Money that can move

Miraia S. Chiou

Survey · Economics

Money that can move

Change the ruler for a price, the fallback for a bargain, or the rules that let promises become payments.

2026-09-08 · 19 min read · 3 interactive demos

The festival has finished, the invoices agree, and nobody can make the first payment.

Three businesses have arranged a circle of promises. A owes B one coin. B owes C one coin. C owes A one coin. Each is owed exactly as much as it owes. Add everyone’s position and everything cancels. Yet their tills are empty.

Would one coin help? Would each invoice require a separate coin? Could changing the rules settle the whole circle without a coin moving at all?

To answer, we need more than a total. We need the relationships between accounts, the times at which payments are possible, and the operations the businesses may perform. Economics makes this habit of Epistemic Physics wonderfully concrete: retain enough structure to know what a system can do next.

Our route begins with a smaller puzzle. Even before money moves, its numerical description can move. Changing the reference for a price can change a probability distribution. Changing the numerical units of utility can move a point through a cone while leaving the bargain untouched. Changing an outside option or a settlement rule can change the available future itself.

A price needs a ruler

Imagine three currencies, called A, B, and C. Let $R_{i j}$ mean the units of currency $j$ received for one unit of currency $i$ . Suppose our invented exchange desk offers

\begin{aligned} R_{A B} & = 2, \\ R_{B C} & = 3, \\ R_{C A} & = 0.17 . \end{aligned}

A round trip multiplies the initial amount by $2 \times 3 \times 0.17 = 1.02$ . One hundred A units would return as 102, if all three exchanges could be executed at those rates.

Now print the accounts in different units. Write $x_{i}^{'} = s_{i} x_{i}$ , where $s_{i} > 0$ is the conversion factor for the account in currency $i$ . Changing dollars into cents is one example. The same exchange must now be labeled

R_{i j}^{'} = \frac{s_{j}}{s_{i}} R_{i j} .

Along a closed route, the factors cancel:

\begin{aligned} \frac{s_{B}}{s_{A}} \frac{s_{C}}{s_{B}} \frac{s_{A}}{s_{C}} & = 1, \\ R_{A B}^{'} R_{B C}^{'} R_{C A}^{'} & = R_{A B} R_{B C} R_{C A} \\ = 1.02 . \end{aligned}

The desk can change its labels without changing that round trip. Taking natural logarithms turns multiplication into addition. Define $a_{i j} = \log R_{i j}$ and $ϕ_{i} = \log s_{i}$ . Then

a_{i j}^{'} = a_{i j} + ϕ_{j} - ϕ_{i} .

This is a gauge transformation: a change in local numerical reference. The loop sum $a_{A B} + a_{B C} + a_{C A}$ survives it. This is the same local-change, closed-loop pattern explored in A world made of little agreements.

In a connected network with reversible, frictionless exchange, zero sum around every loop lets us assign a consistent value level to each currency. Choose a starting currency, add log rates along a path, and the answer does not depend on the path. A nonzero loop exposes a failure of that description.

Execution adds another layer. A 1% fee on each of our three trades changes the round-trip factor to $1.02 (0.99)^{3} \approx 0.9897$ , a loss of about 1.03%. The loop calculation and the route a trader can actually execute belong together.

The ruler can have a future of its own

Currency units are one kind of reference. Another reference can be an asset whose payoff depends on what happens tomorrow.

Take a one-period market with two states: sunshine and rain. A state price is today’s price of one unit paid only in a specified state. Fix the following fictional prices and payoffs:

Quantity	Sunshine	Rain
State price today, $z_{s}$	$0.4$	$0.5$
A sure unit tomorrow, $N_{s}$	$1$	$1$
A sunshine asset, $M_{s}$	$2$	$0.5$
The claim we want to price, $H_{s}$	$6$	$2$

The claim costs $H_{0} = 0.4 (6) + 0.5 (2) = 3.4$ today. The sure unit costs $N_{0} = 0.9$ ; the sunshine asset costs $M_{0} = 1.05$ .

A numeraire is a strictly positive reference asset used to express other prices. Why positive? We will divide by its payoff in every state.

For any such asset $N$ , define

N_{0} = \sum_{s} z_{s} N_{s}, Q_{s}^{N} = \frac{z_{s} N_{s}}{N_{0}} .

