August 17, 2021

Information Geometry (Part 21)

John Baez

Last time I ended with a formula for the 'Gibbs distribution': the probability distribution that maximizes entropy subject to constraints on the expected values of some observables.

This formula is well-known, but I'd like to derive it here. My argument won't be up to the highest standards of rigor: I'll do a bunch of computations, and it would take more work to state conditions under which these computations are justified. But even a nonrigorous approach is worthwhile, since the computations will give us more than the mere formula for the Gibbs distribution.

I'll start by reminding you of what I claimed last time. I'll state it in a way that removes all unnecessary distractions, so go back to Part 20 if you want more explanation.

The Gibbs distribution

$${\displaystyle -{\int }_{\mathrm{\Omega }}\pi (x)\ln \pi (x)d\mu (x)}$$

subject to the requirement that some integrable functions $A^1,\dots ,A^n$ on $\mathrm{\Omega }$ have expected values equal to some chosen list of numbers $q^1,\dots ,q^n$.

(Unlike last time, now I'm writing $A^i$ and $q^i$ with superscripts rather than subscripts, because I'll be using the Einstein summation convention: I'll sum over any repeated index that appears once as a a superscript and once as a subscript.)

Furthermore, suppose $\pi$ depends smoothly on $q\in {\mathbb{R}}^n$. I'll call it ${\pi }_q$ to indicate its dependence on $q$. Then, I claim ${\pi }_q$ is the so-called Gibbs distribution

$${\displaystyle {\pi }_q(x)=\frac{e^{-p_i{A}^i(x)}}{{\int }_{\mathrm{\Omega }}{e}^{-p_i{A}^i(x)}d\mu (x)}}$$

where

$${\displaystyle p_i=\frac{\partial f(q)}{\partial q^i}}$$

and

$${\displaystyle f(q)=-{\int }_{\mathrm{\Omega }}{\pi }_q(x)\ln {\pi }_q(x)d\mu (x)}$$

is the entropy of ${\pi }_q$.

Let's show this is true!

Finding the Gibbs distribution

$${\displaystyle {\int }_{\mathrm{\Omega }}\pi (x)A^i(x)d\mu (x)=q^i}$$

We can solve this problem using Lagrange multipliers. We need one Lagrange multiplier, say ${\beta }_i,$ for each of the above constraints. But it's easiest if we start by letting $\pi$ range over all of $L^1(\mathrm{\Omega }),$ that is, the space of all integrable functions on $\mathrm{\Omega }$. Then, because we want $\pi$ to be a probability distribution, we need to impose one extra constraint

$${\displaystyle {\int }_{\mathrm{\Omega }}\pi (x)d\mu (x)=1}$$

To do this we need an extra Lagrange multiplier, say $\gamma$.

So, that's what we'll do! We'll look for critical points of this function on $L^1(\mathrm{\Omega })$:

$${\displaystyle -\int \pi \ln \pi d\mu -{\beta }_i\int \pi A^{i}d\mu -\gamma \int \pi d\mu }$$

Here I'm using some tricks to keep things short. First, I'm dropping the dummy variable $x$ which appeared in all of the integrals we had: I'm leaving it implicit. Second, all my integrals are over $\mathrm{\Omega }$ so I won't say that. And third, I'm using the Einstein summation convention, so there's a sum over $i$ implicit here.

Okay, now let's do the variational derivative required to find a critical point of this function. When I was a math major taking physics classes, the way physicists did variational derivatives seemed like black magic to me. Then I spent months reading how mathematicians rigorously justified these techniques. I don't feel like a massive digression into this right now, so I'll just do the calculations — and if they seem like black magic, I'm sorry!

