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,

$$m=2q-1,-1\le m\le 1.$$

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

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 supplies a connection between economic incentives and the collective equations familiar from statistical physics.

The half-log-odds coordinate is especially revealing:

$${\rho }_q=\frac{1}{2}\log \frac{q}{1-q}=h+Jm.$$

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:

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?

Choices feed back into choices

A nudge that lasts

Start evenly divided

Thirty-six people are considering a shared technology. Teal means adopt; rose means decline. Click a person to change the starting balance, then let the collective response unfold.

+1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18−19−20−21−22−23−24−25−26−27−28−29−30−31−32−33−34−35−36

The tiles round the continuous adoption share to 36 people. Clicking them sets a new starting state for the displayed adjustment rule.

What happens after the nudge?

Let choices settle Pulse toward adoption Pulse toward declining

Interaction strength J 1.50 Private advantage h 0.00 Remove private advantage

Adopting share (1 + m) / 2

50.0%

Collective direction m

0.000

Stable equilibria at this h

7.1% · 92.9%

Click one person, then let the choices settle.

Where choices can settle

Move the gold handle horizontally, or focus it and use arrow keys, to change h. Teal branches attract nearby states; the dashed rose branch repels them.

The route taken

Collective direction mPrivate advantage h

Adjustment rule: dm/dt = tanh(h + Jm) − m. At J = 1.5 and h = 0, the two stable adoption shares are about 7.1% and 92.9%. The dashed middle equilibrium is unstable.

We can locate the switching points without guessing. At equilibrium,

Differentiate this curve with respect to $m$. For , a fold occurs where its derivative is zero, so a stable and an unstable solution meet:

For $J=1.5$, the private-advantage thresholds are approximately $\pm 0.20755$. 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. Its introduction to the chained index explains the role of expenditure shares.

Choose two goods, fix the second price at one, and write

$$r=\log (p_1/p_2)=\log p_1,s=s_1.$$

Now take a rectangular walk through the price-share plane:

At the end, both prices and both expenditure shares match the beginning. Keep total expenditure at one throughout. Purchased quantities are then

$$q_1=\frac{s}{e^r},q_2=1-s.$$

They return too. Yet the chained index has multiplied by

$${\mathcal{P}}_{\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{p}}=e^{0.6\log 2}=2^{0.6}\approx 1.516.$$

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?

A changing ruler remembers its route

The price index takes a walk

Restore the example loop

The gold point records a journey through price and expenditure share. Drag it to draw your own route, or visit the corners with the buttons. Can the trip change the chained index even after prices and quantities come home?

Clear and draw Undo last leg Reverse the route Close the loop Hold the share fixed

Original price · 80%Double price · 80%Double price · 20%Original price · 20%

Chained index

1.5157×

Direct endpoint index

1.0000×

Accumulated log change ∫ s dr

0.4159

The route is closed: both prices and both quantities match their starting values. The chained index is 1.5157; the direct comparison is 1.

Endpoint audit, with total expenditure fixed at one

Good	Price: start → now	Quantity: start → now
1	1.000 → 1.000	0.800 → 0.800
2	1.000 → 1.000	0.200 → 0.200

Each straight leg uses the average of its endpoint expenditure shares, the two-good Törnqvist link. Good 2 always costs 1; quantities are q₁ = s / exp(r) and q₂ = 1 − s. Reversing the entire route negates its log change.

The geometric calculation is simply

$$\oint sdr=(0.8-0.2)\log 2.$$

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:

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:

$$q_{AB}=q_{BC}=q_{CA}=\frac{2}{3},$$

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

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 $K=\sum _i{D}_i$.

First choose one common, uncertain default propensity,

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:

As $n$ grows, the first contribution shrinks and a variance floor remains:

$$\underset{n\to \mathrm{\infty }}{lim}\mathrm{Var}(K/n)=\frac{\mu (1-\mu )}{\tau +1}.$$

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 $\mathrm{Beta}(1,19)$, 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 $8.41\times {10}^{-11}$. 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?

The joint structure carries the risk

One parent changes the portfolio

Restore the 5% example

Each loan has the same individual default probability in both models. Choose how its uncertain propensity is connected to other loans. Predict the effect on a whole portfolio before drawing a cohort.

One shared PDraw one environment; every loan uses it. Independent PᵢDraw a separate propensity for every loan.

