Markets that learn

Miraia S. Chiou

Survey · Economics

Markets that learn

Pay to move a market quote, spend a business opportunity on learning, and turn an experiment toward what a firm still cannot distinguish.

2026-09-08 · 24 min read · 3 interactive demos

Imagine a little bakery deciding whether to sell a new kind of bread. Its owner can ask customers what they expect, sell advance orders, try a small batch, or change the price and watch what happens. Each action opens a different window onto the future. Some produce money, some produce observations, and some produce both.

Now give the bakery a notebook that remembers how its forecasts were formed. Two entries might both say “50% chance of success.” One comes from no observed trials. The other follows ten successes and ten failures. Those equal answers can lead to different business decisions, because tomorrow’s result can move the two beliefs by different amounts.

Epistemic Physics invites us to follow those differences through an economy. A forecast becomes a trade. A payoff difference becomes a choice. An experiment changes which possibilities the next decision can distinguish. Along the way, we will meet a market register with a hyperbolic curve, a bus whose paint color should not create extra demand by arithmetic alone, and an information matrix living inside a light cone.

The adventure begins with a deceptively practical question: if moving an opinion requires a trade, how much should the move cost?

A cash register with a curved opinion

Suppose there are two mutually exclusive outcomes: the bakery’s new bread succeeds, or it does not. A claim on an outcome pays one unit if that outcome occurs. An automated market maker offers these claims using a cost function. If its outstanding quantities are $Q = (Q_{1}, Q_{2})$ , take

\begin{aligned} C (Q) & = b \log (e^{Q_{1} / b} + e^{Q_{2} / b}), \\ b & > 0. \end{aligned}

The liquidity scale $b$ controls how much trading is required to move the quoted prices. Claims and costs use the normalized unit payout. This construction is the cost-function form of a logarithmic market scoring rule, associated with Robin Hanson’s work on information markets.

The price of a tiny additional claim is the derivative of cost with respect to its quantity:

\begin{aligned} p_{1} & = \frac{\partial C}{\partial Q_{1}} \\ = \frac{e^{Q_{1} / b}}{e^{Q_{1} / b} + e^{Q_{2} / b}}, \\ p_{2} & = 1 - p_{1} . \end{aligned}

Buying a success claim increases $Q_{1}$ and raises its next quoted price. A large purchase therefore cannot be priced by multiplying its entire quantity by the opening quote. The register moves while the customer shops.

Introduce the rapidity, half the natural logarithm of the price odds:

ρ = \frac{Q_{1} - Q_{2}}{2 b} = \frac{1}{2} \log \frac{p_{1}}{p_{2}} .

Factoring the two exponentials gives a familiar pair:

\begin{aligned} C & = \frac{Q_{1} + Q_{2}}{2} \\ + b \log (2 \cosh ρ), \\ 2 p_{1} - 1 & = \tanh ρ . \end{aligned}

The hyperbolic functions obey $\cosh ρ = (e^{ρ} + e^{- ρ}) / 2$ and $\tanh ρ = (e^{ρ} - e^{- ρ}) / (e^{ρ} + e^{- ρ})$ . The same coordinate that lets Epistemic Physics express a binary balance as a velocity now expresses a market quote. Here its physical handle is outstanding claims. The parameter $b$ has an economic calibration; the formula does not identify currency with trial amount.

Before touching the register, make a prediction. If the old quote is 50%, will moving it to 80% cost more or less than pricing the whole purchase at 50%?

The next claim has a different price

Work the market's order ticket

Buy or sell claim packets, then close your ticket when its expected profit looks best. Each claim pays one unit if the event happens. Your belief stays fixed while your trades move the market quote.

Market briefing

Your belief: 80% · Opening quote: 50%

0%100%

Gold dot: market quote · Teal line: your belief

Current quote: 50.0%
Expected profit: 0.00
Net cash paid: 0.00

Your position: 0.00 claims. Negative claims are short sales; negative payment is cash received.

Fine trades, price curve, and mathematical receipt

Every trade uses the integrated logarithmic market cost. Opposite trades undo a position exactly in this fee-free model. The quote is kept between 2% and 98% to avoid infinite endpoint inventories.