We need to find $\pi$ obeying

$${\displaystyle \frac{\delta }{\delta \pi (x)}(-\int \pi \ln \pi d\mu -{\beta }_i\int \pi A^{i}d\mu -\gamma \int \pi d\mu )=0}$$

or in other words

$${\displaystyle \frac{\delta }{\delta \pi (x)}(\int \pi \ln \pi d\mu +{\beta }_i\int \pi A^{i}d\mu +\gamma \int \pi d\mu )=0}$$

First we need to simplify this expression. The only part that takes any work, if you know how to do variational derivatives, is the first term. Since the derivative of $z\ln z$ is $1+\ln z,$ we have

$${\displaystyle \frac{\delta }{\delta \pi (x)}\int \pi \ln \pi d\mu =1+\ln \pi (x)}$$

The second and third terms are easy, so we get

$${\displaystyle \frac{\delta }{\delta \pi (x)}(\int \pi \ln \pi d\mu +{\beta }_i\int \pi A^{i}d\mu +\gamma \int \pi d\mu )=1+\ln \pi (x)+{\beta }_i{A}^i(x)+\gamma }$$

Thus, we need to solve this equation:

$${\displaystyle 1+\ln \pi (x)+{\beta }_i{A}^i(x)+\gamma =0}$$

That's easy to do:

$$\pi (x)=e^{-1-\gamma -{\beta }_i{A}^i(x)}$$

Good! It's starting to look like the Gibbs distribution!

We now need to choose the Lagrange multipliers ${\beta }_i$ and $\gamma$ to make the constraints hold. To satisfy this constraint

$${\displaystyle \int \pi d\mu =1}$$

we must choose $\gamma$ so that

$${\displaystyle \int e^{-1-\gamma -{\beta }_i{A}^i}d\mu =1}$$

or in other words

$${\displaystyle e^{1+\gamma }=\int e^{-{\beta }_i{A}^i}d\mu }$$

Plugging this into our earlier formula

$$\pi (x)=e^{-1-\gamma -{\beta }_i{A}^i(x)}$$

we get this:

$${\displaystyle {\pi }_q(x)=\frac{e^{-{\beta }_i{A}^i(x)}}{\int e^{-{\beta }_i{A}^i}d\mu }}$$

Great! Even more like the Gibbs distribution!

By the way, you must have noticed the "1" that showed up here:

$${\displaystyle \frac{\delta }{\delta \pi (x)}\int \pi \ln \pi d\mu =1+\ln \pi (x)}$$

It buzzed around like an annoying fly in the otherwise beautiful calculation, but eventually went away. This is the same irksome "1" that showed up in Part 19. Someday I'd like to say a bit more about it.

Now, where were we? We were trying to show that

$${\displaystyle {\pi }_q(x)=\frac{e^{-p_i{A}^i(x)}}{\int e^{-p_i{A}^i}d\mu }}$$

minimizes entropy subject to our constraints. So far we've shown

$${\displaystyle \pi (x)=\frac{e^{-{\beta }_i{A}^i(x)}}{\int e^{-{\beta }_i{A}^i}d\mu }}$$

is a critical point. It's clear that

$$\pi (x)\ge 0$$

so $\pi$ really is a probability distribution. We should show it actually maximizes entropy subject to our constraints, but I will skip that. Given that, $\pi$ will be our claimed Gibbs distribution ${\pi }_q$ if we can show

$$p_i={\beta }_i$$

This is interesting! It's saying our Lagrange multipliers ${\beta }_i$ actually equal the so-called conjugate variables $p_i$ given by

$${\displaystyle p_i=\frac{\partial f}{\partial q^i}}$$

where $f(q)$ is the entropy of ${\pi }_q$:

$${\displaystyle f(q)=-{\int }_{\mathrm{\Omega }}{\pi }_q(x)\ln {\pi }_q(x)d\mu (x)}$$

There are two ways to show this: the easy way and the hard way. The easy way is to reflect on the meaning of Lagrange multipliers, and I'll sketch that way first. The hard way is to use brute force: just compute $p_i$ and show it equals ${\beta }_i$. This is a good test of our computational muscle — but more importantly, it will help us rediscover some interesting facts about the Gibbs distribution.

