Decompose/split a single multivariate gauss into random gaussian mixture Say, there is a single $n$-dimensional multivariate Gaussian.
$$Gauss_a(\mu_a,\Sigma_a) $$ 
$\mu_a$ is $1\times n$ vector and $\Sigma_a$ is $n\times n$ matrix.
Is there any easy way to  decompose/split a single gaussian $Gauss_a$ into random multiple $K$ gaussians as  a multivariate Gaussian mixture models where no its Gaussian component is identical to original one, $Gauss_a$.($K$ is given)
$$Gauss_a(\mu_a,\Sigma_a) \approx \sum_{i=1}^Kw_i \cdot Gauss_i(\mu_i, \Sigma_i)$$  where $ Gauss_i(\mu_i, \Sigma_i) \neq Gauss_a , w_i\neq 1$
Thank you.
 A: It's not too hard to show that this isn't possible in general.  For a counterexample, consider the 1-dimensional case with $K=2$ and $Gauss_a(0,1)$ the standard normal, and suppose we had the decomposition
$$Gauss_a(0,1) = w_1 Gauss_1(\mu_1,\sigma_1) + w_2 Gauss_2(\mu_2,\sigma_2)$$
for some parameters $w_i$, $\mu_i$, and $\sigma_i$.  We can calculate the moment generating function of each side and equate them:
$$ \exp\left(\tfrac12t^2\right) =
   w_1\exp\left(\mu_1 t + \tfrac12\sigma_1^2t^2\right)+
   w_2\exp\left(\mu_2 t + \tfrac12\sigma_2^2t^2\right) $$
Note that the moment generating function (or any expectation) of a mixture distribution is easy to calculate -- it's just a weighted sum of the expectations of the mixed distributions.
The nice thing about this equation is that the only way it can hold for all $t$ is if all the coefficients in the power series expansion for $t$ match up.  I used SymPy (a symbolic mathematics library for Python) to (albegraically) solve the system of equations for the first five coefficients:
from sympy import symbols, solve, exp, Eq, diff, N

t,w1,mu1,v1,mu2,v2 = symbols('t,w1,mu1,v1,mu2,v2', real=True)

lhs = exp(t**2/2)
rhs = w1 * exp(mu1 * t + v1*t**2/2) + (1-w1) * exp(mu2 * t + v2*t**2/2)
solve([Eq(diff(lhs, t, k).subs(t,0), diff(rhs, t, k).subs(t,0))
       for k in range(1,5)], check=True)

and it determined that the only exact solution was the trivial one where the components of the mixture are identically distributed:
Out[4]: [{mu1: 0, mu2: 0, v1: 1, v2: 1}]

So, there is no non-trivial solution in the univariate case with $K=2$.  My guess is that this is true in the general multivariate case for any $K>1$ and there are no non-trivial solutions, period, though I'm not sure how to go about proving it.
