Mixing
Find joint CDF given a joint PDF
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Let $X$ and $Y$ have a joint density function given by
$$
f(x, y) =
begincases
1, & textfor 0leq xleq2,;max(0,,x-1)leq yleq min(1,,x), \
0, & textotherwise.
endcases
$$
Determine the joint and marginal distribution functions.
I know that $F_XY(x, y) = int_-infty^xint_-infty^y f(u, v);dudv$. But I have no idea how to apply this fact.
The obvious case that I can find is that if $x > 2$, then $F_XY(x, y) = 1$.
Some other bounds are e.g. $0<x<1$ and $0<y<x$, but how do I find the CDF for this case? I can find some of the bounds, but don't know how to use them for the integrals.
probability probability-distributions marginal-distribution
edited Jul 29 at 10:57
BCLC
6,96621973
asked Jul 28 at 23:07
wednesdaymiko
727
add a comment |Â
up vote
1
down vote
favorite
Let $X$ and $Y$ have a joint density function given by
$$
f(x, y) =
begincases
1, & textfor 0leq xleq2,;max(0,,x-1)leq yleq min(1,,x), \
0, & textotherwise.
endcases
$$
Determine the joint and marginal distribution functions.
I know that $F_XY(x, y) = int_-infty^xint_-infty^y f(u, v);dudv$. But I have no idea how to apply this fact.
The obvious case that I can find is that if $x > 2$, then $F_XY(x, y) = 1$.
Some other bounds are e.g. $0<x<1$ and $0<y<x$, but how do I find the CDF for this case? I can find some of the bounds, but don't know how to use them for the integrals.
probability probability-distributions marginal-distribution
edited Jul 29 at 10:57
BCLC
6,96621973
asked Jul 28 at 23:07
wednesdaymiko
727
For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ and $Y$ have a joint density function given by
$$
f(x, y) =
begincases
1, & textfor 0leq xleq2,;max(0,,x-1)leq yleq min(1,,x), \
0, & textotherwise.
endcases
$$
Determine the joint and marginal distribution functions.
I know that $F_XY(x, y) = int_-infty^xint_-infty^y f(u, v);dudv$. But I have no idea how to apply this fact.
The obvious case that I can find is that if $x > 2$, then $F_XY(x, y) = 1$.
Some other bounds are e.g. $0<x<1$ and $0<y<x$, but how do I find the CDF for this case? I can find some of the bounds, but don't know how to use them for the integrals.
probability probability-distributions marginal-distribution
edited Jul 29 at 10:57
BCLC
6,96621973
asked Jul 28 at 23:07
wednesdaymiko
727
Let $X$ and $Y$ have a joint density function given by
$$
f(x, y) =
begincases
1, & textfor 0leq xleq2,;max(0,,x-1)leq yleq min(1,,x), \
0, & textotherwise.
endcases
$$
Determine the joint and marginal distribution functions.
I know that $F_XY(x, y) = int_-infty^xint_-infty^y f(u, v);dudv$. But I have no idea how to apply this fact.
The obvious case that I can find is that if $x > 2$, then $F_XY(x, y) = 1$.
Some other bounds are e.g. $0<x<1$ and $0<y<x$, but how do I find the CDF for this case? I can find some of the bounds, but don't know how to use them for the integrals.
probability probability-distributions marginal-distribution
edited Jul 29 at 10:57
BCLC
6,96621973
asked Jul 28 at 23:07
wednesdaymiko
727
edited Jul 29 at 10:57
BCLC
6,96621973
edited Jul 29 at 10:57
BCLC
6,96621973
edited Jul 29 at 10:57
BCLC
6,96621973
6,96621973
asked Jul 28 at 23:07
wednesdaymiko
727
asked Jul 28 at 23:07
wednesdaymiko
727
asked Jul 28 at 23:07
wednesdaymiko
727
727
For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24
add a comment |Â
For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24
For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Revised correction, based on conversation with Maxim!
Summary for all $(x,y)$. For $xle 0$ or $yle 0, F(x,y)=0$.
