Yes, I meant to write 'standard Gaussian'. Here's another attempt at what I was trying to say, hopefully clearer: If X and Y are iid random variables such that every combination cos(theta) X + sin(theta) Y has the same distribution as X (or Y), then X = N(0, sigma^2) is a mean-zero Gaussian distribution. The linear combination property you mention is actually implied by this (because you can find k and theta such that a_1 = k cos(theta) and a_2 = k sin(theta)), but is strictly weaker; every distribution in the Levy stable family has this: https://en.wikipedia.org/wiki/Stable_distribution whereas only the Gaussian distribution has an axisymmetric product distribution. Best wishes, Adam P. Goucher
Sent: Thursday, March 14, 2019 at 1:43 AM 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, a drunkard's walkthrough
I suspect X and Y here are random variables.
pedantic interlude: ----- The word Gaussian applies to changes in both scale and location (standard deviation and mean) to the standard N(0,1) density d(x) = sqrt(1/2π) exp(1x^2/2), or in other words applying an affine linear function f(x) = ax + b to a standard Gaussian random variable X:
Y = aX + b
for any real numbers a ≠ 0 and b.) -----
So in fact an "orthogonal linear combination" of independent Gaussians like what Adam suggests will indeed result in a Gaussian. If the original ones are standard, then this careful linear combination will also be standard.
(But it's also true that any affine linear combination of arbitrary Gaussian random variables, X_1 and X_2, like
Y = a_1 X_1 + a_2 X_2 + b
will result in Y's being Gaussian (unless it's degenerate).
—Dan
Adam Goucher écrit: ----- If I'm not mistaken, the Gaussian distribution is unique (up to scale) in that if X and Y are *independently* Gaussian, then every combination cos(theta) X + sin(theta) Y is also Gaussian. (Equivalently,, it's the only axisymmetric product distribution.) -----
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