Why is it necessary X and Y beeing bounded when having same moments to be equal












1















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







share|cite|improve this question





























    1















    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







    share|cite|improve this question



























      1












      1








      1








      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







      share|cite|improve this question
















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 '18 at 13:58

























      asked Nov 27 '18 at 10:20









      Joey Doey

      1104




      1104






















          1 Answer
          1






          active

          oldest

          votes


















          1














          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.






          share|cite|improve this answer





















          • you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
            – Joey Doey
            Nov 27 '18 at 10:29












          • You can use DCT or Funbini.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:32










          • 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.
            – Joey Doey
            Nov 27 '18 at 10:34






          • 1




            Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:35













          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
          });


          }
          });














          draft saved

          draft discarded


















          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









          1














          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.






          share|cite|improve this answer





















          • you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
            – Joey Doey
            Nov 27 '18 at 10:29












          • You can use DCT or Funbini.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:32










          • 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.
            – Joey Doey
            Nov 27 '18 at 10:34






          • 1




            Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:35


















          1














          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.






          share|cite|improve this answer





















          • you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
            – Joey Doey
            Nov 27 '18 at 10:29












          • You can use DCT or Funbini.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:32










          • 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.
            – Joey Doey
            Nov 27 '18 at 10:34






          • 1




            Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:35
















          1












          1








          1






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 10:23









          Kavi Rama Murthy

          50.5k31854




          50.5k31854












          • you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
            – Joey Doey
            Nov 27 '18 at 10:29












          • You can use DCT or Funbini.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:32










          • 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.
            – Joey Doey
            Nov 27 '18 at 10:34






          • 1




            Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:35




















          • you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
            – Joey Doey
            Nov 27 '18 at 10:29












          • You can use DCT or Funbini.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:32










          • 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.
            – Joey Doey
            Nov 27 '18 at 10:34






          • 1




            Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
            – Kavi Rama Murthy
            Nov 27 '18 at 10:35


















          you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
          – Joey Doey
          Nov 27 '18 at 10:29






          you mean fubini? but dom. convergence would work aswell or since $int e^{itx} le 1$ we can change its order by dom. convergence
          – Joey Doey
          Nov 27 '18 at 10:29














          You can use DCT or Funbini.
          – Kavi Rama Murthy
          Nov 27 '18 at 10:32




          You can use DCT or Funbini.
          – Kavi Rama Murthy
          Nov 27 '18 at 10:32












          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.
          – Joey Doey
          Nov 27 '18 at 10:34




          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.
          – Joey Doey
          Nov 27 '18 at 10:34




          1




          1




          Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
          – Kavi Rama Murthy
          Nov 27 '18 at 10:35






          Application of DCT is not that simple. It will not work if $x$ is not restricted to a bounded set.
          – Kavi Rama Murthy
          Nov 27 '18 at 10:35




















          draft saved

          draft discarded




















































          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          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





















































          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







          -

          Popular posts from this blog

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more