Origins of Macroeconomic Fluctuations

1. Introduction

Where do macroeconomic fluctuations come from? Does common variation in firms’ output necessarily come from a single source? In this post, I work through a model which suggests that productivity “factors” might be the result of weak interactions between lots of otherwise independent production decisions. I start with a real world example which has little to do with macroeconomics, then develop the model with this concrete example in mind, and finally conclude by relating this work back to the original questions.

Example (Millenium Bridge):
On June 10th 2000 the Millennium footbridge over the River Thames in London opened to thousands of excited visitors. While engineers carefully designed the bridge to support the total mass of people and survive high wind speeds, the bridge nevertheless started vibrating uncontrollably. The source of this vibration was clear. As shown in the video below at the $37$ second mark, all of the people on the bridge locked step. Of course, the visitors didn’t do this on purpose, so how did this happen? How did many random footsteps turn into a coherent left-right-left-right march?

Here is the rough story:

Oscillators: A person’s center of mass which drifts back and forth from right to left as she walks.
Scale: There were so many people that a small group was really likely to lock step by pure chance.
Weak Interaction: Each step makes the bridge wobble slightly and forces others to compensate by shifting their weight.
Positive Feedback: People respond to an unexpected bridge shift by moving their center of mass in the opposite direction.

With that many people on the bridge, it was really likely that a handful of people standing right next to each other would happen to lock step for a few strides by pure chance. When this happened, the entire bridge shifted a little bit from right to left. This effect was at first imperceptible. However, because the bridge moved a little bit to the left, the people around this initial cluster shifted their weight a bit to the right in the following instant to compensate. Thus, this initial imperceptible shift got amplified until the entire bridge was vibrating violently and everyone on the bridge was marching in lock step.

Strogatz (2005) shows how to study this phenomenon more generally as a group of weakly interacting oscillators using the Kuramoto Model. In Section $2$ , I begin by laying out the basic mathematical framework. Then, in Section $3$ , I solve this model up to a constant parameter. In Section $4$ , I then show how to solve for this constant. Finally, in Section $5$ , I conclude by discussing some potential applications of this idea to macrofinance.

2. Framework

How could I model this idea? In this section, I lay out the mathematical framework for the Kuramoto model which captures these $4$ key elements. Consider a world with $N$ oscillators, and let $\Theta_n(t)$ denote the phase of the $n^{th}$ oscillator at time $t$ in units of radians. For instance, when people walk they shift their weight slightly from left to right and then back again. Thus, the side to side movement in a person’s center of mass while they walk can be thought of as a simple oscillator. $\Theta_t(n) = 0$ might then correspond to person $n$ leaning all the way to the left while $\Theta_t(n) = \pi$ might correspond to person $n$ leaning all the way to the right. The time derivative of oscillator $n$ ‘s phase, $d \Theta_n(t)/dt = \theta_n(t)$ , is known as its frequency and has units of radians per second. I use $\theta_n^*$ to denote the natural frequency at which the phase of the $n^{th}$ oscillator changes in units of radians per second. For example, $\theta_n^*$ corresponds to the rate at which person $n$ would shift her weight from side to side in the absence of any other walkers.

I then look only at the following functional form for the interactions of each oscillator:

(1) $\begin{align*} \theta_n(t) &= \theta_n^* + \sum_{n'=1}^N f\left(\Theta_{n'}(t) - \Theta_n(t) \right) \\ &= \theta_n^* + \frac{K}{N} \cdot \sum_{n'=1}^N \sin\left(\Theta_{n'}(t) - \Theta_n(t) \right) \end{align*}$

This formulation means that the frequency of the $n^{th}$ oscillator at time $t$ equals its natural frequency, $\theta_n^*$ , plus a constant $K \geq 0$ which has units of radians per second times the average of the sine of the distances between the phase of oscillator $n$ and the phase of every other oscillator. Thus, if most of the other $(N-1)$ people on the bridge just got done taking their left step when your right foot hits, then you will slow down your walking speed. Conversely, if most of the other $(N-1)$ people on the bridge just picked up their right foot when yours hits the ground, then you will increase your walking speed. The coupling constant, $K$ , then determines exactly how much you are going to change your walking speed in each of these instances. I assume each oscillator’s natural frequency is drawn from a symmetric unimodal distribution, $\theta_n^* \overset{\scriptscriptstyle \mathrm{iid}}{\sim} g(\theta^*)$ , centered around $\mu_{\theta^*}$ :

