Are uniformly distributed random variables $X$, $X^2$ (in)dependent?












-1












$begingroup$


We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.



However I am not sure how to show that they are (in)dependent.






probability uniform-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to see if the definition of independence holds.
    $endgroup$
    – LoveTooNap29
    Dec 19 '18 at 15:53
















-1












$begingroup$


We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.



However I am not sure how to show that they are (in)dependent.






probability uniform-distribution







share|cite|improve this question











$endgroup$












  • $begingroup$
    Try to see if the definition of independence holds.
    $endgroup$
    – LoveTooNap29
    Dec 19 '18 at 15:53














-1












-1








-1





$begingroup$


We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.



However I am not sure how to show that they are (in)dependent.






probability uniform-distribution







share|cite|improve this question











$endgroup$




We have a random variable $X sim U[-1,1]$. I should find out whether $X, X^2$ are independent or not. I have shown that $textrm{cov}(X,Y) = 0$ and that PDFs are $f_X(x) = frac{1}{2} $ for $x in [-1,1]$ and $0 $ else and $f_{X^2}(x) = frac{1}{2sqrt{x}}$ for $x in [0,1]$ and $0$ else.



However I am not sure how to show that they are (in)dependent.






probability uniform-distribution




probability uniform-distribution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 8:14









Björn Friedrich

2,66961831




2,66961831










asked Dec 19 '18 at 15:36









dwarf_dzdwarf_dz

1




1












  • $begingroup$
    Try to see if the definition of independence holds.
    $endgroup$
    – LoveTooNap29
    Dec 19 '18 at 15:53


















  • $begingroup$
    Try to see if the definition of independence holds.
    $endgroup$
    – LoveTooNap29
    Dec 19 '18 at 15:53
















$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53




$begingroup$
Try to see if the definition of independence holds.
$endgroup$
– LoveTooNap29
Dec 19 '18 at 15:53










3 Answers
3






active

oldest

votes


















1












$begingroup$

Hint:



begin{align}
P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
end{align}






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
    Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      For proving independence?
      $endgroup$
      – NCh
      Dec 20 '18 at 1:11










    • $begingroup$
      @NCh Thanks for attending me. Repaired.
      $endgroup$
      – drhab
      Dec 20 '18 at 7:57



















    0












    $begingroup$

    They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.






    share|cite|improve this answer









    $endgroup$













      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%2f3046516%2fare-uniformly-distributed-random-variables-x-x2-independent%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









      1












      $begingroup$

      Hint:



      begin{align}
      P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
      end{align}






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Hint:



        begin{align}
        P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
        end{align}






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Hint:



          begin{align}
          P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
          end{align}






          share|cite|improve this answer









          $endgroup$



          Hint:



          begin{align}
          P(X^2 le frac14, -frac12 le X le frac12) &= P( -frac12 le X le frac12)= frac12
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 19 '18 at 15:42









          Siong Thye GohSiong Thye Goh

          101k1466118




          101k1466118























              1












              $begingroup$

              For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
              Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                For proving independence?
                $endgroup$
                – NCh
                Dec 20 '18 at 1:11










              • $begingroup$
                @NCh Thanks for attending me. Repaired.
                $endgroup$
                – drhab
                Dec 20 '18 at 7:57
















              1












              $begingroup$

              For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
              Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                For proving independence?
                $endgroup$
                – NCh
                Dec 20 '18 at 1:11










              • $begingroup$
                @NCh Thanks for attending me. Repaired.
                $endgroup$
                – drhab
                Dec 20 '18 at 7:57














              1












              1








              1





              $begingroup$

              For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
              Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.






              share|cite|improve this answer











              $endgroup$



              For proving that $X$ and $X^2$ are not independent it is enough to find Borel set $A,B$ such that:$$P(Xin A,X^2in B)neq P(Xin A)P(X^2in B)$$
              Give it a try for e.g. $A=[frac12,1]$ and $B=[frac14,1]$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Dec 20 '18 at 7:57

























              answered Dec 19 '18 at 15:46









              drhabdrhab

              101k545136




              101k545136












              • $begingroup$
                For proving independence?
                $endgroup$
                – NCh
                Dec 20 '18 at 1:11










              • $begingroup$
                @NCh Thanks for attending me. Repaired.
                $endgroup$
                – drhab
                Dec 20 '18 at 7:57


















              • $begingroup$
                For proving independence?
                $endgroup$
                – NCh
                Dec 20 '18 at 1:11










              • $begingroup$
                @NCh Thanks for attending me. Repaired.
                $endgroup$
                – drhab
                Dec 20 '18 at 7:57
















              $begingroup$
              For proving independence?
              $endgroup$
              – NCh
              Dec 20 '18 at 1:11




              $begingroup$
              For proving independence?
              $endgroup$
              – NCh
              Dec 20 '18 at 1:11












              $begingroup$
              @NCh Thanks for attending me. Repaired.
              $endgroup$
              – drhab
              Dec 20 '18 at 7:57




              $begingroup$
              @NCh Thanks for attending me. Repaired.
              $endgroup$
              – drhab
              Dec 20 '18 at 7:57











              0












              $begingroup$

              They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.






                  share|cite|improve this answer









                  $endgroup$



                  They are dependent, since knowing the value of one variable updates your probability distribution for the other. To know $X$ is to know $X^2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 19 '18 at 15:45









                  J.G.J.G.

                  26.9k22742




                  26.9k22742






























                      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%2f3046516%2fare-uniformly-distributed-random-variables-x-x2-independent%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

                      Ellipse (mathématiques)

                      Image
                      Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
                      Read more

                      Quarter-circle Tiles

                      Image
                      up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
                      Read more

                      Mont Emei

                      Image
                      Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
                      Read more