A “mean” and “variance” for the Cauchy that has neither

Miraia S. Chiou

Field trip · Geometry · Probability

A “mean” and “variance” for the Cauchy that has neither

Wrap deviations into bounded trial directions to find the Cauchy peak and a variance with constant line width. An Epistemic Physics derivation meets McCullagh’s location and scale.

2026-09-07 · 12 min read

The Cauchy distribution has a clear peak and a recognizable width, yet its ordinary population mean, variance, and standard deviation are undefined: the tail integrals do not converge. Can we build useful counterparts from quantities that stay finite?

Imagine a small observatory. Wrap the number line around its circular horizon, let trials contribute arrows, and move the observatory until they balance above and below. That position will be our “mean.” A ratio of trial amounts will give a width; its square will be our “variance.” The mean will sit at the Cauchy peak, and moving it at fixed variance will slide the curve without widening it.

A ruler that travels with the observatory

Choose a tentative center $a$ and a reference length $L > 0$ . Wrap an observation $x$ into the angle

ϕ_{a} (x) = 2 \arctan \frac{x - a}{L}, x = a + L \tan \frac{ϕ_{a}}{2} .

The inverse tangent, $\arctan$ , squeezes the entire line into a finite range of angles. Put angle zero at three o’clock and a cut at nine o’clock. The rightmost point reads $a$ ; the top and bottom read $a + L$ and $a - L$ . Both infinities meet at the cut.

An enormous deviation becomes a finite arrow near nine o’clock. Moving $a$ slides the number markings around the circle, so the ruler travels with the observatory.

How does opening this circle produce a Cauchy?

This map is stereographic projection: a ray from the cut through a rim point meets a straight ruler, which we can turn horizontally to draw the line.

Place on the horizon	Angle $ϕ_{a}$	Line value $x$
Lower approach to the cut	$- π$	$- \infty$
Bottom	$- π / 2$	$a - L$
Rightmost point	$0$	$a$
Top	$+ π / 2$	$a + L$
Upper approach to the cut	$+ π$	$+ \infty$

A uniform horizon has probability per angle $1 / (2 π)$ . Multiply by the stretching factor $| d ϕ_{a} / d x | = 2 L / [L^{2} + (x - a)^{2}]$ to obtain the Cauchy density

f (x) = \frac{L}{π [(x - a)^{2} + L^{2}]} .

Other circular shapes let the width vary independently of $L$ .

Turn deviations into trial amounts

In the compass extension, each trial contributes the length-one arrow $(\cos ϕ, \sin ϕ)$ . Let $A$ be the total trial amount, including the prior, and $E$ an average weighted by the normalized trial distribution.

Write $δ = X - a$ for an observation’s deviation. Split each trial smoothly between the near-center amount $Y$ and far-from-center amount $N$ :

\begin{aligned} Y (a) & = A E [\frac{L^{2}}{L^{2} + δ^{2}}], \\ N (a) & = A E [\frac{δ^{2}}{L^{2} + δ^{2}}] . \end{aligned}

The weights sum to one, so $Y + N = A$ . A trial at $a$ contributes entirely to $Y$ ; a distant trial contributes mostly to $N$ . We also keep the signed vertical balance

T (a) = A E [\frac{2 L δ}{L^{2} + δ^{2}}] .

Positive deviations pull upward; negative deviations pull downward. All three weights stay bounded, even far down a tail.

Find the mean by balancing above and below

Move the observatory until the vertical arrows cancel. Define the balance mean $m$ by

T (m) = 0.

For a Cauchy, this balance has exactly one zero: its peak and median, whose location is conventionally called $μ$ . Thus $m = μ$ . Stand left of the peak and the vertical balance is positive; stand right of it and the sign reverses.

Why does the balance pick out the peak?

A Cauchy with location $μ$ and positive scale $γ$ has density

f (x) = \frac{γ}{π [(x - μ)^{2} + γ^{2}]} .

Averaging our bounded trial weights gives

\frac{Y (a)}{A} = \frac{L (L + γ)}{(μ - a)^{2} + (L + γ)^{2}}, \frac{T (a)}{A} = \frac{2 L (μ - a)}{(μ - a)^{2} + (L + γ)^{2}} .

The denominator stays positive, so $T (a) = 0$ exactly when $a = μ$ .

To evaluate the averages, set $d = μ - a$ and combine them into one integral:

E [\frac{L}{L - i (X - a)}] = \frac{L}{L + γ - i d}, i^{2} = - 1.

