60
$\begingroup$

I'm having trouble deriving the KL divergence formula assuming two multivariate normal distributions. I've done the univariate case fairly easily. However, it's been quite a while since I took math stats, so I'm having some trouble extending it to the multivariate case. I'm sure I'm just missing something simple.

Here's what I have...

Suppose both $p$ and $q$ are the pdfs of normal distributions with means $\mu_1$ and $\mu_2$ and variances $\Sigma_1$ and $\Sigma_2$, respectively. The Kullback-Leibler distance from $q$ to $p$ is:

$\int \left[\log( p(x)) - \log( q(x)) \right]\ p(x)\ dx$, which for two multivariate normals is:

$\frac{1}{2}\left[\log\frac{|\Sigma_2|}{|\Sigma_1|} - d + Tr(\Sigma_2^{-1}\Sigma_1) + (\mu_2 - \mu_1)^T \Sigma_2^{-1}(\mu_2 - \mu_1)\right]$

Following the same logic as this proof, I get to about here before I get stuck:

$=\int \left[ \frac{d}{2} \log\frac{|\Sigma_2|}{|\Sigma_1|} + \frac{1}{2} \left((x-\mu_2)^T\Sigma_2^{-1}(x-\mu_2) - (x-\mu_1)^T\Sigma_2^{-1}(x-\mu_1) \right) \right] \times p(x) dx$

$=\mathbb{E} \left[ \frac{d}{2} \log\frac{|\Sigma_2|}{|\Sigma_1|} + \frac{1}{2} \left((x-\mu_2)^T\Sigma_2^{-1}(x-\mu_2) - (x-\mu_1)^T\Sigma_2^{-1}(x-\mu_1) \right) \right]$

I think I have to implement the trace trick, but I'm just not sure what to do after that. Any helpful hints to put me back on the right track would be appreciated!

$\endgroup$
60
$\begingroup$

Starting with where you began with some slight corrections, we can write

$$ \begin{aligned} KL &= \int \left[ \frac{1}{2} \log\frac{|\Sigma_2|}{|\Sigma_1|} - \frac{1}{2} (x-\mu_1)^T\Sigma_1^{-1}(x-\mu_1) + \frac{1}{2} (x-\mu_2)^T\Sigma_2^{-1}(x-\mu_2) \right] \times p(x) dx \\ &= \frac{1}{2} \log\frac{|\Sigma_2|}{|\Sigma_1|} - \frac{1}{2} \text{tr}\ \left\{E[(x - \mu_1)(x - \mu_1)^T] \ \Sigma_1^{-1} \right\} + \frac{1}{2} E[(x - \mu_2)^T \Sigma_2^{-1} (x - \mu_2)] \\ &= \frac{1}{2} \log\frac{|\Sigma_2|}{|\Sigma_1|} - \frac{1}{2} \text{tr}\ \{I_d \} + \frac{1}{2} (\mu_1 - \mu_2)^T \Sigma_2^{-1} (\mu_1 - \mu_2) + \frac{1}{2} \text{tr} \{ \Sigma_2^{-1} \Sigma_1 \} \\ &= \frac{1}{2}\left[\log\frac{|\Sigma_2|}{|\Sigma_1|} - d + \text{tr} \{ \Sigma_2^{-1}\Sigma_1 \} + (\mu_2 - \mu_1)^T \Sigma_2^{-1}(\mu_2 - \mu_1)\right]. \end{aligned} $$

Note that I have used a couple of properties from Section 8.2 of the Matrix Cookbook.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see you took out the D that I had originally. Wouldn't you have a D term after taking the log of the Gaussian in the first few steps? $\endgroup$ – dmartin Jun 3 '13 at 15:19
  • 1
    $\begingroup$ Consider the scaling factor $(2\pi)^{-d/2} |\Sigma_k|^{-1/2}$, $k = 1,2$ of the multivariate normal density. When computing the log-difference, the $(2\pi)^{-d/2}$ term goes away. There is no $d$ term for the determinants -- simply, a $1/2$, which is factored out. $\endgroup$ – ramhiser Jun 3 '13 at 15:33
  • 1
    $\begingroup$ @acidghost Either one works because we can factor out a negative one from both sides. Multiplying the two negative ones yields a positive one. $\endgroup$ – ramhiser Apr 12 '16 at 0:06
  • 1
    $\begingroup$ Your answer seems wrong, please see the last section of stanford.edu/~jduchi/projects/general_notes.pdf for the correct one. $\endgroup$ – Peng Zhao Apr 19 at 6:40
  • 3
    $\begingroup$ @PengZhao The answers are the same. The primary difference is that we denote the feature dimension with a different letter. $\endgroup$ – ramhiser Apr 21 at 17:24

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.