Finding domain of variables in joint density for marginal density
$begingroup$
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 $0<U-X<1$.
statistics density-function
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
$endgroup$
add a comment |
$begingroup$
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 $0<U-X<1$.
statistics density-function
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
$endgroup$
$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, 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
add a comment |
$begingroup$
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 $0<U-X<1$.
statistics density-function
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
$endgroup$
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
statistics density-function
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
asked Nov 15 '16 at 5:11
ultrainstinctultrainstinct
2,1561725
2,1561725
$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, 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
add a comment |
$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, 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
add a comment |
1 Answer
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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 $(X,X+1)$.
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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oldest
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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 $(X,X+1)$.
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
$endgroup$
add a comment |
$begingroup$
The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
$endgroup$
add a comment |
$begingroup$
The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
$endgroup$
The domain of $(X,U)$ is a parallelogram in $mathbf{R}^2$: $X$ is in $(0,1)$, while $U$ is in $(X,X+1)$.
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
edited Dec 17 '18 at 6:39
Daniele Tampieri
2,1091621
2,1091621
answered Dec 17 '18 at 5:34
james0910james0910
111
answered Dec 17 '18 at 5:34
james0910james0910
111
answered Dec 17 '18 at 5:34
james0910james0910
111
111
add a comment |
add a comment |
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$begingroup$
$0 < U - X < 1 implies X < U < 1+X$, 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