The real part is $Y / A$ ; twice the imaginary part is $T / A$ . A short complex-integration check factors the density as

\frac{1}{2 π i} (\frac{1}{ξ - μ - i γ} - \frac{1}{ξ - μ + i γ}),

where $ξ$ is the complex continuation of $x$ . Close the integration path in the upper half-plane. The factor $L / [L - i (ξ - a)]$ has its pole below the real axis, so the only enclosed pole is $μ + i γ$ . The residue theorem evaluates the factor there, giving the stated result. The large semicircle contributes zero because the full integrand decays as $| ξ |^{- 3}$ .

Give variance a constant line width

At the balanced center, the trial ratio recovers the Cauchy width $γ$ . Define our line variance $V$ as its square:

\begin{aligned} γ & = L \frac{N}{Y}, \\ V & = L^{2} {(\frac{N}{Y})}^{2} = γ^{2} . \end{aligned}

The width $\sqrt{V} = γ$ is half the interquartile range: the interval from $m - \sqrt{V}$ to $m + \sqrt{V}$ contains the middle 50% of the probability. It plays the standard-deviation-like role. If $x$ is in meters, the width is in meters and $V$ is in square meters.

For example, $L = 1$ , $Y = 2.5$ , and $N = 1.5$ give width $0.6$ and variance $0.36$ . Moving the mean from $0.8$ to $3$ slides that middle-half interval from $[0.2, 1.4]$ to $[2.4, 3.6]$ , with the same length.

Check the width calculation

At $a = m = μ$ , the bounded averages simplify to

\frac{Y}{A} = \frac{L}{L + γ}, \frac{N}{A} = \frac{γ}{L + γ} .

Divide the second by the first: $N / Y = γ / L$ . Multiplying by $L$ recovers $γ$ for any choice of reference length. The Cauchy probability formula gives

Pr (m - \sqrt{V} \leq X \leq m + \sqrt{V}) = \frac{1}{2} .

Predict what will move in the demo. The line curve should travel with the rose mean. Will the recentered circular shape or its near/far balance change?

The Epistemic Physics route

Move the Cauchy mean, keep the width

Drag the rose peak: sideways moves the mean; up narrows the curve, down widens it. Watch the circle and line together. Arrow keys work too.

1. The recentered circle

The circular ruler moves with m; both infinities meet at the cut.

2. The same curve on the line

The gold interval holds the middle 50%. Drag the rose peak.

Balance mean / line peakProbability densityMiddle 50% intervalRecentered mean arrow

Precise controls and presets

Balance mean m 0.80 Line variance V 0.36

Balance mean m: 0.800
Line variance V: 0.360
Width √V: 0.600

Middle half: [0.20, 1.40]. Moving m at fixed V translates this interval with length 1.200 throughout.

Explore the recentered trial balance

T (m) = 0, \sqrt{V} = L \frac{N}{Y}, V = L^{2} {(\frac{N}{Y})}^{2}

The recentered trial balance

Blue Y weights near-center deviations; apricot N weights far-from-center deviations. Total amount A = 4 and reference length L = 1 here. Moving the center with the curve preserves both amounts. Changing its width changes their ratio.

Near Y = 2.500 Far N = 1.500

Balance rapidity χ: 0.255
Rest mass M = 2√(YN): 3.873

χ = \frac{1}{2} \log \frac{Y}{N}, \sqrt{V} = L e^{- 2 χ}, V = L^{2} e^{- 4 χ}

The signed rapidity becomes negative when the width exceeds L.

Read the circle and line in detail

Number ticks slide around the circle as m moves. The rightmost point reads m, the top m + 1, and the bottom m − 1. Small ticks mark quarter steps where space allows; both infinities meet at the cut. Darker rim segments carry greater angular density. The blue dot is the recentered mean arrow, with vertical balance zero.

The rose marker stays on the peak as you drag it in either direction. Arrow keys work too. The gold interval contains 50% of the probability. Both axes stay fixed: x runs from −6 to +6, and the density scale from zero to 1.592 per unit x. Tails continue beyond this window.

Compare with McCullagh's location and scale

The values agree exactly: m is McCullagh's location μ, and √V is his scale γ. Our variance V is γ². The trial balance supplies the EP derivation, while A records a separate amount.

