Functions of Characteristic Functions












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Suppose that $phi(t)$ is the characteristic function of a random variable. Show that the functions $phi^2(t)$ and $|phi(t)|^2$ are also characteristic functions.




Attempt: So since $phi(t)$ is a characteristic function it satisfies Bochner's theorem but I was told by user Kavi Rama Murthy that this is not helpful.



So Instead we consider our random variable $X$, and an identically independently distributed variable $Y$, both with the same characteristic function $phi(t)$. The Characteristic function of the random variable $Z=X+Y$ is



$$ phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X+Y)}]=mathbf{E}[e^{itX}e^{itY}]=mathbf{E}[e^{itX}]mathbf{E}[e^{itY}]=phi_X(t)phi_y(t)=phi^2(t) $$
So since $X,Y$ are i.i.d we have that the characteristic function of the random variable $Z=X+Y$ is $phi^2(t)$.




Now I was given a hint, again by Kavi Rama Murthy, to use $X-Y$ but I am slightly confused as to how to continue this is my attempt:



Attempt: Consider the random variable $W=X-Y$ with $X,Y$ i.i.d. then we find the characteristic function of $Z$.



$$phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X-Y)}]=mathbf{E}[e^{itX}]mathbf{E}[e^{-itY}]=phi_X(t)mathbf{E}[e^{-itY}] $$






probability random-variables characteristic-functions







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    0












    $begingroup$



    Suppose that $phi(t)$ is the characteristic function of a random variable. Show that the functions $phi^2(t)$ and $|phi(t)|^2$ are also characteristic functions.




    Attempt: So since $phi(t)$ is a characteristic function it satisfies Bochner's theorem but I was told by user Kavi Rama Murthy that this is not helpful.



    So Instead we consider our random variable $X$, and an identically independently distributed variable $Y$, both with the same characteristic function $phi(t)$. The Characteristic function of the random variable $Z=X+Y$ is



    $$ phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X+Y)}]=mathbf{E}[e^{itX}e^{itY}]=mathbf{E}[e^{itX}]mathbf{E}[e^{itY}]=phi_X(t)phi_y(t)=phi^2(t) $$
    So since $X,Y$ are i.i.d we have that the characteristic function of the random variable $Z=X+Y$ is $phi^2(t)$.




    Now I was given a hint, again by Kavi Rama Murthy, to use $X-Y$ but I am slightly confused as to how to continue this is my attempt:



    Attempt: Consider the random variable $W=X-Y$ with $X,Y$ i.i.d. then we find the characteristic function of $Z$.



    $$phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X-Y)}]=mathbf{E}[e^{itX}]mathbf{E}[e^{-itY}]=phi_X(t)mathbf{E}[e^{-itY}] $$






    probability random-variables characteristic-functions







    share|cite|improve this question











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      0












      0








      0





      $begingroup$



      Suppose that $phi(t)$ is the characteristic function of a random variable. Show that the functions $phi^2(t)$ and $|phi(t)|^2$ are also characteristic functions.




      Attempt: So since $phi(t)$ is a characteristic function it satisfies Bochner's theorem but I was told by user Kavi Rama Murthy that this is not helpful.



      So Instead we consider our random variable $X$, and an identically independently distributed variable $Y$, both with the same characteristic function $phi(t)$. The Characteristic function of the random variable $Z=X+Y$ is



      $$ phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X+Y)}]=mathbf{E}[e^{itX}e^{itY}]=mathbf{E}[e^{itX}]mathbf{E}[e^{itY}]=phi_X(t)phi_y(t)=phi^2(t) $$
      So since $X,Y$ are i.i.d we have that the characteristic function of the random variable $Z=X+Y$ is $phi^2(t)$.




      Now I was given a hint, again by Kavi Rama Murthy, to use $X-Y$ but I am slightly confused as to how to continue this is my attempt:



      Attempt: Consider the random variable $W=X-Y$ with $X,Y$ i.i.d. then we find the characteristic function of $Z$.



      $$phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X-Y)}]=mathbf{E}[e^{itX}]mathbf{E}[e^{-itY}]=phi_X(t)mathbf{E}[e^{-itY}] $$






      probability random-variables characteristic-functions







      share|cite|improve this question











      $endgroup$





      Suppose that $phi(t)$ is the characteristic function of a random variable. Show that the functions $phi^2(t)$ and $|phi(t)|^2$ are also characteristic functions.




      Attempt: So since $phi(t)$ is a characteristic function it satisfies Bochner's theorem but I was told by user Kavi Rama Murthy that this is not helpful.



      So Instead we consider our random variable $X$, and an identically independently distributed variable $Y$, both with the same characteristic function $phi(t)$. The Characteristic function of the random variable $Z=X+Y$ is



      $$ phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X+Y)}]=mathbf{E}[e^{itX}e^{itY}]=mathbf{E}[e^{itX}]mathbf{E}[e^{itY}]=phi_X(t)phi_y(t)=phi^2(t) $$
      So since $X,Y$ are i.i.d we have that the characteristic function of the random variable $Z=X+Y$ is $phi^2(t)$.