C (Q_{1}, Q_{2}) = b \log (e^{Q_{1} / b} + e^{Q_{2} / b}), p_{1} = \frac{e^{Q_{1} / b}}{e^{Q_{1} / b} + e^{Q_{2} / b}}

Liquidity b = 3; the second inventory stays at zero. Cash cost ΔC = 0.00. Pricing the entire net trade at its opening quote gives 0.00. The difference is 0.000.

Δ C - p_{o l d} Δ Q_{1} = b D_{K L} (p_{o l d} ‖ p_{n e w})

Expected profit is rΔQ₁ − ΔC, using your belief r. Moving the quote to r maximizes that expectation. Overshooting it reduces expected profit. Your remaining shortfall is 0.578 = bD(r ∥ p). Natural logarithms are used.

Chosen model: risk-neutral trader, no fees, purchases and short sales allowed, no budget or collateral constraint. This is a toy prediction-market experiment. The market quote and trader's belief are separate probabilities. The rapidity coordinate is half the log odds: currently 0.000.

Two divergences, two different bills

Let a trade move inventory from $Q$ to $Q^{'}$ , with prices changing from $p$ to $p^{'}$ . The actual payment is $Δ C = C (Q^{'}) - C (Q)$ . The old quote would have charged $p \cdot (Q^{'} - Q)$ , where the dot means multiply corresponding entries and add.

Their difference has a precise informational form:

Δ C - p \cdot (Q^{'} - Q) = b D_{KL} (p ‖ p^{'}) .

The Kullback-Leibler divergence is

D_{KL} (p ‖ r) = \sum_{i} p_{i} \log \frac{p_{i}}{r_{i}} .

It measures a directional mismatch between probability distributions. The first argument supplies the weights in the average. Swapping the arguments usually changes the value.

We can check the orientation without trusting the box. From the market price formula,

\log p_{i} = \frac{Q_{i} - C (Q)}{b} .

Subtract the corresponding equation for $p_{i}^{'}$ , multiply by $p_{i}$ , and add over outcomes. The quantities assemble into exactly the extra cost above the old quote. The old distribution belongs first because the tangent bill uses the old prices.

Now stand on the trader’s side. Let $r$ be the trader’s own predictive distribution. For a risk-neutral trader, expected profit from this trade is

\begin{aligned} r \cdot (Q^{'} - Q) - Δ C \\ = b D_{KL} (r ‖ p) \\ - b D_{KL} (r ‖ p^{'}) . \end{aligned}

If arbitrary purchases and short sales are allowed, with no fees or position constraints, the trader maximizes this expectation by choosing $p^{'} = r$ . The remaining divergence vanishes, leaving

Best expected profit = b D_{KL} (r ‖ p) .

For $b = 3$ , an old quote of 50% and trader belief of 80% give about $0.578$ expected units. This expectation is taken under the trader’s belief; the realized outcome can still produce a loss.

The two bills answer different questions. Market curvature compares a trade with its opening tangent. Expected information advantage compares the trader’s belief with the opening market quote. A single reversed divergence would quietly swap those meanings.

There is another geometric clue nearby. For a small price displacement $δ p$ ,

b D_{KL} (p ‖ p + δ p) \approx \frac{b (δ p)^{2}}{2 p (1 - p)} .

The factor $1 / [p (1 - p)]$ is the binary Fisher information in probability coordinates. A fixed absolute move near an endpoint is more costly than the same tiny move near the middle. The market’s curvature has put a price on local distinguishability.

A menu should remember what its boxes mean

The bakery must also choose actions. Should it advertise, lower prices, or develop another recipe? A choice fraction and a belief probability can share mathematics while describing different things.

Write $q_{i}$ for the fraction of attention or probability of choosing action $i$ , and $V_{i}$ for its payoff. Choose a positive reference allocation $q_{i}^{0}$ , summing to one, and a parameter $κ > 0$ that prices departure from that allocation. Define the objective $J (q)$ :

\begin{aligned} J (q) & = \sum_{i} q_{i} V_{i} \\ - κ D_{KL} (q ‖ q^{0}) . \end{aligned}

We maximize $J (q)$ over allocations satisfying $\sum_{i} q_{i} = 1$ . The first term rewards useful actions. The second discourages changing the allocation too far from its reference. Both terms have payoff units because $κ$ converts a natural-log unit of divergence into payoff cost. We are choosing this objective explicitly.

