Physical Statistics

Miraia S. Chiou

Survey · Statistics

Physical Statistics

Connect probability, momentum, and learning through four interactive statistical experiments.

2026-09-08 · 25 min read · 4 interactive demos

Two gardeners test the same kind of seed. Each eventually records 30 germinations and 30 failures. One sees the outcomes alternate. The other gets 30 successes first, becomes increasingly persuaded, and then watches 30 failures arrive.

Their final Beta distributions are identical. Their experiences of learning were quite different.

Could statistics describe that difference using a journey, a clock, and a measure of how much the journey bent? Could disagreement between experiments have an invariant “rest mass”? Could the light cone tell us which changes of belief a finite experiment can possibly achieve?

Physical Statistics is our name for this direction of exploration. Statistical physics uses probability to understand matter. Here, physics suggests new quantities and questions for statistics. Epistemic Physics makes the exchange unusually direct: an evidence state already contains quantities with the algebra of energy, momentum, mass, and velocity.

Let’s use that algebra to build something. Our route passes through statistical position, proper exposure, collision mass, evidence budgets, and the stiffness of a belief. Some steps follow exactly from the existing coordinates; others begin with a new modeling choice. We will make those choices where we need them, then follow their consequences.

Pack a belief for the journey

Let $Y$ be the amount of support for a yes/no proposition and $N$ the amount of opposition. For ordinary binary observations these are counts. Weighted evidence can also use fractional amounts.

A prior describes the probability distribution before these observations. With the Jeffreys prior, which contributes half a unit to each side, the updated belief is

B \sim Beta (α, β), α = Y + \frac{1}{2}, β = N + \frac{1}{2} .

The Beta family describes distributions over probabilities between zero and one, with shape controlled by $α$ and $β$ . Here $B$ is the uncertain probability. Its mean, $μ$ , is the average of that probability under the Beta distribution. The observed trial amount is $t_{E} = Y + N$ ; the amount including the prior is $τ_{E} = t_{E} + 1$ .

Our small dictionary is

\begin{aligned} total amount: & τ_{E} = α + β, \\ signed balance: & x_{E} = α - β = Y - N, \\ mean probability: & μ = \frac{α}{τ_{E}}, \\ velocity: & v_{E} = 2 μ - 1 = \frac{x_{E}}{τ_{E}}, \\ mass: & m_{E} = 2 \sqrt{α β} = \sqrt{τ_{E}^{2} - x_{E}^{2}} . \end{aligned}

A 50% mean has velocity zero. An 80% mean has velocity $0.6$ . Near 100%, the velocity approaches $+ 1$ ; near 0%, it approaches $- 1$ .

In relativistic mechanics with light speed set to one, velocity equals momentum divided by energy. This lets us read the evidence vector $(τ_{E}, x_{E})$ as an energy-momentum state. Its allowed region $τ_{E} \geq | x_{E} |$ is the future cone: total amount cannot be smaller than the magnitude of its signed balance. The prior puts a Beta state strictly inside it.

The following correspondences will do real work:

Relativistic mechanics, with light speed one	Evidence coordinates	Statistical question
Energy $E$	Total amount $τ_{E}$	How large a new input is needed to change this state?
Momentum $P$	Signed balance $x_{E}$	Which direction does the evidence favor?
Velocity $P / E$	Centered mean $v_{E} = 2 μ - 1$	How does a prediction steer a journey?
Rest mass $\sqrt{E^{2} - P^{2}}$	$m_{E} = 2 \sqrt{α β}$	What survives a common change of observer?
Position and elapsed time	A new trajectory $(q, T)$	What does a sequence remember beyond its endpoint?

The usual relativity definitions of energy-momentum, worldlines, and proper time are developed in David Tong’s introductory relativity lectures, §7.4. Our statistical constructions below begin with the evidence dictionary.

Give a prediction a position

If probability is velocity, where does the belief go?

The signed balance $x_{E}$ is already playing the momentum role. Its changes describe incoming evidence: one supporting trial adds $(1, + 1)$ to the energy-momentum vector; one opposing trial adds $(1, - 1)$ . Those increments sit on the cone’s two edges. Their slopes are the incoming signs, even when the current belief’s velocity is near zero.

