August 7, 2021

Information Geometry (Part 19)

John Baez

Last time I figured out the analogue of momentum in probability theory, but I didn't say what it's called. Now I will tell you — thanks to some help from Abel Jansma and Toby Bartels.

SURPRISE: it's called SURPRISAL!

This is a well-known concept in information theory. It's also called 'information content'.

Let's see why. First, let's remember the setup. We have a manifold

Q={qRn:qi>0,i=1nqi=1}

whose points q are nowhere vanishing probability distributions on the set {1,,n}. We have a function

f:QR

called the Shannon entropy, defined by

f(q)=j=1nqjlnqj

For each point qQ we define a cotangent vector pTqQ by

p=(df)q

As mentioned last time, this is the analogue of momentum in probability theory. In the second half of this post I'll say more about exactly why. But first let's compute it and see what it actually equals!

Let's start with a naive calculation, acting as if the probabilities q1,,qn were a coordinate system on the manifold Q. We get

pi=fqi

so using the definition of the Shannon entropy we have

pi=qij=1nqjlnqj=qi(qilnqi)=ln(qi)1

Now, the quantity lnqi is called the surprisal of the probability distribution at i. Intuitively, it's a measure of how surprised you should be if an event of probability qi occurs. For example, if you flip a fair coin and it lands heads up, your surprisal is ln 2. If you flip 100 fair coins and they all land heads up, your surprisal is 100 times ln 2.

Of course 'surprise' is a psychological term, not a term from math or physics, so we shouldn't take it too seriously here. We can derive the concept of surprisal from three axioms:

  1. The surprisal of an event of probability q is some function of q, say F(q).
  2. The less probable an event is, the larger its surprisal is: q1q2F(q1)F(q2)
  3. The surprisal of two independent events is the sum of their surprisals: F(q1q2)=F(q1)+F(q2).

It follows from work on Cauchy's functional equation that F must be of this form:

F(q)=logbq

for some constant b>1. We shall choose b, the base of our logarithms, to be e. We had a similar freedom of choice in defining the Shannon entropy, and we will use base e for both to be consistent. If we chose something else, it would change the surprisal and the Shannon entropy by the same constant factor.

So far, so good. But what about the irksome "-1" in our formula?

pi=ln(qi)1

Luckily it turns out we can just get rid of this! The reason is that the probabilities qi are not really coordinates on the manifold Q. They're not independent: they must sum to 1. So, when we change them a little, the sum of their changes must vanish. Putting it more technically, the tangent space TqQ is not all of Rn, but just the subspace consisting of vectors whose components sum to zero:

TqQ={vRn:j=1nvj=0}

The cotangent space is the dual of the tangent space. The dual of a subspace

SV

is the quotient space

V/{:VR:vS(v)=0}

The cotangent space TqQ thus consists of linear functionals :RnR modulo those that vanish on vectors v obeying the equation

j=1nvj=0

Of course, we can identify the dual of Rn with Rn in the usual way, using the Euclidean inner product: a vector uRn corresponds to the linear functional

(v)=j=1nujvj

From this, you can see that a linear functional vanishes on all vectors v obeying the equation

j=1nvj=0

if and only if its corresponding vector u has

u1==un

So, we get

TqQRn/{uRn:u1==un}

In words: we can describe cotangent vectors to Q as lists of n numbers if we want, but we have to remember that adding the same constant to each number in the list doesn't change the cotangent vector!

This suggests that our naive formula

pi=ln(qi)1

is on the right track, but we're free to get rid of the constant 1 if we want! And that's true.

To check this rigorously, we need to show

p(v)=j=1nln(qi)vi

for all vTqQ. We compute:

p(v)=df(v)=v(f)=j=1nvjfqj=j=1nvj(ln(qi)1)=j=1nln(qi)vi

where in the second to last step we used our earlier calculation:

fqi=qij=1nqjlnqj=ln(qi)1

and in the last step we used

jvj=0

Back to the big picture

Now let's take stock of where we are. We can fill in the question marks in the charts from last time, and combine those charts while we're at it.

 Classical Mechanics   Thermodynamics   Probability Theory 
 q   position extensive variables probabilities
 p  momentum intensive variables surprisals
 S   action entropy Shannon entropy

What's going on here? In classical mechanics, action is minimized (or at least the system finds a critical point of the action). In thermodynamics, entropy is maximized. In the maximum entropy approach to probability, Shannon entropy is maximized. This leads to a mathematical analogy that's quite precise. For classical mechanics and thermodynamics, I explained it here:

These posts may give a more approachable introduction to what I'm doing now: now I'm bringing probability theory into the analogy, with a big emphasis on symplectic and contact geometry.

Let me spell out a bit of the analogy more carefully:

In all three cases, TQ is a symplectic manifold and imposing the constraint p=(df)q picks out a Lagrangian submanifold

Λ={(q,p)TQ:p=(df)q}

There is also a contact manifold TQ×R where the extra dimension comes with an extra coordinate S that means

We can then decree that S=f(q) along with p=(df)q, and these constraints pick out a Legendrian submanifold

Σ={(q,p,S)TQ×R:S=f(q),p=(df)q}

There's a lot more to do with these ideas, and I'll continue next time.


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
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