When evidence takes different roads

Miraia S. Chiou

Survey · Statistics

When evidence takes different roads

Reconstruct beliefs from report fragments, hidden causes, active questions, and complex joint tables.

2026-09-08 · 26 min read · 4 interactive demos

Four headlines announce the same discovery. You feel increasingly sure it happened. Then someone opens the footnotes: every headline copied the same press release.

Now try a different experiment. Ask someone about their happiest memory, then ask whether the world feels welcoming. Reverse those questions tomorrow. The first question might change what the second encounters.

Evidence has routes. It can be copied, hidden behind a common cause, or collected through an interaction that changes the scene. Following those routes leads us from ordinary Bayesian inference into split-complex algebra, positive matrices, and joint tables whose entries point in imaginary directions.

Epistemic Physics gives us a lively question to carry along: what must a belief remember about how its tracelets arrived?

One source can wear four name tags

Let $H$ be a claim, initially assigned probability $1 / 2$ . An original source answers yes or no. Assume it is $80 %$ accurate in either direction:

\begin{aligned} Pr (+ ∣ H) & = 0.8, \\ Pr (+ ∣ \neg H) & = 0.2 . \end{aligned}

The symbol $\neg H$ means that the claim is false. A yes report has likelihood ratio $0.8 / 0.2 = 4$ : it is four times as likely under $H$ as under $\neg H$ .

Odds express a probability $p$ as $p / (1 - p)$ . Bayes’ rule multiplies prior odds by a likelihood ratio. Starting from equal odds, one yes gives

Pr (H ∣ +) = \frac{4}{1 + 4} = \frac{4}{5} .

Suppose four headlines copy that answer perfectly. Conditional on the original answer, their wording supplies no further information about $H$ . The graph contains one informative source event with four appearances. Its posterior remains $4 / 5$ .

Four independently generated yes answers give a different calculation. Here independence means that, once $H$ or $\neg H$ is fixed, the sources’ errors are independent:

\begin{aligned} Pr (H ∣ +, +, +, +) & = \frac{4^{4}}{1 + 4^{4}} \\ = \frac{256}{257} . \end{aligned}

That is approximately $99.61 %$ . The four visible yes answers look identical in both stories. Their source graph determines which calculation belongs to them.

Follow the report back to its origin

Four headlines. How many sources?

Every headline says “yes.” Each original source is 80% accurate in either direction, and the claim starts at 50%. In this model, separate sources are independent once the claim's truth is fixed. Tap a headline to cycle its source, or choose its source below.

Headline 1 Headline 2 Headline 3 Headline 4

Original source factors: 1
Posterior using the source graph: 80.00%
If every headline were independent: 99.61%

All four headlines repeat one original answer. Its likelihood factor enters once.

posterior odds = 4^{m}

Here m is the number of independent original source answers. Copying an answer leaves m unchanged.

A hidden source can also connect two measurements

Let h be a common signal. Each reading adds its own independent noise: x = h + εₓ and y = h + εᵧ. All three input variables have variance one.

hcommon signal

↙ ↘

xy

Independent conditional on a known h

Integrating out h averages over its unknown value. Both readings inherit its variation, which becomes a visible connection between them.

Graph posteriorIndependent-headline calculation

Open the source experiment in the playground.

A source ID answers “which original event was this?” A dependence model answers “how were these different events related?” Separate reporters may share a witness, an instrument, or a training dataset. Distinct names alone establish no independence. The demo explicitly supplies independent source errors; changing that assumption calls for a richer graph.

A factor graph represents a joint model as small pieces called factors. Each factor connects the variables appearing in one local relationship. Summing over hidden variables and multiplying the appropriate factors produces the desired probabilities. This makes source relationships part of the calculation. Kschischang, Frey, and Loeliger, Factor graphs and the sum-product algorithm.

Bayes has a two-channel algebra

Picture two tracks carrying unnormalized weights $w_{+}$ and $w_{-}$ for $H$ and $\neg H$ . Their normalized probability is $w_{+} / (w_{+} + w_{-})$ .

We can package those tracks into a split-complex number. Introduce a unit $j$ satisfying $j^{2} = + 1$ , then define

e_{+} = \frac{1 + j}{2}, e_{-} = \frac{1 - j}{2} .