A multiplier enforcing the sum gives

\begin{aligned} q_{i}^{*} & = \frac{q_{i}^{0} e^{V_{i} / κ}}{Z}, \\ Z & = \sum_{j} q_{j}^{0} e^{V_{j} / κ} . \end{aligned}

One identity both verifies the solution and connects it to the market register:

\begin{aligned} J (q) & = κ \log Z \\ - κ D_{KL} (q ‖ q^{*}) . \end{aligned}

The last term is nonnegative, so the maximum is $κ \log Z$ . Logarithms of sums of exponentials can describe both a market cost and an information-penalized choice. With uniform reference weights, they differ only by the corresponding constant normalization and the meanings of their variables.

For two actions, their choice rapidity obeys

\frac{1}{2} \log \frac{q_{1}^{*}}{q_{2}^{*}} = \frac{1}{2} \log \frac{q_{1}^{0}}{q_{2}^{0}} + \frac{V_{1} - V_{2}}{2 κ} .

A payoff advantage translates half-log-odds. In this equation the result is a decision policy, the distribution of actions the bakery chooses. Its beliefs about demand remain separate.

Matějka and McKay’s rational-inattention model derives conditional choice probabilities with this exponential structure. Their reference frequencies are endogenous: determined inside the optimization and consistent with the resulting unconditional choice frequencies. Our fixed $q^{0}$ problem isolates a useful mathematical piece. It does not solve that whole information-acquisition problem.

Now try a menu trick in your head. A traveler has equally attractive choices: car or bus, each with reference weight $1 / 2$ . We rename the bus “red bus” and “blue bus,” with identical payoffs and availability. What should happen to the total bus fraction?

Representation	Reference weights	Total bus choice at equal payoffs
Car, bus	$1 / 2, 1 / 2$	$1 / 2$
Car, red bus, blue bus; split the bus weight	$1 / 2, 1 / 4, 1 / 4$	$1 / 2$
Car, red bus, blue bus; assign fresh equal weights	$1 / 3, 1 / 3, 1 / 3$	$2 / 3$

Preserving the bus’s total reference mass preserves its total chosen probability, even when bus and car payoffs differ. Both bus terms contain the same exponential, so their divided weights simply add back together.

This is a useful meeting point with Epistemic Physics: refining a description should preserve its total weight when it describes the same underlying possibility. A new departure time, lower fare, or genuinely different route can change the economic option. A second colored box alone has no such content. The choice model needs to know which kind of change happened.

The bakery buys knowledge with a batch of bread

Suppose the bakery has two selling opportunities left before a seasonal stall closes. A familiar recipe succeeds with known probability $s = 0.55$ . A new recipe has a fixed unknown success probability $P$ . A success pays one unit and a failure pays zero. Both recipes have the same cost, and the first result arrives before the second decision.

Represent uncertainty about $P$ by a Beta distribution, a distribution over probabilities between zero and one. Starting from the chosen $Beta (1 / 2, 1 / 2)$ prior, each fresh success adds one to its first shape parameter $α$ ; each fresh failure adds one to its second parameter $β$ .

Compare two possible notebooks:

History	Current distribution	Predictive mean
No observed trials	$Beta (0.5, 0.5)$	$0.5$
Ten successes, ten failures	$Beta (10.5, 10.5)$	$0.5$

The predictive mean $μ = α / (α + β)$ is the probability assigned to the next success. It is not necessarily the distribution’s most likely point. The untried Beta distribution has high density near zero and one, despite its mean of one half.

If only one opportunity remains, familiar bread wins in both notebooks: $0.55 > 0.50$ . With two opportunities, trying the new bread first can also teach us what to do next.

