Marginal PDF from joint PDF











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$$f_{X,Y}(x,y)~=~begin{cases}2 & if &0< x < 1 ~,~ 0< y< x \[1ex] 0 & text{otherwise} end{cases}$$



Now I should calculate the marginal PDF's.



So I use integral and then I get:



$$f_X(x)~=~int_{0}^{1}2 ~mathrm dy~=2$$
$$f_Y(y)~=~int_{0}^{x}2 ~mathrm dx~=4$$



I just feel like I'm not doing this correct. So if any of you could help me and tell me what I am doing wrong and how to get the correct answer.






statistics density-function







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    up vote
    1
    down vote

    favorite












    $$f_{X,Y}(x,y)~=~begin{cases}2 & if &0< x < 1 ~,~ 0< y< x \[1ex] 0 & text{otherwise} end{cases}$$



    Now I should calculate the marginal PDF's.



    So I use integral and then I get:



    $$f_X(x)~=~int_{0}^{1}2 ~mathrm dy~=2$$
    $$f_Y(y)~=~int_{0}^{x}2 ~mathrm dx~=4$$



    I just feel like I'm not doing this correct. So if any of you could help me and tell me what I am doing wrong and how to get the correct answer.






    statistics density-function







    share|cite|improve this question
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      $$f_{X,Y}(x,y)~=~begin{cases}2 & if &0< x < 1 ~,~ 0< y< x \[1ex] 0 & text{otherwise} end{cases}$$



      Now I should calculate the marginal PDF's.



      So I use integral and then I get:



      $$f_X(x)~=~int_{0}^{1}2 ~mathrm dy~=2$$
      $$f_Y(y)~=~int_{0}^{x}2 ~mathrm dx~=4$$



      I just feel like I'm not doing this correct. So if any of you could help me and tell me what I am doing wrong and how to get the correct answer.






      statistics density-function







      share|cite|improve this question













      $$f_{X,Y}(x,y)~=~begin{cases}2 & if &0< x < 1 ~,~ 0< y< x \[1ex] 0 & text{otherwise} end{cases}$$



      Now I should calculate the marginal PDF's.



      So I use integral and then I get:



      $$f_X(x)~=~int_{0}^{1}2 ~mathrm dy~=2$$
      $$f_Y(y)~=~int_{0}^{x}2 ~mathrm dx~=4$$



      I just feel like I'm not doing this correct. So if any of you could help me and tell me what I am doing wrong and how to get the correct answer.






      statistics density-function




      statistics density-function






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      asked yesterday









      MiMaKo

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      588






















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          Observe that $$f_{X,Y}=2mathbf1_{(0,1)}(x)mathbf1_{(0,x)}(y)$$



          For $xin(0,1)$ we get:
          $$f_X(x)=int f_{X,Y}(x,y)dy=int_0^x2dy=2x$$
          For $xnotin(0,1)$ the integrand is $0$ so then $f_X(x)=0$.



          For $yin(0,1)$ we get:
          $$f_Y(y)=int f_{X,Y}(x,y)dx=int_y^12dy=2(1-y)$$
          For $ynotin(0,1)$ the integrand is $0$ so then $f_Y(y)=0$.






          share|cite|improve this answer























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            up vote
            1
            down vote



            accepted










            Observe that $$f_{X,Y}=2mathbf1_{(0,1)}(x)mathbf1_{(0,x)}(y)$$



            For $xin(0,1)$ we get:
            $$f_X(x)=int f_{X,Y}(x,y)dy=int_0^x2dy=2x$$
            For $xnotin(0,1)$ the integrand is $0$ so then $f_X(x)=0$.



            For $yin(0,1)$ we get:
            $$f_Y(y)=int f_{X,Y}(x,y)dx=int_y^12dy=2(1-y)$$
            For $ynotin(0,1)$ the integrand is $0$ so then $f_Y(y)=0$.






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              Observe that $$f_{X,Y}=2mathbf1_{(0,1)}(x)mathbf1_{(0,x)}(y)$$



              For $xin(0,1)$ we get:
              $$f_X(x)=int f_{X,Y}(x,y)dy=int_0^x2dy=2x$$
              For $xnotin(0,1)$ the integrand is $0$ so then $f_X(x)=0$.



              For $yin(0,1)$ we get:
              $$f_Y(y)=int f_{X,Y}(x,y)dx=int_y^12dy=2(1-y)$$
              For $ynotin(0,1)$ the integrand is $0$ so then $f_Y(y)=0$.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Observe that $$f_{X,Y}=2mathbf1_{(0,1)}(x)mathbf1_{(0,x)}(y)$$



                For $xin(0,1)$ we get:
                $$f_X(x)=int f_{X,Y}(x,y)dy=int_0^x2dy=2x$$
                For $xnotin(0,1)$ the integrand is $0$ so then $f_X(x)=0$.



                For $yin(0,1)$ we get:
                $$f_Y(y)=int f_{X,Y}(x,y)dx=int_y^12dy=2(1-y)$$
                For $ynotin(0,1)$ the integrand is $0$ so then $f_Y(y)=0$.






                share|cite|improve this answer














                Observe that $$f_{X,Y}=2mathbf1_{(0,1)}(x)mathbf1_{(0,x)}(y)$$



                For $xin(0,1)$ we get:
                $$f_X(x)=int f_{X,Y}(x,y)dy=int_0^x2dy=2x$$
                For $xnotin(0,1)$ the integrand is $0$ so then $f_X(x)=0$.



                For $yin(0,1)$ we get:
                $$f_Y(y)=int f_{X,Y}(x,y)dx=int_y^12dy=2(1-y)$$
                For $ynotin(0,1)$ the integrand is $0$ so then $f_Y(y)=0$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited yesterday

























                answered yesterday









                drhab

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                93.5k543125






























                     

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