I have the following 3-variate random vector $(X,Y,Z)$ which is distributed as a Gaussian mixture: (with some abuse of notation) $$ f(X,Y,Z)=\underbrace{w_a \mathcal{N}(\mu_a, \Sigma_a)}_{\text{component (a)}}+\underbrace{w_b \mathcal{N}(\mu_b, \Sigma_b)}_{\text{component (b)}} $$ where $f$ denotes the PDF, $(w_a, w_b)$ are the mixture weights, $\mathcal{N}(\mu, \Sigma)$ stays for 3-variate normal distribution with mean $\mu$ and var-cov matrix $\Sigma$.

Suppose that $(X,Y,Z)$ are mutually independent.

I want to understand under which conditions on $(\mu_a,\mu_b)$ $Y-X$ is symmetric around zero.

My attempt (wrong)

Let $\mu_a\equiv (\mu_{a,x},\mu_{a,y},\mu_{a,z})$, $\mu_b\equiv (\mu_{b,x},\mu_{b,y},\mu_{b,z})$, $\Sigma_a\equiv \begin{pmatrix} \sigma^2_{a,x} & 0 & 0\\ 0 & \sigma^2_{a,y} & 0\\ 0 & 0 & \sigma^2_{a,z}\\ \end{pmatrix}$, and $\Sigma_b\equiv \begin{pmatrix} \sigma^2_{b,x} & 0 & 0\\ 0 & \sigma^2_{b,y} & 0\\ 0 & 0 & \sigma^2_{b,z}\\ \end{pmatrix}$

My initial thought was that $$\mu_{a,x}=\mu_{a,y}=\mu_{b,x}=\mu_{b,y}$$ is necessary and sufficient for the distribution of $Y-X$ to be symmetric wrto zero. Let me explain how I arrived to this conclusion.

Following the proof here, $$ f(Y-X)=w_a^2 [\mathcal{N}(\mu_{a,y}, \sigma^2_{a,y})-\mathcal{N}(\mu_{a,x},\sigma^2_{a,x})+w_aw_b [\mathcal{N}(\mu_{a,y}, \sigma^2_{a,y})-\mathcal{N}(\mu_{b,x},\sigma^2_{b,x})]+w_bw_a[\mathcal{N}(\mu_{b,y}, \sigma^2_{b,y})-\mathcal{N}(\mu_{a,x},\sigma^2_{a,x})]+w_b^2[\mathcal{N}(\mu_{b,y}, \sigma^2_{b,y})-\mathcal{N}(\mu_{b,x},\sigma^2_{b,x})] $$

The four components are symmetric wrto zero if and only if the have the mean at zero, which in turn happens if and only if $$\mu_{a,x}=\mu_{a,y}=\mu_{b,x}=\mu_{b,y}$$

Now, this conclusion seems to be contradicted in simulations. For those of you using Matlab, below I consider the case $[\mu_{a,x}=\mu_{a,y}=\mu_{a,z}]\neq [\mu_{b,x}=\mu_{b,y}=\mu_{b,z}]$

rng default

mu_a = [10^6 10^6];
sigma_a = [0.001 0; 0 0.002];
mu_b =[20^6 20^6];
sigma_b= [0.005 0; 0 0.004];
MU = [mu_a;mu_b];
SIGMA = cat(3,sigma_a,sigma_b);
w = [1/3 2/3]; %equal weight 0.5
obj = gmdistribution(MU,SIGMA,w);
values = random(obj,N); 


hold on
hold off
legend('CDF of D', 'CDF of -D')

This is the picture I get

enter image description here

The two CDFs coincide, hence suggesting that $Y-X$ has a distribution symmetric around zero even if $\mu_a\neq \mu_b$. What am I doing wrong?

  • $\begingroup$ Because symmetry around $0$ implies a mean of $0,$ clearly $w_a\mu_a + w_b\mu_b=0.$ But this does not suffice, as you can tell by considering the case $\mu_a = -\mu_b \ne 0$ with different covariance matrices in each component. It should be easy to show that except in special cases, $w_a=w_b,$ so you might want to start there. $\endgroup$
    – whuber
    Nov 1, 2018 at 15:48
  • $\begingroup$ @whuber Thanks. I'm still lost though: 1) from my simulations it seems that as long as $\mu_{a,x}=\mu_{a,y}$ and $\mu_{b,x}=\mu_{b,y}$ (whatever $w_a, w_b, \Sigma_a, \Sigma_b$) we have that $Y-X$ is symmetric wrto zero. 2) This clashes with my proof. 3) Your hint I guess is referred to another way to proceed but I can't see exactly what I should so (sorry I'm just a beginner). $\endgroup$
    – Star
    Nov 1, 2018 at 15:56
  • $\begingroup$ Consider very large values of the mean and small variances: the mixture will have almost all its probability located in one direction far from zero. Obviously your conclusion will be wrong in that case. Since you're using a capable computational and visualization environment, why not plot some of these distributions? Contour plots of the densities ought to make everything clear. BTW, there's little advantage to assuming the covariance matrices are diagonal. Work with the two-dimensional analog of the question, of course! $\endgroup$
    – whuber
    Nov 1, 2018 at 16:01
  • $\begingroup$ @whuber Even with $\mu_a, \mu_b$ very large and variances very small, I still get the the CDFs of $Y-X$ and $X-Y$ coincide (I have added the picture to my question). $\endgroup$
    – Star
    Nov 1, 2018 at 16:11
  • 1
    $\begingroup$ I see that now--I had supposed you were concerned about the symmetry of $f.$ I apologize for going in the wrong direction. You can work out the univariate distribution of $Y-X$ (it's a mixture of at most four distinct Normal distributions) and go on from there. $\endgroup$
    – whuber
    Nov 1, 2018 at 16:24


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.