0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Of course you can approximate it: just set $w_1\approx 1, \mu_1 \approx \mu_a, \Sigma_i\approx \Sigma_a,$ and all other $w_i\approx 0$ for $i \ne 1.$ Might I therefore suggest that you remove the "approximately express" part of this question? $\endgroup$ – whuber Aug 31 '18 at 16:12
  • $\begingroup$ Ahh,, what i mean above is "except" the case identical gauss you just mentioned.. $\endgroup$ – JimSD Aug 31 '18 at 22:24
  • $\begingroup$ Question math.stackexchange.com/questions/3183207/… shows the split methods, but without proof $\endgroup$ – zzqstar Apr 11 at 2:23
1
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ An easier way to demonstrate the impossibility is to examine the asymptotic behavior of the log pdf at infinity: that establishes all the components must have the same values of $\Sigma.$ After that, it's straightforward to show they must all have the same value of $\mu,$ too. $\endgroup$ – whuber Sep 1 '18 at 15:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.