Finding marginal density functions of $X$ and $Y$












0















Let $(X,Y)$ be the coordinates of a point uniformly chosen from a
quadrilateral with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,2)$. Find
the marginal probability density functions of $X$ and $Y$.




Try



Well, these points are vertices of a quadriteral whose area is $frac{2}{3}$. Since it is uniformly chosen the points, then



$$ f_{X,Y}(x,y) = frac{1}{text{Area(S)}} = frac{1}{2/3} = frac{3}{2} $$



Where $(x,y) in S $, the quadrilateral. Hence,



$$ f_X(x) = intlimits_0^2 frac{2}{3} dy = frac{4}{3} $$



and



$$ f_Y(y) = intlimits_0^1 frac{2}{3} = frac{2}{3} $$



Is this correct?






probability







share|cite|improve this question






















  • No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
    – David G. Stork
    Nov 20 at 2:15










  • No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
    – Graham Kemp
    Nov 20 at 2:17
















0















Let $(X,Y)$ be the coordinates of a point uniformly chosen from a
quadrilateral with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,2)$. Find
the marginal probability density functions of $X$ and $Y$.




Try



Well, these points are vertices of a quadriteral whose area is $frac{2}{3}$. Since it is uniformly chosen the points, then



$$ f_{X,Y}(x,y) = frac{1}{text{Area(S)}} = frac{1}{2/3} = frac{3}{2} $$



Where $(x,y) in S $, the quadrilateral. Hence,



$$ f_X(x) = intlimits_0^2 frac{2}{3} dy = frac{4}{3} $$



and



$$ f_Y(y) = intlimits_0^1 frac{2}{3} = frac{2}{3} $$



Is this correct?






probability







share|cite|improve this question






















  • No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
    – David G. Stork
    Nov 20 at 2:15










  • No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
    – Graham Kemp
    Nov 20 at 2:17














0












0








0








Let $(X,Y)$ be the coordinates of a point uniformly chosen from a
quadrilateral with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,2)$. Find
the marginal probability density functions of $X$ and $Y$.




Try



Well, these points are vertices of a quadriteral whose area is $frac{2}{3}$. Since it is uniformly chosen the points, then



$$ f_{X,Y}(x,y) = frac{1}{text{Area(S)}} = frac{1}{2/3} = frac{3}{2} $$



Where $(x,y) in S $, the quadrilateral. Hence,



$$ f_X(x) = intlimits_0^2 frac{2}{3} dy = frac{4}{3} $$



and



$$ f_Y(y) = intlimits_0^1 frac{2}{3} = frac{2}{3} $$



Is this correct?






probability







share|cite|improve this question














Let $(X,Y)$ be the coordinates of a point uniformly chosen from a
quadrilateral with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,2)$. Find
the marginal probability density functions of $X$ and $Y$.




Try



Well, these points are vertices of a quadriteral whose area is $frac{2}{3}$. Since it is uniformly chosen the points, then



$$ f_{X,Y}(x,y) = frac{1}{text{Area(S)}} = frac{1}{2/3} = frac{3}{2} $$



Where $(x,y) in S $, the quadrilateral. Hence,



$$ f_X(x) = intlimits_0^2 frac{2}{3} dy = frac{4}{3} $$



and



$$ f_Y(y) = intlimits_0^1 frac{2}{3} = frac{2}{3} $$



Is this correct?






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 20 at 2:11









Neymar

374113




374113












  • No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
    – David G. Stork
    Nov 20 at 2:15










  • No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
    – Graham Kemp
    Nov 20 at 2:17


















  • No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
    – David G. Stork
    Nov 20 at 2:15










  • No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
    – Graham Kemp
    Nov 20 at 2:17
















No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
– David G. Stork
Nov 20 at 2:15




No. The densities depend upon the variable. So for instance choose a single value for $x$ and find the distribution of possible $y$ values.
– David G. Stork
Nov 20 at 2:15












No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
– Graham Kemp
Nov 20 at 2:17




No. The area is not a rectangle. $X$ and $Y$ are not uniformly distributed in the margins.
– Graham Kemp
Nov 20 at 2:17










2 Answers
2






active

oldest

votes


















2














This is the region:



enter image description here



The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$






share|cite|improve this answer





















  • for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
    – Neymar
    Nov 20 at 2:29










  • That is correct.
    – irchans
    Nov 20 at 2:31










  • yup, the minimal would have taken care of that for us compactly.
    – Siong Thye Goh
    Nov 20 at 2:31



















1














To be correct you need to include the support. This can be expressed in two equivalent ways: $$f_{X,Y}(x,y)~{=tfrac 32mathbf 1_{0le xleq 1, 0leq yleq2-x}\=tfrac 32mathbf 1_{0le yleq 2, 0leq xleq min(1,2-y)}}$$



Then integrate over this support with respect to the appropriate variables.






