--- title: "When an economy remembers" author: "Miraia S. Chiou" creator: "Miraia Chiou" published: "2026-09-08" identifier: "ep_field_92bcd55827654758a0db95e0f5150f1d" source: "https://epistemicphysics.com/fields/economics/when-an-economy-remembers/" citation: "https://epistemicphysics.com/cite/?work=ep_field_92bcd55827654758a0db95e0f5150f1d" --- # When an economy remembers Nudge a town into a new habit, walk a price index around a loop, and discover how shared risks, exchange, and observation clocks select different probability laws. [Read the illustrated article](https://epistemicphysics.com/fields/economics/when-an-economy-remembers/) Imagine a town with a market square, a credit cooperative, and a row of workshops. A temporary discount persuades a few workshops to adopt the same machine. Their neighbors follow because spare parts, training, and compatible suppliers have become easier to find. The discount ends. The machines stay. Across the square, a statistician follows changing shopping baskets. Prices and quantities eventually return to their starting values, yet the chained index in her notebook has moved. At the cooperative, a thousand individually modest risks turn out to share one uncertain economic environment. In the shops, two observers see the same sales but keep different records of their timing. Something consequential survives in each story. Sometimes it is a collective pattern. Sometimes it is a path through changing weights. Sometimes it is a relationship between events, or a piece of the observation record. Epistemic Physics gives us a useful starting question: **what must the state retain so that our next calculation still answers the intended question?** A mean can be enough for one calculation and insufficient for another. A count may need an exposure time. A collection of forecasts may need its shared causes. Let’s tour this town with a notebook that can keep those distinctions. ## A choice can help create its own attraction Suppose workshops choose between adopting a technology and declining it. Write the adopting fraction as $q$, and turn it into a signed collective direction, $$ m=2q-1,\qquad -1\leq m\leq1. $$ An evenly divided town has $m=0$. Universal adoption has $m=1$, and universal rejection has $m=-1$. This is the same change from a probability-like fraction to a centered coordinate that appears throughout Epistemic Physics. Choose a response rule in which the odds of adoption are $$ \frac{q}{1-q}=e^{2(h+Jm)}. $$ Here $h$ is a private payoff advantage, measured in the dimensionless units of this choice model. The nonnegative number $J$ measures how strongly the surrounding adoption pattern changes that advantage. Solving for $q$ gives $$ \begin{aligned} q&=\frac{1+\tanh(h+Jm)}2,\\ m&=\tanh(h+Jm). \end{aligned} $$ The hyperbolic tangent is $\tanh x=(e^x-e^{-x})/(e^x+e^{-x})$. It smoothly turns an unbounded advantage into a bounded direction. The last equation demands **self-consistency**: the average choice must equal the average response to that choice. Brock and Durlauf derive such relationships from discrete choice with social interactions. [Their model](https://hceconomics.uchicago.edu/sites/default/files/pdf/events/Brock_Durlauf_2001_REStud_v68-N2.pdf) supplies a connection between economic incentives and the collective equations familiar from statistical physics. The half-log-odds coordinate is especially revealing: $$ \rho_q=\frac12\log\frac{q}{1-q}=h+Jm. $$ The private advantage and the response to others add in rapidity. Here $\rho_q$ describes a choice fraction. A workshop’s belief about machine reliability would be another variable, with its own trials and observation model. The adoption of neighbors can change both the payoff from buying a machine and the evidence available about it; those are two mechanisms we can connect explicitly. ## Give the town time to answer A self-consistency equation lists possible resting states. To discover which one a nudge reaches, choose an adjustment law: $$ \dot m=\tanh(h+Jm)-m. $$ The dot means change per unit of our model time. The current direction moves toward the response that the current incentives would produce. If the response exceeds $m$, adoption rises. If it falls below $m$, adoption declines. At $h=0$ and $J=0.5$, the only equilibrium is an evenly divided town. At $J=1.5$, there are three: | Collective direction $m$ | Adopting fraction $q$ | Response to a small disturbance | | ------------------------ | ---------------------- | ------------------------------- | | Approximately $-0.85856$ | Approximately $7.1\%$ | Returns toward the equilibrium | | $0$ | $50\%$ | Moves away from the equilibrium | | Approximately $+0.85856$ | Approximately $92.9\%$ | Returns toward the equilibrium | The outer equilibria are **stable**: nearby states return. The middle one is unstable. A small change can select which persistent direction the town takes, even when the private advantage