μ = m: 0.800
γ = √V: 0.600
γ² = V: 0.360

w = m + i \sqrt{V} = μ + i γ

Compare the fixed-circle direction and spread

Keep the wrapping origin at zero instead of moving it with m. Its mean heading and wrapped variance can change during a line translation, because each part of that circle stretches differently. They measure direction and spread on the fixed circle.

Unwrapped circular mean direction: 1.000
Fixed-circle wrapped variance: 0.750

m_{dir} = \tan \frac{\arg z}{2}, σ_{wrap}^{2} = 1 - | z |^{2}

The width stays constant because the construction uses deviations. Shift every observation and the mean together: $X - m$ stays the same, so the wrapped arrows, $Y / N$ , and $V$ stay the same too.

Rapidity turns the ratio into a ruler

The Epistemic Physics rapidity $χ$ expresses the same trial ratio logarithmically:

\begin{aligned} χ & = \frac{1}{2} \log \frac{Y}{N}, \\ \sqrt{V} & = L e^{- 2 χ}, \\ V & = L^{2} e^{- 4 χ} . \end{aligned}

Equal amounts give $χ = 0$ and width $L$ . More near-center amount gives positive rapidity and a narrower curve; more far-from-center amount gives negative rapidity and a wider curve. Equal steps in rapidity multiply the width by equal factors.

The belief and relativity field trip follows this ratio into the light cone.

Follow the light-cone and Doppler connection

The collective spatial arrow is $P = (Y - N, T)$ , and the Epistemic Physics rest mass satisfies

M^{2} = A^{2} - | P |^{2} = 4 Y N - T^{2} .

Here $| P |$ is the arrow’s length. At the balanced center, $T = 0$ , so

A = Y + N, P_{x} = Y - N, M = 2 \sqrt{Y N}, \frac{P_{x}}{A} = \tanh χ .

The hyperbolic tangent is $\tanh χ = (e^{χ} - e^{- χ}) / (e^{χ} + e^{- χ})$ . Write $r = (Y - N) / A$ , allowing either sign. The centered Cauchy wraps into the Poisson density

h_{r} (ϕ) = \frac{1 - r^{2}}{2 π (1 - 2 r \cos ϕ + r^{2})} .

Unwrapping with $x = m + L \tan (ϕ / 2)$ , including the stretching factor, gives

f (x) = \frac{L q}{π [(x - m)^{2} + (L q)^{2}]}, q = \frac{1 - r}{1 + r} = \frac{N}{Y} .

Negative $r$ gathers probability toward the cut and gives a broad line curve. A physical boost of signed rapidity $η$ produces this Poisson family when transformed directions are counted equally, with $r = \tanh (η / 2)$ and hence $η = 2 χ$ . This direction-only law uses one Doppler factor, $D$ ; giving each null trial an additional Doppler amount uses $D^{2}$ and changes the angular law.

Meet McCullagh: the same center and width

McCullagh packages a Cauchy as the complex point $w = μ + i γ$ , where $i^{2} = - 1$ . Our trial construction reaches exactly the same point: $w = m + i \sqrt{V}$ .

Quantity	Epistemic Physics	McCullagh
Center	$m$ solves $T (m) = 0$	$μ = m$
Width	$L N / Y = \sqrt{V}$	$γ = \sqrt{V}$
Squared width	$V = L^{2} (N / Y)^{2}$	$γ^{2} = V$

The numbers agree; the EP route derives them through trial amounts. McCullagh’s 1992 paper develops the complex parametrization, and his 1996 paper develops its circular geometry. The total amount $A$ remains a separate part of the trial ledger, telling us how readily new evidence can move the belief.

Our observatory has found a traveling center and a steady width. The side trips below explore what happens when we change rulers or ask how much trial amount supports the curve.

Further explorations

Change the ruler: try McCullagh's transformation demo

A Möbius transformation changes the ruler by a fraction $(a x + b) / (c x + d)$ , with $a d - b c > 0$ . Here $a, b, c, d$ are transformation coefficients. McCullagh’s rule applies the same fraction to samples and the complex parameter:

x^{'} = \frac{a x + b}{c x + d}, w^{'} = \frac{a w + b}{c w + d} .

A translation gives $m^{'} = m + b$ , $V^{'} = V$ . Scaling by a positive $k$ gives $m^{'} = k m$ , $V^{'} = k^{2} V$ . Try a reciprocal to see a more dramatic change.

McCullagh's transformation rule

Two routes to the same Cauchy curve

Start with balance mean m = 0.8 and line variance V = 0.36, giving McCullagh's parameter w = m + i√V = 0.8 + 0.6i. Shift the curve: its width stays fixed. Then try a scaling or a reciprocal, applying the same transformation to every sample and to w.

