Uniform distribution on unit disk
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Let $(X, Y)$ be a random point chosen according to the uniform distribution in the disk of radius 1 centered at the origin. Compute the densities of $X$ and of $Y$.
I know that the joint density of $X$ and $Y$ is $frac{1}{pi}$ since when we integrate $frac{1}{pi}$ over the unit circle, we get $1$.
So if I wanted to find the density of $X$, I was thinking of finding the cumulative distribution of $X$ and the differentiate it to get its density. In order to get its cumulative distribution function, I was going to use the fact that $P(X<x)=P(X<x, -infty < Y < infty)$, but this integral doesn't seem nice to work with. Am I on the right track or is there a better way?
probability probability-distributions
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Let $(X, Y)$ be a random point chosen according to the uniform distribution in the disk of radius 1 centered at the origin. Compute the densities of $X$ and of $Y$.
I know that the joint density of $X$ and $Y$ is $frac{1}{pi}$ since when we integrate $frac{1}{pi}$ over the unit circle, we get $1$.
So if I wanted to find the density of $X$, I was thinking of finding the cumulative distribution of $X$ and the differentiate it to get its density. In order to get its cumulative distribution function, I was going to use the fact that $P(X<x)=P(X<x, -infty < Y < infty)$, but this integral doesn't seem nice to work with. Am I on the right track or is there a better way?
probability probability-distributions
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up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(X, Y)$ be a random point chosen according to the uniform distribution in the disk of radius 1 centered at the origin. Compute the densities of $X$ and of $Y$.
I know that the joint density of $X$ and $Y$ is $frac{1}{pi}$ since when we integrate $frac{1}{pi}$ over the unit circle, we get $1$.
So if I wanted to find the density of $X$, I was thinking of finding the cumulative distribution of $X$ and the differentiate it to get its density. In order to get its cumulative distribution function, I was going to use the fact that $P(X<x)=P(X<x, -infty < Y < infty)$, but this integral doesn't seem nice to work with. Am I on the right track or is there a better way?
probability probability-distributions
Let $(X, Y)$ be a random point chosen according to the uniform distribution in the disk of radius 1 centered at the origin. Compute the densities of $X$ and of $Y$.
I know that the joint density of $X$ and $Y$ is $frac{1}{pi}$ since when we integrate $frac{1}{pi}$ over the unit circle, we get $1$.
So if I wanted to find the density of $X$, I was thinking of finding the cumulative distribution of $X$ and the differentiate it to get its density. In order to get its cumulative distribution function, I was going to use the fact that $P(X<x)=P(X<x, -infty < Y < infty)$, but this integral doesn't seem nice to work with. Am I on the right track or is there a better way?
probability probability-distributions
probability probability-distributions
edited Apr 11 '15 at 18:30
Did
244k23212451
244k23212451
asked Jan 6 '15 at 4:56
DHH
163110
163110
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By definition, the marginal density of $X$ is simply $$f_X(x) = int_{y=-1}^1 f_{X,Y}(x,y) , dy = int_{y=-sqrt{1-x^2}}^{sqrt{1-x^2}} frac{1}{pi} , dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = frac{1}{pi} mathbb{1}(x^2+y^2 le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.
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$P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
By definition, the marginal density of $X$ is simply $$f_X(x) = int_{y=-1}^1 f_{X,Y}(x,y) , dy = int_{y=-sqrt{1-x^2}}^{sqrt{1-x^2}} frac{1}{pi} , dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = frac{1}{pi} mathbb{1}(x^2+y^2 le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.
add a comment |
up vote
3
down vote
By definition, the marginal density of $X$ is simply $$f_X(x) = int_{y=-1}^1 f_{X,Y}(x,y) , dy = int_{y=-sqrt{1-x^2}}^{sqrt{1-x^2}} frac{1}{pi} , dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = frac{1}{pi} mathbb{1}(x^2+y^2 le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.
add a comment |
up vote
3
down vote
up vote
3
down vote
By definition, the marginal density of $X$ is simply $$f_X(x) = int_{y=-1}^1 f_{X,Y}(x,y) , dy = int_{y=-sqrt{1-x^2}}^{sqrt{1-x^2}} frac{1}{pi} , dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = frac{1}{pi} mathbb{1}(x^2+y^2 le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.
By definition, the marginal density of $X$ is simply $$f_X(x) = int_{y=-1}^1 f_{X,Y}(x,y) , dy = int_{y=-sqrt{1-x^2}}^{sqrt{1-x^2}} frac{1}{pi} , dy.$$ The second equality arises from the fact that $$f_{X,Y}(x,y) = frac{1}{pi} mathbb{1}(x^2+y^2 le 1),$$ from which we see that for a given $X = x$, the support of $Y$ is then $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.
answered Jan 6 '15 at 5:05
heropup
61.9k65997
61.9k65997
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$P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
add a comment |
up vote
1
down vote
$P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
add a comment |
up vote
1
down vote
up vote
1
down vote
$P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.
$P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.
edited Jan 6 '15 at 5:36
answered Jan 6 '15 at 5:22
user1537366
1,519819
1,519819
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
add a comment |
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
@DHH Sorry, I first thought you had a uniform distribution on the circle (only the outline) and not the disk. I've edited for the disk.
– user1537366
Jan 6 '15 at 5:38
add a comment |
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