I need to complete the following problem:

Let $~(X_1, X_2, X_3)~$ be a continuous random vector, with uniform density on the unit sphere $~\{(x_1,~ x_2,~ x_3) ~∈~ \mathbb R^3 ~:~ x_1^2 + x_2^2 + x_3^2 = 1\}~$. What would be the marginal pdf of $~X_1~$?

I have to follow the next steps:

  1. Express the density in spherical coordinates.

  2. Think about the symmetry of the problem. Think about spherical coordinates. Then think how can you integrate the effects of two variables while keeping the other fixed in spherical coordinates.

  3. After 2, you will have the marginal in one angle. Then you need to do a transformation of random variables ("a change of variables in the integral of the density so that the limits become (-1, 1)") to transform the angle to the Cartesian coordinate you are interested in obtaining the marginal.

  4. The marginal pdf of X1 should be one that is intuitive for this problem (yet that intuition will fail if the sphere was a circle or a high-dimensional sphere!).

I have completed the first step by:

$$f(\theta, \varphi) = {\sin(\varphi)\over4\pi}$$

where $\theta \in [0,2\pi]$ and $\varphi \in [0,\pi]$

I'm not sure if the second step is correct by: $$f\varphi(\varphi) = \int_0^{\pi} {\sin(\varphi)\over4\pi}\,d\varphi= {1\over 2\pi}$$ Or the second option: $$f\theta(\theta) = \int_0^{2\pi} {\sin(\varphi)\over4\pi}\,d\theta= {\frac12\sin(\varphi)}$$


1 Answer 1


My take on the problem is that you are supposed to let $x_1=\cos\phi$, $x_2=\sin\phi\cos\theta$, and $x_3=\sin\phi\sin\theta$. Then looking at the integral $$\int_0^{\pi}f(\phi)d\phi=\int_0^{\pi}\frac12\sin\phi\,d\phi=\int_1^{-1}\frac12(-dx_1)=\int_{-1}^1\frac12dx_1=\int_{-1}^1f(x_1)dx_1$$ So that $f(x_1)=\frac12$, i.e. uniformly distributed with respect to $x_1$. In $n$ dimensions the "circumference" of the "circle" at a given $x_1$ is $K_{n-1}(1-x_1^2)^{\frac{n-2}2}$ and the arc length along the surface from $x_1$ to $x_1+dx_1$ is $\frac{dx_1}{\sqrt{1-x_1^2}}$ so $$f(x_1)dx_1=C_n(1-x_1^2)^{\frac{n-3}2}dx_1=\frac{\operatorname{\Gamma}\left(\frac n2\right)(1-x_1^2)^{\frac{n-3}2}}{\sqrt{\pi}\operatorname{\Gamma}\left(\frac{n-1}2\right)}dx_1$$ So you can see that only in $3$ dimensions is it uniform.


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