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







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$endgroup$












  • $begingroup$
    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









    $endgroup$













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      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






























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