Join continuous PDF how to choose interval












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I have the joint pdf
$f_{x,y} (x,y) = 3/4 x $ if $0<x<y<2$



Now if I want to find the marginal pdfs $f_x $ and $ f_y$



How do I choose which interval to use when I integrate?



I know the answers are:



$f_x (x) = 3/4 x (2-x)$ if $0<x<2$



$f_y (y) = 3/8 y^2 $ if $0<y<2$






probability statistics







share|cite|improve this question



























    0














    I have the joint pdf
    $f_{x,y} (x,y) = 3/4 x $ if $0<x<y<2$



    Now if I want to find the marginal pdfs $f_x $ and $ f_y$



    How do I choose which interval to use when I integrate?



    I know the answers are:



    $f_x (x) = 3/4 x (2-x)$ if $0<x<2$



    $f_y (y) = 3/8 y^2 $ if $0<y<2$






    probability statistics







    share|cite|improve this question

























      0












      0








      0







      I have the joint pdf
      $f_{x,y} (x,y) = 3/4 x $ if $0<x<y<2$



      Now if I want to find the marginal pdfs $f_x $ and $ f_y$



      How do I choose which interval to use when I integrate?



      I know the answers are:



      $f_x (x) = 3/4 x (2-x)$ if $0<x<2$



      $f_y (y) = 3/8 y^2 $ if $0<y<2$






      probability statistics







      share|cite|improve this question













      I have the joint pdf
      $f_{x,y} (x,y) = 3/4 x $ if $0<x<y<2$



      Now if I want to find the marginal pdfs $f_x $ and $ f_y$



      How do I choose which interval to use when I integrate?



      I know the answers are:



      $f_x (x) = 3/4 x (2-x)$ if $0<x<2$



      $f_y (y) = 3/8 y^2 $ if $0<y<2$






      probability statistics




      probability statistics






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 17 hours ago









      Winther

      227




      227






















          2 Answers
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          Formally there is no choosing:$$f_X(x)=int_{-infty}^{infty}f_X(x,y);dy$$
          We can substitute $f_{X,Y}=[0<x<y<2]frac34x$ where $[text{condition on }x,y]$ is a function $mathbb R^2tomathbb R$ that takes value $1$ if the condition is satisfied and take value $0$ otherwise.



          Then for fixed $x$:$$f_X(x)=int_{-infty}^{infty}[0<x<y<2]frac34x;dy=frac34xint_{-infty}^{infty}[0<x<y<2];dy$$



          Now we discern cases:



          If $xnotin(0,2)$ then the integrand is $0$ for every $y$ so that $f_X(x)=0$.



          If $xin(0,2)$ then $int_{-infty}^{infty}[0<x<y<2];dy=int_x^2dy=2-x$ so that $f(x)=frac34(1-x)$.



          The first equality is probably the part you label as "choosing which interval".



          It rests on the observation that for $ynotin(x,2)$ the integrand is $0$.




          The same principle works for finding: $$f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx$$for a fixed $y$.



          If $ynotin(0,2)$ then the integrand is $0$ for every $x$ so that $f_Y(y)=0$.



          If $yin(0,2)$ then $f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx=int_0^yfrac34xdx=[frac38x^2]_0^y=frac38y^2$.






          share|cite|improve this answer































            0














            To get the marginal densities from the joint density, you "integrate out" each variable. In this case, the region $0<x<y<2$ is a triangle. To see this, draw the horizontal boundary lines $x=0$, $y=2$, and $y=x$. The triangle is then the region overlapped by $x>0,y<2, x<y$.



            For this region, we see that for a fixed $y$, $x$ varies from $0$ to $y$, hence
            $$f_{Y}(y)=int_{0}^{y}f_{X,Y}(x,y) dx=int_{0}^{y}frac{3}{4}x dx=frac{3}{8}x^{2}biggrvert^{x=y}_{x=0}=frac{3}{8}y^{2}$$
            Similarly, if we fix a value for $x$, $y$ varies from $x$ to $2$
            $$f_{X}(x)=int_{x}^{2}f_{X,Y}(x,y) dy=int_{x}^{2}frac{3}{4}x dy=frac{3}{4}xybiggrvert^{y=2}_{y=x}=frac{3}{4}x(2-x)$$



            If the region is unclear, draw a picture and it should make more sense.






            share|cite|improve this answer





















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              2 Answers
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              2 Answers
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              active

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              2














              Formally there is no choosing:$$f_X(x)=int_{-infty}^{infty}f_X(x,y);dy$$
              We can substitute $f_{X,Y}=[0<x<y<2]frac34x$ where $[text{condition on }x,y]$ is a function $mathbb R^2tomathbb R$ that takes value $1$ if the condition is satisfied and take value $0$ otherwise.



