Note that the probability density function on R^2 where each coordinate is independently uniform [0, 1] is shaped like a box. As such, it seems reasonable to refer to the shape of the probability density function of a bivariate standard Gaussian as a 'muller', so that the Box-Muller transform lives up to its name. -- APG.
Sent: Sunday, March 10, 2019 at 7:18 PM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Statistics for a class of random solids
You can certainly use the "rejection" method for picking points uniformly distributed on a sphere by first choosing n+1 real numbers x_0, ..., x_n in [-1, 1] and using only those sets of points with x_0^2 + ... + x_n^2 <= 1. But as the dimension increases, the fraction rejected becomes very high very fast.
A better method is based on a smart way to choose points from the standard normal distribution in R^2:
1. Pick x, y from the uniform distribution on [0, 1].
2. Let
A = sqrt(-2 ln(x)) cos(2πy)
B = sqrt(-2 ln(x)) sin(2πy).
Then A and B are independent standard normal variates.*
3. Repeating this floor[n/2] times gives the coordinates of a standard normal point in R^n. Dividing by its length gives a uniformly distributed point on the unit sphere S^(n-1).
—Dan ————— * For more on this see <https://en.wikipedia.org/wiki/Box-Muller_transform>
Allan Wechsler: ----- That means that, to answer another question somebody asked here recently, you can produce points uniformly distributed on a sphere by generating N Gaussian coordinates and normalizing the resulting vector. -----
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