      Now I was given a hint, again by Kavi Rama Murthy, to use $X-Y$ but I am slightly confused as to how to continue this is my attempt:



      Attempt: Consider the random variable $W=X-Y$ with $X,Y$ i.i.d. then we find the characteristic function of $Z$.



      $$phi_Z(t)=mathbf{E}[e^{itZ}]=mathbf{E}[e^{it(X-Y)}]=mathbf{E}[e^{itX}]mathbf{E}[e^{-itY}]=phi_X(t)mathbf{E}[e^{-itY}] $$






      probability random-variables characteristic-functions




      probability random-variables characteristic-functions






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      edited Nov 28 '18 at 15:42







      elcharlosmaster

















      asked Nov 27 '18 at 23:11









      elcharlosmasterelcharlosmaster

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          $begingroup$

          If $X$ and $Y$ are i.i.d. with characteristic function $phi$ a simple calculation show that $phi^{2}$ is the characteristic function of $X+Y$ and $|phi|^{2}$ is the characteristic function of $X-Y$. Bochner's Theorem is not useful here.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "simple" seems to be relative...
            $endgroup$
            – elcharlosmaster
            Nov 27 '18 at 23:44










          • $begingroup$
            I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
            $endgroup$
            – elcharlosmaster
            Nov 28 '18 at 15:37










          • $begingroup$
            @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 28 '18 at 23:18











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          $begingroup$

          If $X$ and $Y$ are i.i.d. with characteristic function $phi$ a simple calculation show that $phi^{2}$ is the characteristic function of $X+Y$ and $|phi|^{2}$ is the characteristic function of $X-Y$. Bochner's Theorem is not useful here.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "simple" seems to be relative...
            $endgroup$
            – elcharlosmaster
            Nov 27 '18 at 23:44










          • $begingroup$
            I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
            $endgroup$
            – elcharlosmaster
            Nov 28 '18 at 15:37










          • $begingroup$
            @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 28 '18 at 23:18
















          2












          $begingroup$

          If $X$ and $Y$ are i.i.d. with characteristic function $phi$ a simple calculation show that $phi^{2}$ is the characteristic function of $X+Y$ and $|phi|^{2}$ is the characteristic function of $X-Y$. Bochner's Theorem is not useful here.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            "simple" seems to be relative...
            $endgroup$
            – elcharlosmaster
            Nov 27 '18 at 23:44










          • $begingroup$
            I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
            $endgroup$
            – elcharlosmaster
            Nov 28 '18 at 15:37










          • $begingroup$
            @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 28 '18 at 23:18














          2












          2








          2





          $begingroup$

          If $X$ and $Y$ are i.i.d. with characteristic function $phi$ a simple calculation show that $phi^{2}$ is the characteristic function of $X+Y$ and $|phi|^{2}$ is the characteristic function of $X-Y$. Bochner's Theorem is not useful here.






          share|cite|improve this answer









          $endgroup$



          If $X$ and $Y$ are i.i.d. with characteristic function $phi$ a simple calculation show that $phi^{2}$ is the characteristic function of $X+Y$ and $|phi|^{2}$ is the characteristic function of $X-Y$. Bochner's Theorem is not useful here.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 27 '18 at 23:40









          Kavi Rama MurthyKavi Rama Murthy

          56.6k42159




          56.6k42159












          • $begingroup$
            "simple" seems to be relative...
            $endgroup$
            – elcharlosmaster
            Nov 27 '18 at 23:44










          • $begingroup$
            I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
            $endgroup$
            – elcharlosmaster
            Nov 28 '18 at 15:37










          • $begingroup$
            @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 28 '18 at 23:18


















          • $begingroup$
            "simple" seems to be relative...
            $endgroup$
            – elcharlosmaster
            Nov 27 '18 at 23:44










          • $begingroup$
            I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
            $endgroup$
            – elcharlosmaster
            Nov 28 '18 at 15:37










          • $begingroup$
            @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
            $endgroup$
            – Kavi Rama Murthy
            Nov 28 '18 at 23:18
















          $begingroup$
          "simple" seems to be relative...
          $endgroup$
          – elcharlosmaster
          Nov 27 '18 at 23:44




          $begingroup$
          "simple" seems to be relative...
          $endgroup$
          – elcharlosmaster
          Nov 27 '18 at 23:44












          $begingroup$
          I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
          $endgroup$
          – elcharlosmaster
          Nov 28 '18 at 15:37




          $begingroup$
          I have used your hint to solve the first part but am not sure how to get the negative into an absolute value.
          $endgroup$
          – elcharlosmaster
          Nov 28 '18 at 15:37












          $begingroup$
          @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
          $endgroup$
          – Kavi Rama Murthy
          Nov 28 '18 at 23:18




          $begingroup$
          @elcharlosmaster $Ee^{it(X-Y)}=Ee^{itX}Ee^{-itY}$ and $Ee^{-itY}$ is the complex conjugate of $Ee^{itY}$. For any complex number $z$ we have $|z|^{2}=zoverline {z}$.
          $endgroup$
          – Kavi Rama Murthy
          Nov 28 '18 at 23:18


















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