3
$\begingroup$

$GL_n(\mathbb{R})$ comes equipped with a Haar measure $\mu$, and using it an $f\in L^1(GL_n)$ satisfing $f \geq 0$, and $\int f d\mu = 1 $ is a probability density function for some probability distribution. We can define its entropy $H(f)$ as $\int_{GL_n} log(f(x)) f(x) d \mu(x)$.

According to the "principle of maximum entropy", the $f$ that maximize $H(f)$ given some constraints $C$, are the best choices of prior distributions.

Considering the inclusion $GL_n(R) \to M_n(R)$ as a random variable, or similar, one can compute various 'moments' of the distribution. I'm not sure what conditions $C$ you can impose on $f$ to guarantee that there is a unique entropy maximizer.

It seems reasonable to ask for a mean of the identity matrix and a variance some fixed matrix, because then the entropy maximizer might perhaps be some version of the Gaussian distribution.

Or, as Qiaochu suggests below, perhaps fixing the expectation of the trace turns out to be interesting.

Can someone enlighten me about appropriate conditions $C$ and what the entropy maximizers are?

$\endgroup$
1
$\begingroup$

Moments don't take as input a probability distribution, they take as input a pair consisting of a probability distribution and a random variable. You don't notice this when the sample space is $\mathbb{R}$ because the random variable is implicitly the identity function $X : \mathbb{R} \to \mathbb{R}$, but in this setting you need to pick some variable, say the trace. Now you can impose some conditions using this random variable, say the condition that the expectation of the square of the trace has some fixed value. I don't know anything about existence or uniqueness of entropy maximizers though.

$\endgroup$
  • $\begingroup$ I was thinking of the identity matrix valued random variable in my parenthetical. Will clarify. $\endgroup$ – Lorenzo Mar 17 '17 at 15:57
  • $\begingroup$ I was worried that the moments wouldn't necessarily live in GLn, but as long as they live in Mn I guess it's okay. $\endgroup$ – Lorenzo Mar 17 '17 at 16:32

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.