Noise samples of gaussian mixture distribution

Prashant Kumar answered . 2024-12-21 17:07:30

PDF = sqrt((u.^3)./pi).*exp(-u.*(x.^2)) + sqrt((1-u)^3/pi).*exp(-(1-u).*x.^2)

So we have two terms there.

syms u positive
syms x real

Look at the first term:

pdf1 = sqrt((u.^3)./pi).*exp(-u.*(x.^2));
int(pdf1,x,[-inf,inf])
ans =
u

So, it integrates to u. The second term is similar,

pdf2 = sqrt(((1-u).^3)./pi).*exp(-(1-u).*(x.^2));
int(pdf2,x,[-inf,inf])
ans =
piecewise([u == 1, 0], [1 < u, Inf*1i], [u < 1, 1 - u])

N = 10000000;
u = 0.01;

% random mixture selection
% Here s will be 1 if the element is sampled from first pdf.
%      s will be 0 if the element is sampled from second pdf.
s = rand(1,N) < u;

% sample using randn, with a unit variance initially
x = randn(1,N);;

% now scale each element based on the desired variance
x(s) = x(s)*sqrt(2/u);
x(~s) = x(~s)*sqrt(2/(1-u));

hist(x,1000)

pdf1 = @(x,u) sqrt((u.^3)./pi).*exp(-u.*(x.^2));
pdf2 = @(x,u) sqrt(((1-u).^3)./pi).*exp(-(1-u));

h1 = ezplot(@(x) pdf1(x,0.25),[-10,10]);
hold on
h2 = ezplot(@(x) pdf2(x,0.25),[-10,10]);
title 'pdf1 and pdf2, split apart'

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