--- title: "Markets that learn" author: "Miraia S. Chiou" creator: "Miraia Chiou" published: "2026-09-08" identifier: "ep_field_f8b83ee2d88949878036d5e4e1c5288b" source: "https://epistemicphysics.com/fields/economics/markets-that-learn/" citation: "https://epistemicphysics.com/cite/?work=ep_field_f8b83ee2d88949878036d5e4e1c5288b" --- # 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. [Read the illustrated article](https://epistemicphysics.com/fields/economics/markets-that-learn/) 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 $$ \begin{aligned} C(Q)&=b\log\left(e^{Q_1/b}+e^{Q_2/b}\right),\\ b&>0. \end{aligned} $$ 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](https://hanson.gmu.edu/mktscore.pdf). 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: $$ \rho=\frac{Q_1-Q_2}{2b} =\frac12\log\frac{p_1}{p_2}. $$ Factoring the two exponentials gives a familiar pair: $$ \begin{aligned} C&=\frac{Q_1+Q_2}{2}\\ &\quad+b\log(2\cosh\rho),\\ 2p_1-1&=\tanh\rho. \end{aligned} $$ The hyperbolic functions obey $\cosh\rho=(e^\rho+e^{-\rho})/2$ and $\tanh\rho=(e^\rho-e^{-\rho})/(e^\rho+e^{-\rho})$. 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%? [Work the market's order ticket](https://epistemicphysics.com/fields/economics/markets-that-learn/#market-cost-lab) ## 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 $\Delta C=C(Q')-C(Q)$. 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: $$ \boxed{ \Delta C-p\cdot(Q'-Q)=bD_{\mathrm{KL}}(p\Vert p'). } $$ The **Kullback-Leibler divergence** is $$ D_{\mathrm{KL}}(p\Vert r)=\sum_i p_i\log\frac{p_i}{r_i}. $$ 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 $$ \begin{aligned} &r\cdot(Q'-Q)-\Delta C\\ &\quad=bD_{\mathrm{KL}}(r\Vert p)\\ &\qquad-bD_{\mathrm{KL}}(r\Vert p'). \end{aligned} $$ 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 $$ \boxed{\text{Best expected profit}=bD_{\mathrm{KL}}(r\Vert p).} $$ 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 $\delta p$, $$ bD_{\mathrm{KL}}(p\Vert p+\delta p) \approx\frac{b(\delta p)^2}{2p(1-p)}. $$ 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 $\kappa>0$ that prices departure from that allocation. Define the objective $J(q)$: $$ \begin{aligned} J(q)&=\sum_iq_iV_i\\ &\quad-\kappa D_{\mathrm{KL}}(q\Vert q^0). \end{aligned} $$ We maximize $J(q)$ over allocations satisfying $\sum_iq_i=1$. The first term rewards useful actions. The second discourages changing the allocation too far from its reference. Both terms have payoff units because $\kappa$ converts a natural-log unit of divergence into payoff cost. We are choosing this objective explicitly. A multiplier enforcing the sum gives $$ \begin{aligned} q_i^*&=\frac{q_i^0e^{V_i/\kappa}}{Z},\\ Z&=\sum_jq_j^0e^{V_j/\kappa}. \end{aligned} $$ One identity both verifies the solution and connects it to the market register: $$ \begin{aligned} J(q)&=\kappa\log Z\\ &\quad-\kappa D_{\mathrm{KL}}(q\Vert q^*). \end{aligned} $$ The last term is nonnegative, so the maximum is $\kappa\log Z$. 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 $$ \frac12\log\frac{q_1^*}{q_2^*} =\frac12\log\frac{q_1^0}{q_2^0} +\frac{V_1-V_2}{2\kappa}. $$ 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](https://alisdairmckay.com/pdfs/logit.pdf) 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 $\operatorname{Beta}(1/2,1/2)$ prior, each fresh success adds one to its first shape parameter $\alpha$; each fresh failure adds one to its second parameter $\beta$. Compare two possible notebooks: | History | Current distribution | Predictive mean | | --------------------------- | -------------------------------- | --------------- | | No observed