Minimisation of Kullback-Leibler divergence on an arbitrary parameter

Question

With the definition of the KL divergence between a model probability density function (pdf) and the data pdf

$$D[p|q] = \bigg< \log \frac{p(x)}{q(x)} \bigg>_p = \int_{-\infty}^{\infty} p(x) \log \frac{p(x)}{q(x)}dx$$

taken from this paper*, I'm trying to show that the gradient descent on an arbitrary parameter $x$ is given by

$$\partial_x D[p|q] = \bigg < \bigg (1+\log\frac{p(x)}{q(x)} \bigg) \partial_x \log [p(x)] - \partial_x \log[q(x)] \bigg >_p.$$

The derivative on $x$ gives me

$$\partial_x D[p|q] = \int_{-\infty}^{\infty} \bigg (\partial_xp(x)\log\frac{p(x)}{q(x)}+ p(x)\partial_x \log\frac{p(x)}{q(x)}\bigg)dx $$

but I'm unsure on how to proceed from here as I don't know how to approach the $\partial_xp(x)$ term in the equation above.

Is there a different way to arrive at the result or am I making a mistake here?

* Towards a cross-level theory of neural learning - AJ Bell - AIP Conference Proceedings, 2007

Answer 1

\begin{align*} \partial_x D[p|q] &= \int_{-\infty}^{\infty} \bigg (\partial_xp(x)\log\frac{p(x)}{q(x)}+ p(x)\partial_x \log\frac{p(x)}{q(x)}\bigg)dx \\ &= \int_{-\infty}^{\infty}\partial_xp(x)\log\frac{p(x)}{q(x)}+p(x)\frac{q(x)}{p(x)}\frac{\partial_xp(x)q(x)-p(x)\partial_xq(x) }{q(x)^2} dx \\ &= \int_{-\infty}^{\infty}\partial_xp(x)\log\frac{p(x)}{q(x)}+q(x)\frac{\partial_xp(x)q(x)-p(x)\partial_xq(x) }{q(x)^2} dx \\ &= \int_{-\infty}^{\infty}\partial_xp(x)\log\frac{p(x)}{q(x)}+\partial_xp(x)-\frac{p(x)}{q(x)}\partial_xq(x) dx\\ &= \int_{-\infty}^{\infty} \left[ \bigg (1+\log\frac{p(x)}{q(x)} \bigg) \frac{\partial_x p(x)}{p(x)} - \frac{\partial_xq(x) }{q(x)}\right] p(x) dx\\ &= \bigg < \bigg (1+\log\frac{p(x)}{q(x)} \bigg) \partial_x \log [p(x)] - \partial_x \log[q(x)] \bigg >_p \end{align*}