The numbers $Q_{s}^{N}$ are positive and sum to one, so they form a probability distribution. They are pricing weights. We have constructed them from prices and the reference asset; a weather forecast would be a separate object.

Now divide the claim’s future payoff by the reference’s future payoff, average using these weights, and multiply by the reference’s price today:

\begin{aligned} N_{0} E_{Q^{N}} [H / N] & = N_{0} \sum_{s} \frac{z_{s} N_{s}}{N_{0}} \frac{H_{s}}{N_{s}} \\ = \sum_{s} z_{s} H_{s} = H_{0} . \end{aligned}

Every reference factor cancels, separately in each state. Changing the reference rearranges the calculation without changing the claim being priced. The general method extends to dynamic markets in the work of Geman, El Karoui, and Rochet.

In our example, the sure-unit weights are $(4 / 9, 5 / 9)$ . Using the sunshine asset gives $(16 / 21, 5 / 21)$ . Sunshine receives more weight, while the claim measured in sunshine-asset units changes from $(6, 2)$ to $(3, 4)$ . Together, those changes preserve the price 3.4.

Predict before touching: increase the selected reference asset’s rain payoff. Which pricing weight rises? What has to happen elsewhere in the calculation for the claim’s price to stay fixed?

One price, several pricing measures

Change the reference. Keep the claim.

Tomorrow has two possible states. A fixed claim pays 6 in sunshine and 2 in rain. Choose a positive asset to measure those payments in, then drag its payoff dots. The pricing weights will move. Predict what happens to the claim's price today.

Sunshine asset: payments tomorrow

Sure-unit payoff: 1 in either state

Sunshine payoff

2.00

Rain payoff

0.50

The pricing weights change with it

Sunshine

Sure unit44.4%

Selected76.2%

Rain

Sure unit55.6%

Selected23.8%

State prices stay fixed: 0.4 for sunshine, 0.5 for rain. A state price buys one unit paid only in that state.

Divide the claim by the reference, then price it
State	Claim / reference	Price part today
Sunshine	3.000	2.400
Rain	4.000	1.000

Each last-column entry is N₀ × Q × (claim / reference). Here N₀ is the selected reference asset's price today.

Reference price N₀: 1.050
Expected claim / reference: 3.238
Pricing rapidity ρ: -0.112 → 0.582
Rapidity shift Δρ: 0.693

The same claim costs 3.400 today. 1.050 × 3.238 ≈ 3.400. The displayed factors are rounded; the price identity holds before rounding.

A numeraire is this positive reference asset. Its pricing weights are Qₛ = zₛNₛ/N₀, where zₛ is the state price. Half-log pricing odds are ρ = ½ log(Q₁/Q₂). Relative to the sure unit, Δρ = ½ log(N₁/N₂). These weights describe the pricing calculation, not a weather forecast.

Open the reference-asset experiment in the playground.

There is also a familiar EP coordinate hiding here. For two positive pricing weights, define the rapidity $ρ_{Q} = \frac{1}{2} \log (Q_{1} / Q_{2})$ , half the logarithm of their ratio. Switching reference assets gives

\begin{aligned} Q_{s}^{M} & \propto Q_{s}^{N} \frac{M_{s}}{N_{s}}, \\ ρ_{Q^{M}} - ρ_{Q^{N}} & = \frac{1}{2} \log \frac{M_{1} / N_{1}}{M_{2} / N_{2}} . \end{aligned}

Multiplicative reweighting becomes a translation in rapidity. Bayesian likelihood updates have the same algebraic shape. Here the multiplying factor comes from changing the pricing reference. The equation lets us recognize a common operation while retaining what each weight means.

A better chair at the bargaining table

Two collaborators can earn a total of ten resource units by completing a project together. If talks fail, A can earn one unit elsewhere and B can earn two. These alternatives are their outside options.

Suppose each person’s utility is linear in their resource allocation. Utility is a numerical representation of how they value outcomes. Let $U_{A}, U_{B}$ be the utilities from an agreement and $d_{A}, d_{B}$ their outside-option utilities. Choose positive reference scales $s_{A}, s_{B}$ and measure the gains by

a = \frac{U_{A} - d_{A}}{s_{A}}, c = \frac{U_{B} - d_{B}}{s_{B}} .

An agreement giving both people something beyond their alternatives has $a, c > 0$ . The symmetric Nash bargaining objective chooses a feasible agreement maximizing $a c$ , the product of the gains. Nash’s original bargaining paper develops this solution from assumptions about feasible outcomes and how the solution should behave, including invariance to positive affine changes of utility scales.

