Finding domain of variables in joint density for marginal density












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Let $(X,Y)$ have joint density $f(x,y) = frac{1}{2}(1+x+y)$ for $0<x<1$ and $0<y<1$.



So the joint density of $X$ and $U=X+Y$ is $f_{X,U}(x,u)=frac{1}{2}(1+u)$. Now it is simple to get the domain of $X$ since it is provided in the problem description, but to get the domain of $U$ seems so be trickier since you need to extract the case of $U<1$ and $Ugeq 1$. I need these values in order to get the marginal densities, but I am not quite sure what I need to do after I get $0<U-X<1$.










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$endgroup$












  • $begingroup$
    $0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 5:23












  • $begingroup$
    Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
    $endgroup$
    – chandresh
    Nov 15 '16 at 5:25










  • $begingroup$
    @LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
    $endgroup$
    – ultrainstinct
    Nov 15 '16 at 5:31










  • $begingroup$
    The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 23:46












  • $begingroup$
    The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 16 '16 at 0:03
















2












$begingroup$


Let $(X,Y)$ have joint density $f(x,y) = frac{1}{2}(1+x+y)$ for $0<x<1$ and $0<y<1$.



So the joint density of $X$ and $U=X+Y$ is $f_{X,U}(x,u)=frac{1}{2}(1+u)$. Now it is simple to get the domain of $X$ since it is provided in the problem description, but to get the domain of $U$ seems so be trickier since you need to extract the case of $U<1$ and $Ugeq 1$. I need these values in order to get the marginal densities, but I am not quite sure what I need to do after I get $0<U-X<1$.










share|cite|improve this question









$endgroup$












  • $begingroup$
    $0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 5:23












  • $begingroup$
    Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
    $endgroup$
    – chandresh
    Nov 15 '16 at 5:25










  • $begingroup$
    @LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
    $endgroup$
    – ultrainstinct
    Nov 15 '16 at 5:31










  • $begingroup$
    The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 23:46












  • $begingroup$
    The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 16 '16 at 0:03














2












2








2





$begingroup$


Let $(X,Y)$ have joint density $f(x,y) = frac{1}{2}(1+x+y)$ for $0<x<1$ and $0<y<1$.



So the joint density of $X$ and $U=X+Y$ is $f_{X,U}(x,u)=frac{1}{2}(1+u)$. Now it is simple to get the domain of $X$ since it is provided in the problem description, but to get the domain of $U$ seems so be trickier since you need to extract the case of $U<1$ and $Ugeq 1$. I need these values in order to get the marginal densities, but I am not quite sure what I need to do after I get $0<U-X<1$.










share|cite|improve this question









$endgroup$




Let $(X,Y)$ have joint density $f(x,y) = frac{1}{2}(1+x+y)$ for $0<x<1$ and $0<y<1$.



So the joint density of $X$ and $U=X+Y$ is $f_{X,U}(x,u)=frac{1}{2}(1+u)$. Now it is simple to get the domain of $X$ since it is provided in the problem description, but to get the domain of $U$ seems so be trickier since you need to extract the case of $U<1$ and $Ugeq 1$. I need these values in order to get the marginal densities, but I am not quite sure what I need to do after I get $0<U-X<1$.







statistics density-function






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asked Nov 15 '16 at 5:11









ultrainstinctultrainstinct

2,1611725




2,1611725












  • $begingroup$
    $0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 5:23












  • $begingroup$
    Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
    $endgroup$
    – chandresh
    Nov 15 '16 at 5:25










  • $begingroup$
    @LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
    $endgroup$
    – ultrainstinct
    Nov 15 '16 at 5:31










  • $begingroup$
    The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 23:46












  • $begingroup$
    The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 16 '16 at 0:03


















  • $begingroup$
    $0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 5:23












  • $begingroup$
    Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
    $endgroup$
    – chandresh
    Nov 15 '16 at 5:25










  • $begingroup$
    @LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
    $endgroup$
    – ultrainstinct
    Nov 15 '16 at 5:31










  • $begingroup$
    The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 15 '16 at 23:46












  • $begingroup$
    The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
    $endgroup$
    – Lee David Chung Lin
    Nov 16 '16 at 0:03
















$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
$endgroup$
– Lee David Chung Lin
Nov 15 '16 at 5:23






$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, then $because 0< X < 1 therefore 0 < X < U < 1 + X < 1 + 1 = 2$, which is just the intuitive $0< U < 2$
$endgroup$
– Lee David Chung Lin
Nov 15 '16 at 5:23














$begingroup$
Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
$endgroup$
– chandresh
Nov 15 '16 at 5:25




$begingroup$
Range of $U$ will depend on $X$ values. You can't write $U$ values independent of X.
$endgroup$
– chandresh
Nov 15 '16 at 5:25












$begingroup$
@LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
$endgroup$
– ultrainstinct
Nov 15 '16 at 5:31




$begingroup$
@LeeDavidChungLin So then I got the two cases $U-1 < X < 1$ and $0<X<U$. Are those correct?
$endgroup$
– ultrainstinct
Nov 15 '16 at 5:31












$begingroup$
The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
$endgroup$
– Lee David Chung Lin
Nov 15 '16 at 23:46






$begingroup$
The domain of $U$ is $[0,2]$, it is true and it is the basic understanding of the situation. The joint domain of $f_{XU}$ is a parallelogram above the diagonal with vertices $(x,u)=(0, 0),, (0,1),,(1,2),,(1,0)$. Yes like you said to split into 2 cases, the marginal $f_U$ from $f_{XU}$ is $(1+u)/2 cdot (u-0) = u(1+u)/2$ from $X<U$ and $(1+u)/2 cdot (1- (u-1)) = (2-u)(1+u)/2$ from $U-1 < X < 1$.
$endgroup$
– Lee David Chung Lin
Nov 15 '16 at 23:46














$begingroup$
The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
$endgroup$
– Lee David Chung Lin
Nov 16 '16 at 0:03




$begingroup$
The $(u-0)$ is the horizontal segment length in the lower half of the parallelogram, from the two end points of $0 < X < U$, and the $(1 - (u-1))$ is the horizontal segment length in the upper triangular half of the parallelogram, from the two end points of $U-1 < X < 1$.
$endgroup$
– Lee David Chung Lin
Nov 16 '16 at 0:03










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$begingroup$

The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.






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    1 Answer
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    active

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    1












    $begingroup$

    The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.






        share|cite|improve this answer











        $endgroup$



        The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 17 '18 at 6:39









        Daniele Tampieri

        2,3022922




        2,3022922










        answered Dec 17 '18 at 5:34









        james0910james0910

        111




        111






























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