Empirical estimation of information

Informing yourself from your data how informative your data was

This is an empirical probability metric estimation problem, with especially cruel error properties. There are a few different versions of this problem corresponding to various different information: Mutual information between two variables, KL divergence between two distributions, information of one variable; discrete variables, continuous variable… In the mutual information case this is an independence test.

Say I would like to know the mutual information of the laws of processes generating two streams \(X,Y\) of observations, with weak assumptions on the laws of the generation process. Better, suppose further that each observation from each process is i.i.d. In the case that they have a continuous state space and joint densities \(p_{X,Y}\), marginal densities \(p_{X},p_{Y},\)

\[ \operatorname {I} (X;Y)=\int _{\mathcal {Y}}\int _{\mathcal {X}}{p_{X,Y}X,Y\log {\left({\frac {p_{X,Y}X,Y}{p_{X}(x)\,p_{Y}(y)}}\right)}}\;dx\,dy\]

Information is harder than the many metrics, because observations with low frequency have high influence on that value but are by definition rarely observed. It is easy to get a uselessly biased — or even inconsistent — estimator, especially in the nonparametric case.

Histogram estimator

The obvious one for discrete data. For continuous data where the histogram bins must also be learned, this method is highly sensitive and can be inconsistent if you don’t do it right (Paninski 2003).



Monte Carlo parametric

One case you might want to estimate this value which is one where there is no nonparametric estimation problem per se but the integral to solve it is inconvenient. In which case, we might use a Monte Carlo method.

John Schulmann explicates a good trick for estimating KL divergence in the case that you can simulate from \(x_i\sim q\) and calculate \(p(x)\) and \(q(x_i),\) The following estimator is good despite looking unrelated:

\[\begin{aligned} KL[q, p] &= \int_x q(x) \log \frac{q(x)}{p(x)} \mathrm{d}x\\ &= E_{ x \sim q}\left[\log \frac{q(x)}{p(x)} \right]\\ &\approx \frac1N \sum_{i=1}^N \frac12(\log ⁡p(x)−\log ⁡q(x))^2 \end{aligned}\]

He also introduced a simple debiased one that does even better. The mechanics are interesting. (If you actually want a mutual information this notionally calculates it if we find the KL divergence between joint and product densities; But that is not totally trivial I shall concede.)


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