A position needs a separate rule. Here is a simple choice:

After each observation, move a little traveler at the updated belief velocity for one unit of observation time.

Call this observation clock $T$ . At tick $j$ , the updated mean is $μ_{j}$ and the velocity is $v_{j} = 2 μ_{j} - 1$ . Starting from $q_{0} = 0$ , define

q_{n} = \sum_{j = 1}^{n} v_{j}, T_{n} = n .

The symbol $\sum$ means “add all the terms.” In a continuous version, the same rule reads $d q / d T = v_{E}$ : the slope of the position-time path is the probability-as-velocity.

This position has an immediate statistical reading. Under the usual Beta prediction rule for a binary trial, $μ_{j}$ is the probability of support on the next trial. Thus $q_{n}$ accumulates the signed forecasts made along the way. A long positive excursion records a history spent expecting support, even if later opposition brings the final belief back to 50%.

Imagine using those forecasts to allocate greenhouse space. Two gardeners with the same final estimate might have made very different sequences of planting decisions. Statistical position gives us a compact record of the direction in which their predictions carried them.

A clock carried by the belief

Relativity now suggests a second measurement. Along a small segment, a traveler accumulates proper time according to the segment’s time and position changes. For our chosen journey, define proper exposure $s$ by

Δ s_{j} = \sqrt{Δ T_{j}^{2} - Δ q_{j}^{2}} = \sqrt{1 - v_{j}^{2}} = \frac{m_{j}}{τ_{j}}, s_{n} = \sum_{j = 1}^{n} Δ s_{j} .

Here $m_{j}$ and $τ_{j}$ are the mass and total amount at tick $j$ . “Proper” means that the summed interval survives a common Lorentz change of coordinates along this path. Calling it statistical exposure is our proposed interpretation.

At a balanced mean, one observation tick contributes one unit of proper exposure. At an 80% mean, it contributes $\sqrt{1 - {0.6}^{2}} = 0.8$ . Near either extreme, the contribution becomes small.

So the clock measures time spent with room for both outcomes in the prediction. It distinguishes a history that repeatedly stays near balance from one that spends a long time strongly favoring a side. A well-supported, correct prediction can also have a slow clock; slow proper exposure alone does not diagnose poor learning.

Predict before stepping: will the two gardeners finish with equal proper exposure? What if you reverse the order of the two blocks? What if every tick contributes half a unit to each side?

Position and proper exposure

Same final belief. Different journeys.

Build a route that travels away from home yet finishes with a balanced belief. Tap two cards to swap their order, then launch the traveler. Compare it with the rose alternating route.

Six support and six opposition trials. Same cargo, your choice of order.

Tick 0/12 · position 0.00 · mean 50.0%. The updated belief sets each travel step. Early trials influence many later positions.

Explore 60-trial presets, replay tick by tick, or change observer

Observation tick 0 / 12

Observer rapidity η = 0.000.
Observer velocity w = 0.000.

Chosen historyAlternating referenceEvidence momentum

A journey through position and time

A state in momentum and energy

Coordinate readouts and the travel rule

Mean in original frame: 50.00%
Position q′ / time T′: 0.00 / 0.00
Proper exposure s: 0.00
Invariant mass m_E: 1.00
Journey deficit D: 0.00

Chosen history: Y = 0, N = 0. Its proper exposure is 0.00; the alternating reference has 0.00. Current velocity in this observer frame: 0.000.

The solid path has slope Δq′/ΔT′ equal to the displayed evidence velocity on each segment. Proper exposure sums √(ΔT′² − Δq′²); it and mass stay fixed when you change observer. “Half of each” supplies weighted evidence, with one total unit per tick. Changing the observer transforms both the prior and the observations.

Open the journeys in the playground.

The final Beta retains total support and opposition. The journey also retains a trace of when the predictions changed. We have obtained two candidate statistics, $(q_{n}, s_{n})$ , for questions where the path of learning matters.

There is another useful identity hiding here. A predicted sign $S \in {- 1, + 1}$ with mean $v_{j}$ has variance $1 - v_{j}^{2}$ . Therefore each proper-clock increment equals the standard deviation of the next predicted sign. Standard deviation is the square root of variance. The physical clock has become accumulated predictive spread.

