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Path-dependent conformal pooling

Occupation-time scores remember the start algebraically, overlapping calibration windows remain strongly correlated, and a dependent empirical threshold need not retain its iid rank coverage.

A terminal score forgets its starting regime like $e^{-2\Delta/\varepsilon}$. A score accumulated along the forecast path behaves differently: every early observation remains in the average. In the same two-state chain, its conditional coverage error is generically of order $\varepsilon/\Delta$.

Score and information

Let $Y_t\in\{-1,+1\}$ flip at rate $1/\varepsilon$ in each direction. At forecast origin the information $\mathcal G_0$ determines a posterior $p=\Pr(Y_0=+1\mid\mathcal G_0)$. Define the path average and a noisy upper nonconformity score by

where $Z$ is independent of the chain and $\sigma>0$. The predictor and score map are frozen. This is a population calculation: estimating a threshold from dependent calibration data is a separate error source.

Only the ratio $c=\Delta/\varepsilon$ matters. Write $F_+(q;c)$ and $F_-(q;c)$ for the score CDFs conditional on $Y_0=+1$ and $Y_0=-1$, and $\bar F=(F_++F_-)/2$. The stationary pooled threshold $q_c$ satisfies

It therefore has exact marginal population coverage under a stationary start. Its conditional coverages are $F_+(q_c;c)$ and $F_-(q_c;c)$, which need not equal $1-\alpha$.

An exact occupation-time law

Condition on $N$, the number of flips in the interval. Then $N\sim\operatorname{Poisson}(c)$. If the chain starts at $+1$, the fraction $U$ of the interval spent in state $+1$ has the following law:

• $N=0$: $U=1$;
• $N=2k+1$: $U\sim\operatorname{Beta}(k+1,k+1)$;
• $N=2k\ge2$: $U\sim\operatorname{Beta}(k+1,k)$.

For a negative start replace $U$ by $1-U$. Since $A_\Delta=2U-1$, this gives an exact, rapidly convergent CDF series. With $w_n=e^{-c}c^n/n!$ and $H_q(u)=\Phi((q-\mu(2u-1))/\sigma)$,

and $F_-$ uses $H_q(0)$ in the atom and $\operatorname{Beta}(k,k+1)$ in the even sum. This is not a normal approximation. The certificate evaluates the series by Gauss–Jacobi quadrature and verifies its characteristic function against the independent two-by-two Feynman–Kac matrix exponential.

The crossover theorem

The conditional mean already exposes the changed rate:

For $c$ fixed this memory is nonzero as $\varepsilon\to0$ with $\Delta=c\varepsilon$. For long horizons it falls as $1/(2c)$, not exponentially. The exact characteristic functions are

Expanding this expression under the Gaussian smoothing factor gives the conditional-coverage and quantile asymptotics. If $z_{1-\alpha}=\Phi^{-1}(1-\alpha)$, then

The errors cancel in the stationary average; that is marginal coverage, not conditional coverage. By contrast, the terminal-score error in the companion calculation is exactly a regime-specific constant times $e^{-2c}$.

Unequal switching rates

The equal-and-opposite errors above are a consequence of symmetry, not a general property of path scores. Let the dimensionless generator and its stationary law be

let the physical generator be $mQ_0$, and set $c=m\Delta$. For $y_+=1$, $y_-=-1$, and $\bar y=(b-a)/r$, the initial-memory mean is exact:

For the proof, expand the simple slow eigenvalue and its left and right eigenvectors in the tilted matrix $cQ_0+ik\mu\operatorname{diag}(1,-1)$. Subtracting the stationary start removes the common slow term, while the other eigenmode is $O(e^{-rc})$. Uniformly for bounded $k$,

A bounded-frequency expansion followed by a Gaussian-tail split makes the remainder integrable, so Fourier inversion is termwise valid; the leading term is exactly a translation of the limiting $N(\mu\bar y,\sigma^2)$ law. Evaluating at the stationary quantile (whose displacement from its limit is $O(c^{-1})$) gives, for fixed $a,b,\mu,\sigma>0$,