Six representative loans are shown; the same connection rule applies to all 1,000 loans. Each new cohort redraws its parent propensity or propensities from Beta(1.00, 19.00).

Individual default probability 5% Propensity concentration τ 20

20 loans100 loans1,000 loans

Exact distribution of the number of defaults

One shared PIndependent PᵢSample cohorts / 10% threshold

Chance of at least 100 defaults

13.79%

Variance of default fraction K/n

0.002307

Variance as the portfolio grows

0.002262

Draw a cohortDraw 20 cohorts

Both structures give an expected 50.0 defaults. Under one shared P, increasing the number of loans leaves the common environment uncertain.

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:

The exchange rate is , and the noise strength is . The exchange terms sum to zero. The shocks allow combined wealth to grow or shrink. A Brownian increment $d{B}_i$ has mean zero and variance $dt$; its typical size is proportional to $\sqrt{dt}$. Let the two shocks have correlation , written $d{B}_{1}d{B}_2=\chi dt$.

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. Here the specified two-person Itô rule supplies the mechanism we will calculate.

Return to a familiar coordinate:

$$\rho =\frac{1}{2}\log \frac{w_1}{w_2}.$$

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 ,

The factor $d\rho /dp$ is a Jacobian: it measures how much rapidity interval fits into a small share interval. It changes the height of the density curve while preserving probabilities of corresponding events.

A striking result follows. The rapidity density always peaks at zero. The share density has two equal peaks when ,

$$p_{\pm }=\frac{1\pm \sqrt{1-4k}}{2},$$

and a single central peak when $k\ge 1/4$. The same stationary population can therefore have two density peaks in one coordinate and one in another. The peaks count probability per differently sized intervals. The chance that one owner holds over 90% is the same whichever ruler we use.

Predict the paired pictures: strengthen exchange, then restore it and make the shocks mostly shared. Should both changes pull relative wealth toward equality? Follow the same simulated owners through both histograms.

Exchange, shocks, and the shape of a density

Two fortunes, two rulers

Restore equal starts

Start 192 independent two-person economies at equal wealth. Exchange pulls each pair toward equality; proportional shocks pull their fortunes apart. The same simulated pairs appear in both histograms, measured once by wealth share and once by half-log wealth ratio.

Run the economies Advance 5 time units Advance 20 time units Restart this rule

Exchange rate γ 0.05 Proportional noise σ 1.00

Opposing shocks χ = −0.8Independent shocks χ = 0Mostly shared shocks χ = 0.8

Changing a rate or shock link restarts the cohort at equal wealth. The curve shows the stationary distribution selected by that rule; the gold bars show the current cohort after you advance time.

Person 1: 50.0%Person 2: 50.0%

Share p = w₁ / (w₁ + w₂)

Rapidity ρ = ½ log(w₁ / w₂)

Current cohortStationary exchange law

Model time

0.0

Exchange / relative noise k

0.050

Stationary mean squared imbalance

0.0970

Current cohort squared imbalance

0.0000

Stationary chance one holds over 90%

25.68%

The share density has two equal peaks, at 5.3% and 94.7%. The rapidity density has one peak at zero. Both describe the same stationary pairs and assign the same probability to the over-90% event.

Compare a rule that gives Beta(2,2)

One pair’s recent journey

Simulation uses dρ = −γ sinh(2ρ) dt + σ√((1 − χ)/2) dB, starting at ρ = 0. The blue curve is a density in its labeled coordinate; histogram heights need not agree between coordinates.

The mechanism also tells us which distributional family to expect. Our proportional-wealth shocks give the displayed exponential-in-reciprocal-share kernel. We can choose a different transaction-scale rule that has a Beta stationary law. For a share $p$, take

with restoring rate , target share , and diffusion parameter . This is a Jacobi diffusion. The inward drift and the square-root noise select a zero-current stationary Beta law with

$$\alpha =\frac{\kappa \theta }{D},\beta =\frac{\kappa (1-\theta )}{D}.$$

The optional dashed curve in the lab uses $\kappa =4$, $\theta =1/2$, and $D=1$, giving $\mathrm{Beta}(2,2)$. 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 and 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 $X\sim \mathrm{\Gamma }(a,r_A)$, where is its shape and its rate parameter. Its density is proportional to $x^{a-1}{e}^{-r_{A}x}$, 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,

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 :

Here $B(a,b)=\mathrm{\Gamma }(a)\mathrm{\Gamma }(b)/\mathrm{\Gamma }(a+b)$ is the Beta normalizing function. The Gamma function $\mathrm{\Gamma }(a)$ extends factorials, with $\mathrm{\Gamma }(n)=(n-1)!$ for positive integers. The extra denominator remembers the unequal exposure scales. Setting $\lambda =1$ makes that factor disappear. This ratio construction belongs to the family introduced by Libby and Novick.