(2) $\begin{align*} g(\mu_{\theta^*} + \theta^*) &= g(\mu_{\theta^*} -\theta^*) \end{align*}$

In summary, this section proposes a simple model which captures the $4$ key components of the introductory story. The key state variables, $\Theta_n(t)$ , represent the phases of $N$ oscillators. The rates at which each of the oscillators changes its phase are weakly interacting and governed by a coupling constant, $K$ . The sinusoidal functional form implies a positive feedback effect. “Solving” the model means determining if there are values of the coupling constant $K$ such that, for any collection of $N$ oscillators with natural frequencies selected independently and identically from the distribution $g(\theta^*)$ , all of the oscillators will lump together and take on the similar frequencies. i.e., if there are a bunch of people with different baseline walking speeds on a bridge, will they all end up locking step? This problem is hard because the phase of every single oscillator is pulling on the frequency of every other oscillator at the same time. It is not obvious that a coherent lump will emerge.

3. Solution Strategy

Is this model tractable? In this section, I now show how to “solve” this model up to the determination of a threshold value of the coupling constant $K = \underline{K}$ . i.e., I show that if the bridge is sufficiently wobbly, all of the people on the bridge will lock step; whereas, if it is stiffer than this threshold value, no synchronization will occur. I leave the task of solving for this threshold value to the next section.

The trick is look for a change of variables that simplifies the problem. I do this by introducing the order parameter $Z(t)$ :

(3) $\begin{align*} Z(t) &= \frac{1}{N} \cdot \sum_{n=1}^N e^{i \cdot \Theta_n(t)} = R(t) \cdot e^{i \cdot \Psi(t)} \end{align*}$

What is this equation saying? $\Theta_n(t)$ is a scalar quantity on the $[0,\pi]$ interval. In the first line, the expression $e^{i \cdot \Theta_n(t)}$ then maps this value onto the circumference of the unit circle. Thus, the order parameter $Z(t)$ denotes the average of a group of unit vectors starting. i.e., $Z(t)$ captures the average amount that each person on the bridge has shifted their weight. In the second line, $R(t)$ denotes the phase coherence and $\Psi(t)$ denotes the average phase—i.e., $Z(t)$ ‘s polar coordinates. Where does the name “phase coherence” come from? Intuitively, if all of the $N$ unit vectors are pointing in the same direction, then $R(t) = 1$ ; conversely, if each of the $N$ unit vectors is uniformly spaced around the circle, then $R(t) = 0$ . Thus, for instance, at the $37$ second mark in the video, everyone on the bridge is leaning in the same direction so that $R(37)=1$ . $\Psi(t)$ then denotes the direction which people are leaning towards. By embedding the phases of each of the $N$ oscillators on a unit circle, I can characterize their average phase with a pair of order statistics.

Graphical depiction of order statistics $R$ and $\Psi$ summarizing the average phase of the $N$ oscillators.

I now show how to simplify the problem by using these order statistics and the sinusoidal functional form of the link between each pair of oscillator’s phases. First, I multiply both sides of the equation by $\exp \left\{ - i \cdot \Theta_n(t) \right\}$ to give:

(4) $\begin{align*} R(t) \cdot e^{i \cdot (\Psi(t) - \Theta_n(t) )} &= \frac{1}{N} \cdot \sum_{n'=1}^N e^{i \cdot \left\{ \Theta_{n'}(t) - \Theta_n(t) \right\}} \end{align*}$

Then, I apply Euler’s formula to rewrite the expression:

(5) $\begin{align*} R(t) \cdot &\left\{ \cos\left(\Psi(t) - \Theta_n(t) \right) + i \cdot \sin\left(\Psi(t) - \Theta_n(t) \right) \right\} \\ &= \frac{1}{N} \cdot \sum_{n'=1}^N \left\{ \cos\left(\Theta_{n'}(t) - \Theta_n(t)\right) + i \cdot \sin\left(\Theta_{n'}(t) - \Theta_n(t) \right) \right\} \end{align*}$

Equating the imaginary components then yields:

(6) $\begin{align*} R(t) \cdot \sin\left( \Psi(t) - \Theta_n(t) \right) &= \frac{1}{N} \cdot \sum_{n'=1}^N \sin\left(\Theta_{n'}(t) - \Theta_n(t)\right) \end{align*}$

Notice that the left hand side depends only on the $2$ order statistics, $R(t)$ and $\Psi(t)$ , as well as the phase of oscillator $n$ . Importantly, it does not depend on the phase of any of the other $(N-1)$ oscillators. Thus, I can substitute this simplified form into Equation (1) to get:

