Finding marginal distribution, unit sphere

Question

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?

Answer 1

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 $x^2+y^2\leq1.$

Answer 2

An alternative solution (eschewing spherical coordinates) may be based on the approximation of the surface density by a bulk one as follows. Let $h>0$ (destined to go to zero). Consider the spherical shell $\Sigma_h$ of thickness $h$ about the sphere of radius $1$, i.e., the union of all the spheres of center $0$ and radius ranging between $1-h/2$ and $1+h/2$. Since this is a $3$-dimensional volume you may consider on it a uniform distribution (constant function) $f_{XYZ}^h\equiv c(h)$, with $c(h)$ constant in $x,y,z$ (but depending on $h$) chosen such that the integral of $f_{XYZ}^h$ is $1$. So $c(h)=1/\operatorname{vol}\Sigma_h$, which the following calculation reveals: $$ \operatorname{vol}\Sigma_h = \int_{1-h/2}^{1+h/2}\int_{\partial B_0(\rho)}\operatorname d S\operatorname d\rho = \int_{1-h/2}^{1+h/2}4\pi\rho^2\operatorname d\rho = \frac{4\pi}3\Big(3h+\frac{h^3}4\Big) = 4\pi h+\frac\pi3h^3 . $$ As $h\to0$ the mass of the shell stays $1$, $c(h)\approx1/(4\pi{h}\to\infty$ and the $f_{XYZ}^h$ converges (weakly, or in the sense of distributions, or whatever your mathematical religion prescribes, but intuition is what matters) to $f_{XYZ}$. Thanks to the integral being bounded to compute $f_{XY}$ it is legitimate to interchange limits (here the theory of distribution does come handy, but you can use geometric measure theory,if you prefer) to integrate $f^h_{XYZ}$ in $z$ first and then take $h\to0$. The integration in $z$ rests on some simple geometry, for all $x,y$ such that $x^2+y^2<(1-h/2)^2$, the line passing through $(x,y)$ intersect the shell twice symmetrically about the $(x,y)$-plane. Looking at the upper half (and then doubling the result) the length of the segment is given by $S^+_h(x,y)-S^-_h(x,y)$ where $S_h^\pm$ are functions whose graphs are the outer and inner boundaries of the (upper half of the spherical) shell, respectively. Explicitly $$S^\pm_h(x,y)=\sqrt{(1\pm h/2)^2-x^2-y^2}=:F(\pm h).$$ The auxiliary function $F$'s Taylor expansion of order $1$ about $0$ yields $$ S^+_h(x,y)-S^-_h(x,y) = 2F'(0)h+o(h) = \frac{h+o(h)}{\sqrt{1-x^2-y^2}} , $$ since $$F'(\xi)=\frac{1+\xi/2}{2\sqrt{(1+\xi/2)^2-x^2-y^2}}.$$ Multiplying this by $2$ (to take into account the lower half of the sphere) and the density normalizing constant $c(h)$ we obtain $$ \int_{\mathbb R} f^h_{XYZ}(x,y,z)\operatorname dz = \frac{2h+o(h)}{\sqrt{1-x^2-y^2}(4\pi h+\pi h^3/3)} = \frac{1+o(h^0)}{\sqrt{1-x^2-y^2}(2\pi+\pi h^2/6)} \to \frac1{2\pi\sqrt{1-x^2-y^2}} ,\text{ as }h\to0 . $$