later becomes zero. **Predict before nudging:** begin evenly divided, click one declining workshop into adoption, and let the choices settle. Then reduce $J$ below one and repeat. Which town forgets the nudge? [A nudge that lasts](https://epistemicphysics.com/fields/economics/when-an-economy-remembers/#economic-feedback-lab) We can locate the switching points without guessing. At equilibrium, $$ \begin{aligned} h&=\operatorname{artanh}(m)-Jm,\\ \operatorname{artanh}(m)&=\frac12\log\frac{1+m}{1-m}. \end{aligned} $$ Differentiate this curve with respect to $m$. For $J>1$, a **fold** occurs where its derivative is zero, so a stable and an unstable solution meet: $$ \begin{aligned} m_{\rm fold}&=\pm\sqrt{1-1/J},\\ h_{\rm fold}&=\operatorname{artanh}(m_{\rm fold})\\ &\quad-Jm_{\rm fold}. \end{aligned} $$ For $J=1.5$, the private-advantage thresholds are approximately $\pm0.20755$. Slowly raise $h$ from a town with little adoption, allowing choices to relax along the way. Then lower it again. The switches happen at different advantages on the two trips. This is **hysteresis**, a dependence of the current response on the path used to reach it. The memory lives in the continuing interaction. An initial disturbance changes adoption; adoption changes the attraction of adopting. We could now give each workshop a trial ledger about costs and reliability, and compare three interventions: new information, a temporary subsidy, and a change in the supplier network. They may all increase adoption while leaving different traces after they end. ## Even the measuring ruler can remember Walk from the workshops to the market stalls. Our statistician wants a price index that follows what people actually buy. For each good $i$, let $p_i$ be its price and $s_i$ its share of total expenditure. A continuously updated index $\mathcal P$ obeys $$ d\log\mathcal P=\sum_i s_i\,d\log p_i. $$ The logarithm turns successive multiplicative price changes into additive changes. A **Törnqvist link** replaces each continuously varying share by the average of its values at the two compared endpoints. The U.S. Bureau of Labor Statistics uses this construction in its chained consumer price index and studies the resulting [chain drift](https://www.bls.gov/osmr/research-papers/2017/st170100.htm). Its [introduction to the chained index](https://www.bls.gov/cpi/additional-resources/chained-cpi-introduction.htm) explains the role of expenditure shares. Choose two goods, fix the second price at one, and write $$ r=\log(p_1/p_2)=\log p_1,\qquad s=s_1. $$ Now take a rectangular walk through the price-share plane: | Leg | What changes | Contribution to $\log\mathcal P$ | | -------- | ----------------------------------------------------- | -------------------------------- | | Outward | Double good 1’s price at expenditure share $0.8$ | $0.8\log2$ | | Downward | Lower its expenditure share to $0.2$ at the new price | $0$ | | Homeward | Restore its original price at share $0.2$ | $-0.2\log2$ | | Upward | Restore its original expenditure share | $0$ | At the end, both prices and both expenditure shares match the beginning. Keep total expenditure at one throughout. Purchased quantities are then $$ q_1=\frac{s}{e^r},\qquad q_2=1-s. $$ They return too. Yet the chained index has multiplied by $$ \mathcal P_{\rm loop}=e^{0.6\log2}=2^{0.6}\approx1.516. $$ A direct comparison of the identical endpoints gives one. Each price-changing leg has a constant share, so the displayed result is exact for the four Törnqvist links as well as the continuous integral. **Take the ruler for a walk:** reverse the route before changing any other setting. Should its log change reverse sign? What happens if you flatten the route to a constant expenditure share? [The price index takes a walk](https://epistemicphysics.com/fields/economics/when-an-economy-remembers/#price-index-path-lab) The geometric calculation is simply $$ \oint s\,dr=(0.8-0.2)\log2. $$ The loop symbol means add changes all the way around a closed path. The integral measures the enclosed area with the sign chosen by this traversal. The high-share outward price movement outweighs the low-share return movement. Reversing the route negates the log change, and the two multiplicative index factors become reciprocals. There is also a useful way to restore path independence. Suppose the shares are derivatives of one log expenditure function, at a fixed utility level and under a fixed preference model. Adding their weighted price changes then recovers the change in that function. Such a quantity is a **state function**: its change depends only on the endpoints. Our freely chosen price-share loop instead lets the weighting question change along the route. The town’s spending can come home while the notebook retains a journey. That makes the measurement rule part of the description we need to keep. ## A loop can also live between people At the next town meeting, three equally