Compute $Cov(X,Y)$












0












$begingroup$



Let $X: U(-1,1)$ and $Y=X^2$. Compute $operatorname{Cov}(X,Y)$.




$operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$.



$E[X]= frac{1}{2}(-1+1)=0$ so $operatorname{Cov}(X,Y)=E[XY]$, but how can I calculate $E[XY]$?






probability statistics covariance







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $X: U(-1,1)$ and $Y=X^2$. Compute $operatorname{Cov}(X,Y)$.




    $operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$.



    $E[X]= frac{1}{2}(-1+1)=0$ so $operatorname{Cov}(X,Y)=E[XY]$, but how can I calculate $E[XY]$?






    probability statistics covariance







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $X: U(-1,1)$ and $Y=X^2$. Compute $operatorname{Cov}(X,Y)$.




      $operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$.



      $E[X]= frac{1}{2}(-1+1)=0$ so $operatorname{Cov}(X,Y)=E[XY]$, but how can I calculate $E[XY]$?






      probability statistics covariance







      share|cite|improve this question











      $endgroup$





      Let $X: U(-1,1)$ and $Y=X^2$. Compute $operatorname{Cov}(X,Y)$.




      $operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y]$.



      $E[X]= frac{1}{2}(-1+1)=0$ so $operatorname{Cov}(X,Y)=E[XY]$, but how can I calculate $E[XY]$?






      probability statistics covariance




      probability statistics covariance






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 26 at 16:27









      Clement C.

      50.9k33992




      50.9k33992










      asked Jan 26 at 13:43









      Luke MarciLuke Marci

      856




      856






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          By definition of $Y$,
          $$
          mathbb{E}[XY] = mathbb{E}[X^3],.
          $$
          Now, you could compute this explicitly:
          $$
          mathbb{E}[X^3] = int_{-1}^1 x^3 f_X(x) dx = frac{1}{2}int_{-1}^1 x^3dx ,.
          $$
          However, it suffices to note that $X$ is a symmetric r.v. to see that its odd moments are zero, and thus $mathbb{E}[XY] =0$ (i.e., $X$ and $Y$, while clearly not independent, are uncorrelated).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You are right, thanks a lot!!
            $endgroup$
            – Luke Marci
            Jan 26 at 14:32










          • $begingroup$
            @LukeMarci You're welcome!
            $endgroup$
            – Clement C.
            Jan 26 at 14:33



















          1












          $begingroup$

          begin{align}
          E[XY]=E[X^3] = frac12 int_{-1}^1x^3 , dx = 0
          end{align}






          share|cite|improve this answer









          $endgroup$













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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            By definition of $Y$,
            $$
            mathbb{E}[XY] = mathbb{E}[X^3],.
            $$
            Now, you could compute this explicitly:
            $$
            mathbb{E}[X^3] = int_{-1}^1 x^3 f_X(x) dx = frac{1}{2}int_{-1}^1 x^3dx ,.
            $$
            However, it suffices to note that $X$ is a symmetric r.v. to see that its odd moments are zero, and thus $mathbb{E}[XY] =0$ (i.e., $X$ and $Y$, while clearly not independent, are uncorrelated).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You are right, thanks a lot!!
              $endgroup$
              – Luke Marci
              Jan 26 at 14:32










            • $begingroup$
              @LukeMarci You're welcome!
              $endgroup$
              – Clement C.
              Jan 26 at 14:33
















            1












            $begingroup$

            By definition of $Y$,
            $$
            mathbb{E}[XY] = mathbb{E}[X^3],.
            $$
            Now, you could compute this explicitly:
            $$
            mathbb{E}[X^3] = int_{-1}^1 x^3 f_X(x) dx = frac{1}{2}int_{-1}^1 x^3dx ,.
            $$
            However, it suffices to note that $X$ is a symmetric r.v. to see that its odd moments are zero, and thus $mathbb{E}[XY] =0$ (i.e., $X$ and $Y$, while clearly not independent, are uncorrelated).






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You are right, thanks a lot!!
              $endgroup$
              – Luke Marci
              Jan 26 at 14:32










            • $begingroup$
              @LukeMarci You're welcome!
              $endgroup$
              – Clement C.
              Jan 26 at 14:33














            1












            1








            1





            $begingroup$

            By definition of $Y$,
            $$
            mathbb{E}[XY] = mathbb{E}[X^3],.
            $$
            Now, you could compute this explicitly:
            $$
            mathbb{E}[X^3] = int_{-1}^1 x^3 f_X(x) dx = frac{1}{2}int_{-1}^1 x^3dx ,.
            $$
            However, it suffices to note that $X$ is a symmetric r.v. to see that its odd moments are zero, and thus $mathbb{E}[XY] =0$ (i.e., $X$ and $Y$, while clearly not independent, are uncorrelated).






            share|cite|improve this answer









            $endgroup$



            By definition of $Y$,
            $$
            mathbb{E}[XY] = mathbb{E}[X^3],.
            $$
            Now, you could compute this explicitly:
            $$
            mathbb{E}[X^3] = int_{-1}^1 x^3 f_X(x) dx = frac{1}{2}int_{-1}^1 x^3dx ,.
            $$
            However, it suffices to note that $X$ is a symmetric r.v. to see that its odd moments are zero, and thus $mathbb{E}[XY] =0$ (i.e., $X$ and $Y$, while clearly not independent, are uncorrelated).







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 26 at 13:53









            Clement C.Clement C.

            50.9k33992




            50.9k33992












            • $begingroup$
              You are right, thanks a lot!!
              $endgroup$
              – Luke Marci
              Jan 26 at 14:32










            • $begingroup$
              @LukeMarci You're welcome!
              $endgroup$
              – Clement C.
              Jan 26 at 14:33


















            • $begingroup$
              You are right, thanks a lot!!
              $endgroup$
              – Luke Marci
              Jan 26 at 14:32










            • $begingroup$
              @LukeMarci You're welcome!
              $endgroup$
              – Clement C.
              Jan 26 at 14:33
















            $begingroup$
            You are right, thanks a lot!!
            $endgroup$
            – Luke Marci
            Jan 26 at 14:32




            $begingroup$
            You are right, thanks a lot!!
            $endgroup$
            – Luke Marci
            Jan 26 at 14:32












            $begingroup$
            @LukeMarci You're welcome!
            $endgroup$
            – Clement C.
            Jan 26 at 14:33




            $begingroup$
            @LukeMarci You're welcome!
            $endgroup$
            – Clement C.
            Jan 26 at 14:33











            1












            $begingroup$

            begin{align}
            E[XY]=E[X^3] = frac12 int_{-1}^1x^3 , dx = 0
            end{align}






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              begin{align}
              E[XY]=E[X^3] = frac12 int_{-1}^1x^3 , dx = 0
              end{align}






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                begin{align}
                E[XY]=E[X^3] = frac12 int_{-1}^1x^3 , dx = 0
                end{align}






                share|cite|improve this answer









                $endgroup$



                begin{align}
                E[XY]=E[X^3] = frac12 int_{-1}^1x^3 , dx = 0
                end{align}







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 13:53









                Siong Thye GohSiong Thye Goh

                103k1468119




                103k1468119






























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