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I was reading about generating random unit vectors for n-dimensional space.

I read one method in which the person suggested that we can take n-random samples from standard normal distribution and if we normalize it then we will get a random vector. The reason he gave was that Gaussian distribution is spherically symmetric.

Then I thought of another method , If we pick values from uniform distribution from $-\infty$ to $+\infty$ and then also after normalizing it, We will get a random vector. Since,probability for each point to be getting selected is same then probability for each random vector will be also same.

So ,I was wondering whether what I thought is correct or not? and also what are all distributions for which this will work and if normal distribution is the only distribution for which it will work then what is the reason behind it?

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    $\begingroup$ According to Mark Meckes's comment on MathOverflow, there are no other distributions that will work: "The normal distribution in $\mathbb R^n$ is the unique (up to scaling) rotation-invariant probability measure with independent components." He cites The Normal Distribution: Characterizations with Applications by Bryc for the proof. $\endgroup$ – user856 Aug 12 '16 at 17:39
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Partial answer. There is no uniform distribution from $-\infty$ to $\infty$; thus, what you suggest cannot be implemented (at least not as you describe). The Gaussian will work.

I don't know the conditions on a distribution such that the joint distribution of any number of i.i.d. random variables with that distribution will be spherically (hyperspherically, etc.) symmetric. I'll look into it until someone posts another answer...

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    $\begingroup$ I see Rahul's comment answers in the negative... $\endgroup$ – Brian Tung Aug 12 '16 at 19:10
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Your proposed method clearly will not work, not even in the two-dimensional case, never mind the absurdity of a uniform distribution on an interval of infinite length: suppose $X$ and $Y$ are IID uniform variables. Then $(X,Y)$ will have support on some square; the projection of that square to any circle in the plane will not be uniform on the circumference of that circle.


More concretely, suppose $X, Y \sim \operatorname{Uniform}(-1,1)$. Then the vector-valued random variable $$Z = \frac{(X,Y)}{\sqrt{X^2 + Y^2}}$$ or its angle argument $$\Theta = \tan^{-1} \frac{Y}{X}, \quad -\pi < \Theta \le \pi$$ will have greater density for angles that are odd multiples of $\pi/4$ compared to angles that are even multiples of $\pi/4$. Even if you generalized this to uniform distributions with very large support; e.g., $X, Y \sim \operatorname{Uniform}(-M, M)$ with $M$ some very large number, you can see that this does not in any way rectify the non-uniformity of the density of the angle of the vector $(X,Y)$.

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  • $\begingroup$ Yes, but I was thinking that If I could approximate it to infinity then it would become circle , but I think that is not possible anyways ,but could you add some information on other questions I am stuck with. Like ,Why does it work for normal distribution and what are others for which it will work. $\endgroup$ – Shubham Ugare Aug 12 '16 at 16:26
  • $\begingroup$ @ShubhamUgare: If you are saying that an infinite square covers the whole of $\mathbb{R}^2$ then there is no uniform distribution. If you take the infinite square as the limit of a large square, then a square remains a square no matter how much it is scaled, and does not approach a circle. $\endgroup$ – Henry Aug 12 '16 at 19:00
  • $\begingroup$ Yes, I got that. $\endgroup$ – Shubham Ugare Aug 12 '16 at 19:07

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