sized teams rank projects A, B, and C: | Team | Full ranking | | ----- | ----------------- | | One | A, then B, then C | | Two | B, then C, then A | | Three | C, then A, then B | Each team’s ranking is **transitive**: if it places one project above a second and that second above a third, it also places the first above the third. Yet the pairwise majorities form a **Condorcet cycle**: $$ q_{AB}=q_{BC}=q_{CA}=\frac23, $$ where $q_{AB}$ is the fraction preferring A to B, and similarly for the other pairs. A beats B, B beats C, and C beats A. Try to assign every project one scalar score $V_i$, with comparison odds $$ \log\frac{q_{ij}}{1-q_{ij}}=V_i-V_j. $$ The score differences must add to zero around a closed loop because each score appears once with each sign. The actual comparisons give $$ \begin{aligned} &\log\frac{q_{AB}}{1-q_{AB}}\\ &\quad+\log\frac{q_{BC}}{1-q_{BC}}\\ &\quad+\log\frac{q_{CA}}{1-q_{CA}}=3\log2. \end{aligned} $$ One common score per project therefore loses something. The table of rankings preserves it perfectly: choose a team with probability $1/3$, then use that team’s complete ranking. This is a valid joint model of all three comparisons. It explains the cycle through a mixture of individually consistent views. The nonzero loop has a different origin from the price-index loop. There we accumulated a changing weighted measurement along a route. Here we aggregated people with different priorities. Keeping the team and the question attached to each judgment lets us recover that distinction. It also makes agenda design concrete: which pair gets voted on first can change the surviving proposal. Imagine a planning system that shows the competing rankings and the tradeoffs behind them, then asks which new option could satisfy two teams at once. The disagreement becomes material for design. A single consensus number would have hidden the opening. ## A thousand loans can share one unanswered question The cooperative lends to $n$ workshops. Let $D_i=1$ when loan $i$ defaults and $D_i=0$ otherwise. The total number of defaults is $K=\sum_iD_i$. First choose one common, uncertain default propensity, $$ \begin{aligned} P&\sim\operatorname{Beta}(\alpha,\beta),\\ \mu&=\frac{\alpha}{\alpha+\beta},\\ \tau&=\alpha+\beta. \end{aligned} $$ Both Beta shapes are positive. The mean propensity is $\mu$, and $\tau$ is its concentration. Conditional on the shared $P$, defaults are independent Bernoulli outcomes with probability $P$. **Conditional independence** means that once this common propensity is specified, the remaining random draws are independent. The distinction matters when we average over what remains unknown. Every loan has marginal default probability $\mu$, but two loans share uncertainty about $P$. More loans average away their separate draws while leaving the common question. The law of total variance separates the two contributions: $$ \begin{aligned} \operatorname{Var}(K/n) &=\frac{\mathbb E[P(1-P)]}{n}\\ &\quad+\operatorname{Var}(P)\\ &=\frac{\mu(1-\mu)}{n}\\ &\quad\times\left(1+\frac{n-1}{\tau+1}\right). \end{aligned} $$ As $n$ grows, the first contribution shrinks and a variance floor remains: $$ \lim_{n\to\infty}\operatorname{Var}(K/n) =\frac{\mu(1-\mu)}{\tau+1}. $$ Now give each loan its own independent $P_i$, all drawn from the same Beta law. Every single loan still has default probability $\mu$. This time the aggregate count is binomial, and its fraction variance is $\mu(1-\mu)/n$, tending to zero. For the deliberately simple choice $n=1000$ and $\operatorname{Beta}(1,19)$, both structures give each loan a 5% default probability. Under one shared propensity, the probability of at least 100 defaults is about $0.1379$, or 13.79%. With independent propensities, it is about $8.41\times10^{-11}$. These numbers come from the exact finite count distributions of the two specified models. **Change one relationship:** keep the individual risk fixed and switch the parent structure. Then increase the portfolio size. Which part of the default-fraction spread survives? [One parent changes the portfolio](https://epistemicphysics.com/fields/economics/when-an-economy-remembers/#shared-risk-lab) We must also say what happens tomorrow. A common $P$ could represent a fresh economic environment drawn each period, or a fixed unknown rate that the cooperative keeps learning. Those choices can give the same one-period predictive distribution and different relationships across periods. The demo redraws propensities for each cohort. This is where lineage becomes economic structure. Shared causes, a loss transmitted through obligations, and several reports copied from one source can all create dependence through different routes. A useful ledger records enough of those relationships to calculate the requested collective event. The cooperative might therefore experiment with diversification across environments as well as across loans. It could compare changing