqrt{E[Y^2]}$$
and putting all together $$E[(aX+bY)^2] le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , sqrt{E[X^2]} sqrt{E[Y^2]} \= Big(|a| sqrt{E[X^2]} + |b| sqrt{E[Y^2]}Big)^2$$
which is finite.
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
$endgroup$
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
add a comment |
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%2f2490807%2ffinite-variance-of-a-linear-combination%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
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$You have $$Var[X] = E[X^2] - E[X]^2 le E[X^2] < infty$$ and $$Var[a X + b Y] = a^2 Var[X] + b^2 Var[Y] + 2ab Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$Cov[X, Y]^2 le Var[X] Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$E[(aX+bY)^2] = E[a^2 X^2 + b^2 Y^2 + 2ab , X , Y] \= a^2 E[X^2] + b^2 E[Y^2] + 2ab E[XY] \le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , |E[XY]|$$
From the triangle inequality again you have $$|E[XY]| le |Cov[X,Y]| + |E[X]|,|E[Y]|$$ with the inequalities above and $E[X]^2 le E[X^2]$ that becomes $$|E[XY]| le 2 sqrt{E[X^2]} sqrt{E[Y^2]}$$
and putting all together $$E[(aX+bY)^2] le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , sqrt{E[X^2]} sqrt{E[Y^2]} \= Big(|a| sqrt{E[X^2]} + |b| sqrt{E[Y^2]}Big)^2$$
which is finite.
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
$endgroup$
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$You have $$Var[X] = E[X^2] - E[X]^2 le E[X^2] < infty$$ and $$Var[a X + b Y] = a^2 Var[X] + b^2 Var[Y] + 2ab Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$Cov[X, Y]^2 le Var[X] Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$E[(aX+bY)^2] = E[a^2 X^2 + b^2 Y^2 + 2ab , X , Y] \= a^2 E[X^2] + b^2 E[Y^2] + 2ab E[XY] \le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , |E[XY]|$$
From the triangle inequality again you have $$|E[XY]| le |Cov[X,Y]| + |E[X]|,|E[Y]|$$ with the inequalities above and $E[X]^2 le E[X^2]$ that becomes $$|E[XY]| le 2 sqrt{E[X^2]} sqrt{E[Y^2]}$$
and putting all together $$E[(aX+bY)^2] le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , sqrt{E[X^2]} sqrt{E[Y^2]} \= Big(|a| sqrt{E[X^2]} + |b| sqrt{E[Y^2]}Big)^2$$
which is finite.
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
$endgroup$
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
add a comment |
$begingroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$You have $$Var[X] = E[X^2] - E[X]^2 le E[X^2] < infty$$ and $$Var[a X + b Y] = a^2 Var[X] + b^2 Var[Y] + 2ab Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$Cov[X, Y]^2 le Var[X] Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$E[(aX+bY)^2] = E[a^2 X^2 + b^2 Y^2 + 2ab , X , Y] \= a^2 E[X^2] + b^2 E[Y^2] + 2ab E[XY] \le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , |E[XY]|$$
From the triangle inequality again you have $$|E[XY]| le |Cov[X,Y]| + |E[X]|,|E[Y]|$$ with the inequalities above and $E[X]^2 le E[X^2]$ that becomes $$|E[XY]| le 2 sqrt{E[X^2]} sqrt{E[Y^2]}$$
and putting all together $$E[(aX+bY)^2] le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , sqrt{E[X^2]} sqrt{E[Y^2]} \= Big(|a| sqrt{E[X^2]} + |b| sqrt{E[Y^2]}Big)^2$$
which is finite.
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
$endgroup$
$newcommand{E}{operatorname{E}}newcommand{Var}{operatorname{Var}}newcommand{Cov}{operatorname{Cov}}$You have $$Var[X] = E[X^2] - E[X]^2 le E[X^2] < infty$$ and $$Var[a X + b Y] = a^2 Var[X] + b^2 Var[Y] + 2ab Cov[X,Y]$$ and by the Cauchy-Schwarz inequality, you further have $$Cov[X, Y]^2 le Var[X] Var[Y]$$ and therefore finite variance for the linear combination.
You can also compute the expectation directly: $$E[(aX+bY)^2] = E[a^2 X^2 + b^2 Y^2 + 2ab , X , Y] \= a^2 E[X^2] + b^2 E[Y^2] + 2ab E[XY] \le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , |E[XY]|$$
From the triangle inequality again you have $$|E[XY]| le |Cov[X,Y]| + |E[X]|,|E[Y]|$$ with the inequalities above and $E[X]^2 le E[X^2]$ that becomes $$|E[XY]| le 2 sqrt{E[X^2]} sqrt{E[Y^2]}$$
and putting all together $$E[(aX+bY)^2] le a^2 E[X^2] + b^2 E[Y^2] + 2 , |a| , |b| , sqrt{E[X^2]} sqrt{E[Y^2]} \= Big(|a| sqrt{E[X^2]} + |b| sqrt{E[Y^2]}Big)^2$$
which is finite.
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
edited Jan 29 at 11:40
Martin Sleziak
44.9k10122277
44.9k10122277
answered Oct 26 '17 at 13:40
studentstudent
22628
answered Oct 26 '17 at 13:40
studentstudent
22628
answered Oct 26 '17 at 13:40
studentstudent
22628
22628
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
add a comment |
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
So, does this proof let us conclude that $E[(aX+bY)2]<infty$?
$endgroup$
– mathlife
Oct 26 '17 at 13:58
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
$begingroup$
Yes, and if this is what you are interested in you can get it directly, see the edited answer.
$endgroup$
– student
Oct 26 '17 at 14:20
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%2f2490807%2ffinite-variance-of-a-linear-combination%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
$begingroup$
Your argument is correct. It's somewhat redundant to write $left|operatorname{var}(cdotscdots)right|$ rather than just $operatorname{var}(cdotscdots).$ But with the covariance you do need the absolute value. $qquad$
$endgroup$
– Michael Hardy
Oct 26 '17 at 19:57