The accepted answer here mentions that to sample points from the surface of a hypersphere uniformly, one can generate gaussians and normalize them. Many commentators also say that this is possible only with the normal distribution. My intuition says that any independent distributions should be sufficient. For example, generate n uncorrelated uniforms and divide by their magnitude.

But apparently, we can use only normals for this since it is the only rotationally invariant distribution. How do we prove that normals are the only way to sample uniformly from the surface of a hyper sphere? And if I use uniforms instead, what kind of sampling do I end up with?

EDIT: I realize that the distribution must be radially symmetric (treat all points on the sphere equally). Also, the gaussian is the only radially symmetric and separable function. But there are functions that are radially symmetric but not separable. For example, $e^{e^{-x}}$. Why can't I use a radially symmetric, non-separable function?


I apologize in advance for what is not a complete answer to your question. This might be better suited as a comment, but it certainly won't fit in that space. At any rate, it is at least a partial answer to your question.

The uniform distribution doesn't work in the fashion you described because, simply put, it doesn't place the uniform measure on the sphere. As a small, visualizable example, consider the two-dimensional case:

enter image description here

Choosing a point $(X,Y)$ with $X, Y$ uniformly distributed on $[-1, 1]$ is equivalent to randomly selected a uniformly-distributed point from anywhere in the blue square above. With probability $1$, it will not be at the origin, so we can divide by its modulus to project it to the red circle, as you described. However, this distribution will not be uniform.

To see this, express the circle in angular coordinates and consider the relative probabilities that the point will satisfy $\theta \in [-\pi/8, \pi/8]$ (i.e. between the green and red lines on the right side) and $\theta \in [\pi/8, 3 \pi/8]$ (between the red and purple lines in quadrant I). These events will occur iff the point selected from the square is also in those corresponding regions. It is easy to see without calculation (and not terribly difficult to verify with a calculation) that there is more area in the square between the red and purple lines than there is between the red and green lines. Hence, your point chosen in this fashion is more likely to land in the interval $\theta \in [\pi / 8, 3 \pi /8]$ than it is to land in $\theta \in [-\pi / 8, \pi / 8]$. This is problematic because those two intervals "should" have the same probabilities.

Obviously, I'm leaving unsaid the real answer to your question, which is why the normal distribution is the only one that works in the proper way. I'm not able to prove that off the top of my head, and it's the summer and I have a small baby at home, so the top of my head is the most I can give at this time.

  • $\begingroup$ Thanks, this is at least partly there, +1. $\endgroup$ – Rohit Pandey Jun 24 '18 at 20:34
  • $\begingroup$ One rough reason I can think of looking at your picture is that circles and spheres are quadratic and normal is one way (only way?) to convert a quadratic form into a distribution. $\endgroup$ – Rohit Pandey Jun 24 '18 at 20:36

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