T (x) = x + 1

Original distributionTransformed distribution

Three example samples

Apply the selected ruler

-0.8	0.2
0.4	1.4
1.6	2.6

The parameter takes the same ride

Before: w: 0.8 + 0.6i
After: T(w): 1.800 + 0.600i

X \sim C (w) ⟹ \frac{a X + b}{c X + d} \sim C (\frac{a w + b}{c w + d})

EP statistics and McCullagh's parameters agree

Line statistics before and after the transformation
Quantity	Before	After
Balance mean m = location μ	0.8	1.800
Width √V = scale γ	0.6	0.600
Line variance V = γ²	0.36	0.360

V^{'} = \frac{(a d - b c)^{2} V}{[(c m + d)^{2} + c^{2} V]^{2}}

A shift adds 1 to the mean while preserving variance 0.36 and width 0.6. The two curves have exactly the same shape.

Compare the fixed-circle statistics too

Wrapping line values with φ = 2 arctan x gives a mean vector z. Its heading and wrapped variance use the fixed circle's ruler, so they can change even when line width stays fixed.

Fixed-circle statistics before and after
Circle quantity	Before	After
Unwrapped mean direction	1.000	1.956
Mean vector z	(0.000, 0.500)	(-0.448, 0.621)
Wrapped variance = M²/A²	0.750	0.414

z = \frac{i - w}{i + w}, σ_{wrap}^{2} = 1 - | z |^{2}

These presets have ad − bc > 0, so the parameter stays above the real line. The reciprocal chart has a pole at x = 0; a continuous Cauchy sample hits that exact point with probability zero. The displayed curve window does not include the whole probability mass.

For general coefficients, the parameter rule gives

\begin{aligned} m^{'} & = \frac{(a m + b) (c m + d) + a c V}{(c m + d)^{2} + c^{2} V}, \\ V^{'} & = \frac{(a d - b c)^{2} V}{[(c m + d)^{2} + c^{2} V]^{2}} . \end{aligned}

Compare the fixed circle's direction and spread

Leave the wrapping origin at zero. For its unit arrows $Z = (\cos ϕ, \sin ϕ)$ , let $z = E [Z]$ . The heading $θ$ of this mean vector gives an unwrapped direction $m_{dir} = \tan (θ / 2)$ . Its wrapped variance measures squared straight-line distances inside the circle:

σ_{wrap}^{2} = E | Z - z |^{2} = 1 - | z |^{2} .

Expanding the square and using $| Z |^{2} = 1$ proves the equality. If $z = 0$ , no heading is selected. Compare the two rulers for our starting example:

Ruler	Center statistic	Spread statistic
Line, using recentered trials	Peak $m = 4 / 5$	Squared width $V = 9 / 25$
Fixed unit circle, $z = (0, 1 / 2)$	Unwrapped heading $m_{dir} = 1$	Wrapped variance $3 / 4$

A line translation rearranges the fixed circle’s angles nonuniformly, so its wrapped statistics can change at constant line width. A rigid circle rotation preserves wrapped variance, while its line image can widen near the cut.

Even in the recentered chart, wrapped variance is

σ_{wrap, centered}^{2} = \frac{4 Y N}{(Y + N)^{2}} = \frac{4 L \sqrt{V}}{(L + \sqrt{V})^{2}} .

Widths $L q$ and $L / q$ give the same value. The ratio $N / Y$ distinguishes them because their near/far amounts exchange places.

Decode directly from a fixed-chart trial ledger

Suppose the ledger already uses $ϕ = 2 \arctan x$ . Write its amount as $A$ , its spatial arrow as $(P_{x}, P_{y})$ , and define

M^{2} = A^{2} - P_{x}^{2} - P_{y}^{2}, K = (A + P_{x})^{2} + P_{y}^{2} .

For a Poisson circular belief with positive rest mass, the direct decoder is

m = \frac{2 A P_{y}}{K}, \sqrt{V} = \frac{M^{2}}{K} .

Unwrapping its density gives the denominator $K x^{2} - 4 A P_{y} x + (A - P_{x})^{2} + P_{y}^{2}$ . Complete the square using

K [(A - P_{x})^{2} + P_{y}^{2}] - (2 A P_{y})^{2} = M^{4}

to recover that center and width. In complex notation this is the Cayley transform, from the disk to the upper half-plane:

w = i \frac{A - P}{A + P}, P = P_{x} + i P_{y} .

