Why is it necessary X and Y beeing bounded when having same moments to be equal
Multi tool use
$begingroup$
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
$endgroup$
Let $X,Y$ random variables in $[0,1]$ with $E(X^n)=E(Y^n) ,forall nin mathbb N$. I want to show $Xoverset{d}{=} Y$.
$$E(e^{itX})=intsum frac{(itx)^k}{k!}dF_Xoverset{dom. conv.}{=}sumintfrac{(itx)^k}{k!}dF_x=sumfrac{(it)^k}{k!}int x^kdF_xoverset{EX^n=EY^n}{=}sumfrac{(it)^k}{k!}int x^kdF_Y=dots=E(e^{itY})$$
I am aware of the fact $EX^n=EY^n$ does not imply $Xoverset{d}{=}Y$ in general. I did not use that X and Y are bounded, so what am I missing?
probability-theory proof-verification probability-distributions characteristic-functions
probability-theory proof-verification probability-distributions characteristic-functions
edited Nov 27 '18 at 13:58
Joey Doey
asked Nov 27 '18 at 10:20
Joey DoeyJoey Doey
1155
1155
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
$endgroup$
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
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%2f3015601%2fwhy-is-it-necessary-x-and-y-beeing-bounded-when-having-same-moments-to-be-equal%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
$begingroup$
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
$endgroup$
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
$begingroup$
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
$endgroup$
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
$begingroup$
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
$endgroup$
You have to justify interchange of integral and sum. In this case $int sum |frac {(itx)^{k}} {k!}|d_F(x) leq e^{|t|}$ so we can use Fubini's Theorem.
answered Nov 27 '18 at 10:23
Kavi Rama MurthyKavi Rama Murthy
56k42158
56k42158
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
add a comment |
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
$endgroup$
– Joey Doey
Nov 27 '18 at 10:29
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
You can use DCT or Funbini.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:32
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
$begingroup$
but I still do not understand why I need that X and Y are bounded? What I wrote in the comment above is always true.
$endgroup$
– Joey Doey
Nov 27 '18 at 10:34
1
1
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
$begingroup$
Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
$endgroup$
– Kavi Rama Murthy
Nov 27 '18 at 10:35
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%2f3015601%2fwhy-is-it-necessary-x-and-y-beeing-bounded-when-having-same-moments-to-be-equal%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
wT8A Ruq,irK zOqd9s