For a symmetric $Beta (a, a)$ state, the new predictive means after success and failure are

\begin{aligned} μ_{s u c c e s s} & = \frac{a + 1}{2 a + 1}, \\ μ_{f a i l u r e} & = \frac{a}{2 a + 1} . \end{aligned}

After trying the new recipe, the bakery can select whichever recipe has the better mean. Its expected total payoff is therefore

\begin{aligned} V_{t r y} & = \frac{1}{2} \\ + \frac{1}{2} max (s, μ_{s u c c e s s}) \\ + \frac{1}{2} max (s, μ_{f a i l u r e}) . \end{aligned}

For the untried recipe, one success moves the mean to $0.75$ and one failure moves it to $0.25$ . The calculation is

\begin{aligned} V_{t r y} & = 0.50 + 0.50 (0.75) \\ + 0.50 (0.55) \\ = 1.15 . \end{aligned}

For the experienced notebook, even a success only raises the mean to $11.5 / 22 \approx 0.523$ . Both possible second decisions still favor familiar bread, giving

\begin{aligned} V_{t r y} & = 0.50 + 0.50 (0.55) \\ + 0.50 (0.55) \\ = 1.05 . \end{aligned}

Familiar bread first supplies no information about the new recipe. In either notebook, using it first and then choosing the better recipe gives $1.10$ . The optimal first actions have separated:

Starting notebook	Try new first	Familiar first	Best first action
Untried	$1.15$	$1.10$	Experiment
Twenty balanced trials	$1.05$	$1.10$	Familiar recipe

Predict what happens when you shorten the horizon to one opportunity. Then extend it and make a few decisions yourself.

Earn now, learn for tomorrow

Run the bakery for a few days

Choose today's recipe before opening the sales receipt. Familiar bread succeeds 55% of the time. The new recipe has an unknown, fixed success rate. Each success earns one unit.

Bakery shift

Day 1unopenedDay 2unopened

Starting notebook: no observed trials. Both start at a 50% predictive mean.

What will you bake on day 1?

Days still available: 2
New-recipe predictive mean: 50.0%
Units earned: 0

Decision analysis and the model

These values include the best later choices after each possible outcome.

New recipe first: 1.150
Familiar bread first: 1.100

Expected total units from now. The model favors the new recipe. Actual receipts can differ.

The new recipe uses Beta(0.5, 0.5). The current probability is its predictive mean. With no observations, Beta(0.5,0.5) puts more density near zero and one, despite its 50% mean.

Under the starting notebook, two consecutive new-recipe successes have probability 37.5%. Learning matters when the outcome can change a later action.

One rate is sampled from the starting Beta law and held fixed. Trials are independent conditional on that rate. Familiar outcomes use a known 0.55 rate. Equal costs, no discounting, no recipe improvement, and unique trial lineage are chosen assumptions. Replay uses the same underlying daily random draws, not an extra independent sample.

Learning earns money when it changes a later action

The untried experiment sacrifices $0.05$ expected payoff today. Its expected second-period payoff rises from $0.55$ to $0.65$ , an improvement of $0.10$ . The net advantage is $0.05$ . The extra value comes from the ability to act differently after different results.

Interestingly, the expected updated mean does not rise:

E [μ_{n e x t}] = μ .

The winning move is averaging the best action after observing, rather than selecting one action using the average forecast. The maximum function bends upward, so

E [max (s, μ_{n e x t})] \geq max (s, E [μ_{n e x t}]) .

An observation can have value without making us more optimistic on average. It helps by sorting futures into cases that deserve different responses. For the experienced notebook in this two-period example, neither update changes the chosen response, so that particular learning opportunity has no decision value.

Epistemic Physics calls ease of changing a belief malleability. Here malleability acquires an economic role through a future decision. Its value depends on the payoff threshold, the available alternatives, and how many chances remain to use what is learned.

The longer-horizon calculation follows the same logic. Let $V_{h} (α, β)$ be the best expected payoff with $h$ opportunities left, and set $V_{0} = 0$ . At the current state, write $F_{h}$ for the value of choosing familiar bread first, and $E_{h}$ for the value of experimenting first. With $μ = α / (α + β)$ ,

\begin{aligned} F_{h} & = s + V_{h - 1} (α, β), \\ E_{h} & = μ \\ + μ V_{h - 1} (α + 1, β) \\ + (1 - μ) V_{h - 1} (α, β + 1) . \end{aligned}

Choose the better branch:

V_{h} (α, β) = max {F_{h}, E_{h}} .

This is a small dynamic program: solve the last decision, then use its value to solve earlier decisions. The demo evaluates both branches, including their possible future updates.

Our trial model keeps the new recipe’s $P$ fixed. Conditional on it, fresh trials are independent. Each result has its own lineage, and reopening a sales report does not produce a new trial. Distinct observation IDs alone cannot establish statistical independence; the observation model supplies that additional assumption.