Watch what happens when we express these two positive gains in cone coordinates:

\begin{matrix} T = a + c, X = a - c, \\ M = \sqrt{T^{2} - X^{2}} = 2 \sqrt{a c} . \end{matrix}

The sum $T$ and difference $X$ locate a point inside $T \geq | X |$ . The quantity $M$ is its cone norm. Maximizing the Nash product also maximizes $M$ .

Compare the two constructions directly. Here $Y, N$ denote supporting and opposing trial amounts in an EP evidence description:

Construction	Positive inputs	Sum and difference	Cone norm
Evidence coordinates	$Y, N$	$Y + N, Y - N$	$2 \sqrt{Y N}$
Bargaining gains	$a, c$	$a + c, a - c$	$2 \sqrt{a c}$

The shared geometry follows from keeping both positive inputs. In the bargain, choosing to maximize their product supplies the economic objective. Utilities do not become trial counts through this coordinate change.

We can also choose a bargaining weight $w$ between zero and one, giving A weight $w$ and B weight $1 - w$ . The objective becomes $a^{w} c^{1 - w}$ . Introduce the gain rapidity $ρ = \frac{1}{2} \log (a / c)$ . Since $a = (M / 2) e^{ρ}$ and $c = (M / 2) e^{- ρ}$ ,

\begin{aligned} \log (a^{w} c^{1 - w}) & = \log (M / 2) \\ + (2 w - 1) ρ . \end{aligned}

At equal weights, only the common gain term remains. Other weights also favor a direction of allocation. The cone gives us coordinates for inspecting that choice.

Solve a bargain we can draw

Let $x_{A}$ be A’s allocation from a divisible pie of size $R$ ; B receives $R - x_{A}$ . In this model the gains before rescaling are $x_{A} - d_{A}$ and $R - x_{A} - d_{B}$ . Assume $R > d_{A} + d_{B}$ , so cooperation offers a positive surplus.

To maximize the weighted product, differentiate its logarithm:

\frac{w}{x_{A} - d_{A}} - \frac{1 - w}{R - x_{A} - d_{B}} = 0.

Solving gives

x_{A} = d_{A} + w (R - d_{A} - d_{B}) .

Each person keeps the value of their fallback, then they split the remaining surplus using the chosen weights. For $R = 10$ , $(d_{A}, d_{B}) = (1, 2)$ , and $w = 1 / 2$ , the agreement is $(4.5, 5.5)$ . Their allocations differ, but their gains over the fallbacks are equal.

Try two different interventions: improve A’s fallback by two units, then restore it and merely double the numerical labels on A’s gains while halving B’s. Which intervention changes how much of the resource A receives?

A feasible agreement and its gain cone

Move the fallback. Move the bargain.

A and B divide a resource worth R units. The rose point is what each can get if talks fail. Drag that fallback point, or drag the frontier's R handle to change the pie. Then change the numerical units used for the gains. Which changes move the actual agreement?

Agreements in resource units

FallbackA gainB gain

The same bargain in gain coordinates

Default gain unitsCurrent gain unitsSame cone norm M

A's fallback B's fallback Total resource R

The shaded triangle contains agreements that meet both fallbacks. This sandbox keeps at least 0.5 units of surplus available. Arrow keys move the fallback by 0.25; hold Shift for steps of 1.

Try a different numerical label for the gains

Resource allocated to A: 4.50
Resource allocated to B: 5.50
Labeled gains (a, c): (3.50, 3.50)
Gain rapidity ρ: 0.000

A receives 4.50; B receives 5.50. The chosen weight w = 0.50 gives A that fraction of the 7.00 units left above the two fallbacks. Reciprocal changes of gain units preserve this agreement and M = 7.00.

The weighted objective is aʷc¹⁻ʷ. Its logarithm is log(M/2) + (2w − 1)ρ, with M = 2√(ac) and ρ = ½ log(a/c). Equal weights maximize M; other weights also favor one gain direction.

Open the bargaining experiment in the playground.

Improving A’s fallback from one to three moves the equal-weight agreement to $(5.5, 4.5)$ . A obtains one extra unit from the project. More generally, each extra fallback unit changes A’s allocation by $1 - w$ when the other inputs remain fixed.

