Density function without using convolution












0












$begingroup$


I have the following joint density function:



$$ f_{X,Y}(x,y) = (y-x)e^{-y} (0<x<y) $$



I want to find the density of X + Y.



I find the density of X and Y integrating:



$$f_X(x) = e^{-x} (0<x<infty)$$
$$ f_Y(y) = frac{y^2}{2}e^{-y} (0<y<infty)$$



So I find that X and Y are not indented and thus I cannot find the density of the sum using convolution. My idea is then to find:



$$ F_{X+Y}(t) = mathbb{P}[X + Y leq t] $$ and then integrating to find the density.



I think I made a mistake setting up this double integral though:



$$ int_{0}^{infty} int_{y}^{t-y} (y-x)e^{-y}dxdi $$



Is this right? I am not getting the result I should.



EDIT:
I also tried doing it with this formula:



$$ f_{X+Y}(t) = int f_{X,Y}(x,t-x)dx $$ But I didn't manage to solve this. Plus even here I am not too sure about the integration extremes, I have done the integral from $0$ to $infty$.






probability







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I have the following joint density function:



    $$ f_{X,Y}(x,y) = (y-x)e^{-y} (0<x<y) $$



    I want to find the density of X + Y.



    I find the density of X and Y integrating:



    $$f_X(x) = e^{-x} (0<x<infty)$$
    $$ f_Y(y) = frac{y^2}{2}e^{-y} (0<y<infty)$$



    So I find that X and Y are not indented and thus I cannot find the density of the sum using convolution. My idea is then to find:



    $$ F_{X+Y}(t) = mathbb{P}[X + Y leq t] $$ and then integrating to find the density.



    I think I made a mistake setting up this double integral though:



    $$ int_{0}^{infty} int_{y}^{t-y} (y-x)e^{-y}dxdi $$



    Is this right? I am not getting the result I should.



    EDIT:
    I also tried doing it with this formula:



    $$ f_{X+Y}(t) = int f_{X,Y}(x,t-x)dx $$ But I didn't manage to solve this. Plus even here I am not too sure about the integration extremes, I have done the integral from $0$ to $infty$.






    probability







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I have the following joint density function:



      $$ f_{X,Y}(x,y) = (y-x)e^{-y} (0<x<y) $$



      I want to find the density of X + Y.



      I find the density of X and Y integrating:



      $$f_X(x) = e^{-x} (0<x<infty)$$
      $$ f_Y(y) = frac{y^2}{2}e^{-y} (0<y<infty)$$



      So I find that X and Y are not indented and thus I cannot find the density of the sum using convolution. My idea is then to find:



      $$ F_{X+Y}(t) = mathbb{P}[X + Y leq t] $$ and then integrating to find the density.



      I think I made a mistake setting up this double integral though:



      $$ int_{0}^{infty} int_{y}^{t-y} (y-x)e^{-y}dxdi $$



      Is this right? I am not getting the result I should.



      EDIT:
      I also tried doing it with this formula:



      $$ f_{X+Y}(t) = int f_{X,Y}(x,t-x)dx $$ But I didn't manage to solve this. Plus even here I am not too sure about the integration extremes, I have done the integral from $0$ to $infty$.






      probability







      share|cite|improve this question











      $endgroup$




      I have the following joint density function:



      $$ f_{X,Y}(x,y) = (y-x)e^{-y} (0<x<y) $$



      I want to find the density of X + Y.



      I find the density of X and Y integrating:



      $$f_X(x) = e^{-x} (0<x<infty)$$
      $$ f_Y(y) = frac{y^2}{2}e^{-y} (0<y<infty)$$



      So I find that X and Y are not indented and thus I cannot find the density of the sum using convolution. My idea is then to find:



      $$ F_{X+Y}(t) = mathbb{P}[X + Y leq t] $$ and then integrating to find the density.



      I think I made a mistake setting up this double integral though:



      $$ int_{0}^{infty} int_{y}^{t-y} (y-x)e^{-y}dxdi $$



      Is this right? I am not getting the result I should.