There is also a remarkably simple geometric description. Draw $W\sim \mathrm{Beta}(a,b)$, then set

An unequal-rate Gamma share is a Beta random variable translated along the half-log-odds coordinate! The whole distribution moves through this transformation. Changing only its mean would lose the shape of the cloud. A change of common time unit scales both Gamma rates equally, leaving $\lambda$ and the share law unchanged.

Keep the clock, or keep only the labels?

Now choose how many arrivals to observe. Suppose the first $n$ arrivals contain $s$ sales at A and $f$ at B, with $n=s+f$. One observer keeps the shop labels alone. Their likelihood for $p$ is

$$L_{\mathrm{l}\mathrm{a}\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{s}}(p)\propto p^s(1-p{)}^f.$$

Multiplying it by the Libby-Novick prior gives

$$f(p\mid \text{labels})\propto \frac{p^{a+s-1}(1-p{)}^{b+f-1}}{[1+(\lambda -1)p{]}^{a+b}}.$$

The powers upstairs learn about sales; the denominator keeps its original power. This is a Gauss hypergeometric distribution, often shortened to Gauss hyper. In the standard form

$$f(p)\propto p^{A-1}(1-p{)}^{B-1}(1+zp{)}^{-C},$$

our parameters are $A=a+s$, $B=b+f$, $C=a+b$, and $z=\lambda -1$. Its normalizer is $B(A,B){}_2{F}_1(C,A;A+B;-z)$. The symbol ${}_2{F}_1$ names the Gauss hypergeometric function; here its job is to express the area under that kernel. SciPy’s definition uses exactly this form. Bernoulli observations can keep updating $A$ and $B$ inside the family while $C$ and $z$ stay fixed.

The second observer also keeps the total time $t$ until arrival $n$. The joint likelihood in the intensity variables is

$$L_{\mathrm{c}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{k}}(X,Y)\propto X^s{Y}^f{e}^{-t(X+Y)}.$$

Consequently,

The two updated Gamma variables remain independent under this model, so their share has a new Libby-Novick law, with ${\lambda }_t=(r_A+t)/(r_B+t)$ and denominator power $a+b+n$. The symbol $D$ denotes the observed labels and time together. Labels and waiting time carry different information about the hidden intensities. Keeping the clock changes the likelihood. Equal initial rates make both routes coincide as a Beta update.

Predict before clicking: if the shops have different prior exposure scales, will four rapid sales teach the same thing as four slow sales with the same labels? What if the fourth sale’s receipt is copied three times?

Four receipts, two observers

Investigate the missing market clock

Reset this case

InvestigationStretch the clockCopy a receiptMake the clocks agree

Keep the four shop labels. Tap each receipt’s waiting time until the visit lasts eight hours. Which observer changes their mind?

Shop AReceipt #10.25 htap to change waitShop BReceipt #20.25 htap to change waitShop AReceipt #30.25 htap to change waitShop BReceipt #40.25 htap to change wait

4 unique arrivals · 4 reports · 1.00 hours elapsed

Reads shop labels

22.6%

Mean share of A sales

Keeps the clock too

23.3%

Mean share of A sales

Tap the four waiting-time stamps to build a slower visit. A tap cycles ¼ hour → 1 hour → 2 hours.

Explore freely, compare densities, and inspect the model

Add a fresh A saleAdd a fresh B saleRepeat a report

A prior exposure 0.25 hA prior exposure 1 hA prior exposure 4 h

Rose: labels only. Teal: labels and waiting times. Curves are normalized densities; they are distinct from the mean percentages above.

$$f_{\mathrm{l}\mathrm{a}\mathrm{b}\mathrm{e}\mathrm{l}\mathrm{s}}(p)\propto \frac{p^4(1-p{)}^6}{[1+(3)p{]}^8}$$

With timing, X ∼ Gamma(5, 5.00) and Y ∼ Gamma(7, 2.00), independently. Gamma uses shape/rate notation. The share p = X/(X+Y) has an unequal-rate Gamma-share law (Libby-Novick); labels alone give the Gauss hypergeometric family. Equal rates reduce both to Beta.