Multiplication reveals a delightful arrangement:

\begin{aligned} e_{+}^{2} & = e_{+}, \\ e_{-}^{2} & = e_{-}, \\ e_{+} e_{-} & = 0. \end{aligned}

The two channels reproduce themselves and never cross. Encode the belief weights and a likelihood message as

\begin{aligned} Z & = w_{+} e_{+} + w_{-} e_{-}, \\ L & = ℓ_{+} e_{+} + ℓ_{-} e_{-} . \end{aligned}

Here $ℓ_{+}$ and $ℓ_{-}$ are the likelihoods of the observed data under the two hypotheses. Their product performs the Bayesian update:

Z L = w_{+} ℓ_{+} e_{+} + w_{-} ℓ_{-} e_{-} .

Each message scales its own track. Define rapidity, half the logarithm of the odds, by

r = \frac{1}{2} \log \frac{w_{+}}{w_{-}} .

Multiplication now becomes translation:

r^{'} = r + \frac{1}{2} \log \frac{ℓ_{+}}{ℓ_{-}} .

A prior probability of $1 / 4$ has odds $1 / 3$ . A likelihood ratio of nine changes those odds to three, giving posterior probability $3 / 4$ . In rapidity, the message simply adds $\log 3$ .

Three ways to combine things

How does this multiplication fit with adding fresh trials in a Beta belief? The objects being combined tell us which operation to use.

What arrives?	What we keep	How it combines
Fresh support and opposition	Raw trial amounts $(Y, N)$	Add their amounts
A likelihood message	A function of the unknown state	Multiply the likelihood factors
Two analyses of independent datasets	Posterior densities with a common prior	Multiply, then divide out one shared prior

For a success probability $p$ , $Y$ successes and $N$ failures contribute likelihood $p^{Y} (1 - p)^{N}$ . Multiplying it into a $Beta (α_{0}, β_{0})$ prior gives

p ∣ trials \sim Beta (α_{0} + Y, β_{0} + N) .

Likelihood multiplication adds exponents. That is why fresh trials add to the Beta shapes.

Let $π_{0}$ denote a common prior density. If two disjoint datasets are conditionally independent, with likelihoods $L_{1}, L_{2}$ , their separate posteriors satisfy $π_{1} \propto π_{0} L_{1}$ and $π_{2} \propto π_{0} L_{2}$ . The combined posterior is therefore

π_{12} \propto \frac{π_{1} π_{2}}{π_{0}} .

The symbol $\propto$ means proportional to, with a constant chosen to make the density integrate to one.

Here is an unexpectedly sharp example. Give both analyses the prior $Beta (1 / 2, 1 / 2)$ and two fresh successes each. Each posterior is $Beta (5 / 2, 1 / 2)$ . Their plain product is proportional to

\frac{p^{3}}{1 - p},

whose integral diverges near $p = 1$ . Dividing by the repeated prior restores the proper combined $Beta (9 / 2, 1 / 2)$ distribution. Even a familiar density can become unnormalizable when its shared structure is counted twice.

If the datasets also overlap, each common likelihood factor must occur once. A lineage record gives this algebra somewhere concrete to attach: original events create factors; repeated appearances point back to them.

Reconstruct a law from scattered fragments

FLogic’s tally idea takes us further: join report fragments by lineage, keep the likelihood factors they imply, and reconstruct a distribution from those factors.

Imagine surveying unfamiliar creatures. Let $A$ mean “glows” and $B$ mean “has wings.” One notebook records $A = 1$ ; another records $B = 1$ . Did they describe the same creature?

Choose a joint model with four possible answer combinations, called worlds: $11, 10, 01, 00$ . Their unknown probabilities are $θ_{11}, θ_{10}, θ_{01}, θ_{00}$ and sum to one. Unique sampled events are independent conditional on these probabilities. Observations are direct and error-free; missing fields are ignored by summing over their possible answers.

A Dirichlet distribution places a probability distribution over that four-entry vector. Choose

θ \sim Dirichlet (1 / 4, 1 / 4, 1 / 4, 1 / 4) .

Each world receives a quarter of the prior amount, keeping the total at one. Our question is the unknown frequency $q = θ_{11}$ of creatures possessing both traits.