This suggests an experiment: in a system whose conditions change, compare alarms based on raw observation count with alarms based on accumulated proper exposure. Does the second clock help when a long run of one-sided predictions makes calendar time a misleading measure of continued exploration? The clock’s formula is exact for this construction; its usefulness for detection is a question we can test.

A one-sided journey grows old at a square-root rate

Suppose every observation supports the proposition. After $j$ trials,

τ_{j} = j + 1, x_{j} = j, v_{j} = \frac{j}{j + 1}, m_{j} = \sqrt{2 j + 1} .

The position is

q_{n} = \sum_{j = 1}^{n} \frac{j}{j + 1} = n + 1 - H_{n + 1}, H_{k} = \sum_{j = 1}^{k} \frac{1}{j} .

The quantity $H_{k}$ is a harmonic sum. Our traveler approaches the cone edge, but its distance behind that edge grows slowly, like a logarithm.

The proper exposure is

s_{n} = \sum_{j = 1}^{n} \frac{\sqrt{2 j + 1}}{j + 1} \sim 2 \sqrt{2 n} .

The symbol $\sim$ means that the ratio of the two sides approaches one as $n$ grows. The reason is that a late increment is approximately $\sqrt{2 / j}$ , whose accumulated sum grows like a square root.

By comparison, a stream that adds equal weight to both sides at every tick keeps $v_{j} = 0$ and gives $s_{n} = n$ . The two streams have sharply different clock rates, even though each receives one unit per tick.

These are paths of the predictive mean. A walker moved by the actual realized signs $+ 1$ and $- 1$ would instead have segments on the cone edges, called null segments, and zero proper interval on every step. Taking the mean before constructing the path is a consequential part of our model.

Change the observer, carry the clock

An observer frame here means a common re-expression of the coordinate system. Let its velocity be $w$ , and let $η = artanh w$ be its rapidity, an unbounded coordinate for velocities between $- 1$ and $1$ . The inverse hyperbolic tangent, $artanh$ , undoes $\tanh$ .

The same linear transformation acts on position-time and energy-momentum:

\begin{aligned} T^{'} & = T \cosh η - q \sinh η, & q^{'} & = q \cosh η - T \sinh η, \\ τ_{E}^{'} & = τ_{E} \cosh η - x_{E} \sinh η, & x_{E}^{'} & = x_{E} \cosh η - τ_{E} \sinh η . \end{aligned}

Primes mark the new observer’s coordinates. The hyperbolic functions are $\cosh η = (e^{η} + e^{- η}) / 2$ and $\sinh η = (e^{η} - e^{- η}) / 2$ . They play the role that sine and cosine play in ordinary rotations, while preserving a difference of squares.

Both the trajectory slope and the evidence velocity then obey

v_{E}^{'} = \frac{v_{E} - w}{1 - w v_{E}} .

Coordinate times and energies change; $s$ and $m_{E}$ remain the same. The rest-frame button chooses $w = v_{E}$ at the selected tick, making the current momentum zero and its energy equal to its mass. Earlier parts of the path can still move in that frame.

In support/opposition coordinates, the transformation simply reads $α^{'} = e^{- η} α$ and $β^{'} = e^{η} β$ . It transforms the prior along with the observations. Adding a fresh symmetric prior after every frame change would describe a different operation.

This is a transformation of our mean-state representation and chosen journey. Transforming a random probability $B$ requires its own change-of-variable calculation; a nonlinear transform of a mean generally differs from the mean of the transformed random variable. Nor does choosing an observer generate new data. The attraction is a family of comparable geometric summaries, with an explicit rule for what changes and what stays fixed.

Weigh the disagreement in a collision

Suppose two laboratories each send us 50 trials. Lab A reports 45 support and 5 opposition. Lab B reports 5 support and 45 opposition.

Each laboratory has a strongly directed result. Together they produce a balanced 50/50 tally. How much of the pooled mass comes from this meeting of different directions?

For this calculation, use raw evidence packets, adding no prior to either packet. Packet $i$ has energy $E_{i} = Y_{i} + N_{i}$ , momentum $P_{i} = Y_{i} - N_{i}$ , and mass

m_{i} = \sqrt{E_{i}^{2} - P_{i}^{2}} = 2 \sqrt{Y_{i} N_{i}} .

