Fortune in Flux

The fortune of us that are the moon’s men doth ebb and flow like the sea (Henry IV, Part I)

Wednesday, July 7, 2021 · 7 min read

Surely something will happen — which means that if you integrate over all possible outcomes, the probabilities must add up to 1. This elementary fact remains true, even if the probabilities of the outcomes change for some reason. Are you reminded of something? Fluids behave in the same way: no matter how the particles are distributed, the total quantity of fluid is conserved, which means that if you integrate over all of space, the masses must add up to 1. Indeed, when we talk about probability density functions and probability mass functions, we are borrowing fluidic vocabulary.

I get excited about analogies like this, where the physics of one domain promises to unexpectedly deliver insight into a different domain. Another example of this is how spring constants and capacitances play the same mathematical role in the differential equations that model their respective mechanical and electrical systems. Harry Olson’s wonderful book Dynamical Analogies shows some more beautifully-illustrated examples of mechanical, electrical, and acoustic systems with identical dynamics.

Can we get any mileage out of the probability-fluid analogy? It turns out, yes! We can apply our physical intuition about fluids to design a clever statistical machine learning algorithm. That’s the subject of this piece. By the way, you can undoubtedly find better (and more correct) expositions of this math in other places; my goal here is to emphasize the interaction between physical intuition and probabilistic applications, bridged so beautifully by the fluidic analogy.

A little bit about the problem setup: sometimes, when working with data, you find yourself in a place where you want to sample from a probability distribution $p(x)$. This might not be so bad if you can compute $p(x)$ exactly — for example, you can use rejection sampling or something. Often, however, you can only compute some scaled quantity $q(x)$ such that $p(x) \propto q(x)$ with some unknown constant of proportionality. Commonly, this happens in Bayesian inference, where you may want to sample from $\pi(x \mid y) = \pi(x,y)/\pi(y)$ where $y$ might be data you’ve seen and $x$ might be the parameters to your model. Here, the problem is that the normalizing constant $\pi(y)=\int \pi(x,y)dx$ is hard to compute because that integral is intractable in high dimensions.

One famous way to do this kind of sampling is by the Metropolis-Hastings algorithm, which approximates $p(x)$ with the stable distribution of a Markov chain. The “miracle” of the algorithm is that the transition probabilities between two states only depend on the ratios of the probability densities at those states — the unknown constant simply cancels out!

But now let’s think about this from a fluids perspective. Have unknown normalizing constants popped up in physics anywhere? Yes! The partition function $Z$ in Boltzmann is often unknown. Recall that a system at thermodynamic equilibrium at temperature $T$ has probability $p_i \propto e^{-U_i/k_BT}$ of being in state $i$, where $U_i$ is the potential energy of that state and $k_B$ is a physical constant that you can find, e.g. on Wikipedia. A curious application of this distribution is working out how thin the atmosphere is at altitude $h$ by reasoning about $U_h$ being the gravitational potential energy at altitude $h$: because $U$ is linear in $h$, we can instantly conclude that atmospheric density decreases exponentially in altitude.

But returning to machine learning, here is the insight that the fluidic analogy makes possible: if we set $U(x) = -\log q(x)$ then a particle simulated at some “temperature” $T=1/k_B$ in potential $U$ would eventually have probability $e^{-U(x)}=p(x)$ of being at $x$. So, we can use what we know about the dynamics of particles that are a part of a fluid at thermal equilibrium, to sample from probability distributions that may have nothing at all to do with fluids.

What exactly does it mean to “simulate a particle” of this probability fluid? You can take those words pretty literally, as it turns out. There are two sources of motion at each timestep: $U$ and $T$. Given potential $U(x)$, we can derive the force $-\nabla U$ that acts on our particle, and then apply Newton’s Second Law to infer the velocity. We’ll assume that velocity is roughly independent between timesteps because the particle is repeatedly being bumped around by other particles and having its velocity reset — so, no need to track acceleration. By the way, notice that $\nabla \log q(x)$ doesn’t depend on the unknown normalization constant, because the logarithm turns it into an additive constant, which disappears under the derivative.

As for the effect of $T$, we can model the thermal motion of the particle as Brownian motion: that is, at each time, the particle gets bumped to some random neighboring position drawn independently from a Gaussian that is somehow scaled with respect to “temperature” $T$. (How do we translate $T=1/k_B$ to this un-physical world where “joules” and “kelvins” make no sense? Stay tuned…)

This fluidic analogy leads us to what people call the “Unadjusted Langevin Algorithm.” The algorithm is quite simple to state and implement:

  1. Initialize some $x_0$.
  2. Using some small step size $\Delta t$ integrate $\Delta x = \Delta t \cdot \nabla\log q(x_t) + \sqrt{2\Delta t} \cdot \xi_t$ where you sample $\xi_t \sim N(0,1)$; the Greek letter $\xi$ is chosen for maximum squiggliness, in order to act as a visual metaphor for its function. Importantly, this step only requires you to evaluate the unnormalized density $q$!
  3. That’s it — under some (not too mild but still reasonable) assumptions, snapshots of the trajectory ${x_t}$ are distributed accoding to $p(x)$!

Are you convinced? Here’s a sanity check: without the stochastic component $\sqrt{2\Delta t}\cdot\xi_t$ at each step, $x_t$ deterministically walks towards the maximum likelihood estimator — that is, without the stochastic component, we’d have invented gradient descent! The stochasticity, however, makes $x_t$ a random variable instead.