If the fragments describe one event, they join into a complete $11$ observation. Its likelihood factor is $θ_{11}$ . The exact posterior is

\begin{matrix} q ∣ same event \sim Beta (5 / 4, 3 / 4), \\ E [q] = \frac{5}{8} . \end{matrix}

If they describe different events, each observation leaves one answer missing. The likelihood becomes

\begin{aligned} L & = (θ_{11} + θ_{10}) (θ_{11} + θ_{01}) \\ = θ_{11}^{2} + θ_{11} θ_{01} + θ_{10} θ_{11} + θ_{10} θ_{01} . \end{aligned}

Each term describes a compatible completion of the missing answers. Multiplication by the prior adds that term’s exponents to the four Dirichlet shapes. Projecting onto $q$ gives an exact mixture of Betas:

\begin{matrix} q ∣ different events \sim \\ \begin{aligned} \frac{1}{8} Beta (1 / 4, 11 / 4) \\ + \frac{1}{4} Beta (5 / 4, 7 / 4) \\ + \frac{5}{8} Beta (9 / 4, 3 / 4) . \end{aligned} \end{matrix}

The weights specify how much each component contributes to the resulting distribution. They include Dirichlet integrals, the prior averages of the likelihood terms. The squared term averages to $5 / 32$ ; each cross term averages to $1 / 32$ . Normalize those four contributions, then combine the two completions containing exactly one $11$ event. Equal polynomial coefficients have produced unequal posterior weights.

This mixture has mean $7 / 12$ , about $58.33 %$ , compared with $62.5 %$ for the joined event. The notebooks contain the same written yes answers. Knowing which facts belong together changes what they teach us about the conjunction.

FLogic's tally idea, with lineage kept visible

Two fragments, one reconstructed belief

Two fragments say A = 1 and B = 1. Should the archive join them into one event, or retain two? Inspect their event stamps, then choose. Matching answers alone cannot identify their lineage.

Read the stamps before deciding. The answer fields alone leave the event identity unresolved.

Report A Event stamp covered

A = 1

B = ?

Report B Event stamp covered

A = ?

B = 1

Open the archive workbench and posterior

No extra report copies. A question mark means the answer was not recorded.

Join the fragments, then multiply the event factors

Event 1

θ_{11}

Exact mixture for q = θ₁₁

q is the chance that A and B are both one. Each row contributes a Beta distribution with the displayed weight and shapes α, β.

Weight

α

β

15/43/4

Report appearances: 2
Unique observed events: 1
Posterior mean of q: 5/8 = 62.50%
Variance of q: 0.0781

The two fragments complete one A = 1, B = 1 observation. Their shared event contributes one complete factor. Copies and arrival order leave this law unchanged.

Compare the entire distributions

A cumulative distribution gives the probability that q is at most a threshold t. Drag the gold marker along the horizontal axis, use its arrow keys, or choose the buttons below.

Same eventDifferent eventsMove the threshold

At t = 0.50, the selected posterior places 33.04% of its probability at q ≤ t.

Open the likelihood expansion

\begin{aligned} θ_{11} \end{aligned}

Each monomial adds its exponents to the four Dirichlet shapes. Its mixture weight includes the monomial's coefficient and its integral under the prior. Equal polynomial coefficients need not give equal posterior weights.

Term 1: coefficient 1; prior integral 1/4; normalized weight 1.

Model: independent event draws conditional on four unknown joint frequencies, direct answers, and ignored missing fields. Each category starts with prior weight 1/4. Matching event identity allows joining; it alone does not establish independence between distinct events.

Open the reconstruction experiment in the playground.

Copy a report and the reconstructed law stays fixed. Shuffle the reports and it stays fixed again: these likelihood factors commute. Changing which fragments refer to the same event can change the law, because it changes the factors themselves. Revealing a missing field supplies another way to refine a factor.

After joining the fragments, a tally of distinct compatible-world factors preserves the likelihood for this chosen joint model. The current calculation can use those counts; retained lineage can still support later joins and duplicate checks. This offers a compact architecture for a belief engine that carries a whole distribution forward through a changing network of reports.

Hide a cause, and a connection appears

Imagine two sensors measuring the same fluctuating room temperature. Write

\begin{aligned} h & \sim N (0, 1), \\ x & = h + ϵ_{x}, \\ y & = h + ϵ_{y}, \end{aligned}

where $N (0, 1)$ is a Gaussian distribution with mean zero and variance one. The sensor noises $ϵ_{x}, ϵ_{y}$ are independent copies of that distribution, also independent of the shared temperature signal $h$ .