What if making the recipe improves the baker’s skill, customers recommend it to friends, or a rival changes prices? Then doing the experiment also changes the world being estimated. We could give skill, reputation, and demand their own evolving states. The experiment’s value would include both learning and changing those states, a promising next step beyond this fixed-rate bakery.

Turn the experiment toward a question the data cannot answer

A successful batch suggests that something worked. It may leave the bakery unsure which part worked. Perhaps it lowered the price and increased advertising together. Repeating that combination can estimate the combined response while leaving the separate coefficients unresolved.

Choose standardized units for price changes, advertising changes, and response. For this new problem, write $β = (β_{1}, β_{2})$ for the two regression coefficients. A controlled observation follows

\begin{aligned} y_{i} & = x_{i}^{T} β + ε_{i}, \\ x_{i} & = (\cos θ_{i}, \sin θ_{i}), \\ ε_{i} & \sim N (0, 1 / w_{i}) . \end{aligned}

The superscript $T$ turns a column into a row, so the product adds the two feature-times-coefficient contributions. The angle $θ_{i}$ selects an intervention direction. The weight $w_{i}$ is precision, inverse noise variance. Noise is independent across observations, the baseline is controlled, and the coefficients remain fixed.

For Gaussian observations, likelihood curvature is described by the Fisher information matrix

F = \sum_{i} w_{i} x_{i} x_{i}^{T} .

Its quadratic form says how sharply the likelihood distinguishes nearby coefficient values. When the design has full rank, the coefficient estimator has covariance $F^{- 1}$ . Boyd and Vandenberghe’s treatment of experiment design develops the connection between selecting observations and controlling this error covariance in Section 7.5.

Now expand the trigonometric squares. Define

\begin{aligned} A & = \sum_{i} w_{i}, \\ X & = \sum_{i} w_{i} \cos 2 θ_{i}, \\ Y & = \sum_{i} w_{i} \sin 2 θ_{i} . \end{aligned}

Then

\begin{matrix} F = \frac{1}{2} (\begin{matrix} A + X & Y \\ Y & A - X \end{matrix}), \\ 4 det F = A^{2} - X^{2} - Y^{2} . \end{matrix}

A positive semidefinite matrix has no negative quadratic directions. In these coordinates that condition becomes $A \geq \sqrt{X^{2} + Y^{2}}$ : the future Lorentz cone. Each single observation contributes a point on its boundary. Combining different information axes can carry the sum inside.

The doubled angle has a simple job. Reversing both input changes sends $x_{i}$ to $- x_{i}$ , but leaves $x_{i} x_{i}^{T}$ unchanged. Information sees an axis, which returns to itself after $180^{\circ}$ ; its cone direction makes a full turn.

A tiny experiment workshop

Build a test that sees both directions

A shop has one price-only trial. Spend the remaining test slots to separate price from advertising. Turn the teal test, then place it. Aim for a compact, nearly round error region.

Turn your next test

30°

One direction is still open

0°empty

Fresh trials placed: 1 / 2
Weakest-direction precision: 0.00 / 1

Repeating this same axis tightens one combination but leaves another completely unresolved. Try a genuinely different direction.

Free exploration, equations, and model

Each line is a unit feature vector in standardized price/advertising coordinates. Each fresh trial has independent Gaussian noise of variance one and its own lineage. Reversing both changes gives the same information axis. The baseline is controlled and coefficients stay fixed.

F = \sum_{i} x_{i} x_{i}^{T}, 4 det F = A^{2} - X^{2} - Y^{2}

A = 1, X = 1.000, Y = 0.000; determinant = 0.000. The smallest eigenvalue is the weakest-direction precision. At fixed A its maximum is A/2.

The plotted region is dᵀFd ≤ 1 in coefficient-error coordinates. When bounded, it encloses 39.3% of estimation errors under this Gaussian model and has area unbounded. The fixed plot window can clip long regions. This matrix/cone correspondence is algebraic; causal claims require the intervention assumptions too. Copying a report would add no fresh precision.

Drag or touch the teal handle, focus it for arrow keys, or use the turn buttons. The rose lines are placed tests; the teal dashed line is only a proposal.

