Finding the density function given the conditional.












-1












$begingroup$


I have the following conditional density function:



$f_{Y vert X = x}(y) sim Normal(0,frac{1}{x})$



I also know that $ X sim Gamma(frac{3}{2}, frac{1}{2})$.



I want to find the density of just $f_Y$.



My problem is that I do not know the joint density of X and Y, so I do not know formulas to solve this. Any hints?






probability







share|cite|improve this question









$endgroup$

















    -1












    $begingroup$


    I have the following conditional density function:



    $f_{Y vert X = x}(y) sim Normal(0,frac{1}{x})$



    I also know that $ X sim Gamma(frac{3}{2}, frac{1}{2})$.



    I want to find the density of just $f_Y$.



    My problem is that I do not know the joint density of X and Y, so I do not know formulas to solve this. Any hints?






    probability







    share|cite|improve this question









    $endgroup$















      -1












      -1








      -1





      $begingroup$


      I have the following conditional density function:



      $f_{Y vert X = x}(y) sim Normal(0,frac{1}{x})$



      I also know that $ X sim Gamma(frac{3}{2}, frac{1}{2})$.



      I want to find the density of just $f_Y$.



      My problem is that I do not know the joint density of X and Y, so I do not know formulas to solve this. Any hints?






      probability







      share|cite|improve this question









      $endgroup$




      I have the following conditional density function:



      $f_{Y vert X = x}(y) sim Normal(0,frac{1}{x})$



      I also know that $ X sim Gamma(frac{3}{2}, frac{1}{2})$.



      I want to find the density of just $f_Y$.



      My problem is that I do not know the joint density of X and Y, so I do not know formulas to solve this. Any hints?






      probability




      probability






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 23 at 13:06









      qcc101qcc101

      627213




      627213






















          1 Answer
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          $begingroup$

          Hint:
          $$
          f_Y(y)=int f_{X,Y}(x,y) mathrm{d}x=int f_{Y|X}(y|x)f_X(x) mathrm{d}x.
          $$



          By this formula, we get
          $$
          f_{Y|X}(y|x)f_X(x)=frac{x^{1/2}}{sqrt{2pi}}e^{-frac{xy^2}{2}}frac{2^{3/2}}{Gamma(3/2)}x^{1/2}e^{-2x},
          $$           for $x>0$. Integrating over $x$ yields
          $$
          f_Y(y)=frac{16}{pi(y^2+4)^2}.
          $$          Identities $xGamma(x)=Gamma(x+1)$ and $Gamma(1/2)=sqrt{pi}$ is used for simplification.






          share|cite|improve this answer











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            $begingroup$

            Hint:
            $$
            f_Y(y)=int f_{X,Y}(x,y) mathrm{d}x=int f_{Y|X}(y|x)f_X(x) mathrm{d}x.
            $$



            By this formula, we get
            $$
            f_{Y|X}(y|x)f_X(x)=frac{x^{1/2}}{sqrt{2pi}}e^{-frac{xy^2}{2}}frac{2^{3/2}}{Gamma(3/2)}x^{1/2}e^{-2x},
            $$             for $x>0$. Integrating over $x$ yields
            $$
            f_Y(y)=frac{16}{pi(y^2+4)^2}.
            $$            Identities $xGamma(x)=Gamma(x+1)$ and $Gamma(1/2)=sqrt{pi}$ is used for simplification.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Hint:
              $$
              f_Y(y)=int f_{X,Y}(x,y) mathrm{d}x=int f_{Y|X}(y|x)f_X(x) mathrm{d}x.
              $$



              By this formula, we get
              $$
              f_{Y|X}(y|x)f_X(x)=frac{x^{1/2}}{sqrt{2pi}}e^{-frac{xy^2}{2}}frac{2^{3/2}}{Gamma(3/2)}x^{1/2}e^{-2x},
              $$               for $x>0$. Integrating over $x$ yields
              $$
              f_Y(y)=frac{16}{pi(y^2+4)^2}.
              $$              Identities $xGamma(x)=Gamma(x+1)$ and $Gamma(1/2)=sqrt{pi}$ is used for simplification.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Hint:
                $$
                f_Y(y)=int f_{X,Y}(x,y) mathrm{d}x=int f_{Y|X}(y|x)f_X(x) mathrm{d}x.
                $$



                By this formula, we get
                $$
                f_{Y|X}(y|x)f_X(x)=frac{x^{1/2}}{sqrt{2pi}}e^{-frac{xy^2}{2}}frac{2^{3/2}}{Gamma(3/2)}x^{1/2}e^{-2x},
                $$                 for $x>0$. Integrating over $x$ yields
                $$
                f_Y(y)=frac{16}{pi(y^2+4)^2}.
                $$                Identities $xGamma(x)=Gamma(x+1)$ and $Gamma(1/2)=sqrt{pi}$ is used for simplification.






                share|cite|improve this answer











                $endgroup$



                Hint:
                $$
                f_Y(y)=int f_{X,Y}(x,y) mathrm{d}x=int f_{Y|X}(y|x)f_X(x) mathrm{d}x.
                $$



                By this formula, we get
                $$
                f_{Y|X}(y|x)f_X(x)=frac{x^{1/2}}{sqrt{2pi}}e^{-frac{xy^2}{2}}frac{2^{3/2}}{Gamma(3/2)}x^{1/2}e^{-2x},
                $$                 for $x>0$. Integrating over $x$ yields
                $$
                f_Y(y)=frac{16}{pi(y^2+4)^2}.
                $$                Identities $xGamma(x)=Gamma(x+1)$ and $Gamma(1/2)=sqrt{pi}$ is used for simplification.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 23 at 18:04

























                answered Jan 23 at 13:10









                SongSong

                17.3k21246




                17.3k21246






























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