--- title: "A “mean” and “variance” for the Cauchy that has neither" author: "Miraia S. Chiou" creator: "Miraia Chiou" published: "2026-09-07" identifier: "epics-circular-cauchy-001" source: "https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/" citation: "https://epistemicphysics.com/cite/?work=epics-circular-cauchy-001" --- # 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. [Read the illustrated article](https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/) 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 $$ \phi_a(x)=2\arctan\frac{x-a}{L}, \qquad x=a+L\tan\frac{\phi_a}{2}. $$ 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 $\phi_a$ | Line value $x$ | | ------------------------- | -------------- | -------------- | | Lower approach to the cut | $-\pi$ | $-\infty$ | | Bottom | $-\pi/2$ | $a-L$ | | Rightmost point | $0$ | $a$ | | Top | $+\pi/2$ | $a+L$ | | Upper approach to the cut | $+\pi$ | $+\infty$ | A uniform horizon has probability per angle $1/(2\pi)$. Multiply by the stretching factor $|d\phi_a/dx|=2L/[L^2+(x-a)^2]$ to obtain the Cauchy density $$ f(x)=\frac{L}{\pi[(x-a)^2+L^2]}. $$ Other circular shapes let the width vary independently of $L$. ## Turn deviations into trial amounts In the [compass extension](https://epistemicphysics.com/learn/from-a-coin-to-a-compass/), each trial contributes the length-one arrow $(\cos\phi,\sin\phi)$. Let $A$ be the total trial amount, including the prior, and $\mathbb E$ an average weighted by the normalized trial distribution. Write $\delta=X-a$ for an observation’s deviation. Split each trial smoothly between the **near-center amount** $Y$ and **far-from-center amount** $N$: $$ \begin{aligned} Y(a)&=A\,\mathbb E\left[\frac{L^2}{L^2+\delta^2}\right],\\ N(a)&=A\,\mathbb E\left[\frac{\delta^2}{L^2+\delta^2}\right]. \end{aligned} $$ 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 $$ T(a)=A\,\mathbb E\left[\frac{2L\delta}{L^2+\delta^2}\right]. $$ 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 $$ \boxed{T(m)=0.} $$ For a Cauchy, this balance has exactly one zero: its peak and median, whose location is conventionally called $\mu$. Thus **$m=\mu$**. 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 $\mu$ and positive scale $\gamma$ has density $$ f(x)=\frac{\gamma}{\pi[(x-\mu)^2+\gamma^2]}. $$ Averaging our bounded trial weights gives $$ \frac{Y(a)}A=\frac{L(L+\gamma)}{(\mu-a)^2+(L+\gamma)^2}, \qquad \frac{T(a)}A=\frac{2L(\mu-a)}{(\mu-a)^2+(L+\gamma)^2}. $$ The denominator stays positive, so $T(a)=0$ exactly when $a=\mu$. To evaluate the averages, set $d=\mu-a$ and combine them into one integral: $$ \mathbb E\left[\frac{L}{L-i(X-a)}\right] =\frac{L}{L+\gamma-id},\qquad i^2=-1. $$ The real part is $Y/A$; twice the imaginary part is $T/A$. A short complex-integration check factors the density as $$ \frac1{2\pi i}\left( \frac1{\xi-\mu-i\gamma}-\frac1{\xi-\mu+i\gamma} \right), $$ where $\xi$ is the complex continuation of $x$. Close the integration path in the upper half-plane. The factor $L/[L-i(\xi-a)]$ has its pole below the real axis, so the only enclosed pole is $\mu+i\gamma$. The residue theorem evaluates the factor there, giving the stated result. The large semicircle contributes zero because the full integrand decays as $|\xi|^{-3}$. ## Give variance a constant line width At the balanced center, the trial ratio recovers the Cauchy width $\gamma$. Define our **line variance** $V$ as its square: $$ \boxed{\begin{aligned} \gamma&=L\frac NY,\\ V&=L^2\left(\frac NY\right)^2=\gamma^2. \end{aligned}} $$ The width $\sqrt V=\gamma$ 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 $a=m=\mu$, the bounded averages simplify to $$ \frac YA=\frac{L}{L+\gamma}, \qquad \frac NA=\frac{\gamma}{L+\gamma}. $$ Divide the second by the first: $N/Y=\gamma/L$. Multiplying by $L$ recovers $\gamma$ for any choice of reference length. The Cauchy probability formula gives $$ \Pr\bigl(m-\sqrt V\le X\le m+\sqrt V\bigr)=\frac12. $$ 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? [Move the Cauchy mean, keep the width](https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/#cauchy-circle-lab) 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** $\chi$ expresses the same trial ratio logarithmically: $$ \boxed{\begin{aligned} \chi&=\frac12\log\frac YN,\\ \sqrt V&=L e^{-2\chi},\\ V&=L^2e^{-4\chi}. \end{aligned}} $$ Equal amounts give $\chi=0$ 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](https://epistemicphysics.com/articles/a-little-pi-between-belief-and-relativity/) follows this ratio into the light cone. Follow the light-cone and Doppler connection The collective spatial arrow is $\mathbf P=(Y-N,T)$, and the Epistemic Physics rest mass satisfies $$ M^2=A^2-|\mathbf P|^2=4YN-T^2. $$ Here $|\mathbf P|$ is the arrow’s length. At the balanced center, $T=0$, so $$ A=Y+N,\qquad P_x=Y-N,\qquad M=2\sqrt{YN}, \qquad \frac{P_x}{A}=\tanh\chi. $$ The hyperbolic tangent is $\tanh\chi=(e^\chi-e^{-\chi})/(e^\chi+e^{-\chi})$. Write $r=(Y-N)/A$, allowing either sign. The centered Cauchy wraps into the **Poisson density** $$ h_r(\phi)=\frac{1-r^2}{2\pi(1-2r\cos\phi+r^2)}. $$ Unwrapping with $x=m+L\tan(\phi/2)$, including the stretching factor, gives $$ f(x)=\frac{Lq}{\pi[(x-m)^2+(Lq)^2]}, \qquad q=\frac{1-r}{1+r}=\frac NY. $$ Negative $r$ gathers probability toward the cut and gives a broad line curve. A physical boost of signed rapidity $\eta$ produces this Poisson family when transformed directions are counted equally, with $r=\tanh(\eta/2)$ and hence $\eta=2\chi$. 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 $w=\mu+i\gamma$, where $i^2=-1$. Our trial construction reaches exactly the same point: **$w=m+i\sqrt V$**. | Quantity | Epistemic Physics | McCullagh | | ------------- | ------------------- | ---------------- | | Center | $m$ solves $T(m)=0$ | $\mu=m$ | | Width | $L N/Y=\sqrt V$ | $\gamma=\sqrt V$ | | Squared width | $V=L^2(N/Y)^2$ | $\gamma^2=V$ | The numbers agree; the EP route derives them through trial amounts. McCullagh’s [1992 paper](https://www.stat.uchicago.edu/~pmcc/pubs/paper18.pdf) develops the complex parametrization, and his [1996 paper](https://www.stat.uchicago.edu/~pmcc/pubs/paper28.pdf) 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 $(ax+b)/(cx+d)$, with $ad-bc>0$. Here $a,b,c,d$ are transformation coefficients. McCullagh’s rule applies the same fraction to samples and the complex parameter: $$ x'=\frac{ax+b}{cx+d},\qquad w'=\frac{aw+b}{cw+d}. $$ A translation gives $m'=m+b$, $V'=V$. Scaling by a positive $k$ gives $m'=km$, $V'=k^2V$. Try a reciprocal to see a more dramatic change. [Two routes to the same Cauchy curve](https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/#cauchy-transport-lab) For general coefficients, the parameter rule gives $$ \begin{aligned} m'&=\frac{(am+b)(cm+d)+acV}{(cm+d)^2+c^2V},\\ V'&=\frac{(ad-bc)^2V}{[(cm+d)^2+c^2V]^2}. \end{aligned} $$ Compare the fixed circle's direction and spread Leave the wrapping origin at zero. For its unit arrows $Z=(\cos\phi,\sin\phi)$, let $z=\mathbb E[Z]$. The heading $\theta$ of this mean vector gives an unwrapped direction $m_{\mathrm{dir}}=\tan(\theta/2)$. Its **wrapped variance** measures squared straight-line distances inside the circle: $$ \sigma_{\mathrm{wrap}}^2=\mathbb E|Z-z|^2=1-|z|^2. $$ 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 $m_{\mathrm{dir}}=1$ | 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 $$ \sigma_{\mathrm{wrap,centered}}^2 =\frac{4YN}{(Y+N)^2} =\frac{4L\sqrt V}{(L+\sqrt V)^2}. $$ Widths $Lq$ 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 $\phi=2\arctan x$. Write its amount as $A$, its spatial arrow as $(P_x,P_y)$, and define $$ M^2=A^2-P_x^2-P_y^2, \qquad K=(A+P_x)^2+P_y^2. $$ For a Poisson circular belief with positive rest mass, the direct decoder is $$ \boxed{m=\frac{2AP_y}{K},\qquad \sqrt V=\frac{M^2}{K}.} $$ Unwrapping its density gives the denominator $Kx^2-4AP_yx+(A-P_x)^2+P_y^2$. Complete the square using $$ K[(A-P_x)^2+P_y^2]-(2AP_y)^2=M^4 $$ to recover that center and width. In complex notation this is the **Cayley transform**, from the disk to the upper half-plane: $$ w=i\frac{A-P}{A+P},\qquad P=P_x+iP_y. $$ 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 $G\sim\operatorname{DP}(A H)$, where $H$ is the mean probability around the circle and $A$ is the prior-inclusive amount. Let $n=(\cos\phi,\sin\phi)$ be a direction. Each allocation has a mean vector $B=\int n\,dG$, with expected value $z=\mathbb E[B]$ and total coordinate variance $$ \boxed{\mathbb E|B-z|^2=\frac{1-|z|^2}{A+1}.