Re: Uniform distribution on a hyperannulus

Ray Koopman wrote:
Let u_1,...,u_n be independent Uniform[0,1].

z_i = Z(u_i), i = 1,...,n, where Z is the standard normal inverse cdf.

y = sum_i z_i^2.

r = (F_n(y)*(r_max^n - r_min^n) + r_min^n)^(1/n), where F_n

is the cdf of a chi-square variable with n degrees of freedom.

[x_1,...,x_n] = [z_1,...,z_n]*r/sqrt(y)

are the coordinates of a point whose distribution is uniform on the
hyperannulus.

Thank you, Ray! That is exactly what I was searching for.

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http://kaba.hilvi.org
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