# Picking a random point on the surface of a unit sphere

It is mentioned in many places (e.g., in Wikipedia) that if $X_1, X_2,\dots, X_n$ are independent standard Gaussian random variables, then the vector $$Y:=\frac{1}{\sqrt {X_1^2+X^2_2+\dots+X_n^2}} (X_1,X_2,\dots X_n)$$ is uniformly distributed on the unit sphere.

How can I prove this result?

Thanks a lot!

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Look at the joint distribution of $(X_1,X_2,\ldots,X_n)$. Note that this distribution is spherically symmetric. The proposition follows immediately from this observation. –  Will Nelson Feb 26 '14 at 6:12
@WillNelson Thanks for your comment. I agree that the distribution of this vector is very symmetric. Could you please add some more details on how this fact implies the proposition? Thanks! –  Oleg Feb 26 '14 at 10:26
Since the distribution is spherically symmetric, the probability density for any two directions must be the same. To see this, just think what "spherically symmetric" means. It means that if you rotate the distribution by any rotation, the distribution remains unchanged. Doesn't that mean the probability density for any two directions must be exactly the same, given that one can find a rotation that transforms one direction into the other? –  Will Nelson Feb 26 '14 at 10:58