Independence of ratios of independent variates












5












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If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
$x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
respectively, are $X$ & $Y$ independent?




I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?










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    5












    $begingroup$



    If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
    $x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
    respectively, are $X$ & $Y$ independent?




    I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?










    share|cite|improve this question











    $endgroup$















      5












      5








      5





      $begingroup$



      If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
      $x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
      respectively, are $X$ & $Y$ independent?




      I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?










      share|cite|improve this question











      $endgroup$





      If $X= x_1/(x_1+x_2)$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ where
      $x_1,x_2,x_3$ independent chi-square variates with d.f $n_1,n_2,n_3$
      respectively, are $X$ & $Y$ independent?




      I know the condition for independence of two random variables and the usual method followed i.e using joint distribution . But is there any short way to establish independence in the above question?







      distributions chi-squared independence gamma-distribution






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      edited Jan 13 at 15:28









      Xi'an

      55.5k792353




      55.5k792353










      asked Jan 13 at 14:28









      Jor_ElJor_El

      544




      544






















          2 Answers
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          6












          $begingroup$

          It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
          begin{align*}
          x_1&={varrho sin(theta)}^2\
          x_1&={varrho cos(theta)}^2\
          end{align*}

          we get that
          $$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
          and
          $$Y=frac{varrho^2}{varrho^2+x_3}$$
          are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            A geometrical interpretation/intuition



            You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points



            In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.



            The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".





            You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.



            See here a similar (but much simpler) calculation.






            share|cite|improve this answer











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              2 Answers
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              6












              $begingroup$

              It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
              begin{align*}
              x_1&={varrho sin(theta)}^2\
              x_1&={varrho cos(theta)}^2\
              end{align*}

              we get that
              $$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
              and
              $$Y=frac{varrho^2}{varrho^2+x_3}$$
              are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).






              share|cite|improve this answer











              $endgroup$


















                6












                $begingroup$

                It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
                begin{align*}
                x_1&={varrho sin(theta)}^2\
                x_1&={varrho cos(theta)}^2\
                end{align*}

                we get that
                $$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
                and
                $$Y=frac{varrho^2}{varrho^2+x_3}$$
                are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).






                share|cite|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
                  begin{align*}
                  x_1&={varrho sin(theta)}^2\
                  x_1&={varrho cos(theta)}^2\
                  end{align*}

                  we get that
                  $$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
                  and
                  $$Y=frac{varrho^2}{varrho^2+x_3}$$
                  are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).






                  share|cite|improve this answer











                  $endgroup$



                  It is a "well-known" property of the Gamma distributions that $x_1/(x_1+x_2)$ and $(x_1+x_2)$ are independent, and that $x_1+x_2+x_3$ and $Y= (x_1+x_2)/(x_1+x_2+x_3)$ are independent. For instance, writing
                  begin{align*}
                  x_1&={varrho sin(theta)}^2\
                  x_1&={varrho cos(theta)}^2\
                  end{align*}

                  we get that
                  $$X=frac{{varrho sin(theta)}^2}{{varrho sin(theta)}^2+{varrho cos(theta)}^2}=sin(theta)^2$$
                  and
                  $$Y=frac{varrho^2}{varrho^2+x_3}$$
                  are indeed functions of different variates (although the independence between $varrho$ and $theta$ has to be established).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 14 at 7:08

























                  answered Jan 13 at 15:31









                  Xi'anXi'an

                  55.5k792353




                  55.5k792353

























                      1












                      $begingroup$

                      A geometrical interpretation/intuition



                      You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points



                      In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.



                      The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".





                      You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.



                      See here a similar (but much simpler) calculation.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        A geometrical interpretation/intuition



                        You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points



                        In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.



                        The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".





                        You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.



                        See here a similar (but much simpler) calculation.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          A geometrical interpretation/intuition



                          You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points



                          In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.



                          The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".





                          You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.



                          See here a similar (but much simpler) calculation.






                          share|cite|improve this answer











                          $endgroup$



                          A geometrical interpretation/intuition



                          You could view the chi-squared variables $x_1,x_2,x_3$ as relating to independent standard normal distributed variables which in it's turn relates to uniformly distributed variables on a n-sphere https://en.wikipedia.org/wiki/N-sphere#Generating_random_points



                          In the same way as you can cut up a regular sphere into circles of different sizes you can cut up the hyper-sphere into hyper-spheres of lower dimension.



                          The distribution of $frac{x_1}{x_1+x_2}$ "the point/angle on a circle" is independent from $frac{x_1+x_2}{x_1+x_2+x_3}$ "the (relative) squared radius of the circle" or $x_1+x_2+x_3$ "the squared radius of the sphere in which that circle is embedded".





                          You could take a n-sphere (embedded in dimension $n_1+n_2+n_3$) and compute explicitly the value for $f_X(x)$ by computing the relative ratio of the areas of sub-sphere, the $(n_1-1)$-sphere with radius $sqrt{x_1}$ and the $(n_1-n_2-1)$-sphere with radius $sqrt{x_1+x_2}$. The result should only depend on the fraction $X=frac{x_1}{x_1+x_2}$ and be independent from $x_1+x_2$ or $x_1+x_2+x_3$.



                          See here a similar (but much simpler) calculation.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 15 at 15:09

























                          answered Jan 15 at 14:33









                          Martijn WeteringsMartijn Weterings

                          13.4k1659




                          13.4k1659






























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