The easy way

$$f:{\mathbb{R}}^2\to \mathbb{R}$$

subject to the constraint

$$g=c$$

for some smooth function

$$g:{\mathbb{R}}^2\to \mathbb{R}$$

and constant $c$. (The function $f$ here has nothing to do with the $f$ in the previous sections, which stood for entropy.) To answer this we introduce a Lagrange multiplier $\lambda$ and seek points where

$$\mathrm{\nabla }(f-\lambda g)=0$$

This works because the above equation says

$$\mathrm{\nabla }f=\lambda \mathrm{\nabla }g$$

Geometrically this means we're at a point where the gradient of $f$ points at right angles to the level surface of $g$:

Thus, to first order we can't change $f$ by moving along the level surface of $g$.

But also, if we start at a point where

$$\mathrm{\nabla }f=\lambda \mathrm{\nabla }g$$

and we begin moving in any direction, the function $f$ will change at a rate equal to $\lambda$ times the rate of change of $g$. That's just what the equation says! And this fact gives a conceptual meaning to the Lagrange multiplier $\lambda$.

Our situation is more complicated, since our functions are defined on the infinite-dimensional space $L^1(\mathrm{\Omega }),$ and we have an $n$-tuple of constraints with an $n$-tuple of Lagrange multipliers. But the same principle holds.

So, when we are at a solution ${\pi }_q$ of our constrained entropy-maximization problem, and we start moving the point ${\pi }_q$ by changing the value of the $i$th constraint, namely $q^i,$ the rate at which the entropy changes will be ${\beta }_i$ times the rate of change of $q^i$. So, we have

$${\displaystyle \frac{\partial f}{\partial q^i}={\beta }_i}$$

But this is just what we needed to show!

The hard way

$${\displaystyle \frac{\partial f}{\partial q^i}={\beta }_i}$$

We start by solving our constrained entropy-maximization problem using Lagrange multipliers. As already shown, we get

$${\displaystyle {\pi }_q(x)=\frac{e^{-{\beta }_i{A}^i(x)}}{\int e^{-{\beta }_i{A}^i}d\mu }}$$

Then we'll compute the entropy

$$f(q)=-\int {\pi }_q\ln {\pi }_{q}d\mu$$

Then we'll differentiate this with respect to $q_i$ and show we get ${\beta }_i$.

Let's try it! The calculation is a bit heavy, so let's write $Z(q)$ for the so-called partition function

$${\displaystyle Z(q)=\int e^{-{\beta }_i{A}^i}d\mu }$$

so that

$${\displaystyle {\pi }_q(x)=\frac{e^{-{\beta }_i{A}^i(x)}}{Z(q)}}$$

and the entropy is

\[ $$\begin{array}{ccl}f(q)& =& -{\displaystyle \int {\pi }_q\ln \left(\frac{e^{-{\beta }_k{A}^k}}{Z(q)}\right)d\mu }\\ \\ & =& {\displaystyle \int {\pi }_q({\beta }_k{A}^k+\ln Z(q))d\mu }\\ \end{array}$$ \)

This is the sum of two terms. The first term

$${\displaystyle \int {\pi }_q{\beta }_k{A}^{k}d\mu ={\beta }_k\int {\pi }_q{A}^{k}d\mu }$$

is ${\beta }_k$ times the expected value of $A^k$ with respect to the probability distribution ${\pi }_q,$ all summed over $k$. But the expected value of $A^k$ is $q^k,$ so we get

$${\displaystyle \int {\pi }_q{\beta }_k{A}^{k}d\mu ={\beta }_k{q}^k}$$

The second term is easier:

$${\displaystyle {\int }_{\mathrm{\Omega }}{\pi }_q\ln Z(q)d\mu =\ln Z(q)}$$

since ${\pi }_q(x)$ integrates to 1 and the partition function $Z(q)$ doesn't depend on $x\in \mathrm{\Omega }$.

Putting together these two terms we get an interesting formula for the entropy:

$$f(q)={\beta }_k{q}^k+\ln Z(q)$$

This formula is one reason this brute-force approach is actually worthwhile! I'll say more about it later.