(7) $\begin{align*} \theta_n(t) &= \theta_n^* + K \cdot R(t) \cdot \sin\left(\Psi(t) - \Theta_n(t)\right) \end{align*}$

This new equation says that rather than depending on the phase of every single other oscillator, the frequency of oscillator $n$ depends on its natural frequency, its current phase, and order statistics characterizing the average phase of all of the other oscillators. In the language of the Millenium bridge example, this equation says that my walking speed is determined by my natural walking speed, my current weight distribution, and the average weight distribution of everyone else on the bridge. I don’t have to worry about whether Bob from Hertfordshire who is standing $100{\scriptstyle \mathrm{ft}}$ in front of me happens to be leaning a bit more to his left. All I care about is the average weight distribution for everyone on the bridge.

I then look for solutions where $R(t)=R^*$ and $\Psi(t)=\mu_{\theta^*} \cdot t$ . By choosing the right reference frame, I can then set $\mu_{\theta^*} = 0$ without loss of generality to get:

(8) $\begin{align*} \theta_n(t) &= \theta_n^* - K \cdot R^* \cdot \sin\left( \Theta_n(t) \right) \end{align*}$

These $2$ assumptions are necessary to solve the model in closed form, but are somewhat at odds with the introductory example. $R(t) = R^*$ constant means that people on the bridge are not transitioning from incoherent steps (start of the video) to synchrony (the $37$ second mark). Rather, their level of phase coherence is constant. $\Psi(t) = \mu_{\theta^*} \cdot t$ means that their average weight distribution is evolving at the mean natural frequency. There will be some fast walkers who shift their weight quickly from left to right; there will be some slow walkers who shift their weight gradually from left to right; but, the average location of everyone’s weight must change according to the average natural walking speed. In this setting, the $N$ oscillators can be split into $2$ groups: locked and drifting. Oscillators in the first group are phase locked at the original frequency $\Psi^* = \mu_{\theta^*} \cdot t$ . Oscillators in the second group, on the other hand, drift around the unit circle in a non-uniform way.

(9) $\begin{align*} \begin{cases} |\theta_n^*| \leq K \cdot R^* &\text{ locked} \\ |\theta_n^*| > K \cdot R^* &\text{ drifting} \end{cases} \end{align*}$

The conditions above show why I refer to the constant $K$ as the model’s coupling strength. As $K$ increases, more and more of the oscillators will become locked and rotate around the circle at the natural frequency $\Psi^* = \mu_{\theta^*} \cdot t$ . i.e., as the bridge becomes more wobbly, more and more people will start to deviate from the natural walking speed and synchronize their steps with the rest of the people on the bridge.

I then look only for solutions where the drifting oscillators form a stationary distribution on the circle. This stationarity condition means that the probability that a fast walker is leaning left or right does not change over time. Let $\rho(\Theta,\theta^*) \cdot d \theta$ denote the fraction of oscillators with the natural frequency $\theta^*$ that lie in the interval $[\Theta,\Theta + d\Theta)$ . Then, stationarity implies that:

(10) $\begin{align*} \rho(\Theta,\theta^*) &= \frac{C(\theta^*)}{\left\vert \theta^* - K \cdot r \cdot \sin(\Theta) \right\vert} \end{align*}$

Click image to view animation. Simulation parameters: $N = 100$ , $K=\pi/4$ , $\mu_{\theta^*} = \pi/4$ , and $\sigma_{\theta^*}=\pi/8$ .

In summary, in this section solves this model up to the determination of the coupling constant $K$ . I first propose a pair of order parameters $R(t)$ and $\Psi(t)$ , and then show that sinusoidal link between the frequencies of each pair of oscillators means that each oscillator’s frequency only depends on these $2$ order statistics. The only free parameter that I have not used yet is the coupling constant, $K$ . I tie my hands and propose that any solution must be such that both order parameters are constants, $R(t) = R^*$ and $\Psi(t) = 0$ , and also that the distribution of frequencies, $\rho(\Theta,\theta^*)$ , must be stationary.

4. Coupling Threshold

In this section, I ask the question: “Does there exist a $K$ which solves the model above subject to these $3$ constraints?” I do this in $3$ steps. First, I solve for the stationary density. Then, I enforce the conditions $R(t) = R^*$ and $\Psi(t) = 0$ . Finally, I look for satisfactory solutions of $K$ .