These amounts belong to the fixed chart; the ratio $\sqrt{V} = L N / Y$ uses the chart recentered at $m$ . Both decode the same curve.

Same curve, different trial amounts: explore uncertainty

Balances $2.5 : 1.5$ and $25 : 15$ give the same width, but a new unit trial changes the smaller ledger more easily. It is more malleable.

A Dirichlet process, or DP, models random allocations of weight around the circle. Write $G \sim DP (A H)$ , where $H$ is the mean probability around the circle and $A$ is the prior-inclusive amount. Let $n = (\cos ϕ, \sin ϕ)$ be a direction. Each allocation has a mean vector $B = \int n d G$ , with expected value $z = E [B]$ and total coordinate variance

E | B - z |^{2} = \frac{1 - | z |^{2}}{A + 1} .

Increase $A$ while holding the curve fixed: the cloud of possible wrapped balances tightens. The demo shows uncertainty in $B$ ; uncertainty about a decoded line peak requires a further nonlinear calculation.

Uncertainty about the wrapped balance

Same Cauchy curve, different trial amounts

Keep one Cauchy curve; change how much trial weight carries it. Choose the right-hand amount and compare the two interior clouds. Drag either rose point to move both rims together.

Reference: A = 4

Wrapped variance on the rim: 0.750
Variance of wrapped balances: 0.1500

Your comparison: A = 16

Wrapped variance on the rim: 0.750
Variance of wrapped balances: 0.0441

Both curves have line mean 0.800 and line variance 0.360. Their possible wrapped balances have total coordinate variances 0.1500 and 0.0441. More amount tightens uncertainty about the wrapped balance.

Why the rim stays fixed while its balance cloud tightens

\underset{directions’ spread}{\underset{⏟}{σ_{wrap}^{2} = \frac{M^{2}}{A^{2}}}}, \underset{wrapped balances’ spread}{\underset{⏟}{E | B - z |^{2} = \frac{σ_{wrap}^{2}}{A + 1}}}

A includes the prior and is the concentration of a Dirichlet process, a probability model for random weights around the circle. B is the mean vector of one allocation; z is its expected value, the rose point. Each coordinate variance is half the displayed total in this family. The density's peak may lie away from the rose point. These values describe uncertainty in B; uncertainty about a decoded line peak requires a further nonlinear calculation.

01.897.35

The interiors share one color scale for probability density per unit area, using log(1 + density). Both rims share a separate scale for directional density. At A = 1 with a uniform horizon, the possible wrapped balance is uniform throughout the disk.

The exact density drawn here

f_{A, z} (v) = \frac{A (1 - | z |^{2})}{π} \frac{(1 - | v |^{2})^{A - 1}}{(1 + | z |^{2} - 2 z \cdot v)^{A + 1}}, | v | < 1

Each pixel evaluates this continuous-circle formula. Changing A preserves the chosen Poisson base with mean z; adding particular observed trials can change that base too.

The probability model and its exact density

For $K$ equal angular bins, the unit uniform prior gives each bin $1 / K$ . Counts $n_{k}$ give Dirichlet parameters $α_{k} = 1 / K + n_{k}$ , preserving total prior amount one under refinement. The DP extends these random allocations to the continuous circle. Ferguson’s original construction.

The demo uses the Poisson kernel, or circular Cauchy, as its base density:

h_{z} (ϕ) = \frac{1 - | z |^{2}}{2 π | n (ϕ) - z |^{2}} .

Its mean is $z$ . With $τ = A$ , the exact area density of a possible balance $b$ is

f_{τ, z} (b) = \frac{τ (1 - | z |^{2})}{π} \frac{(1 - | b |^{2})^{τ - 1}}{(1 + | z |^{2} - 2 z \cdot b)^{τ + 1}}, | b | < 1.

Its covariance matrix is $(1 - | z |^{2}) I_{2} / [2 (τ + 1)]$ , where $I_{2}$ is the two-by-two identity matrix: the coordinate variances are equal and the cross-covariance is zero. The density peak can lie away from the mean.

For the uniform horizon, $z = 0$ , this reduces to

f_{τ} (x, y) = \frac{τ}{π} (1 - x^{2} - y^{2})^{τ - 1}, | B |^{2} \sim Beta (1, τ) .

The angle is uniform and independent of the radius, with $Pr (| B | \leq r) = 1 - (1 - r^{2})^{τ}$ . At $τ = 1$ , the balance is uniform throughout the disk.

