So, have a system that estimates a covariance matrix. The scenario occurs where at time step $t$, I will either have a Gaussian with covariance matrix $\Sigma_A$ with probability $p$ or $\Sigma_B$ with probability $1-p$. The mean is the same in both cases.
This scenario seems odd, but essentially, I have an estimator, and I am predicting if observation will be made or not. I know the probability $p$ of such an observation occurring.
Is this simply just $\Sigma_t = p\Sigma_A + (1-p)\Sigma_B$?
One interpretation is $$X=Y\Sigma^{1/2}_AZ+(1-Y)\Sigma^{1/2}_BZ,\quad (Y,Z)\sim \mathrm{Bern}(p)\otimes\mathcal N(0,I_p)$$Hence
$$E(\exp(i\langle u, X\rangle)\mid Y) = \exp(-\langle (Y\Sigma^{1/2}_A+(1-Y)\Sigma^{1/2}_B)^2u ,u\rangle/2)$$
Taking expectations shows that $X$ is not gaussian.
On the other hand, a simple calculation shows $$\mathrm{Cov} X = p\Sigma_A +q\Sigma_B$$
There is another interpretation that $$X=Y\Sigma_A^{1/2}Z_1+(1-Y)\Sigma_B^{1/2}Z_2,\quad (Z_1,Z_2)\sim \mathcal N(0,I_p)\otimes \mathcal N(0,I_p)$$ A similar calculation yields the same answer as before.