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?










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    up vote
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    down vote

    favorite
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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?










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      up vote
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      down vote

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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?










      share|cite|improve this question
















      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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      edited Apr 11 '15 at 18:30









      Did

      244k23212451




      244k23212451










      asked Jan 6 '15 at 4:56









      DHH

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      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 $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.






            share|cite|improve this answer























            • @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
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            active

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            2 Answers
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            active

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            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 $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.






            share|cite|improve this answer

























              up vote
              3
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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 $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.






              share|cite|improve this answer























                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 $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.






                share|cite|improve this answer












                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 $-sqrt{1-x^2} le Y le sqrt{1-x^2}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 6 '15 at 5:05









                heropup

                61.9k65997




                61.9k65997






















                    up vote
                    1
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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 $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.






                    share|cite|improve this answer























                    • @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

















                    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 $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.






                    share|cite|improve this answer























                    • @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















                    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 $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.






                    share|cite|improve this answer














                    $P(X<x)=int_{-1}^xint_{-sqrt{1-x^2}}^{sqrt{1-x^2}}frac{1}{pi}text{d}ytext{ d}x=int_{-1}^xfrac{1}{pi}2sqrt{1-x^2}text{d}x$, so differentiating you just get back the inside $f_X(x)=frac{1}{pi}2sqrt{1-x^2}$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    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




















                    • @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




















                     

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