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

    votes









    2














    This is the region:



    enter image description here



    The area should be $frac32$.
    For $x in (0,1)$,
    $$f_X(x) = int_0^{2-x} frac23 , dy$$
    For $y in (0,2)$,
    $$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$






    share|cite|improve this answer





















    • for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
      – Neymar
      Nov 20 at 2:29










    • That is correct.
      – irchans
      Nov 20 at 2:31










    • yup, the minimal would have taken care of that for us compactly.
      – Siong Thye Goh
      Nov 20 at 2:31
















    2














    This is the region:



    enter image description here



    The area should be $frac32$.
    For $x in (0,1)$,
    $$f_X(x) = int_0^{2-x} frac23 , dy$$
    For $y in (0,2)$,
    $$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$






    share|cite|improve this answer





















    • for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
      – Neymar
      Nov 20 at 2:29










    • That is correct.
      – irchans
      Nov 20 at 2:31










    • yup, the minimal would have taken care of that for us compactly.
      – Siong Thye Goh
      Nov 20 at 2:31














    2












    2








    2






    This is the region:



    enter image description here



    The area should be $frac32$.
    For $x in (0,1)$,
    $$f_X(x) = int_0^{2-x} frac23 , dy$$
    For $y in (0,2)$,
    $$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$






    share|cite|improve this answer












    This is the region:



    enter image description here



    The area should be $frac32$.
    For $x in (0,1)$,
    $$f_X(x) = int_0^{2-x} frac23 , dy$$
    For $y in (0,2)$,
    $$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 20 at 2:25









    Siong Thye Goh

    98.9k1464116




    98.9k1464116












    • for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
      – Neymar
      Nov 20 at 2:29










    • That is correct.
      – irchans
      Nov 20 at 2:31










    • yup, the minimal would have taken care of that for us compactly.
      – Siong Thye Goh
      Nov 20 at 2:31


















    • for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
      – Neymar
      Nov 20 at 2:29










    • That is correct.
      – irchans
      Nov 20 at 2:31










    • yup, the minimal would have taken care of that for us compactly.
      – Siong Thye Goh
      Nov 20 at 2:31
















    for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
    – Neymar
    Nov 20 at 2:29




    for marginal for $Y$ can you just integrate from $0$ to $1$ if $0<y<1$ and from $1<y<2$ we integrate from $1$ to $2-y$, correct?
    – Neymar
    Nov 20 at 2:29












    That is correct.
    – irchans
    Nov 20 at 2:31




    That is correct.
    – irchans
    Nov 20 at 2:31












    yup, the minimal would have taken care of that for us compactly.
    – Siong Thye Goh
    Nov 20 at 2:31




    yup, the minimal would have taken care of that for us compactly.
    – Siong Thye Goh
    Nov 20 at 2:31











    1














    To be correct you need to include the support. This can be expressed in two equivalent ways: $$f_{X,Y}(x,y)~{=tfrac 32mathbf 1_{0le xleq 1, 0leq yleq2-x}\=tfrac 32mathbf 1_{0le yleq 2, 0leq xleq min(1,2-y)}}$$



    Then integrate over this support with respect to the appropriate variables.






    share|cite|improve this answer


























      1














      To be correct you need to include the support. This can be expressed in two equivalent ways: $$f_{X,Y}(x,y)~{=tfrac 32mathbf 1_{0le xleq 1, 0leq yleq2-x}\=tfrac 32mathbf 1_{0le yleq 2, 0leq xleq min(1,2-y)}}$$



      Then integrate over this support with respect to the appropriate variables.






      share|cite|improve this answer
























        1












        1








        1






        To be correct you need to include the support. This can be expressed in two equivalent ways: $$f_{X,Y}(x,y)~{=tfrac 32mathbf 1_{0le xleq 1, 0leq yleq2-x}\=tfrac 32mathbf 1_{0le yleq 2, 0leq xleq min(1,2-y)}}$$



        Then integrate over this support with respect to the appropriate variables.






        share|cite|improve this answer












        To be correct you need to include the support. This can be expressed in two equivalent ways: $$f_{X,Y}(x,y)~{=tfrac 32mathbf 1_{0le xleq 1, 0leq yleq2-x}\=tfrac 32mathbf 1_{0le yleq 2, 0leq xleq min(1,2-y)}}$$



        Then integrate over this support with respect to the appropriate variables.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 20 at 2:28









        Graham Kemp

        84.7k43378




        84.7k43378






























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