For $0lt xle 1, 0le yle 1$, two parts:
1) $0lt yle x, F(x,y)=xy-fracy^22$,
2) $xlt yle 1, F(x,y)=fracx^22$,
For $1lt x, 0le yle 1$, two parts: [part 2) vacuous for $xgt 2$]
1) $0le yle x-1, F(x,y)=y$,
2) $x-1lt yle1, F(x,y)=xy-fracy^22-frac(x-1)^22$,
For $1lt y, F(x,y)=F(x,1)$.
answered Jul 29 at 0:05
herb steinberg
93529
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
 |Â
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up vote
1
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Let
$$f_X, Y(x, y) = [0 leq x leq 2 land max(0, x-1) leq y leq min(1, x)],$$ where $[P]$ is $1$ if $P$ is true and $0$ otherwise. Consider the cases $0 leq x leq 1$ and $1 < x leq 2$ separately to get rid of $min$ and $max$ and verify that
$$f_X, Y(x, y) = [0 leq y leq 1 land y leq x leq y + 1].$$
Integrating out the variable $x$ gives $1$ if the horizontal line intersects the parallelogram and $0$ otherwise:
$$f_Y(y) = int_-infty^infty f_X, Y(x, y) dx = [0 leq y leq 1],$$
from which the marginal cdf of $Y$ is
$$F_Y(y) = y ,[0 leq y leq 1] + [1 < y].$$
The marginal pdf of $X$ is symmetric wrt the line $x = 1$:
$$f_X(x) = x ,[0 leq x leq 1] + (2 - x) ,[1 < x leq 2], \
F_X(x) = frac x^2 2 ,[0 leq x leq 1] +
left( 2 x - frac x^2 2 - 1 right) [1 < x leq 2] +
[2 < x].$$
To find $F_X, Y(x, y)$, first take $(x, y)$ inside the parallelogram:
$$G(x, y) =
int_-infty^x int_-infty^y f_X, Y(u, v) dv du = \
left( x y - frac y^2 2 right) ,[x leq 1] +
left( x - frac (x - y)^2 + 1 2 right) [1 < x], \
f_X, Y(x, y) = 1.$$
Note that this is different from $int_-infty^x int_-infty^y f_X, Y(u, v) du dv$.
For $(x, y)$ outside of the parallelogram, project the point onto the boundary:
$$F_X, Y(x, y) =
G(min(x, y + 1, 2), min(y, x, 1)) ,[0 leq x land 0 leq y].$$
It can be verified that
$$F_X(x) = F_X, Y(x, infty), ;F_Y(y) = F_X, Y(infty, y).$$
answered Jul 31 at 16:17
Maxim
2,035112
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Revised correction, based on conversation with Maxim!
Summary for all $(x,y)$. For $xle 0$ or $yle 0, F(x,y)=0$.
For $0lt xle 1, 0le yle 1$, two parts:
1) $0lt yle x, F(x,y)=xy-fracy^22$,
2) $xlt yle 1, F(x,y)=fracx^22$,
For $1lt x, 0le yle 1$, two parts: [part 2) vacuous for $xgt 2$]
1) $0le yle x-1, F(x,y)=y$,
2) $x-1lt yle1, F(x,y)=xy-fracy^22-frac(x-1)^22$,
For $1lt y, F(x,y)=F(x,1)$.
answered Jul 29 at 0:05
herb steinberg
93529
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
 |Â
show 12 more comments
up vote
2
down vote
accepted
Revised correction, based on conversation with Maxim!
Summary for all $(x,y)$. For $xle 0$ or $yle 0, F(x,y)=0$.
For $0lt xle 1, 0le yle 1$, two parts:
1) $0lt yle x, F(x,y)=xy-fracy^22$,
2) $xlt yle 1, F(x,y)=fracx^22$,
For $1lt x, 0le yle 1$, two parts: [part 2) vacuous for $xgt 2$]
1) $0le yle x-1, F(x,y)=y$,
2) $x-1lt yle1, F(x,y)=xy-fracy^22-frac(x-1)^22$,
For $1lt y, F(x,y)=F(x,1)$.
answered Jul 29 at 0:05
herb steinberg
93529
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
 |Â
show 12 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Revised correction, based on conversation with Maxim!
Summary for all $(x,y)$. For $xle 0$ or $yle 0, F(x,y)=0$.