Step 1: What does $\rho(\Theta,\theta^*)$ look like? $C(\theta^*)$ is pinned down by the fact that all oscillators have to lie somewhere on the circle:

(11) $\begin{align*} 1 &= \int_{-\pi}^{\pi} \rho(\Theta,\theta^*) \cdot d\Theta = \int_{-\pi}^{\pi} \left( \frac{C(\theta^*)}{\left\vert \theta^* - K \cdot R^* \cdot \sin \Theta \right\vert} \right) \cdot d\Theta = \frac{(2 \cdot \pi) \cdot C(\theta^*)}{\sqrt{\left(\theta^*\right)^2 - \left( K \cdot R^*\right)^2}} \end{align*}$

where $\left(\theta^*\right)^2 > \left( K \cdot R^*\right)^2$ by definition for all of the drifting oscillators. This constraint basically says that everyone who is walking on the bridge has to be leaning in some direction. It may be right; it may be left; it may be right down the center; but, it is somewhere. $C(\theta^*)$ is then given by:

(12) $\begin{align*} C(\theta^*) &= \frac{1}{2 \cdot \pi} \cdot \sqrt{\left(\theta^*\right)^2 - \left( K \cdot R^*\right)^2} \end{align*}$

Step 2: What are the consequences of imposing $R(t) = R^*$ and $\Psi(t) = 0$ ? To answer this question, I return to Equation (3):

(13) $\begin{align*} R(t) \cdot e^{i \cdot \Psi(t)} &= \sum_{n=1}^N e^{i \cdot \Theta_n(t)} = \sum_{n \in \mathbf{N}_{\mathrm{Lock}}} e^{i \cdot \Theta_n(t)} + \sum_{n \in \mathbf{N}_{\mathrm{Drift}}} e^{i \cdot \Theta_n(t)} \end{align*}$

Since $\Psi(t) = 0$ , then in order to have $R(t) = R^*$ be constant it must be the case that:

(14) $\begin{align*} R^* &= \sum_{n \in \mathbf{N}_{\mathrm{Lock}}} e^{i \cdot \Theta_n(t)} + \sum_{n \in \mathbf{N}_{\mathrm{Drift}}} e^{i \cdot \Theta_n(t)} \end{align*}$

So, in order to $R^*$ to be constant, either both of these summations have to be constant, or their changes have to exactly offset each other as time marches forward. Let’s examine each of these summations in turn. Looking first at the locked state, I have that for all $n$ such that $|\theta_n^*| \leq K \cdot R^*$ :

(15) $\begin{align*} \sin (\Theta_n(t)) &= \frac{\theta_n^*}{K \cdot R^*} \end{align*}$

Thus for all locked oscillators $\Theta_n(t)$ is an implicit function of $\theta_n^*$ rather than $t$ . Since the distribution of locked phases is symmetric about $0$ by assumption, I have that:

(16) $\begin{align*} \sum_{n \in \mathbf{N}_{\mathrm{Lock}}} e^{i \cdot \Theta_n(\theta^*)} &= \sum_{n \in \mathbf{N}_{\mathrm{Lock}}}\left\{ \cos\left(\Theta_n(\theta^*)\right) + i \cdot \sin\left(\Theta_n(\theta^*)\right) \right\} \\ &= \sum_{n \in \mathbf{N}_{\mathrm{Lock}}}\left\{ \cos\left(\Theta_n(\theta^*)\right) \right\} \\ &= \int_{-K \cdot R^*}^{K \cdot R^*} \cos(\Theta_n(\theta^*)) \cdot g(\theta^*) \cdot d\theta^* \\ &= K \cdot R^* \cdot \int_{-\pi/2}^{\pi/2} \cos(\Theta)^2 \cdot g(K \cdot R^* \cdot \sin(\Theta)) \cdot d\Theta \end{align*}$

Looking at the drifting state, I have that:

(17) $\begin{align*} \sum_{n \in \mathbf{N}_{\mathrm{Drift}}} e^{i \cdot \Theta_n(t)} &= \int_{-\pi}^{\pi} \int_{|\theta^*| > K \cdot R^*} e^{i \cdot \Theta} \cdot \rho(\Theta,\theta^*) \cdot g(\theta^*) \cdot d\theta^* \cdot d\Theta \end{align*}$

However, this integral has to vanish since $g(\theta^*) = g(-\theta^*)$ and $\rho(\Theta + \pi,-\theta^*) = \rho(\Theta,\theta^*)$ . Thus, in order for $R(t) = R^*$ to be a constant, I have to have that:

(18) $\begin{align*} R^* &= K \cdot R^* \cdot \int_{-\pi/2}^{\pi/2} \cos(\Theta)^2 \cdot g(K \cdot R^* \cdot \sin(\Theta)) \cdot d\Theta \end{align*}$