An index $i$ labels a packet. Pooling adds the energies and momenta. The pooled mass is

M = \sqrt{{(\sum_{i} E_{i})}^{2} - {(\sum_{i} P_{i})}^{2}} .

Now define the collision surplus

Δ m = M - \sum_{i} m_{i} .

For positive evidence, this surplus is nonnegative. It is zero when all packets point in the same direction. Each separate mass and the pooled mass survive a common Lorentz transformation, so their difference does too.

For our laboratories, each separate mass is $2 \sqrt{45 \cdot 5} = 30$ . The pooled mass is $100$ . Thus $Δ m = 100 - 60 = 40$ : forty units of rest mass appear through pooling different directions.

Try to break the interpretation: what happens when both labs agree at 90%? When they are individually unanimous but opposite? When you change the observer?

Invariant disagreement

Let two bodies of evidence collide.

Each colored packet contains 50 units of evidence. Drag its point sideways to change its support/opposition balance. The gold point is their sum. How much pooled mass comes from the difference between their directions?

Packet APacket BPooled state

A: Y = 45, N = 5 B: Y = 5, N = 45

Sum of separate masses: 60.00
Pooled mass M: 100.00
Mass surplus Δm: 40.00
Fraction from pooling: 40.0%

Pooled energy in this frame: 100.00. Mass surplus stays 40.00 under a common change of observer. The different directions contribute extra rest mass.

These are raw packets, with no prior added to either. The surplus depends on how you divided the data into packets. When the points overlap, use Tab to select either packet and arrow keys to separate them.

Open the collision in the playground.

This offers a candidate measure of heterogeneity, meaning variation between groups: the amount of pooled rest mass attributable to their different directions. We could use it to compare batches of a manufacturing process, regions in a survey, or habitats in the seed experiment.

A normalized version, $Δ m / M$ , describes the fraction of pooled mass contributed by the collision, when $M > 0$ . Its meaning is quite specific. Two individually unanimous, opposing packets give a fraction of one even if one packet is much smaller. The absolute surplus still records their scale. When every packet is unanimous in the same direction, $M = 0$ and the fraction is undefined.

The packet division is part of the statistic. Treat all 100 trials as one packet and the surplus is zero. Separate them by laboratory and it can reveal a difference. This makes the choice of grouping scientifically meaningful: we are asking which distinctions explain how pooled mass was assembled. A reference distribution for deciding whether a surplus exceeds ordinary sampling variation would require a sampling model for those groups.

Why does disagreement create extra mass?

For packets with positive mass, write

E_{i} = m_{i} \cosh ρ_{i}, P_{i} = m_{i} \sinh ρ_{i}, ρ_{i} = artanh (P_{i} / E_{i}) .

For two packets, expanding the pooled difference of squares gives

M^{2} = m_{1}^{2} + m_{2}^{2} + 2 m_{1} m_{2} \cosh (ρ_{1} - ρ_{2}) .

Since $\cosh z \geq 1$ , we have $M \geq m_{1} + m_{2}$ . Relative rapidity $ρ_{1} - ρ_{2}$ measures their directional separation and stays unchanged when both rapidities shift by the same amount.

Move to the pooled rest frame, whose rapidity is $ρ_{pool}$ . There, the total energy is $M$ , and

Δ m = \sum_{i} m_{i} [\cosh (ρ_{i} - ρ_{pool}) - 1] .

The right side is exactly the packets’ total kinetic energy, their energy of motion, in that frame, in units with light speed one. Relative motion becomes the extra rest mass of the combined state. Packets of zero mass can be handled directly with their $(E_{i}, P_{i})$ coordinates or by a limit.

There is also a statistical form. Let $p_{i} = Y_{i} / E_{i}$ , let $E = \sum_{i} E_{i}$ , and let $w_{i} = E_{i} / E$ be each packet’s fraction of the total amount. Define $g (p) = 2 \sqrt{p (1 - p)}$ and the pooled fraction $\bar{p} = \sum_{i} w_{i} p_{i}$ . Then

\frac{Δ m}{E} = g (\bar{p}) - \sum_{i} w_{i} g (p_{i}) .

The function $g$ bends downward: averaging its inputs gives at least as large a value as averaging its outputs. This difference is called a Jensen gap. The energy-momentum geometry has selected a particular gap for comparing groups.

