1
$\begingroup$

From the perspective of information theory, I understand how D(P|Q) is non-negative and why the KL divergence is asymmetric, i.e. $D(P|Q) \neq D(Q|P)$, given two gaussian univariate gaussian distributions.

I also know that $D(P|Q) = ln\frac{\sigma_q}{\sigma_p} + \frac{\sigma_p^2 + (\mu_p - \mu_q)^2}{2\sigma_q^2} - \frac{1}{2}$

My question is that if the mean of both the divergence is equal, i.e. $\mu_p = \mu_q$, Which of the divergence would be larger, D(P|Q) or D(Q|P)?

$\endgroup$
1
  • 1
    $\begingroup$ Hi Inderpartap, welcome! Do you have an idea of how to check the sign of $D(P|Q) - D(Q|P)$? $\endgroup$ – Konstantin Nov 12 '19 at 10:28
1
$\begingroup$

It depends on the ratio of deviations. Let $x=\sigma_p/\sigma_q$ When $\mu_p=\mu_q$, $$D(P|Q)=-\ln x+x^2/2-1/2, D(Q|P)=\ln x+{1 \over 2x^2}-1/2$$ As you might guess, these two are equal when $x=1$ and there is no other solution because if $x=a$ is a solution, so as $x=1/a$; and for $x>1$, this expression is greater than $0$, i.e. there is no zero-crossing.

To sum up, $D(P|Q)$ is larger when $\sigma_p>\sigma_q$.

$\endgroup$

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.