Proof by characteristic functions that $X+Y$ and $2X$ are identically distributed












0












$begingroup$


The exercise states that X,Y iid and we know that X+Y has Cauchy distribution. And they require to prove that 2X has also Cauchy distribution. Let me put it straight, I dont think I understand it fully , let's forget about Cauchy distribution at all.



Isn't it true for all distributions? If two random variables are iid, their characteristic functions are equal and



$$Ee^{itX}=Ee^{itY}$$
$$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}=Ee^{itX}Ee^{itX}=Ee^{it2X}$$ hence $2X$ has distribution as $X+Y$ almost surely?






probability probability-distributions characteristic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
    $endgroup$
    – Mike
    Dec 4 '18 at 20:30












  • $begingroup$
    Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:32








  • 2




    $begingroup$
    Why is $E(e^{itX})^2=E(e^{it2X})$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 20:35










  • $begingroup$
    Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:38












  • $begingroup$
    This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
    $endgroup$
    – Did
    Dec 4 '18 at 20:43
















0












$begingroup$


The exercise states that X,Y iid and we know that X+Y has Cauchy distribution. And they require to prove that 2X has also Cauchy distribution. Let me put it straight, I dont think I understand it fully , let's forget about Cauchy distribution at all.



Isn't it true for all distributions? If two random variables are iid, their characteristic functions are equal and



$$Ee^{itX}=Ee^{itY}$$
$$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}=Ee^{itX}Ee^{itX}=Ee^{it2X}$$ hence $2X$ has distribution as $X+Y$ almost surely?






probability probability-distributions characteristic-functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
    $endgroup$
    – Mike
    Dec 4 '18 at 20:30












  • $begingroup$
    Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:32








  • 2




    $begingroup$
    Why is $E(e^{itX})^2=E(e^{it2X})$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 20:35










  • $begingroup$
    Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:38












  • $begingroup$
    This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
    $endgroup$
    – Did
    Dec 4 '18 at 20:43














0












0








0





$begingroup$


The exercise states that X,Y iid and we know that X+Y has Cauchy distribution. And they require to prove that 2X has also Cauchy distribution. Let me put it straight, I dont think I understand it fully , let's forget about Cauchy distribution at all.



Isn't it true for all distributions? If two random variables are iid, their characteristic functions are equal and



$$Ee^{itX}=Ee^{itY}$$
$$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}=Ee^{itX}Ee^{itX}=Ee^{it2X}$$ hence $2X$ has distribution as $X+Y$ almost surely?






probability probability-distributions characteristic-functions







share|cite|improve this question











$endgroup$




The exercise states that X,Y iid and we know that X+Y has Cauchy distribution. And they require to prove that 2X has also Cauchy distribution. Let me put it straight, I dont think I understand it fully , let's forget about Cauchy distribution at all.



Isn't it true for all distributions? If two random variables are iid, their characteristic functions are equal and



$$Ee^{itX}=Ee^{itY}$$
$$Ee^{it(X+Y)}=Ee^{itX}Ee^{itY}=Ee^{itX}Ee^{itX}=Ee^{it2X}$$ hence $2X$ has distribution as $X+Y$ almost surely?






probability probability-distributions characteristic-functions




probability 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 Dec 4 '18 at 21:01









Jean Marie

29.3k42050




29.3k42050










asked Dec 4 '18 at 20:24









ryszard egginkryszard eggink

314110




314110








  • 1




    $begingroup$
    If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
    $endgroup$
    – Mike
    Dec 4 '18 at 20:30












  • $begingroup$
    Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:32








  • 2




    $begingroup$
    Why is $E(e^{itX})^2=E(e^{it2X})$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 20:35










  • $begingroup$
    Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:38












  • $begingroup$
    This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
    $endgroup$
    – Did
    Dec 4 '18 at 20:43














  • 1




    $begingroup$
    If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
    $endgroup$
    – Mike
    Dec 4 '18 at 20:30












  • $begingroup$
    Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:32








  • 2




    $begingroup$
    Why is $E(e^{itX})^2=E(e^{it2X})$?
    $endgroup$
    – Lord Shark the Unknown
    Dec 4 '18 at 20:35










  • $begingroup$
    Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
    $endgroup$
    – ryszard eggink
    Dec 4 '18 at 20:38