borrowers, changing the shared production exposure, and acquiring a more informative observation of the common state. Each intervention changes a different term in the model. ## Let exchange choose a distribution Follow two owners after the market closes. Let their positive wealths be $w_1$ and $w_2$. Choose a small model with transfers toward equality and proportional external gains or losses: $$ \begin{aligned} dw_1&=\gamma(w_2-w_1)dt+\sigma w_1dB_1,\\ dw_2&=\gamma(w_1-w_2)dt+\sigma w_2dB_2. \end{aligned} $$ The exchange rate is $\gamma>0$, and the noise strength is $\sigma>0$. The exchange terms sum to zero. The shocks allow combined wealth to grow or shrink. A Brownian increment $dB_i$ has mean zero and variance $dt$; its typical size is proportional to $\sqrt{dt}$. Let the two shocks have correlation $-1\leq\chi<1$, written $dB_1dB_2=\chi\,dt$. We interpret these as **Itô stochastic differential equations**, where coefficients of a random increment use the state at the beginning of the small interval. Exchange with multiplicative random growth has a substantial econophysics history, including [Bouchaud and Mézard’s model](https://arxiv.org/abs/cond-mat/0002374). Here the specified two-person Itô rule supplies the mechanism we will calculate. Return to a familiar coordinate: $$ \rho=\frac12\log\frac{w_1}{w_2}. $$ Positive rapidity means owner 1 holds more; zero means equal wealth. Applying Itô’s change-of-variable rule gives $$ d\rho=-\gamma\sinh(2\rho)dt +\sigma\sqrt{\frac{1-\chi}{2}}\,dB, $$ where $\sinh x=(e^x-e^{-x})/2$. To see the cancellation, the drift of $\log w_1$ contains $\gamma(w_2/w_1-1)-\sigma^2/2$, and the other log wealth has the corresponding reversed ratio. Subtracting the two logs cancels their equal Itô corrections. The noise difference has variance proportional to $1-\chi$: shocks shared equally by both owners cancel from relative wealth. The inward pull grows strongly at extreme ratios. Define the dimensionless exchange-to-relative-noise ratio $$ k=\frac{\gamma}{\sigma^2(1-\chi)}. $$ At a stationary distribution, the probability flow from the drift balances the flow from diffusion. Setting the net current to zero gives $$ \frac{d}{d\rho}\log f_\rho(\rho) =-4k\sinh(2\rho), $$ and integrating produces $$ f_\rho(\rho)\propto e^{-2k\cosh(2\rho)}. $$ Here $\cosh x=(e^x+e^{-x})/2$, and $f_\rho$ is density per unit rapidity. The proportionality symbol means divide by the total integral to make the density integrate to one. ## The same owners, measured with two rulers The wealth share is $$ p=\frac{w_1}{w_1+w_2}=\frac{1+\tanh\rho}{2}. $$ Write $z=p(1-p)$. Then $\cosh(2\rho)=1/(2z)-1$ and $d\rho/dp=1/(2z)$. Preserving probability under this coordinate change gives, for $00$, target share $0<\theta<1$, and diffusion parameter $D>0$. This is a **Jacobi diffusion**. The inward drift and the square-root noise select a zero-current stationary Beta law with $$ \alpha=\frac{\kappa\theta}{D},\qquad \beta=\frac{\kappa(1-\theta)}{D}. $$ The optional dashed curve in the lab uses $\kappa=4$, $\theta=1/2$, and $D=1$, giving $\operatorname{Beta}(2,2)$. It is a comparison between two specified mechanisms. Asking which transactions could generate each drift and noise law turns the family resemblance into an economic modeling program. There is another route to larger families: keep the mechanisms of observation and the structure of the evidence itself. For that, we need to return to the shops and inspect their clocks. ## Let the observation record choose a larger family ### Ask the Gamma parents The wealth experiment has opened a trapdoor: a variable between zero and one can have many possible laws. To follow it down, visit two shops. Let $X>0$ and $Y>0$ be their unknown sales intensities, measured in sales per hour. Assume each shop runs a Poisson process: during a short interval $dt$, its chance of one sale is approximately its intensity times $dt$. Conditional on the two intensities, the processes run independently. A **Gamma distribution** describes a positive quantity. Write $X\sim\Gamma(a,r_A)$, where $a>0$ is its shape and $r_A>0$ its **rate parameter**. Its density is proportional to $x^{a-1}e^{-r_Ax}$, and its mean is $a/r_A$. The Gamma rate is the reciprocal of its scale. Here it has units of hours, because its product with a sales intensity must be dimensionless. Think of shape as a sales tally and rate as exposure time in a chosen prior model. Multiplying an intensity’s density by a Poisson likelihood adds observed sales to shape and elapsed time to rate. Give the shops independent Gamma beliefs, $$ \begin{aligned} X&\sim\Gamma(a,r_A),\\ Y&\sim\Gamma(b,r_B),\\ p&=\frac{X}{X+Y},\\ \lambda&=\frac{r_A}{r_B}. \end{aligned} $$ The next combined arrival comes from A with probability $p$. When $r_A=r_B$, the share $p$ is Beta. When the rates differ, the same ratio construction gives the **Libby-Novick distribution**, a generalized Beta law for $0