If we know $h$ , the two remaining sensor errors are independent. If we integrate over its unknown value, both readings inherit its variation:

\begin{aligned} Var (x) & = Var (y) = 2, \\ Cov (x, y) & = 1, \\ Corr (x, y) & = \frac{1}{2} . \end{aligned}

Covariance measures their shared fluctuation. Correlation divides covariance by the two standard deviations, producing a dimensionless quantity.

The full Gaussian density has exponent

- \frac{1}{2} [h^{2} + (x - h)^{2} + (y - h)^{2}] .

Complete the square in $h$ :

\begin{aligned} h^{2} + (x - h)^{2} + (y - h)^{2} \\ = 3 {(h - \frac{x + y}{3})}^{2} \\ + x^{2} + y^{2} - \frac{(x + y)^{2}}{3} . \end{aligned}

Integrating the first term’s Gaussian leaves a constant. The remaining expression couples $x$ to $y$ . Hiding the common node has created a visible connection between its children.

This is an example of a Schur complement. A Gaussian’s precision matrix is the inverse of its covariance matrix. Partition a precision matrix into retained variables, hidden variables, and their connections:

J = (\begin{matrix} A & B \\ B^{T} & C \end{matrix}) .

The superscript $T$ means transpose. After integrating out the hidden variables, the retained precision is

J_{effective} = A - B C^{- 1} B^{T} .

For our sensors, $A = I$ , the two-by-two identity; $B = (- 1, - 1)^{T}$ ; and $C = 3$ . We obtain

\begin{aligned} J_{x y} & = (\begin{matrix} 2 / 3 & - 1 / 3 \\ - 1 / 3 & 2 / 3 \end{matrix}), \\ Σ_{x y} = J_{x y}^{- 1} & = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix}) . \end{aligned}

Notice another familiar shape. For any real positive two-by-two precision matrix $J = (\begin{matrix} a & c \\ c & b \end{matrix})$ , define $τ = a + b$ , $P_{z} = a - b$ , and $P_{x} = 2 c$ . Then

τ^{2} - P_{z}^{2} - P_{x}^{2} = 4 det J .

An ordinary Gaussian source graph contains the same cone identity we encounter in matrix evidence geometry. Here the coordinates have precision units; raw trial amounts retain their own units.

This suggests a useful kind of memory compression. A detailed graph can retain who measured what, while a compact graph retains the dependence created when hidden causes are averaged away. The two descriptions can agree exactly for the chosen Gaussian model. Zooming out changes which connections are explicit.

A question can move the thing it measures

Now imagine an interview, a learning session, or a sequence of selective filters. Let us model each successful interaction as actively changing a positive matrix $R$ :

R ⟼ K R K^{†} .

The dagger means conjugate transpose. Normalize by the trace, the sum of diagonal entries, to obtain the state $ρ = R / tr R$ .

A two-by-two state can be written

ρ = \frac{1}{2} (\begin{matrix} 1 + v_{z} & v_{x} - i v_{y} \\ v_{x} + i v_{y} & 1 - v_{z} \end{matrix}),

where $i^{2} = - 1$ is the ordinary imaginary unit. The real vector $v = (v_{x}, v_{y}, v_{z})$ lies inside the unit ball. Its length is at most one, ensuring nonnegative state eigenvalues $(1 \pm | v |) / 2$ . These three coordinates summarize directional biases. A horizontal question has yes probability $(1 + v_{x}) / 2$ . When a question changes the shape of uncertainty explores how an observer can learn a distribution over such states.

Choose two filters:

K_{z} (η) = (\begin{matrix} e^{η / 2} & 0 \\ 0 & e^{- η / 2} \end{matrix}),

K_{x} (ξ) = (\begin{matrix} \cosh (ξ / 2) & \sinh (ξ / 2) \\ \sinh (ξ / 2) & \cosh (ξ / 2) \end{matrix}) .

The rapidities $η$ and $ξ$ set their strengths. Here $\cosh t = (e^{t} + e^{- t}) / 2$ and $\sinh t = (e^{t} - e^{- t}) / 2$ . The first filter favors the vertical direction; the second favors the horizontal direction. Each has determinant one.

