If $f$ is the pdf of a random variable, show that $g(x, y) = f(x+y)/(x + y)$ is a density function in the...












1












$begingroup$


Let $f$ be the pdf of a positive random variable and write



$$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$



Show that $g$ is the density function in the plane.




Clearly, $g(x, y) geq 0$, but I need help evaluating the integral



$$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$



One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.



Does anyone have any suggestions?






probability probability-theory probability-distributions







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    1












    $begingroup$


    Let $f$ be the pdf of a positive random variable and write



    $$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$



    Show that $g$ is the density function in the plane.




    Clearly, $g(x, y) geq 0$, but I need help evaluating the integral



    $$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$



    One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.



    Does anyone have any suggestions?






    probability probability-theory probability-distributions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f$ be the pdf of a positive random variable and write



      $$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$



      Show that $g$ is the density function in the plane.




      Clearly, $g(x, y) geq 0$, but I need help evaluating the integral



      $$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$



      One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.



      Does anyone have any suggestions?






      probability probability-theory probability-distributions







      share|cite|improve this question









      $endgroup$




      Let $f$ be the pdf of a positive random variable and write



      $$g(x, y) = frac{f(x + y)}{x + y} , text{ if } x, y > 0.$$



      Show that $g$ is the density function in the plane.




      Clearly, $g(x, y) geq 0$, but I need help evaluating the integral



      $$int_{0}^{infty}int_{0}^{infty}frac{f(x + y)}{x + y} mathop{dx dy}. $$



      One substitution that looks pretty clear is $u = x + y$. Then, we get the $f(u)$ term, and I know $int_{0}^{infty} f(u) mathop{du} = 1$, but I don't know when I would utilize that.



      Does anyone have any suggestions?






      probability probability-theory probability-distributions




      probability probability-theory probability-distributions






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      share|cite|improve this question











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      asked Dec 9 '18 at 7:27









      josephjoseph

      491111




      491111






















          1 Answer
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          $begingroup$

          You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
          $$dv = dy,$$ and hence
          $$dudv = dxdy.
          $$           Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
          $$
          int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
          $$
          $textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
          $$left( {begin{array}{c}
          du \
          dv\
          end{array} } right) =
          left( {begin{array}{cc}
          1 & 1 \
          0 & 1 \
          end{array} } right)left( {begin{array}{c}
          dx \
          dy \
          end{array} } right),
          $$           and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            don't we need to do a jacobian ?
            $endgroup$
            – joseph
            Dec 9 '18 at 19:07











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
          $$dv = dy,$$ and hence
          $$dudv = dxdy.
          $$           Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
          $$
          int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
          $$
          $textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
          $$left( {begin{array}{c}
          du \
          dv\
          end{array} } right) =
          left( {begin{array}{cc}
          1 & 1 \
          0 & 1 \
          end{array} } right)left( {begin{array}{c}
          dx \
          dy \
          end{array} } right),
          $$           and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            don't we need to do a jacobian ?
            $endgroup$
            – joseph
            Dec 9 '18 at 19:07
















          2












          $begingroup$

          You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
          $$dv = dy,$$ and hence
          $$dudv = dxdy.
          $$           Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
          $$
          int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
          $$
          $textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
          $$left( {begin{array}{c}
          du \
          dv\
          end{array} } right) =
          left( {begin{array}{cc}
          1 & 1 \
          0 & 1 \
          end{array} } right)left( {begin{array}{c}
          dx \
          dy \
          end{array} } right),
          $$           and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            don't we need to do a jacobian ?
            $endgroup$
            – joseph
            Dec 9 '18 at 19:07














          2












          2








          2





          $begingroup$

          You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
          $$dv = dy,$$ and hence
          $$dudv = dxdy.
          $$           Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
          $$
          int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
          $$
          $textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
          $$left( {begin{array}{c}
          du \
          dv\
          end{array} } right) =
          left( {begin{array}{cc}
          1 & 1 \
          0 & 1 \
          end{array} } right)left( {begin{array}{c}
          dx \
          dy \
          end{array} } right),
          $$           and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.






          share|cite|improve this answer











          $endgroup$



          You're in the right direction, but recall that change of variables in dimension 2 requires one more variable $v = v(x,y)$ as well as $u = x+y$. One of the simplest choice is $v=y$. Then, $$du = dx+dy,$$
          $$dv = dy,$$ and hence
          $$dudv = dxdy.
          $$           Also, the support of the integral is mapped to $x=u-v>0, y=v>0$. Therefore, the integral has value
          $$
          int_0^inftyint_0^infty frac{f(x+y)}{x+y}dxdy =iint_{0<v<u} frac{f(u)}{u}dudv = int_0^inftyleft(int_0^u frac{f(u)}{u}dvright)du = int_0^infty f(u)du =1.
          $$
          $textbf{Note:}$ A jacobian matrix is involved in the calculation. What $du = dx+dy, ,dv=dy $ means is exactly
          $$left( {begin{array}{c}
          du \
          dv\
          end{array} } right) =
          left( {begin{array}{cc}
          1 & 1 \
          0 & 1 \
          end{array} } right)left( {begin{array}{c}
          dx \
          dy \
          end{array} } right),
          $$           and $frac{dudv}{dxdy} = 1$ is a jacobian determinant.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 23:02

























          answered Dec 9 '18 at 8:13









          SongSong

          13.6k633




          13.6k633












          • $begingroup$
            don't we need to do a jacobian ?
            $endgroup$
            – joseph
            Dec 9 '18 at 19:07


















          • $begingroup$
            don't we need to do a jacobian ?
            $endgroup$
            – joseph
            Dec 9 '18 at 19:07
















          $begingroup$
          don't we need to do a jacobian ?
          $endgroup$
          – joseph
          Dec 9 '18 at 19:07




          $begingroup$
          don't we need to do a jacobian ?
          $endgroup$
          – joseph
          Dec 9 '18 at 19:07


















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