How to find the pdf of difference of r.v How do I calculate the pdf for the following case?  In general, if we have 2 r.v. $x,y$ which are normal, then the pdf of the difference of 2 r.v. which are Gaussian will also be Gaussian, I think with mean $\mu_Z = \mu_x - \mu_y$ and variance $\sigma^2_Z = \sigma^2_x + \sigma^2_y$.
Based on this premise, how to find the pdf from a Gaussian Mixture model (GMM). The time series $Z$ has the pdf $f_Z$ which is GMM distribution. The time series contains 2 r.v $x,y$.  So, both the r.v. together constitute a GMM. Considering that there are only 2 mixtures.
I have observations of multivariate time series $Z_i = {[x_i,y_i]}_{i=1}^n$ where $x,y$ are the random variables. The pdf of $Z$ is Gaussian mixture model (GMM). The parameters of the GMM model are learnt through Expectation Maximization.   How to get the functional form for the pdf $f(d_i) = f(x_i-y_i)$ where $d_ i = x_i-y_i$. Thank you for help.


How do I calculate the pdf for the following case? The pdf of the difference of 2 r.v. which are Gaussian will also be Gaussian, I think with mean $\mu_D =\mu_X −\mu_Y$   and variance $\sigma^2_D =\sigma^2_X +\sigma^2_Y\;$.

Yes.   Now you have a Gaussian random variable with given mean and variance.
$$D\sim \mathcal{N}(\mu_X −\mu_Y, \sigma_X^2+\sigma_Y^2)$$
So the pmf of that Normal Distribution is: $$\color{blue}{\boxed{\color{black}{f_D(d)=\dfrac {\mathsf {\large e}^{\displaystyle-(d-\mu_D+\mu_D)/(2\sigma_X^2+2\sigma_Y^2)}}{\sqrt{2\pi(\sigma_X^2+\sigma_Y^2)}}}}}$$

By substitution into:  $Z\sim\mathcal{N}(\mu_D, \sigma_D) \iff f_D(d)=\dfrac{\mathsf e^{-(d-\mu_D)/2\sigma_D^2}}{\sigma_D\sqrt{2\pi}}$
