Pdf of $|X-Y|$ when $X,Y$ are independent Uniform $[0,a]$ variables












0














Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
As it is symmetrical I have taken one part of triangle.
Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
Not getting expected answer.
Could any help.



Answer:
$f_{z}(z)=frac{2}{a}(1-frac{z}{a})$










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    0














    Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
    As it is symmetrical I have taken one part of triangle.
    Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
    Not getting expected answer.
    Could any help.



    Answer:
    $f_{z}(z)=frac{2}{a}(1-frac{z}{a})$










    share|cite|improve this question



























      0












      0








      0


      1





      Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
      As it is symmetrical I have taken one part of triangle.
      Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
      Not getting expected answer.
      Could any help.



      Answer:
      $f_{z}(z)=frac{2}{a}(1-frac{z}{a})$










      share|cite|improve this question















      Need to find pdf of $ |X-Y| $ .I little confused and not getting answer after taking below limits.
      As it is symmetrical I have taken one part of triangle.
      Considering lower triangle limits I have taken is x from $ y+z $ to $ a $ and outer limit : y from $ 0 $ to $ a-z $
      Not getting expected answer.
      Could any help.



      Answer:
      $f_{z}(z)=frac{2}{a}(1-frac{z}{a})$







      probability probability-distributions random-variables uniform-distribution






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      edited Dec 22 '18 at 19:58









      StubbornAtom

      5,36411138




      5,36411138










      asked Nov 22 '18 at 12:35









      Pramod_achar

      13




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          2 Answers
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          |X-Y| can be written as below ,



          $ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $



          The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.



          $ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $



          After differentiating w.r.t z ,



          $ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $



          $ =frac{2}{a^{2}}(a-z) $






          share|cite|improve this answer





























            0














            Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$



            So for $0< z<a$,



            begin{align}
            P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
            \&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
            \&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
            \&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
            \&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
            \&=frac{2}{a^2}left(az-frac{z^2}{2}right)
            end{align}



            Hence the pdf of $Z=|X-Y|$ is



            $$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$



            Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.



            Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)



            enter image description here






            share|cite|improve this answer























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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              0














              |X-Y| can be written as below ,



              $ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $



              The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.



              $ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $



              After differentiating w.r.t z ,



              $ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $



              $ =frac{2}{a^{2}}(a-z) $






              share|cite|improve this answer


























                0














                |X-Y| can be written as below ,



                $ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $



                The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.



                $ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $



                After differentiating w.r.t z ,



                $ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $



                $ =frac{2}{a^{2}}(a-z) $






                share|cite|improve this answer
























                  0












                  0








                  0






                  |X-Y| can be written as below ,



                  $ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $



                  The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.



                  $ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $



                  After differentiating w.r.t z ,



                  $ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $



                  $ =frac{2}{a^{2}}(a-z) $






                  share|cite|improve this answer












                  |X-Y| can be written as below ,



                  $ P(|X-Y| leq z )= P(X-Y leq z ,X ge Y )+P(Y-X leq z ,Y >X ) \ $



                  The limits can be visualized by drawing rectangle and |x-y| and take area of the other sides,which would be symmetry hence multiple 2.



                  $ F_Z(z)=1-2int_{y=0}^{y=a-z} int_{y+z}^{a}f(x,y) dxdy \ $



                  After differentiating w.r.t z ,



                  $ f_Z(z)=0+2 int_{0}^{a-z} f(y+z,y)dy \ $



                  $ =frac{2}{a^{2}}(a-z) $







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 22 '18 at 17:32









                  Pramod_achar

                  13




                  13























                      0














                      Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$



                      So for $0< z<a$,



                      begin{align}
                      P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
                      \&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
                      \&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
                      \&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
                      \&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
                      \&=frac{2}{a^2}left(az-frac{z^2}{2}right)
                      end{align}



                      Hence the pdf of $Z=|X-Y|$ is



                      $$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$



                      Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.



                      Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)



                      enter image description here






                      share|cite|improve this answer




























                        0














                        Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$



                        So for $0< z<a$,



                        begin{align}
                        P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
                        \&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
                        \&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
                        \&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
                        \&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
                        \&=frac{2}{a^2}left(az-frac{z^2}{2}right)
                        end{align}



                        Hence the pdf of $Z=|X-Y|$ is



                        $$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$



                        Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.



                        Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)



                        enter image description here






                        share|cite|improve this answer


























                          0












                          0








                          0






                          Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$



                          So for $0< z<a$,



                          begin{align}
                          P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
                          \&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
                          \&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
                          \&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
                          \&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
                          \&=frac{2}{a^2}left(az-frac{z^2}{2}right)
                          end{align}



                          Hence the pdf of $Z=|X-Y|$ is



                          $$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$



                          Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.



                          Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)



                          enter image description here






                          share|cite|improve this answer














                          Since $X$ and $Y$ are identically distributed, $P(X-Yle z,Xge Y)=P(Y-Xle z,Yge X)$



                          So for $0< z<a$,



                          begin{align}
                          P(|X-Y|le z)&=2times P(X-Yle z,Xge Y)
                          \&=2 int P(X-yle z,Xge ymid Y=y)f_Y(y),dy
                          \&=2int P(yle Xle z+y)frac{mathbf1_{0<y<a}}{a},dy
                          \&=frac{2}{a}int_0^a int_y^{min(z+y,,,a)}frac{1}{a},dx,dy
                          \&=frac{2}{a^2}left[int_0^{a-z}int_y^{z+y},dx,dy+int_{a-z}^aint_y^a,dx,dyright]
                          \&=frac{2}{a^2}left(az-frac{z^2}{2}right)
                          end{align}



                          Hence the pdf of $Z=|X-Y|$ is



                          $$f_Z(z)=frac{2}{a^2}(a-z)mathbf1_{0<z<a}$$



                          Needless to say, the above algebra for the CDF is not as simple as drawing a picture of the region ${(x,y)in[0,a]^2:|x-y|le z}$ and finding its area.



                          Here is a picture for $z=0.6$ and $a=3$: (Also see the case $a=1$ discussed in this post)



                          enter image description here







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Dec 22 '18 at 19:54

























                          answered Dec 22 '18 at 19:19









                          StubbornAtom

                          5,36411138




                          5,36411138






























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