The rest of this piece is dedicated to answering the question: where does that stochastic coefficient $\sqrt{2\Delta t}$ come from? You might have already worked out that it’s the magnitude of the thermal motion, which depends on $T$. If that coefficient were too small, then, as we just discussed above, we simply have gradient descent: our particle is analogous to a marble rolling down a (syrupy — overdamped!) hill. If the coefficient were too large, then thermal motion would dominate and the “signal” from the potential $U$ would stop mattering. So, clearly, the threshold $\sqrt{2\Delta t}$ is “special” in some way.

First let me convince you that we want this coefficient to indeed be proportional to $\sqrt{\Delta t}$, even though it looks dimensionally wrong when placed in a sum right next to the $\Delta t\cdot\nabla \log q$ term. What’s going on? It turns out that this “lots of little Gaussian bumps” motion doesn’t have an intuitive integral calculus. In particular, for Gaussians, it’s the variance that adds, not the standard deviation. So in time $t$ the variance of the total distance traveled would be $\Sigma\sqrt{\Delta t}^2 = \Sigma \Delta t = t$, and therefore the standard deviation would be $\sqrt{t}$, which is exactly what we want for the algorithm to be invariant to how small $\Delta t$ is. (For more on this, you could read about “Wiener processes.”)

Now for the “2.” To derive this value, let’s return to our fluidic analogy. Suppose we wanted to compute the change in probability density at a point $x$ over time, that is, ${\partial p(x,t)}/{\partial t}$, with the understanding that we will set this time derivative to zero later to reason about the system’s behavior at equilibrium. By following the fluidic analogy, we can hallucinate a “probability flow” $\vec{J}(x, t)$ and assert, by Gauss’ Law, that ${\partial p}/{\partial t} = -\nabla \cdot \vec{J}$.

To measure $\vec{J}(x, t)$, we should think about what’s flowing; that is, about “particles,” whose dynamics is given by the Langevin update rule for $\Delta x_t$ I wrote out above. The deterministic part is easy to deal with: from a tiny region of volume $\Delta V$, a mass $p\Delta V$ of particles leaves with velocity $\nabla\log q$, so the flux is $p\nabla \log q$.

For the stochastic part, notice that the dynamics of the thermal motion are spatially invariant, i.e. independent of $x$. So, whatever amount diffuses out of $\Delta V$ is proportional to $p$, and whatever amount diffuses in is proportional to the density in the neighborhood of $\Delta V$: this means the net flow is proportional to $-\nabla p$. When we talk about fluids, this observation is also known as Fick’s Law, and the constant of proportionality is called the diffusion coefficient $D$.

How can we compute $D$? Looking at only the stochastic part, we have $\partial p/\partial t = -D\nabla^2 p$, which we recognize as the heat equation. That is, for this brief aside, we’re going to think of $p$ representing — not probability over Bayesian model parameters or fluid density in a potential well, but — heat in a block of metal — a bonus analogy!

Thankfully, the solution to the heat equation turns out to be well-known: it is a time-varying Gaussian. And it better be! After all, the stochastic diffusion is just a sum of lots of independent little Gaussian-sampled steps of variance $\Delta(\sigma^2) = {2\Delta t}$, which together form a big Gaussian of variance $\sigma^2 = 2t$, as discussed above. But let’s say we didn’t know the coefficient was 2. Let’s set it to $a$ instead. Now, if we know that $p(x,t) = e^{-x^2/2(a t)}/\sqrt{2\pi(at)}$, then we can literally take the partial derivatives on both sides of the heat equation and expand to solve for $D$. A miracle occurs, and many things cancel, leaving us with simply $-D=a/2$. It’s not worth writing out the algebra here: I encourage you to try it yourself, since it is that rare combination of straightforward and satisfying. But furthermore, it gives some mechanical (if not physical) insight into the “2” that appears in the denominator of the result: algebraically, the $-1/2$ power of $at$ in the normalization constant $1/\sqrt{2\pi(at)}$ kicks out some factors of two that are not balanced by the $-1$ power of $t$ in the exponential.

In any case, though, we conclude from the above that $\Delta p = -(a/2)\Delta t \nabla^2 p$, which means the stochastic flux is simply $\nabla p$. Perhaps it’s all clear to you by now, but let’s take this over the finish line. Putting these two components together, $U$ and $T$, we have $\partial p/\partial t = \nabla\cdot (p\nabla\log q-(a/2)\nabla p)$. This nice identity is known as the Fokker-Planck equation (the devoted reader might note that this is the second reference to Fokker on this blog).

Now, finally, as promised many paragraphs ago, we will now reason about this equation at equilibrium. At equilibrium the distribution is static in time, so we have $0 = \nabla \cdot (p\nabla \log q -(a/2) \nabla p)$. Assuming suitable boundary conditions, we can assume the argument of the divergence is also zero, so $\nabla \log q = (a/2)\nabla p / p$. Integrating in all directions, we have $\log q + c = (a/2)\log p$, or rather $p \propto q^{2/a}$. This makes sense: just as expected, for large $a$ the effect of $q$ is wiped out, and for small $a$ all but the maximum value get squashed to zero. Crucially, however, if we want $p \propto q$, then we must set $a=2$. Aha! So that’s why the thermal motion needs to be scaled by $\sqrt{2\Delta t}$. (By the way, that constant of integration above, $c$, oddly enough encodes that pesky normalization constant that started off this whole discussion…)

Okay. I’m glad I got that out of my system, it’s been rattling around for weeks, now! How wonderful it is, that we can start with the simplest axiom of probability — “something must happen” — and, reasoning only with physical intuition, reach a sampling algorithm that solves a real problem.

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