              Then for fixed $x$:$$f_X(x)=int_{-infty}^{infty}[0<x<y<2]frac34x;dy=frac34xint_{-infty}^{infty}[0<x<y<2];dy$$



              Now we discern cases:



              If $xnotin(0,2)$ then the integrand is $0$ for every $y$ so that $f_X(x)=0$.



              If $xin(0,2)$ then $int_{-infty}^{infty}[0<x<y<2];dy=int_x^2dy=2-x$ so that $f(x)=frac34(1-x)$.



              The first equality is probably the part you label as "choosing which interval".



              It rests on the observation that for $ynotin(x,2)$ the integrand is $0$.




              The same principle works for finding: $$f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx$$for a fixed $y$.



              If $ynotin(0,2)$ then the integrand is $0$ for every $x$ so that $f_Y(y)=0$.



              If $yin(0,2)$ then $f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx=int_0^yfrac34xdx=[frac38x^2]_0^y=frac38y^2$.






              share|cite|improve this answer




























                2














                Formally there is no choosing:$$f_X(x)=int_{-infty}^{infty}f_X(x,y);dy$$
                We can substitute $f_{X,Y}=[0<x<y<2]frac34x$ where $[text{condition on }x,y]$ is a function $mathbb R^2tomathbb R$ that takes value $1$ if the condition is satisfied and take value $0$ otherwise.



                Then for fixed $x$:$$f_X(x)=int_{-infty}^{infty}[0<x<y<2]frac34x;dy=frac34xint_{-infty}^{infty}[0<x<y<2];dy$$



                Now we discern cases:



                If $xnotin(0,2)$ then the integrand is $0$ for every $y$ so that $f_X(x)=0$.



                If $xin(0,2)$ then $int_{-infty}^{infty}[0<x<y<2];dy=int_x^2dy=2-x$ so that $f(x)=frac34(1-x)$.



                The first equality is probably the part you label as "choosing which interval".



                It rests on the observation that for $ynotin(x,2)$ the integrand is $0$.




                The same principle works for finding: $$f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx$$for a fixed $y$.



                If $ynotin(0,2)$ then the integrand is $0$ for every $x$ so that $f_Y(y)=0$.



                If $yin(0,2)$ then $f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx=int_0^yfrac34xdx=[frac38x^2]_0^y=frac38y^2$.






                share|cite|improve this answer


























                  2












                  2








                  2






                  Formally there is no choosing:$$f_X(x)=int_{-infty}^{infty}f_X(x,y);dy$$
                  We can substitute $f_{X,Y}=[0<x<y<2]frac34x$ where $[text{condition on }x,y]$ is a function $mathbb R^2tomathbb R$ that takes value $1$ if the condition is satisfied and take value $0$ otherwise.



                  Then for fixed $x$:$$f_X(x)=int_{-infty}^{infty}[0<x<y<2]frac34x;dy=frac34xint_{-infty}^{infty}[0<x<y<2];dy$$



                  Now we discern cases:



                  If $xnotin(0,2)$ then the integrand is $0$ for every $y$ so that $f_X(x)=0$.



                  If $xin(0,2)$ then $int_{-infty}^{infty}[0<x<y<2];dy=int_x^2dy=2-x$ so that $f(x)=frac34(1-x)$.



                  The first equality is probably the part you label as "choosing which interval".



                  It rests on the observation that for $ynotin(x,2)$ the integrand is $0$.




                  The same principle works for finding: $$f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx$$for a fixed $y$.



                  If $ynotin(0,2)$ then the integrand is $0$ for every $x$ so that $f_Y(y)=0$.



                  If $yin(0,2)$ then $f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx=int_0^yfrac34xdx=[frac38x^2]_0^y=frac38y^2$.






                  share|cite|improve this answer














                  Formally there is no choosing:$$f_X(x)=int_{-infty}^{infty}f_X(x,y);dy$$
                  We can substitute $f_{X,Y}=[0<x<y<2]frac34x$ where $[text{condition on }x,y]$ is a function $mathbb R^2tomathbb R$ that takes value $1$ if the condition is satisfied and take value $0$ otherwise.



                  Then for fixed $x$:$$f_X(x)=int_{-infty}^{infty}[0<x<y<2]frac34x;dy=frac34xint_{-infty}^{infty}[0<x<y<2];dy$$



                  Now we discern cases:



                  If $xnotin(0,2)$ then the integrand is $0$ for every $y$ so that $f_X(x)=0$.



                  If $xin(0,2)$ then $int_{-infty}^{infty}[0<x<y<2];dy=int_x^2dy=2-x$ so that $f(x)=frac34(1-x)$.



                  The first equality is probably the part you label as "choosing which interval".



                  It rests on the observation that for $ynotin(x,2)$ the integrand is $0$.




                  The same principle works for finding: $$f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx$$for a fixed $y$.



