Mixing
without computing it, show that $I = mathbb{E}[e^{XY} |X] geq 1$
$begingroup$
$X, Y$ are two independent $mathcal{N}(0,1)$ random variables
this question was a follow up question of this one
I honestly don't know if that series of equivalences could be of any help.
my reasoning was as follows :
$ mathbb{E}[I] = mathbb{E}[e^{frac{X^{2}}{2}}] = +infty$
now let's assume that $I<1$ because $e^{XY} geq0$
then $0leq I<1 $ this gives $0leq mathbb{E}[I]<1 $
contradiction !
is this it ?
probability-theory proof-verification conditional-expectation
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
$endgroup$
add a comment |
$begingroup$
$X, Y$ are two independent $mathcal{N}(0,1)$ random variables
this question was a follow up question of this one
I honestly don't know if that series of equivalences could be of any help.
my reasoning was as follows :
$ mathbb{E}[I] = mathbb{E}[e^{frac{X^{2}}{2}}] = +infty$
now let's assume that $I<1$ because $e^{XY} geq0$
then $0leq I<1 $ this gives $0leq mathbb{E}[I]<1 $
contradiction !
is this it ?
probability-theory proof-verification conditional-expectation
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
$endgroup$
add a comment |
$begingroup$
$X, Y$ are two independent $mathcal{N}(0,1)$ random variables
this question was a follow up question of this one
I honestly don't know if that series of equivalences could be of any help.
my reasoning was as follows :
$ mathbb{E}[I] = mathbb{E}[e^{frac{X^{2}}{2}}] = +infty$
now let's assume that $I<1$ because $e^{XY} geq0$
then $0leq I<1 $ this gives $0leq mathbb{E}[I]<1 $
contradiction !
is this it ?
probability-theory proof-verification conditional-expectation
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
$endgroup$
$X, Y$ are two independent $mathcal{N}(0,1)$ random variables
this question was a follow up question of this one
I honestly don't know if that series of equivalences could be of any help.
my reasoning was as follows :
$ mathbb{E}[I] = mathbb{E}[e^{frac{X^{2}}{2}}] = +infty$
now let's assume that $I<1$ because $e^{XY} geq0$
then $0leq I<1 $ this gives $0leq mathbb{E}[I]<1 $
contradiction !
is this it ?
probability-theory proof-verification conditional-expectation
probability-theory proof-verification conditional-expectation
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
asked Feb 1 at 21:18
rapidracimrapidracim
1,7441419
1,7441419
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}mid X)geqslant1$ almost surely.
The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.
Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Zgeqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)ne0$. And the hypothesis that $P(Z<1)ne0$ does not imply that $E(Z)<1$.
Edit: If one is forbidden to compute $E(e^{XY}mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}mid X)geqslant1$ almost surely.
This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.
answered Feb 1 at 21:23
DidDid
249k23228466
$endgroup$
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
add a comment |
$begingroup$
$1+E[XY|X]leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!!
So we have the desired inequality without any computing
answered Feb 1 at 23:06
mathexmathex
1138
$endgroup$
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
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%2f3096741%2fwithout-computing-it-show-that-i-mathbbeexy-x-geq-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}mid X)geqslant1$ almost surely.
The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.
Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Zgeqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)ne0$. And the hypothesis that $P(Z<1)ne0$ does not imply that $E(Z)<1$.
Edit: If one is forbidden to compute $E(e^{XY}mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}mid X)geqslant1$ almost surely.
This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.
answered Feb 1 at 21:23
DidDid
249k23228466
$endgroup$
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
add a comment |
$begingroup$
Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}mid X)geqslant1$ almost surely.
The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.
Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Zgeqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)ne0$. And the hypothesis that $P(Z<1)ne0$ does not imply that $E(Z)<1$.
Edit: If one is forbidden to compute $E(e^{XY}mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}mid X)geqslant1$ almost surely.
This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.
answered Feb 1 at 21:23
DidDid
249k23228466
$endgroup$
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
add a comment |
$begingroup$
Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}mid X)geqslant1$ almost surely.
The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.
Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Zgeqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)ne0$. And the hypothesis that $P(Z<1)ne0$ does not imply that $E(Z)<1$.
Edit: If one is forbidden to compute $E(e^{XY}mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}mid X)geqslant1$ almost surely.
This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.
answered Feb 1 at 21:23
DidDid
249k23228466
$endgroup$
Since $Y$ is standard normal, $E(e^{xY})=e^{x^2/2}$ for every $x$, hence, by independence of $X$ and $Y$, $E(e^{XY}mid X)=e^{X^2/2}$ almost surely. Thus, $E(e^{XY}mid X)geqslant1$ almost surely.
The proof does not use the distribution of $X$, only the distribution of $Y$ and the independence of $X$ and $Y$.
Your approach does not yield a contradiction: you are asked to show that some random variable $Z$ is such that $Zgeqslant1$ almost surely. The negation of this property is not that $Z<1$ almost surely, but that $P(Z<1)ne0$. And the hypothesis that $P(Z<1)ne0$ does not imply that $E(Z)<1$.
Edit: If one is forbidden to compute $E(e^{XY}mid X)$, one can note that, by Jensen convexity inequality, for every $x$, $E(e^{xY})geqslant e^{xE(Y)}=1$, hence, again by independence, $E(e^{XY}mid X)geqslant1$ almost surely.
This uses only that $E(Y)=0$ and the independence of $X$ and $Y$.
answered Feb 1 at 21:23
DidDid
249k23228466
edited Feb 1 at 21:33
answered Feb 1 at 21:23
DidDid
249k23228466
answered Feb 1 at 21:23
DidDid
249k23228466
answered Feb 1 at 21:23
DidDid
249k23228466
249k23228466
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
add a comment |
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
$begingroup$
I see now my mistake, btw you still computed $I$, as in you got an explicit formula for it. usually I understand 'without computing it' as 'don't use an explicitfinal formula of it'. but maybe then they meant don't use the standard approach, thanks for the answer anyways.
$endgroup$
– rapidracim
Feb 1 at 21:27
add a comment |
$begingroup$
$1+E[XY|X]leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!!
So we have the desired inequality without any computing
answered Feb 1 at 23:06
mathexmathex
1138
$endgroup$
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
add a comment |
$begingroup$
$1+E[XY|X]leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!!
So we have the desired inequality without any computing
answered Feb 1 at 23:06
mathexmathex
1138
$endgroup$
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
add a comment |
$begingroup$
$1+E[XY|X]leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!!
So we have the desired inequality without any computing
answered Feb 1 at 23:06
mathexmathex
1138
$endgroup$
$1+E[XY|X]leq E[e^{XY}|X]$ a.s and $E[XY|X]=XE[Y]=0$ by independence!!
So we have the desired inequality without any computing
answered Feb 1 at 23:06
mathexmathex
1138
answered Feb 1 at 23:06
mathexmathex
1138
answered Feb 1 at 23:06
mathexmathex
1138
answered Feb 1 at 23:06
mathexmathex
1138
1138
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
add a comment |
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
very nice observation !
$endgroup$
– rapidracim
Feb 1 at 23:10
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
Did you solve the last equivalence here:math.stackexchange.com/questions/3096530/…
$endgroup$
– mathex
Feb 1 at 23:12
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
$begingroup$
I have added my attempt at the last equivalence in the post. btw all 3 claims are technically false as none of those random variables are integrable.
$endgroup$
– rapidracim
Feb 1 at 23:16
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%2f3096741%2fwithout-computing-it-show-that-i-mathbbeexy-x-geq-1%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
