I'm asked to find the marginal distribution of $(X,Y)$ as $(X,Y,Z)$ is a point chosen uniformly on the unit sphere.
I've worked out that the joint density function $f_{XYZ}(x,y,z) = \frac{3}{4\pi}$ for $x^2+y^2+z^2 \leq 1.$
I know how to find the marginal distribution when dealing with 2-dimensions, but I'm not sure how I should set up the integrals in this case.
Would this be the correct way to go,
$$F_{X,Y}(x,y) = \int_{-\infty}^x \int_{-\infty}^y \int_{z} f_{XYZ}(x,y,z) dzdydx,$$
or am I really computing the marginal distribution of $F_X(x)$ if I do that?
Also, I'm having some problems understanding the difference between finding the density function and distribution function in cases like this. To me, it seems to be the same method for both functions?
Best Answer
A uniform distribution on the sphere does not have a density function in three variables, but the marginal distribution for two of the three variables does have a density. It is obtained by expressing the area element of the sphere $\sin\theta d\varphi\wedge d\theta$ in new coordinates $x$ and $y$ and then normalizing.
$$\eqalign{ \sin^2\theta&=x^2+y^2\\ 2\sin\theta\cos\theta d\theta&=2xdx+2ydy\\ \sin\theta d\theta&=\frac{xdx+ydy}{\sqrt{1-x^2-y^2}}\\ \varphi&=\arctan\left(\frac y x\right)\\ d\varphi&=\frac1{1+\frac{y^2}{x^2}}\left(\frac1xdy-\frac{y}{x^2}dx\right)\\ \sin\theta d\varphi\wedge d\theta&=\frac1{\sqrt{1-x^2-y^2}}dx\wedge dy\\ }$$
This gives the density function
$$\frac1{2\pi\sqrt{1-x^2-y^2}}$$
on the disk $x^2+y^2\leq1.$