Start from $R = I$ and set $η = ξ = \log 2$ . Multiplying the matrices in their actual order gives

\begin{aligned} z then x & : v = (3 / 5, 0, 12 / 25), \\ x then z & : v = (12 / 25, 0, 3 / 5) . \end{aligned}

The final horizontal yes probabilities are $4 / 5$ and $37 / 50$ : $80 %$ versus $74 %$ . Both unnormalized matrices have trace $25 / 8$ and determinant one. Their directions remember something those two scalar summaries forget.

An interaction can change the next answer

Build a route through two questions

Choose which filter acts first, then apply them one at a time. Each successful filter moves the state. Before comparing routes, predict which order leaves the stronger horizontal “yes.”

Balanced start. Choose a route and apply its first filter.

z firstz secondTurn this filter

Change filter strengths and inspect the mathematics

z filter factor 2.00 Second filter factor 2.00

Compare passive Bayes messages

A second control: observe an unchanged hypothesis

Ordinary fixed likelihood messages multiply in either order. Starting from equal odds, factors 4 and 9 both give posterior probability 36/37. The active filters above change the state between questions.

L_{z} L_{x} = L_{x} L_{z}

Open the question-order experiment in the playground.

A prediction about gentle interactions

For arbitrary perpendicular strengths, direct calculation gives

\begin{aligned} v_{z \to x} & = (\tanh ξ, 0, \tanh η / \cosh ξ), \\ v_{x \to z} & = (\tanh ξ / \cosh η, 0, \tanh η) . \end{aligned}

Here $\tanh t = \sinh t / \cosh t$ . The difference in horizontal yes probability is

Δ p_{x} = \frac{1}{2} \tanh ξ (1 - \frac{1}{\cosh η}) .

When both strengths are small, $Δ p_{x} \approx η^{2} ξ / 4$ . The effect begins at cubic order in these strengths. Under this model, making two gentle interactions twice as strong can make their order difference roughly eight times larger. That is a specific prediction to explore in simulated learning or fitted interview dynamics.

We also have two controls. Filters aligned with the same axis commute, meaning their order leaves the endpoint unchanged. Fixed classical likelihood factors about an unchanged hypothesis commute as well: $π L_{1} L_{2} = π L_{2} L_{1}$ . The proposed matrix interaction supplies state change between questions.

For a physical success/failure experiment at the chosen strength, use $M = K / \sqrt{2}$ as the success operator. Then $M^{†} M ⪯ I$ : every success probability is at most one. A complementary operator $\sqrt{I - M^{†} M}$ supplies the failure outcome. Starting from $I / 2$ , both successful filters occur with probability $25 / 64$ in either order, and conditioning on those successes gives the two states above.

A table can remember an imaginary direction

The previous experiment changed the state actively. We can also build a joint representation that compares two questions against the same initial state.

A projector $Π$ represents one answer to a sharp matrix question; applying it twice has the same effect as once, $Π^{2} = Π$ . For vertical and horizontal questions, label answers by $a, b \in {- 1, + 1}$ and use

\begin{aligned} Π_{a}^{z} & = \frac{1}{2} (\begin{matrix} 1 + a & 0 \\ 0 & 1 - a \end{matrix}), \\ Π_{b}^{x} & = \frac{1}{2} (\begin{matrix} 1 & b \\ b & 1 \end{matrix}) . \end{aligned}

The Kirkwood-Dirac quasiprobability is

q_{z, x} (a, b) = tr (Π_{b}^{x} Π_{a}^{z} ρ) .

It is a probability-like representation whose entries may be complex or negative. Summing a row or column gives an ordinary probability for the corresponding question asked on its own. This representation has applications ranging from quantum measurement to thermodynamics. Arvidsson-Shukur and coauthors, Properties and applications of the Kirkwood-Dirac distribution.

Multiply our little matrices:

q_{z, x} (a, b) = \frac{1 + a v_{z} + b v_{x} - i a b v_{y}}{4} .

Every imaginary part cancels in each marginal:

\begin{aligned} \sum_{b} q_{z, x} (a, b) & = \frac{1 + a v_{z}}{2}, \\ \sum_{a} q_{z, x} (a, b) & = \frac{1 + b v_{x}}{2} . \end{aligned}

For $v = (0, 1, 0)$ , all single-question answers are balanced, while the joint entries are $(1 + i) / 4$ or $(1 - i) / 4$ . Imagine four arrows leaning upward and downward. Their vertical components cancel when we add any row or column.

Reverse the operator order. Each entry becomes its complex conjugate, meaning the imaginary part changes sign:

q_{x, z} (a, b) = q_{z, x} (a, b)^{*} .