For closely spaced interior fractions, a quadratic approximation gives

\frac{Δ m}{E} \approx \frac{{Var}_{w} (p_{i})}{4 [\bar{p} (1 - \bar{p})]^{3 / 2}}, {Var}_{w} (p_{i}) = \sum_{i} w_{i} (p_{i} - \bar{p})^{2} .

Thus the candidate emphasizes a given small variation more strongly near the probability boundaries. That is a testable design choice: does it help identify meaningful differences among mostly successful systems, or overreact to rare failures? The exact surplus lets us investigate without relying on the approximation near a boundary.

The collision also measures a journey’s bending

Here is a connection between our first two tools. Regard each journey segment $(Δ T_{j}, Δ q_{j})$ as a little future-cone vector. Its mass is its proper exposure $Δ s_{j}$ . The sum of the segments joins the journey’s endpoints.

The same surplus formula becomes

D = \sqrt{T_{n}^{2} - q_{n}^{2}} - s_{n} \geq 0.

A straight journey with constant predictive velocity has $D = 0$ . Changing velocity produces a positive deficit between the straight path’s proper interval and the actual accumulated clock. We can interpret $D$ as a measure of variation in predictive direction across a journey, and it is invariant under a common observer change.

The same geometry now compares disagreement between experiments and variation along a sequence of predictions. This seems a promising place to look for a unified description of changing conditions and conflicting populations.

Ask what a finite evidence budget can reach

A proposed experiment will collect 100 more trials. Can it move a mean from 50% to 80%?

The answer depends on the evidence already present. Physics suggests drawing the set of permitted incoming energy-momenta, then asking whether it reaches the target direction.

Let $a \geq 0$ be the new evidence amount and $u \in [- 1, 1]$ its signed direction. All support means $u = + 1$ ; all opposition means $u = - 1$ . Adding this packet gives

τ_{E}^{'} = τ_{E} + a, x_{E}^{'} = x_{E} + a u, v_{E}^{'} = \frac{τ_{E} v_{E} + a u}{τ_{E} + a} .

Because the incoming direction is bounded, the reachable mean probabilities lie in the interval

\frac{α}{τ_{E} + a} \leq μ^{'} \leq \frac{α + a}{τ_{E} + a} .

The endpoints spend the entire budget on opposition or support. Fractional weights fill the interval. If all observations must be whole binary trials, only the corresponding discrete means are attainable at a fixed integer budget.

We can also solve backward. Let $μ_{*}$ be the target mean and $v_{*} = 2 μ_{*} - 1$ its velocity. For a different target, write $d = sign (v_{*} - v_{E})$ , the direction toward it: $+ 1$ for an increase and $- 1$ for a decrease. The least possible positive evidence amount is

a_{min} = \frac{τ_{E} | v_{*} - v_{E} |}{1 - d v_{*}} .

If the target equals the current mean, the required amount is zero. At either exact probability boundary, the required amount is infinite for a Beta state with positive weight on both sides.

A speed limit for belief revision

Can that much evidence get you there?

Drag the rose target along the probability axis. The teal band contains every mean reachable with the chosen amount of new, positive evidence, including a best-case run entirely on one side.

New evidence budget a 100 units

Current meanReachable meansTarget mean

Current mean, μ: 50.00%
Target mean, μ*: 80%
Minimum positive amount: 151.50
Whole trials to reach or pass: 152

The target is beyond this budget. The reachable interval is 25.12% to 74.88%. The minimum assumes every added unit favors support.

Starting evidence: Y = 50, N = 50, plus the Jeffreys prior once. A fractional amount can hit the target exactly; whole trials may pass it. The band allows fractional support/opposition weights. Actual observations determine which reachable mean you obtain.

Open the evidence budget in the playground.

Starting with $Y = N = 50$ , we have $τ_{E} = 101$ . To reach $μ_{*} = 0.8$ , the target velocity is $0.6$ , so

a_{min} = \frac{101 \cdot 0.6}{1 - 0.6} = 151.5 .

It takes 151.5 units of pure support to hit 80% exactly, or 152 whole supporting trials to reach or pass it. Even an all-success experiment of 100 trials cannot get there. Starting with $Y = N = 1$ instead requires only 4.5 units. Both beliefs start at 50%; their malleability, or ease of changing, is very different.

