fint joint and marginal distributions of two uniformy distributed variables over a specified region
$begingroup$
This exercise (partially) comes from Rice 3.9
Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.
Find the marginal densities of X and Y.
Find the joint density.
Find the two conditional densities.
$X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$
then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$
I am not sure if this is correct? I am confused with how to find the joint distribution also.
probability probability-theory statistics probability-distributions uniform-distribution
asked Dec 11 '18 at 12:45
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
This exercise (partially) comes from Rice 3.9
Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.
Find the marginal densities of X and Y.
Find the joint density.
Find the two conditional densities.
$X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$
then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$
I am not sure if this is correct? I am confused with how to find the joint distribution also.
probability probability-theory statistics probability-distributions uniform-distribution
asked Dec 11 '18 at 12:45
user1607user1607
1718
$endgroup$
add a comment |
$begingroup$
This exercise (partially) comes from Rice 3.9
Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.
Find the marginal densities of X and Y.
Find the joint density.
Find the two conditional densities.
$X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$
then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$
I am not sure if this is correct? I am confused with how to find the joint distribution also.
probability probability-theory statistics probability-distributions uniform-distribution
asked Dec 11 '18 at 12:45
user1607user1607
1718
$endgroup$
This exercise (partially) comes from Rice 3.9
Suppose that (X, Y ) is uniformly distributed over the region defined
by $0 ≤ y ≤ 1 − x^2$ and $−1 ≤ x ≤ 1$.
Find the marginal densities of X and Y.
Find the joint density.
Find the two conditional densities.
$X sim U[-1,1]$ and $Y∼U[0, 1-x^2]$
then $F_x(x)=x+1$ for $−1 ≤ x ≤ 1$ and $F_Y(y)=frac{y}{1-x^2}$ for $0 ≤ y ≤ 1 − x^2$
I am not sure if this is correct? I am confused with how to find the joint distribution also.
probability probability-theory statistics probability-distributions uniform-distribution
probability probability-theory statistics probability-distributions uniform-distribution
asked Dec 11 '18 at 12:45
user1607user1607
1718
asked Dec 11 '18 at 12:45
user1607user1607
1718
asked Dec 11 '18 at 12:45
user1607user1607
1718
asked Dec 11 '18 at 12:45
user1607user1607
1718
asked Dec 11 '18 at 12:45
user1607user1607
1718
1718
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
$endgroup$
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
add a comment |
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1 Answer
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$begingroup$
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
$endgroup$
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
add a comment |
$begingroup$
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
$endgroup$
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
add a comment |
$begingroup$
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
$endgroup$
$X$ is not uniform on $[-1,1]$. Even if it were, the cumulative distribution function would be $F_X(x)=frac12(x+1)$ on that interval. But this would not be the marginal density
Hints for the question:
Since the distribution is uniform in the region, call it a constant multiplied by an indicator function $f_{X,Y}(x,y)=k, I[-1 le x le 1, 0 le y le 1-x^2$. You can find the value of $k$ from $intlimits_{x=-infty}^inftyintlimits_{y=-infty}^{infty} f_{X,Y}(x,y) , dy , dx = intlimits_{x=-1}^1intlimits_{y=0}^{1-x^2} k , dy , dx = 1$
You can do the same sort of thing for the conditional probabilities: in each case it will be constant integrating to $1$ over a particular interval, which will determine the density
For the marginal density for $X$ you have $f_X(x)=intlimits_{y=0}^{1-x^2} k , dy$ using the $k$ you found earlier
For the marginal density for $Y$ you will find it easier to rewrite the region inequalities $-1 le x le 1, 0 le y le 1-x^2$ as $0 le y le 1, -sqrt{1-y} le x le sqrt{1-y}$ and you then want $f_Y(y)=intlimits_{x=-sqrt{1-y}}^{sqrt{1-y}} k , dx$ again using the $k$ you found earlier
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
edited Dec 11 '18 at 23:57
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
answered Dec 11 '18 at 14:32
HenryHenry
101k482169
101k482169
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
add a comment |
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
1
1
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
thanks. I have obtained $k=frac34$ and also the marginal distributions: $f_X(x)= frac34 (a-x^2)$ for $-1leq x leq 1$ and $f_Y(y)= frac32 sqrt{1-y}$ for $0leq y leq 1$. However, I am not sure how to obtain the conditional distributions? The only approach I know is via division such that : $f_{x,y}(x|y)=frac{f_{x,y}(x,y)}{f_{y}(y)}$. Ist this the fastest/simplest way to find the conditional distributions?
$endgroup$
– user1607
Dec 11 '18 at 15:02
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
$begingroup$
Inside the region you have the conditionally uniform densities $f(y mid x) = frac1{1-x^2}$ for $0 le y le 1-x^2$ and $f(x mid y) = frac1{2sqrt{1-y}}$ for $-sqrt{1-y} le x le sqrt{1-y}$
$endgroup$
– Henry
Dec 11 '18 at 15:31
add a comment |
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