Step 3: So, what values of $K$ solve this equation? An obvious solution is that $R^*=0$ . This solution corresponds to the completely incoherent state with $\rho(\Theta,\theta^*) = 1/2 \cdot \pi$ . A second set of solutions comes from the solution to the equation:

(19) $\begin{align*} 1 &= K \cdot \int_{-\pi/2}^{\pi/2} \cos(\Theta)^2 \cdot g(K \cdot R^* \cdot \sin(\Theta)) \cdot d\Theta \end{align*}$

This equation bifurcates continuously from $R^* = 0$ at a value $K = \underline{K}$ obtained by letting $R^* \searrow 0$ . Thus, there is a critical value, $\underline{K}$ , such that for all $K \geq \underline{K}$ the $N$ drifting oscillators lump together. I can compute $\underline{K}$ as follows:

(20) $\begin{align*} \underline{K} &= \frac{2}{\pi \cdot g(0)} \end{align*}$

In other words, when the bridge is sufficiently stiff there is only one outcome: incoherence. People with randomly selected natural walking speeds will place their steps at random. However, at some critical level of “wobbliness” there suddenly exists a second possibility. As a result of the weak interaction of each person’s leaning from left to right as they walk, people on the bridge might synchronize their steps. An aggregate behavior might emerge from seemingly independent private decisions.

Relationship between phase coherence, $R^*$ , and the coupling constant, $K$ , which contains a bifurcation point at $\underline{K}$ .

5. Conclusion

What does any of this have to do with macroeconomics? Instead of $N$ people leaning left to right on a bridge as they walk, think about $N$ firms making lumpy investment decisions between $2$ complimentary inputs to their production technology. For instance, think about a firm hiring people (labor) and purchasing equipment (capital). Because these are complimentary inputs, firms want to always have them in roughly equal proportions. Thus, they oscillate between increasing their stock of labor and capital.

First, think about a world where the firms’ decisions to invest are completely independent. Suppose that on average $10{\scriptstyle \%}$ of all firms increase either their labor or their capital stock in any given period. If each addition adds $\delta$ to the firm’s output $y_{n,t}$ , then growth in aggregate output, $Y_t$ , is given by:

(21) $\begin{align*} \frac{\Delta Y_{t+1}}{Y_t} &= \frac{1}{Y_t} \cdot \sum_{n=1}^N \Delta y_{n,t+1} = \sum_{n=1}^N \frac{(1 + \delta \cdot \varepsilon_{n,t+1}) \cdot y_{n,t}}{Y_t} \end{align*}$

where $\varepsilon_{n,t+1}$ is an indicator variable if a firm makes an investment in either labor or capital. Thus, the standard deviation of output growth would be:

(22) $\begin{align*} \sigma_{\mathrm{GDP}} &= \left( \mathrm{Var}\left[ \frac{\Delta Y_{t+1}}{Y_t} \right] \right)^{1/2} = \frac{3 \cdot \delta}{10 \cdot \sqrt{N}} \end{align*}$

Thus, if firm volatility is on the order of $0.30 \cdot \delta = 12{\scriptstyle \%}$ per year and there are $10^3$ firms in the economy, the aggregate volatility of the economy should be on the order of $0.012{\scriptstyle \%}$ per year as discussed in Gabaix (2011).

With this benchmark in mind, now think about a different world where each firm’s investment decision’s weakly interact. If I am running a gas station and you are running a convenience store across the street, I will be quicker to add an extra pump to my station in the months after you add a few more aisles to your store. After all, this means that more people will be parked out front of your store, and each of these cars needs gas to run. This weak interaction will exist even if I don’t know about your extra capacity. I don’t need to understand exactly why more drivers are stopping by my station.

In this world, suppose that a fraction $\phi$ of all firms are phase locked to make investments at the same time. Then, the standard deviation of output growth would be:

(23) $\begin{align*} \sigma_{\mathrm{GDP}} &\approx 0.30 \cdot \delta \cdot \sqrt{\phi} \end{align*}$

where the approximation holds for $N$ grows large. The idea is that, as the number of firms grows large, the dominant component of aggregate volatility will come from common investment decisions made by the $\phi \cdot N$ phase locked firms. These decisions will happen in $1$ out of every $10$ periods. Nevertheless, none of these phase locked firms need to be actively trying to coordinate investment decisions in the exact same way that none of the people on Millenium bridge was trying to march in step.