0
$\begingroup$

I was trying to understand the mathematical proof of KL-Divergence when using the chain rule:

$D(p(y|x)||q(y|x)) = D(p(x)||q(x)) + D(p(y|x)||q(y|x))$

And I'm a bit lost in the last step (https://www.cs.princeton.edu/courses/archive/fall11/cos597D/L03.pdf, 1.3 Conditional Divergence). What I don't understand is why is this true?

$D(p(x)||q(x)) = \sum_x \sum_y p(x, y) log \frac{p(x)}{q(x)}$

The definition says something slightly different:

$D(p(x)||q(x)) = \sum_x p(x) log \frac{p(x)}{q(x)}$

I have also seen this written in another way that I still don't understand (https://homes.cs.washington.edu/~anuprao/pubs/CSE533Autumn2010/lecture3.pdf, 2.3 Conditional Divergence):

$D(p(x)||q(x)) = \sum_x p(x) log \frac{p(x)}{q(x)} \sum_y p(y|x)$

Why is that last part $\sum_y p(y|x)$ also absorbed into the KL definition?

$\endgroup$
2
$\begingroup$

First of all, $\sum_y{p(x,y)}=p(x)$, because you're fixing $x$ and summing over all possible $y$. Therefore, $$\begin{align}D(p||q) &= \sum_x \sum_y p(x, y) log \frac{p(x)}{q(x)}\\ &= \sum_x log \frac{p(x)}{q(x)} \sum_y p(x, y) \\ &= \sum_x log \frac{p(x)}{q(x)} p(x)\end{align}$$, which is what the definition says.

For the latter part, from the definition of joint PMF, we have $p(x,y)=p(x)p(y|x)$. So, $$\sum_y{p(x,y)}=\sum_y p(x)p(y|x)=p(x)\sum_y p(y|x)$$, which explains the factorization in the KL formulation.

$\endgroup$
  • $\begingroup$ It was p(y|x), sorry. $\endgroup$ – kuonb Feb 21 at 13:12
  • $\begingroup$ OK, I have also commented on the last part. $\endgroup$ – gunes Feb 21 at 13:17
  • $\begingroup$ And, @user20160's comment on $\sum_y p(y|x)=1$ is another good point by the way. $\endgroup$ – gunes Feb 21 at 13:23
1
$\begingroup$

What I don't understand is why is this true? $$D(p(x) \parallel q(x)) = \sum_x \sum_y p(x, y) \log \frac{p(x)}{q(x)}$$

$\log \frac{p(x)}{q(x)}$ doesn't depend on $y$, so it can be moved out of the inner sum:

$$D(p(x) \parallel q(x)) = \sum_x \log \frac{p(x)}{q(x)} \sum_y p(x, y)$$

$\sum_y p(x, y) = p(x)$ by the definition of marginal probability. Plugging this in gives the canonical expression for KL divergence:

$$D(p(x) \parallel q(x)) = \sum_x p(x) \log \frac{p(x)}{q(x)}$$

Why is that last part $\sum_y p(y \mid y)$ also absorbed into the KL definition?

I don't see that in the notes you linked. Maybe a typo? Assuming you meant $\sum_y p(y \mid x)$, then this is equal to one, since a probability distribution must sum to one by definition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.