trials | $\operatorname{Beta}(0.5,0.5)$ | $0.5$ | | Ten successes, ten failures | $\operatorname{Beta}(10.5,10.5)$ | $0.5$ | The **predictive mean** $\mu=\alpha/(\alpha+\beta)$ 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 $\operatorname{Beta}(a,a)$ state, the new predictive means after success and failure are $$ \begin{aligned} \mu_{\rm success}&=\frac{a+1}{2a+1},\\ \mu_{\rm failure}&=\frac{a}{2a+1}. \end{aligned} $$ After trying the new recipe, the bakery can select whichever recipe has the better mean. Its expected total payoff is therefore $$ \begin{aligned} V_{\rm try}&=\frac12\\ &\quad+\frac12\max(s,\mu_{\rm success})\\ &\quad+\frac12\max(s,\mu_{\rm failure}). \end{aligned} $$ For the untried recipe, one success moves the mean to $0.75$ and one failure moves it to $0.25$. The calculation is $$ \begin{aligned} V_{\rm try}&=0.50+0.50(0.75)\\ &\quad+0.50(0.55)\\ &=1.15. \end{aligned} $$ For the experienced notebook, even a success only raises the mean to $11.5/22\approx0.523$. Both possible second decisions still favor familiar bread, giving $$ \begin{aligned} V_{\rm try}&=0.50+0.50(0.55)\\ &\quad+0.50(0.55)\\ &=1.05. \end{aligned} $$ 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. [Run the bakery for a few days](https://epistemicphysics.com/fields/economics/markets-that-learn/#business-experiment-lab) ## 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: $$ \mathbb E[\mu_{\rm next}]=\mu. $$ 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 $$ \mathbb E[\max(s,\mu_{\rm next})] \ge\max(s,\mathbb E[\mu_{\rm next}]). $$ 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 $V_h(\alpha,\beta)$ 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 $\mu=\alpha/(\alpha+\beta)$, $$ \begin{aligned} F_h&=s+V_{h-1}(\alpha,\beta),\\ E_h&=\mu\\ &\quad+\mu V_{h-1}(\alpha+1,\beta)\\ &\quad+(1-\mu)V_{h-1}(\alpha,\beta+1). \end{aligned} $$ Choose the better branch: $$ V_h(\alpha,\beta)=\max\{F_h,E_h\}. $$ 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 $\boldsymbol\beta=(\beta_1,\beta_2)$ for the two regression coefficients. A controlled observation follows $$ \begin{aligned} y_i&=\mathbf x_i^{\mathsf T}\boldsymbol\beta+\varepsilon_i,\\ \mathbf x_i&=(\cos\theta_i,\sin\theta_i),\\ \varepsilon_i&\sim\mathcal N(0,1/w_i). \end{aligned} $$ The superscript $\mathsf T$ turns a column into a row, so the product adds the two feature-times-coefficient contributions. The angle $\theta_i$ 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** $$ F=\sum_iw_i\mathbf x_i\mathbf x_i^{\mathsf T}. $$ 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](https://web.stanford.edu/~boyd/cvxbook/bv_cvxbook.pdf) develops the connection between selecting observations and controlling this error covariance in Section 7.5. Now expand the trigonometric squares. Define $$ \begin{aligned} A&=\sum_iw_i,\\ X&=\sum_iw_i\cos2\theta_i,\\ Y&=\sum_iw_i\sin2\theta_i. \end{aligned} $$ Then $$ \begin{gathered} F=\frac12\begin{pmatrix}A+X&Y\\Y&A-X\end{pmatrix},\\ \boxed{4\det F=A^2-X^2-Y^2.} \end{gathered} $$ A positive semidefinite matrix has no negative quadratic directions. In these coordinates that condition becomes $A\ge\sqrt{X^2+Y^2}$: 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 $\mathbf x_i$ to $-\mathbf x_i$, but leaves $\mathbf x_i\mathbf x_i^{\mathsf T}$ unchanged. Information sees an axis, which returns to itself after $180^\circ$; its cone direction makes a full turn. [Build a test that sees both directions](https://epistemicphysics.com/fields/economics/markets-that-learn/#economic-design-lab) ## Diversity of directions becomes an area The determinant can also be written $$ \det F=\sum_{i