# Uniform distribution on unit disk

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?

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

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

• @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