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?

  • $\begingroup$ You're doing the right job. The distribution of the random variable $(X,Y)$ (which is a marginal in this case) must be a function of two real variables, say $(x,y)$; you must therefore "zap" the $z$ from $f_{XYZ}$ by averaging it out. So marginal density is $$f_{XY}(x,y)=\int_{\mathbb R}f_{XYZ}(x,y,z)\operatorname dz,$$ whence (upon averagin in $x$ and $y$) your formula for $(X,Y)$'s distribution. $\endgroup$ Jan 25, 2016 at 9:27
  • $\begingroup$ Note that "=" should read "$\leq$" in $x^2+y^2+z^2\leq1$. I advise newbies to learn the Iverson--Knuth brackets that make the algebra much easier for this type of exercises. $\endgroup$ Jan 25, 2016 at 9:33
  • $\begingroup$ About the "=", you are of course right! Will edit. But if I would like to find $f_X(x)$, wouldn't I essentially compute the exact same thing? And is the density function and the distribution function the same thing in this case? $\endgroup$ Jan 25, 2016 at 9:45
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    $\begingroup$ There is a big difference between a ball ($\leq1$) and a sphere ($=1$). If the original question is about the unit sphere then there cannot exist a 3-dimensional density function. $\endgroup$ Jan 25, 2016 at 9:47
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    $\begingroup$ Yes you follow the formula to integrate and obtain the joint pdf of $(X, Y)$: $\displaystyle \int_{-\sqrt{1-x^2-y^2}}^{\sqrt{1-x^2-y^2}} \frac {3} {4\pi} dz = \frac {3} {2\pi} \sqrt{1 - x^2 - y^2}, x^2 + y^2 \leq 1$ Note that they are not uniformly distributed inside the unit disk. $\endgroup$
    – BGM
    Jan 25, 2016 at 10:52

2 Answers 2


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


on the disk $x^2+y^2\leq1.$

  • $\begingroup$ I am not sure I follow, would you like to specify how you reached that result? $\endgroup$ Jan 25, 2016 at 12:06
  • $\begingroup$ There was an error in my original volume element (missing $\sin\theta$); does this one look better? $\endgroup$ Jan 25, 2016 at 12:23
  • $\begingroup$ That was very helpful, thank you! $\endgroup$ Jan 25, 2016 at 12:30
  • $\begingroup$ @Justpassingby Very impressive! Could you give a reference which presents the proof details? In multivariate statistics, we know that if $x$ has a density, then any marginal distribution of $x$ has a density. However, the fact that any marginal distribution of $x$ has a density can not guarantee the density existence of $x$; uniform distribution in the surface of the unit is such a case (counterexample), but I don't know how to proof it in details? $\endgroup$
    – John Stone
    Mar 9 at 3:55

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 $$ \int_{\mathbb R} f^h_{XYZ}(x,y,z)\operatorname dz = \frac{2h+o(h)}{\sqrt{1-x^2-y^2}(4\pi h+\pi h^3/3)} = \frac{1+o(h^0)}{\sqrt{1-x^2-y^2}(2\pi+\pi h^2/6)} \to \frac1{2\pi\sqrt{1-x^2-y^2}} ,\text{ as }h\to0 . $$


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