This could become a useful planning tool. Before discussing expected outcomes, first ask whether a proposed study can reach a decision threshold under its most favorable possible observations. If its budget is too small, the threshold is geometrically out of reach. If it is large enough, a sampling model can tell us how likely the required outcomes are.

Find the target by intersecting a ray with the future cone

The target is the ray $x^{'} = v_{*} τ^{'}$ . The boundary of the available positive input points in direction $d$ . Substituting $x^{'} = τ_{E} v_{E} + d a$ and $τ^{'} = τ_{E} + a$ gives

v_{*} (τ_{E} + a) = τ_{E} v_{E} + d a .

Rearrange to obtain the minimum above. Any less directed packet needs at least as much amount because its $u$ lies between $- 1$ and $1$ .

There is a differential version too. During a smooth stream of direction $u$ ,

\frac{d v_{E}}{d τ_{E}} = \frac{u - v_{E}}{τ_{E}} .

Existing amount appears in the denominator of the response. The same incoming direction bends a small state more quickly than a large one. This is a precise sense in which trial amount produces resistance to changing the mean.

The budget assumes we keep the existing evidence and add nonnegative input. Forgetting old observations, changing their weights, or admitting signed tracelets changes the allowed operations and therefore the reachable set.

Connect the clock to the shape of a belief

We have built a journey from the mean of a Beta distribution. Can its proper-clock rate tell us anything about the full distribution?

Introduce the random rapidity

R = \frac{1}{2} \ln \frac{B}{1 - B} .

The natural logarithm $\ln$ turns ratios into additive distances. Rapidity stretches the probability boundaries out to $- \infty$ and $+ \infty$ .

A probability density tells us how probability is distributed per unit of a coordinate. When we change from $B$ to $R$ , that unit changes too. Including the required change-of-variable factor gives

f_{R} (r) \propto \frac{e^{x_{E} r}}{(2 \cosh r)^{τ_{E}}} .

Here $f_{R}$ is the density in rapidity and $\propto$ means “equal up to a positive normalization constant.” Its peak is at

ρ_{E} = \frac{1}{2} \ln \frac{α}{β}, \tanh ρ_{E} = v_{E} .

The peak in rapidity maps back to the Beta mean $μ$ . It is generally different from the mode, or location of a peak, of the density in probability coordinates. Density peaks depend on the coordinate used.

Now imagine the negative logarithm of this density as a bowl. Moving away from its bottom costs log density. The bowl’s curvature, its second derivative at the bottom, measures its local stiffness:

K = τ_{E} (1 - v_{E}^{2}) = \frac{m_{E}^{2}}{τ_{E}} .

Compare this with the clock we built earlier:

K = τ_{E} {(\frac{d s}{d T})}^{2} .

The same factor that slows the proper clock also reduces density stiffness per unit of total evidence on the rapidity scale. This is an exact connection between a chosen mean trajectory and the local shape of its Beta state.

Mass, clocks, and curvature

How stiff is a belief in rapidity?

Try 50 support / 50 opposition, then 99 / 1. Both contain 100 observed trials. Drag the gold probe along the same half-log-odds axis. How quickly does the curve climb away from its minimum?

Y = 50 · N = 50 Total observed trials

Showing the posterior density in r, including the change-of-variable factor. Its peak maps to the Beta mean.

Selected evidenceBalanced reference, same totalLocal spring approximation

The vertical scale stays fixed. Curves above 14 continue beyond the window. The blue and teal curves coincide at balance.

Total including prior, τ_E: 101
Epistemic mass, m_E: 101.000
Stiffness at the peak: 101.000

Probe r = 0.50 means p = 73.11%. Log drop: 12.132. Relative density in r: 0.00000539.

ρ_{E} = \frac{1}{2} \ln \frac{α}{β}, K = \frac{m_{E}^{2}}{τ_{E}} = 101.000

Y and N count observed support and opposition. The Jeffreys option uses α = Y + ½ and β = N + ½. K is the curvature of the negative log posterior density in r. I_r is data likelihood curvature at its fit. The gold parabola agrees locally at a finite minimum; farther away, the exact blue curve can differ.

Open the density landscape in the playground.

