Proving that if $varphi(t)$ is an infinitely divisible characteristic function then $|varphi(t)|$ as well












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Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.



My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
$$varphi(t) = big(varphi_1(t) big)^n.$$
I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
$$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
$$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
I got stuck here. What can I do next? Is my attempt a correct one?






probability characteristic-functions







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    $begingroup$


    Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.



    My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
    $$varphi(t) = big(varphi_1(t) big)^n.$$
    I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
    $$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
    And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
    $$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
    I got stuck here. What can I do next? Is my attempt a correct one?






    probability characteristic-functions







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.



      My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
      $$varphi(t) = big(varphi_1(t) big)^n.$$
      I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
      $$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
      And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
      $$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
      I got stuck here. What can I do next? Is my attempt a correct one?






      probability characteristic-functions







      share|cite|improve this question









      $endgroup$




      Problem. I am given an infinitely divisible characteristic function $varphi(t)$. My task it to prove that $|varphi(t)|$ is infinitely divisible too.



      My attempt. Because $varphi$ is infinitely divisible then for any $n in mathbb{N}$ there exists characteristic functions $varphi_1(t)= varphi_2(t) = ldots varphi_n(t)$ such that
      $$varphi(t) = big(varphi_1(t) big)^n.$$
      I noticed that $|varphi(t)|^2$ is an infinitely divisible characteristic function either because
      $$|varphi(t)|^2 = varphi(t) overlinevarphi(t).$$
      And it's easy to show that a product of a finite amount of infinitely divisible characteristic functions is a infinitely divisible characteristic function too. $overlinevarphi(t)$ is infinitely divisible characteristic function because
      $$varphi(t) = overlinevarphi(-t) = big(varphi_1(t) big)^n.$$
      I got stuck here. What can I do next? Is my attempt a correct one?






      probability characteristic-functions




      probability characteristic-functions






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      asked Jan 8 at 15:51









      HendrraHendrra

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          Let $varphi_j(t)$ be the characteristic function such that
          $$
          varphi(t) = [varphi_j(t)]^j
          $$           holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
          [phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
          $$           and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.






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          • $begingroup$
            Thank you. Very smart solution :)
            $endgroup$
            – Hendrra
            Jan 9 at 14:17











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          $begingroup$

          Let $varphi_j(t)$ be the characteristic function such that
          $$
          varphi(t) = [varphi_j(t)]^j
          $$           holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
          [phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
          $$           and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. Very smart solution :)
            $endgroup$
            – Hendrra
            Jan 9 at 14:17
















          1












          $begingroup$

          Let $varphi_j(t)$ be the characteristic function such that
          $$
          varphi(t) = [varphi_j(t)]^j
          $$           holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
          [phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
          $$           and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. Very smart solution :)
            $endgroup$
            – Hendrra
            Jan 9 at 14:17














          1












          1








          1





          $begingroup$

          Let $varphi_j(t)$ be the characteristic function such that
          $$
          varphi(t) = [varphi_j(t)]^j
          $$           holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
          [phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
          $$           and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.






          share|cite|improve this answer









          $endgroup$



          Let $varphi_j(t)$ be the characteristic function such that
          $$
          varphi(t) = [varphi_j(t)]^j
          $$           holds. Let $varphi_2(t) =phi(t)$. We claim that $phi$ is infinitely divisible and $phi_j(t) = varphi_{2j}(t)$. Note that $$
          [phi(t)]^{2}=varphi(t)=[varphi_{2j}(t)]^{2j},
          $$           and hence $frac{phi(t)}{ (varphi_{2j}(t))^j}=omega(t)$ where $omega(t)in {-1,1}$ for all $tinmathbb{R}$ (note that infinitely divisible characteristic function $varphi$ nowhere vanishes and hence neither does $varphi_{2j}$.) This gives by continuity of $omega(t)$ that $omega equiv 1$ and $phi(t) =[varphi_{2j}(t)]^j$. This shows $phi$ is infinitely divisible. Now, since $phi(t)$ is infinitely divisible, so does $overline{phi}(t)$ and $phi(t)overline{phi}(t)=|phi(t)|^2=|varphi(t)|$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 8 at 17:05









          SongSong

          11.2k628




          11.2k628












          • $begingroup$
            Thank you. Very smart solution :)
            $endgroup$
            – Hendrra
            Jan 9 at 14:17


















          • $begingroup$
            Thank you. Very smart solution :)
            $endgroup$
            – Hendrra
            Jan 9 at 14:17
















          $begingroup$
          Thank you. Very smart solution :)
          $endgroup$
          – Hendrra
          Jan 9 at 14:17




          $begingroup$
          Thank you. Very smart solution :)
          $endgroup$
          – Hendrra
          Jan 9 at 14:17


















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