Noise samples of gaussian mixture distribution

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Harsha Vats - 2022-05-23T13:24:13+00:00
Question: Noise samples of gaussian mixture distribution

I'm new to matlab. I want to generate noise samples of Gaussian mixture with PDF= sqrt((u.^3)./pi).*exp(-u.*(x.^2)) + sqrt((1-u)^3/pi).*exp(-(1-u).*x.^2);   The length of noise sample is (1,200) Please help me out.

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    • Expert Answer

    Profile picture of Prashant Kumar Prashant Kumar answered . 2025-11-20

    Ok, so you have formulated what seems to me to be a rather strange mixture model.
     
    You have two modes, with variances that are related to each other, the mean of both terms is zero.
     
    So, if the PDF is:
     
    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])
    For u <= 1, it integrates to 1-u.
     
    So the integral of the sum is indeed 1, therefore it is a valid PDF.
     
    Essentially, you have two Gaussian modes, such that the variances are 2/u and 2/(1-u) respectively. You have chosen the mixture parameter as the inverse of those variances. Effectively, when u is small (or very near 1), you have a mixture distribution that rarely, you will generate large outliers. When u is exactly 1/2, this reduces to a standard normal distribution, with a unit variance.
     
    I'm not sure why, but I suppose that is not really important. Normally, one would choose a mixture parameter that is independent of the variances.
     
    The solution is simple.
     
    1. Pick some value for u. u determines both variances for each Gaussian mode, as well as the mixture coefficient.
    2. For a sample size of N, pick N uniformly distributed random numbers. These will determine which mode you will sample from. Then just use a classical tool like randn to do the sampling.
     
    For a sample size of 1e7, and some arbitrary value for u, say 1/100. I've picked a tiny value for u to make the result clear. I've also chosen a large sample to make it easier to see as a histogram.
     
     
    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)
    So, for this fairly tiny value of u, see that the histogram looks much like a traditional Gaussian, BUT with very wide tails.
    I'll next plot the separate pieces of the PDF, split into two parts. Here I've chosen u as 0.25.
     
    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'
    So, the sampling was rather simple. No need to do anything strenuous. I still have no idea why you want this distribution, but whatever floats your boat is a good thing.


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