For a general DP base $H$ , the covariance uses its full second directional moment:

Cov (B) = \frac{\int n n^{⊤} d H - z z^{⊤}}{A + 1},

where $⊤$ means transpose. Its coordinate variances still sum to $(1 - | z |^{2}) / (A + 1)$ , but the full distribution depends on $H$ .

Why does this density describe the DP mean?

The Markov-Krein transform records all the moments of a DP mean. James develops the identity and its Gamma-process interpretation. We can verify the density by recovering its fingerprint exactly. Throughout this check, $τ > 0$ , $| z | < 1$ , and $| t | < 1$ .

The general DP identity is

E (1 - t \cdot B)^{- τ} = \exp [- τ \int \log (1 - t \cdot n) d H (n)] .

It follows by representing Dirichlet weights as normalized independent Gamma amounts. Their Laplace transforms multiply; taking logarithms turns that product into the integral. For the circular-Cauchy base, the Poisson kernel’s harmonic averaging property evaluates the logarithmic integral and gives

E (1 - t \cdot B)^{- τ} = {[\frac{1 + | z |^{2} + (1 - | z |^{2}) \sqrt{1 - | t |^{2}}}{2} - z \cdot t]}^{- τ} .

Here “harmonic averaging” means that the kernel reconstructs an interior harmonic function from its values on the rim.

Now check the proposed disk density. A useful tool is the Beta integral, defined by

B (p, q) = \int_{0}^{1} s^{p - 1} (1 - s)^{q - 1} d s .

For positive numbers $U, W$ , it combines two denominators:

\frac{1}{U^{τ + 1} W^{τ}} = \frac{1}{B (τ + 1, τ)} \int_{0}^{1} \frac{s^{τ} (1 - s)^{τ - 1}}{[s U + (1 - s) W]^{2 τ + 1}} d s .

Use $U = 1 + | z |^{2} - 2 z \cdot b$ and $W = 1 - t \cdot b$ . The disk integral then has a standard form: for a scalar $q_{0}$ and a vector $q$ with $q_{0} > | q |$ ,

\int_{| b | < 1} \frac{(1 - | b |^{2})^{τ - 1}}{(q_{0} - q \cdot b)^{2 τ + 1}} d^{2} b = \frac{π}{τ} (q_{0}^{2} - | q |^{2})^{- τ - 1 / 2} .

The notation $d^{2} b$ means ordinary area. To verify this identity, rotate $q$ onto the horizontal axis, expand the denominator, and integrate the even powers using the Beta integral; their sum is the binomial series on the right.

Define three positive quantities

P = (1 - | z |^{2})^{2}, Q = 1 - | t |^{2}, R = 1 + | z |^{2} - 2 z \cdot t .

Substituting $u = s / (1 - s)$ leaves the one-dimensional integral

\int_{0}^{\infty} \frac{u^{τ}}{(P u^{2} + 2 R u + Q)^{τ + 1 / 2}} d u = \frac{B (τ + 1, τ)}{\sqrt{P}} {(\frac{R + \sqrt{P Q}}{2})}^{- τ} .

This is another Beta integral. One route is to set $u = \sqrt{Q / P} e^{s}$ , pair the terms at $s$ and $- s$ , and then use $y = \sinh (s / 2)$ . Here $\sinh r = (e^{r} - e^{- r}) / 2$ is the hyperbolic sine.

Multiplying back the density’s prefactor cancels the Beta factor and $\sqrt{P} = 1 - | z |^{2}$ . What remains is exactly the displayed DP transform. At $t = 0$ , the answer is one, confirming normalization. Its other coefficients give every moment. A probability law on the bounded disk is determined by these moments, completing the check.

What changes beyond the Cauchy family?

For observed trial weights, the same near, far, and vertical sums produce an empirical ledger. A general dataset may have several balance centers. The unique peak and exact scale identities above are population results for the Cauchy family.

A full ledger can retain multiple peaks and observed atoms. Replacing its mean measure by a Poisson kernel keeps the mean vector and total wrapped variance while selecting a particular shape. East-west and north-south ledgers can share those summaries and still retain different information. The compass chapter explores this distinction.

Transporting the entire random measure, keeping every direction’s weight, preserves trial amounts and lineage. An atom at the cut can retain the point at infinity, or we can choose another cut. Adding ledgers pools evidence; it does not generally multiply their decoded Cauchy curves.