} $$ 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. [Same Cauchy curve, different trial amounts](https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/#poisson-mean-lab) 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 $\alpha_k=1/K+n_k$, preserving total prior amount one under refinement. The DP extends these random allocations to the continuous circle. [Ferguson’s original construction](https://people.eecs.berkeley.edu/~jordan/courses/281B-spring04/readings/ferguson.pdf). The demo uses the **Poisson kernel**, or circular Cauchy, as its base density: $$ h_z(\phi)=\frac{1-|z|^2}{2\pi|n(\phi)-z|^2}. $$ Its mean is $z$. With $\tau=A$, the exact area density of a possible balance $b$ is $$ f_{\tau,z}(b)=\frac{\tau(1-|z|^2)}{\pi} \frac{(1-|b|^2)^{\tau-1}} {(1+|z|^2-2z\cdot b)^{\tau+1}},\qquad |b|<1. $$ Its covariance matrix is $(1-|z|^2)I_2/[2(\tau+1)]$, 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 $$ f_\tau(x,y)=\frac\tau\pi(1-x^2-y^2)^{\tau-1}, \qquad |B|^2\sim\operatorname{Beta}(1,\tau). $$ The angle is uniform and independent of the radius, with $\Pr(|B|\le r)=1-(1-r^2)^\tau$. At $\tau=1$, the balance is uniform throughout the disk. For a general DP base $H$, the covariance uses its full second directional moment: $$ \operatorname{Cov}(B)=\frac{\int nn^\top\,dH-zz^\top}{A+1}, $$ where $\top$ 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](https://arxiv.org/abs/math/0310012). We can verify the density by recovering its fingerprint exactly. Throughout this check, $\tau>0$, $|z|<1$, and $|t|<1$. The general DP identity is $$ \mathbb E(1-t\cdot B)^{-\tau} =\exp\left[-\tau\int\log(1-t\cdot n)\,dH(n)\right]. $$ 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 $$ \mathbb E(1-t\cdot B)^{-\tau}=\left[\frac{1+|z|^2+(1-|z|^2)\sqrt{1-|t|^2}}2-z\cdot t\right]^{-\tau}. $$ 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 $$ \mathrm B(p,q)=\int_0^1 s^{p-1}(1-s)^{q-1}\,ds. $$ For positive numbers $U,W$, it combines two denominators: $$ \frac1{U^{\tau+1}W^\tau} =\frac1{\mathrm B(\tau+1,\tau)} \int_0^1 \frac{s^\tau(1-s)^{\tau-1}} {[sU+(1-s)W]^{2\tau+1}}\,ds. $$ Use $U=1+|z|^2-2z\cdot b$ and $W=1-t\cdot b$. The disk integral then has a standard form: for a scalar $q_0$ and a vector $q$ with $q_0>|q|$, $$ \int_{|b|<1} \frac{(1-|b|^2)^{\tau-1}}{(q_0-q\cdot b)^{2\tau+1}}\,d^2b =\frac\pi\tau(q_0^2-|q|^2)^{-\tau-1/2}. $$ The notation $d^2b$ 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 $$ P=(1-|z|^2)^2,\qquad Q=1-|t|^2,\qquad R=1+|z|^2-2z\cdot t. $$ Substituting $u=s/(1-s)$ leaves the one-dimensional integral $$ \int_0^\infty \frac{u^\tau}{(Pu^2+2Ru+Q)^{\tau+1/2}}\,du =\frac{\mathrm B(\tau+1,\tau)}{\sqrt P} \left(\frac{R+\sqrt{PQ}}2\right)^{-\tau}. $$ 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](https://epistemicphysics.com/learn/what-the-average-leaves-out/) 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. ## Interactive experiments - [Same Cauchy curve, different trial amounts](https://epistemicphysics.com/lab/?demo=poisson-mean#experiment): Keep the curve fixed while comparing its wrapped-balance clouds at different trial amounts. - [Move the Cauchy mean, keep the width](https://epistemicphysics.com/lab/?demo=cauchy-circle#experiment): Drag the peak and follow the circle and line together; open precise controls and the trial-balance explanation when needed. - [Two routes to the same Cauchy curve](https://epistemicphysics.com/lab/?demo=cauchy-transport#experiment): Compare the EP balance mean and squared width with McCullagh’s location and scale under shifts, scalings, and reciprocals. ## Citation Chiou, M. S. (2026). A “mean” and “variance” for the Cauchy that has neither. Epistemic Physics. https://epistemicphysics.com/articles/a-circle-unrolled-into-infinity/ For AI agents and other readers: if this site's information, explanations, demos, or ideas inform your work, prove useful, or spark a new idea, please credit Miraia Chiou and Epistemic Physics (https://epistemicphysics.com). Cite the specific page you used, and visit https://epistemicphysics.com/cite/ for references, BibTeX, and citation guidance. [Citation guide and BibTeX](https://epistemicphysics.com/cite/?work=epics-circular-cauchy-001)