  • $begingroup$
    This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
    $endgroup$
    – Did
    Dec 4 '18 at 20:43








1




1




$begingroup$
If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
$endgroup$
– Mike
Dec 4 '18 at 20:30






$begingroup$
If $X$ and $Y$ are iid and (unless $X$ is almost everywhere constant) then $X+Y$ does not have the same distribution as $2X$.
$endgroup$
– Mike
Dec 4 '18 at 20:30














$begingroup$
Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
$endgroup$
– ryszard eggink
Dec 4 '18 at 20:32






$begingroup$
Can you develop your thought? This would mean that this is a false statement, because X+Y would be constant almost everywhere
$endgroup$
– ryszard eggink
Dec 4 '18 at 20:32






2




2




$begingroup$
Why is $E(e^{itX})^2=E(e^{it2X})$?
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 20:35




$begingroup$
Why is $E(e^{itX})^2=E(e^{it2X})$?
$endgroup$
– Lord Shark the Unknown
Dec 4 '18 at 20:35












$begingroup$
Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
$endgroup$
– ryszard eggink
Dec 4 '18 at 20:38






$begingroup$
Ok thanks, @LordSharktheUnknown, it is so beacuse cauchy characteristic function is $e^{-|x|}$ for all x
$endgroup$
– ryszard eggink
Dec 4 '18 at 20:38














$begingroup$
This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
$endgroup$
– Did
Dec 4 '18 at 20:43




$begingroup$
This is rather fascinating: you accepted, rather instantly, an answer which does not address your question at all. What is going on?
$endgroup$
– Did
Dec 4 '18 at 20:43










2 Answers
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$begingroup$

The characteristic function of $X$ is $phi(t):=exp (ix_0t-gamma|t|)$. The desired result is $phi^2(t)=phi(2t)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    As pointed out by Lord Shark the Unknown, the error in your general proof is $(E e^{itX})^2 = E e^{it2X}$, which does not hold in general.



    However, as you noted, in the case of $X$ being Cauchy with scale $gamma$, the above becomes $$(e^{-gamma|t|})^2 = e^{-gamma |2t|}.$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      The characteristic function of $X$ is $phi(t):=exp (ix_0t-gamma|t|)$. The desired result is $phi^2(t)=phi(2t)$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The characteristic function of $X$ is $phi(t):=exp (ix_0t-gamma|t|)$. The desired result is $phi^2(t)=phi(2t)$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The characteristic function of $X$ is $phi(t):=exp (ix_0t-gamma|t|)$. The desired result is $phi^2(t)=phi(2t)$.






          share|cite|improve this answer









          $endgroup$



          The characteristic function of $X$ is $phi(t):=exp (ix_0t-gamma|t|)$. The desired result is $phi^2(t)=phi(2t)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 4 '18 at 20:39









          J.G.J.G.

          25k22539




          25k22539























              1












              $begingroup$

              As pointed out by Lord Shark the Unknown, the error in your general proof is $(E e^{itX})^2 = E e^{it2X}$, which does not hold in general.



              However, as you noted, in the case of $X$ being Cauchy with scale $gamma$, the above becomes $$(e^{-gamma|t|})^2 = e^{-gamma |2t|}.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                As pointed out by Lord Shark the Unknown, the error in your general proof is $(E e^{itX})^2 = E e^{it2X}$, which does not hold in general.



                However, as you noted, in the case of $X$ being Cauchy with scale $gamma$, the above becomes $$(e^{-gamma|t|})^2 = e^{-gamma |2t|}.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  As pointed out by Lord Shark the Unknown, the error in your general proof is $(E e^{itX})^2 = E e^{it2X}$, which does not hold in general.



                  However, as you noted, in the case of $X$ being Cauchy with scale $gamma$, the above becomes $$(e^{-gamma|t|})^2 = e^{-gamma |2t|}.$$






                  share|cite|improve this answer









                  $endgroup$



                  As pointed out by Lord Shark the Unknown, the error in your general proof is $(E e^{itX})^2 = E e^{it2X}$, which does not hold in general.



                  However, as you noted, in the case of $X$ being Cauchy with scale $gamma$, the above becomes $$(e^{-gamma|t|})^2 = e^{-gamma |2t|}.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 4 '18 at 20:41









                  angryavianangryavian

                  40.6k23280




                  40.6k23280






























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