Distribution of $(XY)^Z$ if $(X,Y,Z)$ is i.i.d. uniform on $[0,1]$












6












$begingroup$


$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$



How is random variable $(XY)^Z$ distributed?



I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.






probability-theory probability-distributions







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$endgroup$












  • $begingroup$
    Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
    $endgroup$
    – Dilip Sarwate
    Dec 18 '12 at 23:45
















6












$begingroup$


$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$



How is random variable $(XY)^Z$ distributed?



I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.






probability-theory probability-distributions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
    $endgroup$
    – Dilip Sarwate
    Dec 18 '12 at 23:45














6












6








6


0



$begingroup$


$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$



How is random variable $(XY)^Z$ distributed?



I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.






probability-theory probability-distributions







share|cite|improve this question











$endgroup$




$X,Y$ and $Z$ are independent uniformly distributed on $[0,1]$



How is random variable $(XY)^Z$ distributed?



I had an idea to logarithm this and use convolution integral for the sum, but I'm not sure it's possible.






probability-theory probability-distributions




probability-theory probability-distributions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 1 '17 at 21:08









Did

247k23223460




247k23223460










asked Dec 18 '12 at 23:31









XxxXxx

354110




354110












  • $begingroup$
    Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
    $endgroup$
    – Dilip Sarwate
    Dec 18 '12 at 23:45


















  • $begingroup$
    Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
    $endgroup$
    – Dilip Sarwate
    Dec 18 '12 at 23:45
















$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45




$begingroup$
Why do you need to know? Also, does $(XY)^{frac{1}{2}}$ mean $sqrt{XY}$ or $-sqrt{XY}$.
$endgroup$
– Dilip Sarwate
Dec 18 '12 at 23:45










3 Answers
3






active

oldest

votes


















9












$begingroup$

Hints:




  • The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.


  • If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.


  • If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.



Conclusion:




  • If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
    $endgroup$
    – Fabian
    Dec 19 '12 at 0:02





















5












$begingroup$

Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.



The distribution of the random variable $W=(XY)^Z$ is given by:
$$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
&= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
&=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
&=-int_0^w!deta log eta (1-log_{eta} w)\
&=w.
end{align}$$
with $eta=xy$.



Thus the variable $W$ is also uniformly distributed (between 0 and 1).






share|cite|improve this answer









$endgroup$





















    -1












    $begingroup$

    Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You can format your answer using latex.
      $endgroup$
      – user60610
      Apr 18 '13 at 14:17










    • $begingroup$
      Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
      $endgroup$
      – Did
      Nov 27 '18 at 19:00











    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    9












    $begingroup$

    Hints:




    • The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.


    • If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.


    • If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.



    Conclusion:




    • If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
      $endgroup$
      – Fabian
      Dec 19 '12 at 0:02


















    9












    $begingroup$

    Hints:




    • The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.


    • If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.


    • If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.



    Conclusion:




    • If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
      $endgroup$
      – Fabian
      Dec 19 '12 at 0:02
















    9












    9








    9





    $begingroup$

    Hints:




    • The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.


    • If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.


    • If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.



    Conclusion:




    • If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.






    share|cite|improve this answer









    $endgroup$



    Hints:




    • The random variable $X$ is uniform on $(0,1)$ if and only if $-log X$ is exponential with parameter $1$.


    • If $U$ and $V$ are independent and exponential with parameter $1$, then $U+V$ is gamma distributed $(2,1)$, that is, with density $wmapsto wmathrm e^{-w}mathbf 1_{wgt0}$.


    • If $W$ is gamma distributed $(2,1)$ and $T$ is uniform on $(0,1)$ and independent of $W$, then $WU$ is exponential with parameter $1$.



    Conclusion:




    • If $X$, $Y$ and $Z$ are independent and uniform on $(0,1)$, then $(XY)^Z$ is uniform on $(0,1)$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 19 '12 at 0:01









    DidDid

    247k23223460




    247k23223460












    • $begingroup$
      +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
      $endgroup$
      – Fabian
      Dec 19 '12 at 0:02




















    • $begingroup$
      +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
      $endgroup$
      – Fabian
      Dec 19 '12 at 0:02


















    $begingroup$
    +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
    $endgroup$
    – Fabian
    Dec 19 '12 at 0:02






    $begingroup$
    +1: this is a nicer argument than my calculation... Even though they rely on fact which the OP might or might not have already proven.
    $endgroup$
    – Fabian
    Dec 19 '12 at 0:02













    5












    $begingroup$

    Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.



