Finding marginal density functions of $X$ and $Y$
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
asked Nov 20 at 2:11
Neymar
374113
add a comment |
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
asked Nov 20 at 2:11
Neymar
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
add a comment |
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
asked Nov 20 at 2:11
Neymar
374113
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
asked Nov 20 at 2:11
Neymar
374113
asked Nov 20 at 2:11
Neymar
374113
asked Nov 20 at 2:11
Neymar
374113
asked Nov 20 at 2:11
Neymar
374113
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
This is the region:
The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$
answered Nov 20 at 2:25
Siong Thye Goh
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
add a comment |
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.
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is the region:
The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$
answered Nov 20 at 2:25
Siong Thye Goh
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
add a comment |
This is the region:
The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$
answered Nov 20 at 2:25
Siong Thye Goh
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
add a comment |
This is the region:
The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$
answered Nov 20 at 2:25
Siong Thye Goh
98.9k1464116
This is the region:
The area should be $frac32$.
For $x in (0,1)$,
$$f_X(x) = int_0^{2-x} frac23 , dy$$
For $y in (0,2)$,
$$f_Y(y) = int_0^{min(1,2-y)} frac23 , dx$$
answered Nov 20 at 2:25
Siong Thye Goh
98.9k1464116
answered Nov 20 at 2:25
Siong Thye Goh
98.9k1464116
answered Nov 20 at 2:25
Siong Thye Goh
98.9k1464116
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
add a comment |
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
add a comment |
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.
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
add a comment |
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.
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
add a comment |
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.
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
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.
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
answered Nov 20 at 2:28
Graham Kemp
84.7k43378
84.7k43378
add a comment |
add a comment |
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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