                  If $ynotin(0,2)$ then the integrand is $0$ for every $x$ so that $f_Y(y)=0$.



                  If $yin(0,2)$ then $f_Y(y)=int_{-infty}^{infty}[0<x<y<2]frac34x;dx=int_0^yfrac34xdx=[frac38x^2]_0^y=frac38y^2$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 17 hours ago

























                  answered 17 hours ago









                  drhab

                  98.1k544129




                  98.1k544129























                      0














                      To get the marginal densities from the joint density, you "integrate out" each variable. In this case, the region $0<x<y<2$ is a triangle. To see this, draw the horizontal boundary lines $x=0$, $y=2$, and $y=x$. The triangle is then the region overlapped by $x>0,y<2, x<y$.



                      For this region, we see that for a fixed $y$, $x$ varies from $0$ to $y$, hence
                      $$f_{Y}(y)=int_{0}^{y}f_{X,Y}(x,y) dx=int_{0}^{y}frac{3}{4}x dx=frac{3}{8}x^{2}biggrvert^{x=y}_{x=0}=frac{3}{8}y^{2}$$
                      Similarly, if we fix a value for $x$, $y$ varies from $x$ to $2$
                      $$f_{X}(x)=int_{x}^{2}f_{X,Y}(x,y) dy=int_{x}^{2}frac{3}{4}x dy=frac{3}{4}xybiggrvert^{y=2}_{y=x}=frac{3}{4}x(2-x)$$



                      If the region is unclear, draw a picture and it should make more sense.






                      share|cite|improve this answer


























                        0














                        To get the marginal densities from the joint density, you "integrate out" each variable. In this case, the region $0<x<y<2$ is a triangle. To see this, draw the horizontal boundary lines $x=0$, $y=2$, and $y=x$. The triangle is then the region overlapped by $x>0,y<2, x<y$.



                        For this region, we see that for a fixed $y$, $x$ varies from $0$ to $y$, hence
                        $$f_{Y}(y)=int_{0}^{y}f_{X,Y}(x,y) dx=int_{0}^{y}frac{3}{4}x dx=frac{3}{8}x^{2}biggrvert^{x=y}_{x=0}=frac{3}{8}y^{2}$$
                        Similarly, if we fix a value for $x$, $y$ varies from $x$ to $2$
                        $$f_{X}(x)=int_{x}^{2}f_{X,Y}(x,y) dy=int_{x}^{2}frac{3}{4}x dy=frac{3}{4}xybiggrvert^{y=2}_{y=x}=frac{3}{4}x(2-x)$$



                        If the region is unclear, draw a picture and it should make more sense.






                        share|cite|improve this answer
























                          0












                          0








                          0






                          To get the marginal densities from the joint density, you "integrate out" each variable. In this case, the region $0<x<y<2$ is a triangle. To see this, draw the horizontal boundary lines $x=0$, $y=2$, and $y=x$. The triangle is then the region overlapped by $x>0,y<2, x<y$.



                          For this region, we see that for a fixed $y$, $x$ varies from $0$ to $y$, hence
                          $$f_{Y}(y)=int_{0}^{y}f_{X,Y}(x,y) dx=int_{0}^{y}frac{3}{4}x dx=frac{3}{8}x^{2}biggrvert^{x=y}_{x=0}=frac{3}{8}y^{2}$$
                          Similarly, if we fix a value for $x$, $y$ varies from $x$ to $2$
                          $$f_{X}(x)=int_{x}^{2}f_{X,Y}(x,y) dy=int_{x}^{2}frac{3}{4}x dy=frac{3}{4}xybiggrvert^{y=2}_{y=x}=frac{3}{4}x(2-x)$$



                          If the region is unclear, draw a picture and it should make more sense.






                          share|cite|improve this answer












                          To get the marginal densities from the joint density, you "integrate out" each variable. In this case, the region $0<x<y<2$ is a triangle. To see this, draw the horizontal boundary lines $x=0$, $y=2$, and $y=x$. The triangle is then the region overlapped by $x>0,y<2, x<y$.



                          For this region, we see that for a fixed $y$, $x$ varies from $0$ to $y$, hence
                          $$f_{Y}(y)=int_{0}^{y}f_{X,Y}(x,y) dx=int_{0}^{y}frac{3}{4}x dx=frac{3}{8}x^{2}biggrvert^{x=y}_{x=0}=frac{3}{8}y^{2}$$
                          Similarly, if we fix a value for $x$, $y$ varies from $x$ to $2$
                          $$f_{X}(x)=int_{x}^{2}f_{X,Y}(x,y) dy=int_{x}^{2}frac{3}{4}x dy=frac{3}{4}xybiggrvert^{y=2}_{y=x}=frac{3}{4}x(2-x)$$



                          If the region is unclear, draw a picture and it should make more sense.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 17 hours ago









                          pwerth

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                          1,747411






























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