Chosen model: independent Poisson streams with fixed unknown intensities, observed from time zero until a chosen number of consecutive arrivals. Prior shapes are 3 and 5; B's prior exposure is 1 h. Gamma rate parameters measure hours, while the latent intensities measure sales/hour. Editing a waiting time edits every copy of that event. Copies supply no extra independent trials or elapsed time.

Lineage answers the receipt question. A repeated report points to the same arrival and contributes no additional sale or waiting interval. Distinct events enter under the stated process model. The richer tally logic in Money that can move follows the same idea when separate reports describe partial aspects of a transaction. Unresolved combinations can produce mixtures because several complete event histories still fit the observations.

From shares to fortunes

The shops’ intensity ratio $R=X/Y$ lives on the positive line. Write $G_a$ and $G_b$ for independent unit-rate Gamma variables with shapes $a$ and $b$. Then

$$R=\frac{r_B}{r_A}\frac{G_a}{G_b}.$$

This is a scaled Beta prime law: the odds transform of the Libby-Novick share. More broadly, choose a positive scale $q$ and power , and define

$$Z=q{\left(\frac{G_a}{G_b}\right)}^{1/h}.$$

Its density, for , is the generalized Beta prime, also called generalized Beta of the second kind, or GB2:

For large $z$, the density falls as $z^{-hb-1}$; the probability of exceeding $z$ falls as $z^{-hb}$. The Gamma ratio and a nonlinear measurement scale together create an adjustable power-law tail. Such positive distributions have been explored as income and wealth models; this stochastic wealth study connects Beta prime laws to an explicit exchange-and-noise mechanism.

Place the routes beside one another:

Each row specifies a generative or observational recipe. The earlier wealth process has a different recipe: redistribution continually pulls extreme wealth ratios inward while multiplicative shocks shake the log ratio. Its stationary odds have a generalized inverse Gaussian (GIG) distribution, with density for ,

$$f_R(R)\propto R^{-1}\exp [-k(R+R^{-1})].$$

The factor $R^{-1}$ comes from $d\rho /dR=1/(2R)$. The exponential factor penalizes both huge ratios and their reciprocals.

We can even manufacture an evidence interpretation of that last factor. Choose the scale-uniform reference weight $dR/R$, and imagine two independent Poisson counters with expected counts $kR$ and $k/R$. Observing zero in both contributes likelihood $e^{-kR}{e}^{-k/R}$, giving the same normalized stationary law for . The reference weight alone has infinite total mass; the two zero-count observations make the integral finite. This is a constructed observation model for the same density, and it suggests a question: could reciprocal exposure mechanisms explain the tendency of some exchange networks to suppress both dominance and disappearance?

The expedition now has a larger vocabulary. Beta describes one especially elegant balance. Gamma parents, unequal scales, omitted clocks, partial transaction records, and collective dynamics tell us which larger family a particular economic story needs.

The next town could learn from its own interventions

We now have the ingredients for a small economy whose history remains available to its next decision. Workshops could keep uncertain beliefs about production methods. Households could choose products through explicit payoff and attention rules. The cooperative could retain shared environments. Shops could record arrivals with identities and clocks. Contracts could specify which future actions become possible after a payment.

The adventurous step is to let these pieces change one another. Adoption changes which trials get observed. Credit changes which experiments can be attempted. A new public service changes feasible choices. Published results change expectations, which can change the outcomes those expectations predict.

Several comparisons would reveal different mechanisms:

One particularly fertile experiment would let a town choose between buying another machine and funding a shared measurement service. Machines change production capacity. Measurements change what everyone can learn about a common state. Their values depend on who receives the results, which actions become available, and whether the resulting choices change the state being measured.

A second experiment could compare organizations that immediately compress every disagreement to one score with organizations that keep a small mixture of coherent perspectives. Could the latter discover a new project that the former never proposes? The rankings table gives that question a concrete starting point.

The town’s memory would live in the relationships among trials, receipts, clocks, choices, and exchanges. A useful state would keep the pieces that let its next question be answered, even when the question changes.