Changing utility units has another effect. Under reciprocal rescaling,

\begin{matrix} a^{'} = e^{η} a, c^{'} = e^{- η} c, \\ M^{'} = M, ρ^{'} = ρ + η . \end{matrix}

The gain point receives a Lorentz-shaped transformation, while the maximizing agreement stays put. Any positive rescaling of the two utility gains multiplies the bargaining objective by a positive constant independent of the agreement. Reciprocal rescaling additionally preserves $M$ itself.

A shared tool could enlarge the pie. Access to childcare could improve someone’s fallback or open a previously impossible working arrangement. A bargaining rule could change the weights. These are different ways institutions can act on the negotiation diagram. A policy conversation becomes sharper when we identify which part of the model an intervention changes.

One relationship, two appearances

A bargain creates promises, and promises acquire records.

If A owes B one unit under contract AB-1, A records a liability and B records a receivable, an amount it is owed. These are two appearances of the same contractual relationship. Adding both appearances as though they were two independent claims would lose that relationship.

Bank lending makes the distinction vivid. When a bank grants a loan by crediting a deposit, it obtains a loan asset and incurs a deposit liability. The borrower has a deposit asset and a loan liability. The Bank of England explains this process in Money creation in the modern economy.

There are two linked relationships here:

Relationship	Bank’s appearance	Borrower’s appearance
Loan `L-1`	Loan asset	Loan liability
Deposit `D-1`	Deposit liability	Deposit asset

One origination event can establish both relationships. Keeping its lineage connects the records without collapsing the loan and deposit into one contract. The deposit can serve as a spendable asset even though the internal financial assets and liabilities cancel when we consolidate the two parties.

Reports must meet their sale

The same identity question changes what a business can learn. Suppose each sale has two properties: shipped or unshipped, paid or unpaid. Represent the unknown probabilities of those four joint states with a $Dirichlet (1, 1, 1, 1)$ prior, placing one initial weight on each state. Assume sales are independent draws conditional on that unknown probability vector.

Ask whether a sale shipped and whether a sale was paid. Both answers are yes. If the answers refer to the same sale ID, they describe one completed sale. The posterior distribution for the rate of completed sales is $Beta (2, 3)$ .

If the IDs differ, each sale is only partially described. The exact posterior becomes the mixture

\begin{aligned} \frac{1}{5} Beta (1, 5) \\ + \frac{2}{5} Beta (2, 4) \\ + \frac{2}{5} Beta (3, 3) . \end{aligned}

The components allow zero, one, or two completed sales. The four compatible completions have relative weights $1, 1, 1, 2$ after integrating over the shared Dirichlet model; the two middle cases combine. Merely assigning distinct IDs would not establish the conditional-independence assumption. That assumption belongs to the sale model we specified.

A copied report of the same fact about the same sale adds another appearance, rather than another sale trial. Lineage determines which facts can join into one observed event and which unresolved combinations the belief must retain.

A balanced account can still wait for cash

Return to the festival’s three contracts. Initially, every business has one receivable and one liability, with no cash. At their stated values, each balance sheet has zero equity: assets minus liabilities equals zero.

We can put those positive balance-sheet amounts into cone coordinates too. Call assets $A$ and liabilities $L$ :

\begin{matrix} T = A + L, X = A - L, \\ M^{2} = 4 A L . \end{matrix}

If $A = L > 0$ , then $X = 0$ and $T > 0$ . Zero equity lies on the central axis inside the cone. Its boundary occurs when $A L = 0$ . In our initial network, each firm sits at $(T, X) = (2, 0)$ , yet cannot make a cash payment. This geometry organizes the two balance-sheet amounts; the next event also depends on what those assets are and when they can be used.

Give the network the following rules: payments are sequential, each contract requires a full unit of cash, no overdrafts are available, and a contract becomes payable at its due time. An unpaid due contract remains payable later. These are the rules of our experiment.

Place one external coin at A. If all three contracts are due, the cash locations can follow

\begin{aligned} (1, 0, 0) & \overset{AB-1}{\to} (0, 1, 0) \\ \overset{BC-1}{\to} (0, 0, 1) \\ \overset{CA-1}{\to} (1, 0, 0) . \end{aligned}

Three unit payments occur using one conserved coin. Each arrow settles a different contract. The coin returns to A, while all three obligations have disappeared from the outstanding balances.