      EDIT:
      I also tried doing it with this formula:



      $$ f_{X+Y}(t) = int f_{X,Y}(x,t-x)dx $$ But I didn't manage to solve this. Plus even here I am not too sure about the integration extremes, I have done the integral from $0$ to $infty$.






      probability




      probability






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      edited Jan 26 at 13:52







      qcc101

















      asked Jan 26 at 13:43









      qcc101qcc101

      627213




      627213






















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          $begingroup$

          If you're familiar with the Dirac delta, we can obtain the density of $Z=g(X,,Y)$ as the definite integral $f_Z(z):=int f_{X,,Y}(x,,y)delta(g(x,,y)-z)dxdy$, where the integration range is the support of the joint density $f_{X,,Y}$. In the case at hand, with $z=x+yge 2x$,$$f_Z(z)=int (y-x)e^{-y}delta(x+y-z)dxdy,$$where again I haven't shown the integration range. Where you need to know your Dirac delta onions is in integrating out $y$:$$f_Z(z)=int_0^{z/2}(z-2x)e^{x-z}dx=e^{-z}(2e^{z/2}-2-z).$$






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            $begingroup$

            If you're familiar with the Dirac delta, we can obtain the density of $Z=g(X,,Y)$ as the definite integral $f_Z(z):=int f_{X,,Y}(x,,y)delta(g(x,,y)-z)dxdy$, where the integration range is the support of the joint density $f_{X,,Y}$. In the case at hand, with $z=x+yge 2x$,$$f_Z(z)=int (y-x)e^{-y}delta(x+y-z)dxdy,$$where again I haven't shown the integration range. Where you need to know your Dirac delta onions is in integrating out $y$:$$f_Z(z)=int_0^{z/2}(z-2x)e^{x-z}dx=e^{-z}(2e^{z/2}-2-z).$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              If you're familiar with the Dirac delta, we can obtain the density of $Z=g(X,,Y)$ as the definite integral $f_Z(z):=int f_{X,,Y}(x,,y)delta(g(x,,y)-z)dxdy$, where the integration range is the support of the joint density $f_{X,,Y}$. In the case at hand, with $z=x+yge 2x$,$$f_Z(z)=int (y-x)e^{-y}delta(x+y-z)dxdy,$$where again I haven't shown the integration range. Where you need to know your Dirac delta onions is in integrating out $y$:$$f_Z(z)=int_0^{z/2}(z-2x)e^{x-z}dx=e^{-z}(2e^{z/2}-2-z).$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                If you're familiar with the Dirac delta, we can obtain the density of $Z=g(X,,Y)$ as the definite integral $f_Z(z):=int f_{X,,Y}(x,,y)delta(g(x,,y)-z)dxdy$, where the integration range is the support of the joint density $f_{X,,Y}$. In the case at hand, with $z=x+yge 2x$,$$f_Z(z)=int (y-x)e^{-y}delta(x+y-z)dxdy,$$where again I haven't shown the integration range. Where you need to know your Dirac delta onions is in integrating out $y$:$$f_Z(z)=int_0^{z/2}(z-2x)e^{x-z}dx=e^{-z}(2e^{z/2}-2-z).$$






                share|cite|improve this answer









                $endgroup$



                If you're familiar with the Dirac delta, we can obtain the density of $Z=g(X,,Y)$ as the definite integral $f_Z(z):=int f_{X,,Y}(x,,y)delta(g(x,,y)-z)dxdy$, where the integration range is the support of the joint density $f_{X,,Y}$. In the case at hand, with $z=x+yge 2x$,$$f_Z(z)=int (y-x)e^{-y}delta(x+y-z)dxdy,$$where again I haven't shown the integration range. Where you need to know your Dirac delta onions is in integrating out $y$:$$f_Z(z)=int_0^{z/2}(z-2x)e^{x-z}dx=e^{-z}(2e^{z/2}-2-z).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 26 at 14:12









                J.G.J.G.

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