For $0lt xle 1, 0le yle 1$, two parts:
1) $0lt yle x, F(x,y)=xy-fracy^22$,
2) $xlt yle 1, F(x,y)=fracx^22$,
For $1lt x, 0le yle 1$, two parts: [part 2) vacuous for $xgt 2$]
1) $0le yle x-1, F(x,y)=y$,
2) $x-1lt yle1, F(x,y)=xy-fracy^22-frac(x-1)^22$,
For $1lt y, F(x,y)=F(x,1)$.
answered Jul 29 at 0:05
herb steinberg
93529
Revised correction, based on conversation with Maxim!
Summary for all $(x,y)$. For $xle 0$ or $yle 0, F(x,y)=0$.
For $0lt xle 1, 0le yle 1$, two parts:
1) $0lt yle x, F(x,y)=xy-fracy^22$,
2) $xlt yle 1, F(x,y)=fracx^22$,
For $1lt x, 0le yle 1$, two parts: [part 2) vacuous for $xgt 2$]
1) $0le yle x-1, F(x,y)=y$,
2) $x-1lt yle1, F(x,y)=xy-fracy^22-frac(x-1)^22$,
For $1lt y, F(x,y)=F(x,1)$.
answered Jul 29 at 0:05
herb steinberg
93529
edited Jul 31 at 19:13
answered Jul 29 at 0:05
herb steinberg
93529
answered Jul 29 at 0:05
herb steinberg
93529
answered Jul 29 at 0:05
herb steinberg
93529
93529
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
 |Â
show 12 more comments
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
I seem to get different results for the first integral. I get $int_0^xint_x^y,dudv = int_0^x (y - x);dv = (yv - xv)|_0^x = xy - x^2$. Should the integral be in $dxdy$ instead, as that do give me the right solution. But in all the definitions of the CDF, it's stated as other variables like $dudv$. I think I'm potentially misunderstanding something?
– wednesdaymiko
Jul 29 at 0:54
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
@wednesdaymike Your integrand $(y-x)$ should be $(y-v)$. As a check put in the upper limit, the value should be $frac12$. Your expression gives 0.
– herb steinberg
Jul 29 at 2:24
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To see it more clearly, diagram $f(x,y)=1$. It consists of two triangles. The first has $(x,y)=(0,0),(0,1),(1,1)$ as vertices. The second has $(x,y)=(1,0),(1,1),(2,1)$ as vertices.
– herb steinberg
Jul 29 at 2:37
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
To get cdf from the diagram, set $(x,y)$ at a point inside a triangle and draw straight lines from that point to the axes, making a rectangle with the axes. The area inside the rectangle within the triangle(s) is the cdf at that point.
– herb steinberg
Jul 29 at 3:54
1
1
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
If $f$ is $f_X,Y$, then your first integral gives $P(X < y, Y < x)$. The second integral is incorrect (you probably shouldn't use the same name for a free and a bound variable). Also, this can't be the complete answer because a cdf has to be between $0$ and $1$. The full result is $$int_-infty^x int_-infty^y f(t_1, t_2) dt_2 dt_1 = \ cases 0 & $x < 0 lor y < 0$ \ 1 & $2 < x land 1 < y$ \ x min(x, y) - frac 1 2 min(x, y)^2 & $0 leq x leq 1$ \ -frac 1 2 (min(x, 1 + y) - min(1, y))^2 + min(x, 1 + y) - frac 1 2 & $1 < x land lnot (2 < x land 1 < y)$.$$
– Maxim
Jul 29 at 13:52
 |Â
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up vote
1
down vote
Let
$$f_X, Y(x, y) = [0 leq x leq 2 land max(0, x-1) leq y leq min(1, x)],$$ where $[P]$ is $1$ if $P$ is true and $0$ otherwise. Consider the cases $0 leq x leq 1$ and $1 < x leq 2$ separately to get rid of $min$ and $max$ and verify that
$$f_X, Y(x, y) = [0 leq y leq 1 land y leq x leq y + 1].$$
Integrating out the variable $x$ gives $1$ if the horizontal line intersects the parallelogram and $0$ otherwise:
$$f_Y(y) = int_-infty^infty f_X, Y(x, y) dx = [0 leq y leq 1],$$
from which the marginal cdf of $Y$ is
$$F_Y(y) = y ,[0 leq y leq 1] + [1 < y].$$
The marginal pdf of $X$ is symmetric wrt the line $x = 1$:
$$f_X(x) = x ,[0 leq x leq 1] + (2 - x) ,[1 < x leq 2], \
F_X(x) = frac x^2 2 ,[0 leq x leq 1] +
left( 2 x - frac x^2 2 - 1 right) [1 < x leq 2] +
[2 < x].$$
To find $F_X, Y(x, y)$, first take $(x, y)$ inside the parallelogram:
$$G(x, y) =
int_-infty^x int_-infty^y f_X, Y(u, v) dv du = \
left( x y - frac y^2 2 right) ,[x leq 1] +
left( x - frac (x - y)^2 + 1 2 right) [1 < x], \
f_X, Y(x, y) = 1.$$
Note that this is different from $int_-infty^x int_-infty^y f_X, Y(u, v) du dv$.