But for now, let's use this formula to show what we're trying to show, namely

$${\displaystyle \frac{\partial f}{\partial q^i}={\beta }_i}$$

For starters,

$$\begin{array}{ccl}{\displaystyle \frac{\partial f}{\partial q^i}}& =& {\displaystyle \frac{\partial }{\partial q^i}({\beta }_k{q}^k+\ln Z(q))}\\ \\ & =& {\displaystyle \frac{\partial {\beta }_k}{\partial q^i}{q}^k+{\beta }_k\frac{\partial q^k}{\partial q^i}+\frac{\partial }{\partial q^i}\ln Z(q)}\\ \\ & =& {\displaystyle \frac{\partial {\beta }_k}{\partial q^i}{q}^k+{\beta }_k{\delta }_i^k+\frac{\partial }{\partial q^i}\ln Z(q)}\\ \\ & =& {\displaystyle \frac{\partial {\beta }_k}{\partial q^i}{q}^k+{\beta }_i+\frac{\partial }{\partial q^i}\ln Z(q)}\end{array}$$

where we played a little Kronecker delta game with the second term.

Now we just need to compute the third term:

$$\begin{array}{ccl}{\displaystyle \frac{\partial }{\partial q^i}\ln Z(q)}& =& {\displaystyle \frac{1}{Z(q)}\frac{\partial }{\partial q^i}Z(q)}\\ \\ & =& {\displaystyle \frac{1}{Z(q)}\frac{\partial }{\partial q^i}\int e^{-{\beta }_j{A}^j}d\mu }\\ \\ & =& {\displaystyle \frac{1}{Z(q)}\int \frac{\partial }{\partial q^i}\left(e^{-{\beta }_j{A}^j}\right)d\mu }\\ \\ & =& {\displaystyle \frac{1}{Z(q)}\int \frac{\partial }{\partial q^i}(-{\beta }_k{A}^k)e^{-{\beta }_j{A}^j}d\mu }\\ \\ & =& {\displaystyle -\frac{1}{Z(q)}\int \frac{\partial {\beta }_k}{\partial q^i}{A}^k{e}^{-{\beta }_j{A}^j}d\mu }\\ \\ & =& {\displaystyle -\frac{\partial {\beta }_k}{\partial q^i}\frac{1}{Z(q)}\int A^k{e}^{-{\beta }_j{A}^j}d\mu }\\ \\ & =& {\displaystyle -\frac{\partial {\beta }_k}{\partial q^i}{q}^k}\end{array}$$

Ah, you don't know how good it feels, after years of category theory, to be doing calculations like this again!

Now we can finish the job we started:

$$\begin{array}{ccl}{\displaystyle \frac{\partial f}{\partial q^i}}& =& {\displaystyle \frac{\partial {\beta }_k}{\partial q^i}{q}^k+{\beta }_i+\frac{\partial }{\partial q^i}\ln Z(q)}\\ \\ & =& {\displaystyle \frac{\partial {\beta }_k}{\partial q^i}{q}^k+{\beta }_i-\frac{\partial {\beta }_k}{\partial q^i}{q}^k}\\ \\ & =& {\beta }_i\end{array}$$

Voilà!

Conclusions

$${\displaystyle p_i=\frac{\partial f}{\partial q^i}}$$

Thus, they are rich in meaning. From what we've seen earlier, they are 'surprisals'. They are analogous to momentum in classical mechanics and have the meaning of intensive variables in thermodynamics:

$$f(q)=p_i{q}^i+\ln Z(q)$$

(We proved this formula with ${\beta }_i$ replacing $p_i,$ but now we know those are equal.)

This formula suggests that the logarithm of the partition function is important — and it is! It's closely related to the concept of free energy — even though 'energy', free or otherwise, doesn't show up at the level of generality we're working at now.

This formula should also remind you of the tautological 1-form on the cotangent bundle $T^{\ast }Q,$ namely

$$\theta =p_{i}d{q}^i$$

It should remind you even more of the contact 1-form on the contact manifold $T^{\ast }Q\times \mathbb{R},$ namely

$$\alpha =-dS+p_{i}d{q}^i$$

Here $S$ is a coordinate on the contact manifold that's a kind of abstract stand-in for our entropy function $f$.

So, it's clear there's a lot more to say: we're seeing hints of things here and there, but not yet the full picture.

You can read a discussion of this article on Azimuth, and make your own comments or ask questions there!

© 2021 John Baez
baez@math.removethis.ucr.andthis.edu
home