Diversity of directions becomes an area

The determinant can also be written

det F = \sum_{i < j} w_{i} w_{j} \sin^{2} (θ_{i} - θ_{j}) .

Every pair contributes its weights multiplied by the squared sine of its angular separation. Parallel pairs contribute zero. Perpendicular pairs contribute their full weight product. The sine is the area made by two unit vectors; its square measures how effectively their two questions span a plane.

Two unit-weight tests along one axis give $det F = 0$ . Two perpendicular unit-weight tests give $det F = 1$ . Under the Gaussian model, a fixed quadratic error contour has area proportional to $1 / \sqrt{det F}$ . Maximizing this determinant is called D-optimal design.

For two perpendicular tests with fixed total precision $A = w_{1} + w_{2}$ , the product $w_{1} w_{2}$ is largest when their weights are equal. Equal precision in complementary directions produces the smallest ellipse under this budget and these available tests. Different noise levels, experiment costs, or a particular coefficient of interest can change the useful design objective.

Try collecting ten fresh repetitions along a single direction. The likelihood becomes sharper across its strip, while remaining flat along the unmeasured combination. Turn one test sideways and a finite ellipse appears. A copied report would not produce the first change; a genuinely new observation supplies another contribution under the independent-noise model.

Fresh trials can still share disturbances. Two branches of the bakery might face the same storm or advertising campaign. If their errors have a known, nonsingular covariance matrix $Σ$ , the appropriate information is

F = X_{d a t a}^{T} Σ^{- 1} X_{d a t a},

where the rows of $X_{d a t a}$ are the feature vectors. This replaces the independent weighting rule with one that accounts for the shared fluctuations. A duplicate appearance of one observation should first be recognized through its lineage; it is not a second noisy measurement that deserves another independent term.

The coordinate scales matter too. Measuring price in cents instead of dollars changes the numerical coefficient and its information matrix together. We compare designs after fixing those scales. The physical question is how well the observations distinguish the effects the bakery cares about, under the experiments it can actually afford. A beautiful right angle on a diagram may require an unavailable intervention; then the design problem becomes choosing the most useful feasible combination.

The cone and the ellipse occupy different spaces. The cone describes possible information matrices. The ellipse describes errors in estimated coefficients. This correspondence does not turn the positive Fisher metric on coefficients into a Lorentzian metric. Instead, the determinant connects those two pictures through a shared algebraic structure.

Nor can matrix rank establish causality. If price and advertising respond to an unrecorded holiday rush, the assumed observation model may fail. A causal interpretation needs an intervention design and assumptions that connect coefficient changes to what would happen under the interventions. Directional diversity answers which combinations the fitted model can distinguish.

A second claim can ask a second-order question

Return to the market with our two bakery notebooks. A claim paying on the next new-recipe success has the same predictive value, $1 / 2$ , under both. Could a different claim reveal something the first price concealed?

Suppose a claim pays one unit only if two fresh new-recipe trials both succeed. Under the fixed- $P$ , conditionally independent model, its predictive probability is $E [P^{2}]$ :

\begin{aligned} E [P^{2}] & = μ^{2} + Var (P) \\ = \frac{α (α + 1)}{(α + β) (α + β + 1)} . \end{aligned}

The untried notebook gives $0.375$ . The twenty-trial notebook gives approximately $0.2614$ . They agree about one future result and disagree about the pair. Their different uncertainty about a shared underlying rate changes the dependence between future outcomes.

This suggests an intriguing market design: offer claims that ask different orders of predictive questions. Under the chosen Beta family and successful elicitation of these probabilities, a mean and a second moment can recover the family’s concentration. Without that family assumption, the same two moments leave many distributions possible. If trials share additional shocks or the recipe changes between them, the claim asks a different model-dependent question again.

A learning economy could connect such markets to firms that choose experiments, with budgets determining which observations become possible. The market rewards specified predictive distinctions; the firm decides whether producing them can improve future actions. Lineage keeps each result attached to the trial that generated it. Experimental direction determines what the collection can resolve.

Our bakery’s next breakthrough might come from a more informative claim, a longer window for learning, or a test pointed away from yesterday’s favorite direction. These are three concrete ways to change what an economy can discover, even when today’s headline forecast stays exactly the same.