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







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












$begingroup$


$X,Y$ 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





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f261783%2fdistribution-of-xyz-if-x-y-z-is-i-i-d-uniform-on-0-1%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f261783%2fdistribution-of-xyz-if-x-y-z-is-i-i-d-uniform-on-0-1%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
            Read more