The connection to order is exact. Define the commutator by $[B, A] = B A - A B$ . Then

Im q_{z, x} (a, b) = \frac{1}{2 i} tr ([Π_{b}^{x}, Π_{a}^{z}] ρ) .

The imaginary component records the expectation of this operator-order difference in the chosen state. At the balanced center, the same noncommuting question operators produce a real, positive table. The state matters too.

Real answers, complex relationships

A joint table with turning arrows

A complex number has a horizontal real part and a vertical imaginary part. Each arrow below represents one entry of the Kirkwood-Dirac table for z and x questions. Reverse the question order, then compare the table with frequencies from actual successive sharp measurements.

Choose the state

State coordinate vᵧ 1.00

The three coordinates obey vₓ² + vᵧ² + v_z² ≤ 1. At vᵧ = ±1 the allowed disk shrinks to a point; move vᵧ inward to open it.

Kirkwood-Dirac entries

z+, x+ 0.250− 0.250i

z+, x− 0.250+ 0.250i

z−, x+ 0.250+ 0.250i

z−, x− 0.250− 0.250i

Rows fix the z answer; columns fix the x answer. Horizontal arrow components give real parts; upward components give positive imaginary parts. Row and column sums cancel the imaginary parts.

z yes, asked on its own: 50.0%
x yes, asked on its own: 50.0%
Smallest state eigenvalue: 0.000
Negative real weight in the KD table: 0.000

The state matrix remains positive throughout this experiment. Reversing the operator order keeps every real part and flips every imaginary part. The ordinary single-question probabilities stay the same.

\begin{aligned} q_{z, x} (a, b) & = \frac{1 + a v_{z} + b v_{x}}{4} \\ - \frac{i a b v_{y}}{4} \\ q_{x, z} & = q_{z, x}^{*} \end{aligned}

State pointNonnegative real componentNegative real componentImaginary value

Open the complex-table experiment in the playground.

A positive state with a negative corner

Turn the state toward the diagonal:

v = \frac{1}{\sqrt{2}} (1, 0, 1) .

Its state eigenvalues are one and zero, both nonnegative. Yet

q_{z, x} (-, -) = \frac{1 - \sqrt{2}}{4} < 0.

The two separate yes probabilities are each $(1 + 1 / \sqrt{2}) / 2$ , approximately $85.36 %$ . Their chosen joint representation includes a negative corner to reproduce the matrix relationships.

What would an actual sequence of sharp measurements record? It includes the state change caused by the first question:

\begin{aligned} p_{z \to x} (a, b) & = tr (Π_{b}^{x} Π_{a}^{z} ρ Π_{a}^{z}) \\ = \frac{1 + a v_{z}}{4} . \end{aligned}

These are ordinary positive frequencies. After a sharp vertical answer, the horizontal answer is fair for our perpendicular axes. Reversing the experiment instead gives $p_{x \to z} (a, b) = (1 + b v_{x}) / 4$ . The demo lets you switch between the complex representation and this explicitly defined experiment.

The imaginary coordinate $v_{y}$ remains invisible to the separate $z$ and $x$ marginals. A third question along $y$ reads it through probability $(1 + v_{y}) / 2$ . That suggests a practical strategy: collect enough question directions to reconstruct a state, then use that state to predict each specified interaction and readout.

Build a belief that remembers its journey

Let us imagine a research assistant whose internal evidence lives in a graph. A paper, instrument reading, or interview creates an original source event. Reports and summaries carry tracelets pointing back to that event. Likelihood messages propagate through the graph, with each factor represented once. Hidden causes can be integrated out when a smaller dependence model is useful.

Now give some interactions an active matrix transformation. A question can bring a memory into focus; a rehearsal can strengthen one direction; a selective observation can change what remains available. Model those operations explicitly and retain their order. A final answer is a specified readout of the resulting state.

There are several scales of ambition here. A small implementation could repair copied-source Bayesian pooling. A larger one could combine source graphs with stateful interview models. An adventurous one could learn both the latent state and the transformations produced by different questions, then choose a question for the path it opens through future states.

Such a system would have an unusual form of curiosity. It could ask whether another report adds an origin, whether a hidden cause explains an apparent connection, or whether a question will change the experiment that follows. Its memory would preserve routes through evidence as well as the answers found along them.

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