Near its peak, a displacement $δ r = r - ρ_{E}$ produces approximately

- \ln \frac{f_{R} (r)}{f_{R} (ρ_{E})} \approx \frac{1}{2} K (δ r)^{2} .

That is the shape of a spring’s energy. It suggests choosing small numerical search steps on the scale $1 / \sqrt{K}$ : wider steps where a rapidity landscape is locally shallow, smaller steps where it is steep. The demo lets us inspect where the quadratic approximation stops following the exact curve.

There is an intriguing surprise as trials accumulate. With $n$ supporting trials and no opposition, $m_{E}^{2} = 2 n + 1$ and $τ_{E} = n + 1$ , so $K \to 2$ . The posterior keeps concentrating near probability one, yet its local stiffness in rapidity approaches a finite value. The moving peak travels toward infinity while the rapidity scale keeps resolving multiplicative changes in the increasingly small failure probability.

Proper exposure and mass are observer invariants of the constructions above. $K = m_{E}^{2} / τ_{E}$ depends on energy and the chosen density coordinate. Its role is a local shape measurement, with its coordinate stated.

Derive the density and its stiffness

A Beta density is proportional to $p^{α - 1} (1 - p)^{β - 1}$ . The inverse rapidity map and its derivative are

p = \frac{1 + \tanh r}{2}, \frac{d p}{d r} = 2 p (1 - p) .

Multiplying the original density by $d p / d r$ gives a rapidity density proportional to $p^{α} (1 - p)^{β}$ . Using $p = e^{r} / (2 \cosh r)$ and $1 - p = e^{- r} / (2 \cosh r)$ produces the displayed formula.

Write its negative logarithm as $U (r)$ , ignoring the additive constant:

U (r) = τ_{E} \ln (2 \cosh r) - x_{E} r .

Differentiate:

U^{'} (r) = τ_{E} \tanh r - x_{E}, U^{″} (r) = τ_{E} {sech}^{2} r .

The hyperbolic secant is $sech r = 1 / \cosh r$ . Setting the first derivative to zero locates the peak; evaluating the second there gives $K$ .

There are two different operations to keep track of here. Relabeling the same random rapidity as $R^{'} = R - η$ translates its density and preserves its curvature. Constructing a new Beta distribution from the boosted evidence shapes $α^{'} = e^{- η} α$ and $β^{'} = e^{η} β$ instead gives $K^{'} = m_{E}^{2} / τ_{E}^{'}$ . The observer demo transforms the evidence vector and mean trajectory; it does not equate these two operations on the full probability law.

The demo’s observed-trials option uses the likelihood, the probability of the data regarded as a function of a candidate parameter. Its corresponding expression has $Y + N$ in place of $τ_{E}$ . If both counts are positive, the fitted curvature is $I_{r} = 4 Y N / (Y + N)$ . With unanimous data, its best fit lies at infinite rapidity, so there is no finite bottom. The prior-inclusive density retains a finite peak for every finite nonnegative tally.

A statistical laboratory built from the cone

We now have a collection of tools that ask different questions of the same evidence:

Proposed tool	What it measures	A place to try it
Position $q = \sum_{j} v_{j}$	Accumulated direction of sequential predictions	Compare histories of decisions with the same final estimate
Proper exposure $s = \sum_{j} \sqrt{1 - v_{j}^{2}}$	Accumulated predictive spread on a chosen observation clock	Schedule checks during long stretches of one-sided prediction
Collision surplus $Δ m$	Pooled mass arising from differences between groups	Compare batches, habitats, or experimental conditions
Journey deficit $D$	Variation in predictive direction along a path	Compare steady and changing prediction histories
Minimum budget $a_{min}$	Least positive evidence that can reach a target mean	Check whether a proposed study can cross a threshold
Rapidity stiffness $K$	Local curvature of the full belief density in rapidity	Scale numerical steps and inspect boundary behavior

One especially tempting direction is to organize evidence streams by their own proper exposure. Another is to treat changes between experimental conditions as collisions, asking how much new structure is hiding in the extra mass. A third is to design an experiment as a navigation problem: choose feasible evidence inputs, follow the evolving momentum, and reach a target with the smallest cost.

The cone already supplies exact relations among these quantities. The next adventure is to discover which of them explain statistical behavior that our usual summaries overlook.

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