KL-Divergence of two GMMs

Question

I have two GMMs that I used to fit two different sets of data in the same space, and I would like to calculate the KL-divergence between them.

Currently I am using the GMMs defined in sklearn ( http://scikit-learn.org/stable/modules/generated/sklearn.mixture.GMM.html ) and the SciPy implementation of KL-divergence ( http://docs.scipy.org/doc/scipy-dev/reference/generated/scipy.stats.entropy.html )

How would I go about doing this? Do I want to just create tons of random points, get their probabilities on each of the two models (call them P and Q) and then use those probabilities as my input? Or is there some more canonical way to do this within the SciPy/SKLearn environment?

Answer 1

There's no closed form for the KL divergence between GMMs. You can easily do Monte Carlo, though. Recall that KL(p||q) = \\int p(x) log(p(x) / q(x)) dx = E_p[ log(p(x) / q(x)) . So:

def gmm_kl(gmm_p, gmm_q, n_samples=10**5):
    X = gmm_p.sample(n_samples)
    log_p_X, _ = gmm_p.score_samples(X)
    log_q_X, _ = gmm_q.score_samples(X)
    return log_p_X.mean() - log_q_X.mean()

( mean(log(p(x) / q(x))) = mean(log(p(x)) - log(q(x))) = mean(log(p(x))) - mean(log(q(x))) is somewhat cheaper computationally.)

You don't want to use scipy.stats.entropy ; that's for discrete distributions.

If you want the symmetrized and smoothed Jensen-Shannon divergence KL(p||(p+q)/2) + KL(q||(p+q)/2) instead, it's pretty similar:

def gmm_js(gmm_p, gmm_q, n_samples=10**5):
    X = gmm_p.sample(n_samples)
    log_p_X, _ = gmm_p.score_samples(X)
    log_q_X, _ = gmm_q.score_samples(X)
    log_mix_X = np.logaddexp(log_p_X, log_q_X)

    Y = gmm_q.sample(n_samples)
    log_p_Y, _ = gmm_p.score_samples(Y)
    log_q_Y, _ = gmm_q.score_samples(Y)
    log_mix_Y = np.logaddexp(log_p_Y, log_q_Y)

    return (log_p_X.mean() - (log_mix_X.mean() - np.log(2))
            + log_q_Y.mean() - (log_mix_Y.mean() - np.log(2))) / 2

( log_mix_X / log_mix_Y are actually the log of twice the mixture densities; pulling that out of the mean operation saves some flops.)