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Given two uniformly random directions on a hemisphere, n0 and n1, is the normalized sum of these vectors also a uniformly random direction on the same hemisphere?

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up vote 4 down vote accepted

No - average directions away from the edge are more likely than average directions near the edge of the hemisphere

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@Henry gave you a good answer, I just wanted to supplement it with some visuals. Using normalized standard Gaussian 3D vector to produce points, uniformly distributed on $S^2$ (see near the end of this article on MathWorld) in Mathematica:

enter image description here

The above visualizes distribution for the normalized sum of $n=1$, $n=2$, $n=4$ and $n=8$ vectors uniformly distributed on a hemisphere. Computation of the probability for the $z$-component of such a random point to be above $1/2$ then follows and shows greater concentration of points near then north pole as $n$ increases.

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Here is a suggestion: add a graphics for the sum of, say, 4 or 5 uniform unit vectors, the concentration towards the pole should be more visible. – Did May 24 '12 at 18:52
@Dider Thanks for the suggestion. I have just updated the graphics. – Sasha May 24 '12 at 19:03
Magnifique. :-) – Did May 24 '12 at 19:07

If you roll one die, the data is uniformly distributed. How about with 2 dice? It's not uniform anymore, because options closer to the centre of the data can be reached in more ways. Convolution is your friend.

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Yes, but in some cases the sum is uniform. If n0 and n1 had been on a sphere their normalized sum would have been uniformly distributed. – Andreas Brinck May 24 '12 at 18:59
Yes, but the case of the complete sphere amounts to considering the sum of the two dice modulo 6 (which is indeed uniform). – Did May 24 '12 at 19:09
I agree that the sum can be uniform in some cases, but I that's why I didn't draw the analogy with them. I was hoping that the bit about reaching some data in more ways would give the intuition as to the reason. – Robert Mastragostino May 25 '12 at 1:58

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