For $(x, y)$ outside of the parallelogram, project the point onto the boundary:
$$F_X, Y(x, y) =
G(min(x, y + 1, 2), min(y, x, 1)) ,[0 leq x land 0 leq y].$$
It can be verified that
$$F_X(x) = F_X, Y(x, infty), ;F_Y(y) = F_X, Y(infty, y).$$
answered Jul 31 at 16:17
Maxim
2,035112
add a comment |Â
up vote
1
down vote
Let
$$f_X, Y(x, y) = [0 leq x leq 2 land max(0, x-1) leq y leq min(1, x)],$$ where $[P]$ is $1$ if $P$ is true and $0$ otherwise. Consider the cases $0 leq x leq 1$ and $1 < x leq 2$ separately to get rid of $min$ and $max$ and verify that
$$f_X, Y(x, y) = [0 leq y leq 1 land y leq x leq y + 1].$$
Integrating out the variable $x$ gives $1$ if the horizontal line intersects the parallelogram and $0$ otherwise:
$$f_Y(y) = int_-infty^infty f_X, Y(x, y) dx = [0 leq y leq 1],$$
from which the marginal cdf of $Y$ is
$$F_Y(y) = y ,[0 leq y leq 1] + [1 < y].$$
The marginal pdf of $X$ is symmetric wrt the line $x = 1$:
$$f_X(x) = x ,[0 leq x leq 1] + (2 - x) ,[1 < x leq 2], \
F_X(x) = frac x^2 2 ,[0 leq x leq 1] +
left( 2 x - frac x^2 2 - 1 right) [1 < x leq 2] +
[2 < x].$$
To find $F_X, Y(x, y)$, first take $(x, y)$ inside the parallelogram:
$$G(x, y) =
int_-infty^x int_-infty^y f_X, Y(u, v) dv du = \
left( x y - frac y^2 2 right) ,[x leq 1] +
left( x - frac (x - y)^2 + 1 2 right) [1 < x], \
f_X, Y(x, y) = 1.$$
Note that this is different from $int_-infty^x int_-infty^y f_X, Y(u, v) du dv$.
For $(x, y)$ outside of the parallelogram, project the point onto the boundary:
$$F_X, Y(x, y) =
G(min(x, y + 1, 2), min(y, x, 1)) ,[0 leq x land 0 leq y].$$
It can be verified that
$$F_X(x) = F_X, Y(x, infty), ;F_Y(y) = F_X, Y(infty, y).$$
answered Jul 31 at 16:17
Maxim
2,035112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let
$$f_X, Y(x, y) = [0 leq x leq 2 land max(0, x-1) leq y leq min(1, x)],$$ where $[P]$ is $1$ if $P$ is true and $0$ otherwise. Consider the cases $0 leq x leq 1$ and $1 < x leq 2$ separately to get rid of $min$ and $max$ and verify that
$$f_X, Y(x, y) = [0 leq y leq 1 land y leq x leq y + 1].$$
Integrating out the variable $x$ gives $1$ if the horizontal line intersects the parallelogram and $0$ otherwise:
$$f_Y(y) = int_-infty^infty f_X, Y(x, y) dx = [0 leq y leq 1],$$
from which the marginal cdf of $Y$ is
$$F_Y(y) = y ,[0 leq y leq 1] + [1 < y].$$
The marginal pdf of $X$ is symmetric wrt the line $x = 1$:
$$f_X(x) = x ,[0 leq x leq 1] + (2 - x) ,[1 < x leq 2], \
F_X(x) = frac x^2 2 ,[0 leq x leq 1] +
left( 2 x - frac x^2 2 - 1 right) [1 < x leq 2] +
[2 < x].$$
To find $F_X, Y(x, y)$, first take $(x, y)$ inside the parallelogram:
$$G(x, y) =
int_-infty^x int_-infty^y f_X, Y(u, v) dv du = \
left( x y - frac y^2 2 right) ,[x leq 1] +
left( x - frac (x - y)^2 + 1 2 right) [1 < x], \
f_X, Y(x, y) = 1.$$
Note that this is different from $int_-infty^x int_-infty^y f_X, Y(u, v) du dv$.
