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Given two gaussian mixture models (GMMs) with different degrees of freedom, is there a way to determine the probability that one is generated from the other? That is, can we give a probability to the hypothesis that the two distributions are actually the same?

My current approach is to use a correlation score, which is easy to calculate but is not really a probability:

$$ C(p_1,p_2) = -\log \left[ \frac{\int p_1(x)p_2(x)dx}{\int p_1^2(x)+p_2^2(x)dx}\right] $$ For two GMMs $p_1(\boldsymbol r)=\sum_i\pi_i\phi_i(\boldsymbol r|\boldsymbol \mu_i,\boldsymbol \Sigma_i)$ and $p_2(\boldsymbol r)=\sum_j\pi_j\phi_j(\boldsymbol r|\boldsymbol \mu_j,\boldsymbol \Sigma_j)$ we have: $$ \int p_1p_2 = \sum_{i,j} \pi_i\pi_j\int\phi_i(\boldsymbol r)\phi_j(\boldsymbol r)d\boldsymbol r $$ The final integral is straightforward to compute because the product of two gaussians is another gaussian. I like this function because it's analytic, but it really doesn't tell us a probability. One additional thought is to use a multivariate t-test but this seems to only be useful for comparing two normal distributions. I need to compare two mixtures with unequal degrees of freedom.

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You're comparing two distributions, not two random variables, so I'm not sure where "probability" would come into play. The closest thing to what you describe is the Kullback-Leibler divergence, which measures the amount of extra information one would need to encode with one distribution a sample produced by the other.

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  • $\begingroup$ Yeah, it's a bit confusing for me. Doesn't the two-way t-test, for example, give you the probability that two distributions are the same thing? $\endgroup$ – cgreen Oct 30 '14 at 21:06
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    $\begingroup$ No, it compares two samples. $\endgroup$ – Arthur B. Oct 30 '14 at 21:09
  • $\begingroup$ OK I'll give you the answer, that was very helpful. I think I need to rework my question :) $\endgroup$ – cgreen Nov 2 '14 at 19:54

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