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












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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







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    2












    $begingroup$



    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







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$



      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







      share|cite|improve this question











      $endgroup$





      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






      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







      Joey Doey

















      asked Nov 27 '18 at 10:20









      Joey DoeyJoey Doey

      1155




      1155






















          1 Answer
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          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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













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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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


















          2












          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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




















          • $begingroup$
            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




















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