Or CLT and Walsh Hadamard transform. Note CLT applies as equally to differences as additions.

https://archive.org/details/bitsavers_mitreESDTe69266ANewMethodofGeneratingGaussianRando_2706065 ]]>

Greg – Neat generalization.

]]>…and somehow, this is really all about biology!

https://johncarlosbaez.wordpress.com/2014/01/22/relative-entropy-in-evolutionary-dynamics/

Thinking about this a bit more I suspected that there would be a relationship between Dirichlet distributions on the unit simplex and uniform distributions on spheres. Google found this

for me, and Lemma 2.3 is a more general version of what I guessed. Those Rademacher distributed epsilons sound scary but they are just random + or – signs. The case p=2 is the one we’re talking about here.

Sorry for doubting you about the Beta distribution.

Squared distances are what I use in Part 1 and also Random points on a group. Basically, polynomials are good but square roots suck—a lesson I always emphasize in my calculus classes!

]]>I think everything is easier if you look at squared Euclidean distances. For example: Consider one point at (1,0,0…0), and a random point at (x1,x2,…xn). Let x = x1 and D be the sum of squares of x2,…xn. The squared distance is (1-x)^2 + D = 1-2x+x^2+D = 2-2x. The squared distance to the ‘mirror’ point (-x1,x2,…xn) is (1+x)^2 + D = 2+2x. So the squared distance has a symmetric distribution in [0,4]. In particular its mean is 2 for all dimensions. (Greg’s formula shows this too of course.) As for the factors of 4, just look at spheres with diameter 1 instead of radius 1!

]]>The th moment of the (random) area of the triangle whose vertices are three independent, uniformly distributed random points on the unit circle appears to be . Can anyone prove this? Better yet, can anyone give a conceptual explanation for why this moment should be rational? (If this observation is not new, references would be appreciated.)

Iosif Pinelis said that for every natural number the th moment is

]]>Grahamâ€™s right, you can tweak the parameter to account for the derivative and match the probability distribution, so the relevant Beta distribution becomes .

There is one remaining catch, though: the change of variables means that the original moments now come from powers, not of , but of .

So the moments of these Beta distributions need to be multiplied by powers of 4 to make them integral. For example, to get the Catalan numbers, you need to multiply the mth moments of the Beta(3/2,3/2) distribution by .

]]>so it still has the form of a beta distribution. No?

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