For $(x, y)$ outside of the parallelogram, project the point onto the boundary:
$$F_X, Y(x, y) =
G(min(x, y + 1, 2), min(y, x, 1)) ,[0 leq x land 0 leq y].$$
It can be verified that
$$F_X(x) = F_X, Y(x, infty), ;F_Y(y) = F_X, Y(infty, y).$$
answered Jul 31 at 16:17
Maxim
2,035112
Let
$$f_X, Y(x, y) = [0 leq x leq 2 land max(0, x-1) leq y leq min(1, x)],$$ where $[P]$ is $1$ if $P$ is true and $0$ otherwise. Consider the cases $0 leq x leq 1$ and $1 < x leq 2$ separately to get rid of $min$ and $max$ and verify that
$$f_X, Y(x, y) = [0 leq y leq 1 land y leq x leq y + 1].$$
Integrating out the variable $x$ gives $1$ if the horizontal line intersects the parallelogram and $0$ otherwise:
$$f_Y(y) = int_-infty^infty f_X, Y(x, y) dx = [0 leq y leq 1],$$
from which the marginal cdf of $Y$ is
$$F_Y(y) = y ,[0 leq y leq 1] + [1 < y].$$
The marginal pdf of $X$ is symmetric wrt the line $x = 1$:
$$f_X(x) = x ,[0 leq x leq 1] + (2 - x) ,[1 < x leq 2], \
F_X(x) = frac x^2 2 ,[0 leq x leq 1] +
left( 2 x - frac x^2 2 - 1 right) [1 < x leq 2] +
[2 < x].$$
To find $F_X, Y(x, y)$, first take $(x, y)$ inside the parallelogram:
$$G(x, y) =
int_-infty^x int_-infty^y f_X, Y(u, v) dv du = \
left( x y - frac y^2 2 right) ,[x leq 1] +
left( x - frac (x - y)^2 + 1 2 right) [1 < x], \
f_X, Y(x, y) = 1.$$
Note that this is different from $int_-infty^x int_-infty^y f_X, Y(u, v) du dv$.
For $(x, y)$ outside of the parallelogram, project the point onto the boundary:
$$F_X, Y(x, y) =
G(min(x, y + 1, 2), min(y, x, 1)) ,[0 leq x land 0 leq y].$$
It can be verified that
$$F_X(x) = F_X, Y(x, infty), ;F_Y(y) = F_X, Y(infty, y).$$
answered Jul 31 at 16:17
Maxim
2,035112
edited Jul 31 at 16:57
answered Jul 31 at 16:17
Maxim
2,035112
answered Jul 31 at 16:17
Maxim
2,035112
answered Jul 31 at 16:17
Maxim
2,035112
2,035112
add a comment |Â
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For these integrals, the lower limit is 0 not $-infty$.
– herb steinberg
Jul 28 at 23:10
Yeah, I've kind of understood that part. But how do I set up the rest of the integral? Just like this $int_0^x int_0^y f(x, y);dudv = int_0^x int_0^y 1;dudv$? Though, with the bounds I've given above, I end up with wrong solution for some reason.
– wednesdaymiko
Jul 28 at 23:12
For $0leq xleq 1$, we get $0leq yleq x$, but not sure how to find the bounds when $1leq x leq 2$. My guess was to plug in $x$ into $max(0, x-1) leq y leq min(1, x)$, but end up with the wrong results. E.g. $max(0, 2 - 1) leq y leq min(1, 2) = 1 leq y leq 1$ which doesn't make sense.
– wednesdaymiko
Jul 28 at 23:24