### Variational free energies

##### 2015-10-27

Say we’re really interested in some distribution $p(x;J) \propto e^{-\beta E(x;J)}$ on vectors $x$ and parameterized by $J$, where for the sake of discussion we have assumed an exponential model. It is often the case that evaluating sums or integrals over $x$ is intractable, i,e. we can’t evaluate things like $\langle O\rangle = \sum_x O(x) p(x;j)$. A common way to proceed is to propose a distribution you can work with, say $q(x;\theta)$, now paramterized by $\theta$ and then to fiddle with $\theta$ until $q$ looks like $p$. Formally, we want to minimize:

At first glance evaluating this function looks just as hard as the problem we can’t solve. If we expand $D_{\text{KL}}$ using the definition $p$:

Where $Z = \sum_x p(x;J)$ and we used the fact that $\sum_x q(x;\theta)=1$ and the definition of the free energy $F=-\beta^{-1}\log Z$. Notice that $D_{\text{KL}}(p||q)\geq 0$ and equal to 0 only when $p(x)=q(x)$, so that minimizing the equation on the left will have found an upper bound (and in the case of a tight bound) an approximation to the free energy of our original model! This motivates the definition of the variational free energy as:

Where we then want to minimize

over $\theta$. This is also called variational bayes in the literature, and is useful beyond just exponential distributions. Now we just have to choose a distribution $q(x)$ that is easy to evaluate, and most of the time people choose product distributions, so that if $x\in\mathcal{R}^m$, $q(x) = \prod_i q_i(x_i)$, where each individual $q_i(x)$ is integrable. Then, since we should be able to evaluate $e^{-\beta E}$, we have a well-defined and tractable objective.

### Notes:

We can also write

To really make it look like an entropy. Arxiv:1505.0542 proposes a simple scheme for stochastic gradient descent on this quantity in the context of neural networks.