Predict the interruption: suppose A’s payment becomes due only at time two. Does placing the coin at A still start the route immediately? Try placing it at B instead, after resetting the run.

A settlement puzzle

Route one coin through three promises

Settlement route

Place one coin at any firm, then route it through all three promises. Tap the firm holding the coin to pay its next neighbor.

Choose a firm to place the coin, or try the permitted netting rule.

External coins: 0
Promises settled: 0 / 3

Button alternatives, contract receipts, and rules

A owes B one unit, B owes C one, and C owes A one. Cash payments require one full coin and a due contract. No overdrafts or partial payments; a due promise remains payable in later rounds. The gold arrow is ready; dashed arrows are due later.

AB-1: A liability ↔ B receivable. Both one unit; due round 0.
BC-1: B liability ↔ C receivable. Both one unit; due round 0.
CA-1: C liability ↔ A receivable. Both one unit; due round 0.

D0, round 0: Due times advance to round 0.

Netting is an additional chosen institution. It permits the complete equal cycle to settle together only while all original contracts are unpaid and due. One receipt links three identities; no cash transfer occurs. Final balances alone do not show this event sequence.

Compare simultaneous clearing

F(0.5, 0.5, 0.5) = (0.5, 0.5, 0.5), where F(pₐ,pᵦ,p꜀) = (min(1,p꜀),min(1,pₐ),min(1,pᵦ)). Every constant from zero to one solves this zero-cash clearing equation. Choosing a fixed point does not execute an enabled first payment in the dated protocol above.

Open the dated-payment experiment in the playground.

The final balances alone cannot reveal the route. The event sequence can: P1 settles a named contract, which changes two linked ledger appearances and enables another payment. A due-time event changes eligibility without moving cash. The experiment keeps these operations distinct.

Clearing an equation and executing a sequence

There is another possible rule: permit multilateral netting, offsetting obligations across the participating firms. Once the entire equal cycle is due, our chosen rule can settle all three together. The balances vanish in one linked event, and no coin needs to circulate.

A simultaneous clearing model provides a useful neighboring viewpoint. Let $p_{i}$ be firm $i$ ’s total payment, ${\bar{p}}_{i}$ its promised total, $e_{i}$ its external assets available for payment, and $Π_{j i}$ the fraction of firm $j$ ’s payments directed to $i$ . The clearing equation is

p_{i} = min ({\bar{p}}_{i}, e_{i} + \sum_{j} Π_{j i} p_{j}) .

It asks for payments consistent with resources received from one another. Eisenberg and Noe study the existence of such fixed points, values reproduced by the equation, and conditions that ensure uniqueness in Systemic Risk in Financial Systems.

For our zero-cash unit cycle, the equation reduces to

\begin{aligned} p_{A} & = min (1, p_{C}), \\ p_{B} & = min (1, p_{A}), \\ p_{C} & = min (1, p_{B}) . \end{aligned}

Every vector $(c, c, c)$ with $0 \leq c \leq 1$ is a fixed point. Both “nobody pays” and “everyone pays in full” satisfy the simultaneous equation. This closed cycle does not provide a unique solution.

The sequential protocol resolves a different question: what event can happen first, given the current cash and permissions? With zero cash and only cash-payment events, none can. A permitted netting event changes the available operations. An external coin opens a cash-payment route. We can understand all three descriptions once we specify the institution that connects obligations to events.

Institutions change the available future

This suggests a way to explore economic possibility through an event graph. A vertex is a state containing cash locations, outstanding dated contracts, and permissions. An edge is an allowed event: a payment, an advance of time, a netting operation, or another action explicitly included in the model.

Two vertices can have identical total cash and identical net positions while admitting different next steps. Two histories can reach identical final balances through different routes. Aggregating the accounts therefore need not preserve the graph of possible futures.

We could build a planning model on this graph: assign values to completed deliveries, costs to temporary funding, and penalties to specified delays. Then search for the most valuable allowed path. Another version could ask which single rule change makes a previously unreachable delivery possible. These are concrete ways to investigate institutions as arrangements that create, remove, or coordinate options.

The three experiments give us distinct interventions to try. A numeraire change rewrites a valuation. A utility-unit change relabels gains. A better fallback changes the bargaining problem. A settlement mechanism changes which events can occur. A lineage correction changes which observations belong together.

Carry that last distinction back to the empty tills. Before deciding that the town needs three more coins, inspect its promises, its clocks, and its permitted next move.