    The distribution of the random variable $W=(XY)^Z$ is given by:
    $$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
    &= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
    &=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
    &=-int_0^w!deta log eta (1-log_{eta} w)\
    &=w.
    end{align}$$
    with $eta=xy$.



    Thus the variable $W$ is also uniformly distributed (between 0 and 1).






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.



      The distribution of the random variable $W=(XY)^Z$ is given by:
      $$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
      &= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
      &=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
      &=-int_0^w!deta log eta (1-log_{eta} w)\
      &=w.
      end{align}$$
      with $eta=xy$.



      Thus the variable $W$ is also uniformly distributed (between 0 and 1).






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.



        The distribution of the random variable $W=(XY)^Z$ is given by:
        $$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
        &= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
        &=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
        &=-int_0^w!deta log eta (1-log_{eta} w)\
        &=w.
        end{align}$$
        with $eta=xy$.



        Thus the variable $W$ is also uniformly distributed (between 0 and 1).






        share|cite|improve this answer









        $endgroup$



        Given the simplicity of the result there must be a nice short way to obtain it. However, I did not find one so I present the long and complicated calculation.



        The distribution of the random variable $W=(XY)^Z$ is given by:
        $$begin{align}P(wgeq W) &= int_0^1!dxint_0^1!dyint_0^1!dz, theta(w-(xy)^z)\
        &= int_0^1!dxint_0^1!dy max{1-log_{xy} w,0}\
        &=int_0^1!detaint_eta^1!frac{dx}{x}max{1-log_{eta} w,0} \
        &=-int_0^w!deta log eta (1-log_{eta} w)\
        &=w.
        end{align}$$
        with $eta=xy$.



        Thus the variable $W$ is also uniformly distributed (between 0 and 1).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '12 at 0:00









        FabianFabian

        19.7k3674




        19.7k3674























            -1












            $begingroup$

            Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You can format your answer using latex.
              $endgroup$
              – user60610
              Apr 18 '13 at 14:17










            • $begingroup$
              Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
              $endgroup$
              – Did
              Nov 27 '18 at 19:00
















            -1












            $begingroup$

            Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You can format your answer using latex.
              $endgroup$
              – user60610
              Apr 18 '13 at 14:17










            • $begingroup$
              Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
              $endgroup$
              – Did
              Nov 27 '18 at 19:00














            -1












            -1








            -1





            $begingroup$

            Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.






            share|cite|improve this answer











            $endgroup$



            Using the definition of weak convergence, it is so easy. First, for any positive integer $kge 0$ we have $E{W^k}=1/(k+1)=EU^k$. Hence, for any polynomial $f(x)$, we have $Ef(W)=Ef(U)$. For any bounded and continuous function $g(cdot)$, we can find a polynomial function $f(cdot)$ such that $f$ can approximate $g$ uniformly by Weierstrass's theorem. Thus, $Eg(W)=Eg(U)$. So $Wsim U(0,1)$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 21 at 19:03









            Mars Plastic

            865




            865










            answered Apr 18 '13 at 13:57









            user73223user73223

            1




            1












            • $begingroup$
              You can format your answer using latex.
              $endgroup$
              – user60610
              Apr 18 '13 at 14:17










            • $begingroup$
              Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
              $endgroup$
              – Did
              Nov 27 '18 at 19:00


















            • $begingroup$
              You can format your answer using latex.
              $endgroup$
              – user60610
              Apr 18 '13 at 14:17










            • $begingroup$
              Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
              $endgroup$
              – Did
              Nov 27 '18 at 19:00
















            $begingroup$
            You can format your answer using latex.
            $endgroup$
            – user60610
            Apr 18 '13 at 14:17




            $begingroup$
            You can format your answer using latex.
            $endgroup$
            – user60610
            Apr 18 '13 at 14:17












            $begingroup$
            Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
            $endgroup$
            – Did
            Nov 27 '18 at 19:00




            $begingroup$
            Hmmm... By the way, is this so obvious that E{W^k}=1/(k+1)?
            $endgroup$
            – Did
            Nov 27 '18 at 19:00


















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