# Uniform distribution on the surface of unit sphere

It is known that given $X=(X_1, X_2, \ldots, X_n)$ iid $\sim N(0,1)$, then $X/\sqrt{X_1^2+\cdots+X_n^2}$ is uniformly distributed on the surface of unit sphere.

Intuitively, I know that that's because the probability of $X/\sqrt{X_1^2+\cdots+X_n^2}$ belonging to any region with the same area on the surface should be the same. But how can I prove it mathematically?

• See the references in the article. – user64494 Jul 16 '13 at 5:30
• @user64494 i saw that page before but the references there only mentioned the method, not a detailed proof. – Julie Jul 16 '13 at 12:35
• Have you tried to use spherical coordinates? – Davide Giraudo Jul 16 '13 at 17:20
• – cardinal Jul 16 '13 at 23:25
• @cardinal Thanks a lot! – Julie Jul 17 '13 at 3:14

Suppose $$X_1$$ and $$X=(X_1, X_2, \ldots, X_n)$$ iid $$\sim N(0,1)$$, then $$X \sim N(0, I_n)$$, where $$N(0, I_n)$$ is the multivariate normal distribution with zero-mean and identity covariance matrix. From that it follows, that if $$O$$ is an orthogonal matrix, that $$OX$$ is identically distributed with $$X$$. From that it follows, that $$Y = \frac{X}{||X||_2}$$ is identically distributed with $$\frac{OX}{||OX||_2} = \frac{OX}{||X||_2}$$. From that we can conclude, that $$Y$$ is invariant under rotations and belongs to the unit sphere. There is only one probability distribution, that satisfies both those conditions at the same time: that is the uniform distribution on the unit sphere.