3
$\begingroup$

If I use a latent profile analysis (Gaussian Mixture Model) to model my observed multivariate probability distribution as a mixture (K-classes) of conditionally-independent normal pdfs, does this model imply that each within-class multivariate pdf is multivariate normal?

Can you show my why, or why not, this is true?

If it is true, can tests for multivariate normallity be used to help evaluate the model?

$\endgroup$
2
$\begingroup$

Except in degenerate cases, it won't be multivariate Normal. The easiest way to see that is in the $\mathbb R^2$ case. The probability density function of a GMM is a sum over Normal densities $p(x) = \sum_i \pi_i \cdot \mathcal N(x|\mu_i,\Sigma_i)$. Since a multivariate Normal distribution is unimodal with its only a single maximum at the mean, the mixture should also be unimodal. However, if $\mu_i\not=\mu_j$ and $\pi_i\not=0$, you get several "bumps" since there is a Gaussian sitting on each mean $\mu_i$. Therefore, the distribution you can't be multivariate Normal.

$\endgroup$
3
  • $\begingroup$ You are right, but I don't think I've put my question well. What about the within class pdf when each estimated normal describes a different random variable? Will this imply a mvnorm? $\endgroup$
    – D L Dahly
    Dec 28 '13 at 10:02
  • $\begingroup$ @DLDahly It's not clear what you're asking, but in any case, could you clarify in your actual question? $\endgroup$
    – Glen_b
    Dec 28 '13 at 10:07
  • 1
    $\begingroup$ Hmm, I try to guess what you mean. The sampling model for a mixture of Normals would be: randomly draw a normal distribution from the $K$ you got according to $\pi_1, ...,\pi_K$. Then draw a sample from that Normal distribution. This means that conditioned on the component your data point is multivariate Normal. $\endgroup$
    – fabee
    Dec 28 '13 at 10:51

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.