Equivalently, the leading coefficients for starts $+$ and $-$ are $-2\mu a\phi(z_{1-\alpha})/(\sigma r^2)$ and $+2\mu b\phi(z_{1-\alpha})/(\sigma r^2)$. Their magnitudes have ratio $a/b$; they are equal and opposite only when $a=b$. What remains exact is the stationary-weighted cancellation

Thus an exact marginal guarantee can conceal conditional errors of different sizes. A threshold adapted to an initial distribution $\nu$ instead obeys $q_\nu(c)-q_c=\mu(\nu y-\bar y)/(rc)+O(c^{-2})$. This remains a population, model-conditional result rather than feature-conditional conformal validity.

The certificate evaluates the exact characteristic functions

by matrix exponentiation and independent Fourier inversion. This follows the Feynman–Kac treatment of telegraph occupation functionals in Bogachev and Ratanov (2011).

Here $a=1.7$, $b=0.4$, $\mu=1$, $\sigma=0.5$, and the target coverage is $0.9$. The predicted limits are $-0.27060967$ and $0.06367286$; both residuals converge with measured order $1.9665$, consistent with the stated $O(c^{-2})$ remainder.

Any finite irreducible chain: two Poisson-equation coefficients

The two-state coefficient is not special. Let $Q$ be the row generator of a fixed finite irreducible continuous-time chain, let $\pi Q=0$, and let $f$ be a real state observable. Write $\bar f=\pi f$, $\widetilde f=f-\bar f\mathbf1$, and define the group inverse $Q^\#$ by

Put $h=Q^\#\widetilde f$, define the Green–Kubo variance rate and a second Poisson solution by

and write $a=\mu/\sigma$ and $z=z_{1-\alpha}$. For $A_c=c^{-1}\int_0^c f(Y_s)\,ds$, $S_c=\mu A_c+\sigma Z$, and the stationary pooled threshold $q_c$ satisfying $\sum_i\pi_i\Pr_i(S_c\le q_c)=1-\alpha$, the fixed-start gap obeys

The pooled threshold itself has the expansion

This is a fixed-$Q$, long-horizon statement: the state space, generator, observable, $\mu$, $\sigma>0$, and interior target level are held fixed as $c\to\infty$. It does not claim a remainder uniform over chains whose spectral gap degenerates.

The semigroup identity first gives the exact conditional mean correction

for any rate $\gamma$ below the spectral decay rate. For $F=\operatorname{diag}(\widetilde f)$, let $\lambda(t)$ and $P(t)$ denote the analytic dominant eigenvalue and spectral projector of $Q+tF$ near zero. Differentiating the eigenvalue and projector equations and applying $Q^\#$ gives

where the scalar $s(t)$ is common to every initial state. Substituting $t=ik\mu/c$ in the finite-dimensional Feynman–Kac transform, expanding under the integrable Gaussian factor, and solving the stationary quantile equation gives the two displayed formulas.

At second order the common projector terms and the third Perron cumulant cancel when the stationary CDF is subtracted; the remaining variance correction is exactly $-a^3\mathcal V h_i\phi(z)/2$. This uses the classical Feynman–Kac framework of Kac (1949) and the finite-chain group-inverse formalism of Meyer (1975).

Because $\pi Q^\#=0$, both displayed coefficient vectors have stationary weighted sum zero. More strongly, the weighted coverage gaps cancel exactly at every $c$ by the definition of $q_c$.

This remains a model-conditional statement indexed by the initial latent state, not distribution-free conditional conformal coverage. For the two-state generator above, $Q^\#\widetilde f=-\widetilde f/r$, so the formula reduces to the unequal-rate coefficients already displayed.

The certificate uses the nonreversible generator $Q=\left(\begin{smallmatrix}-1.1&.8&.3\\.4&-1.4&1\\.7&.5&-1.2\end{smallmatrix}\right)$, $f=(-1,.35,1.4)$, $\mu=1$, and $\sigma=.5$. Exact matrix Feynman–Kac inversion predicts first coefficients $(0.28660839,-0.08388809,-0.20088698)$ and second coefficients $(-0.85550337,0.45880750,0.41001447)$.

At $c=1024$, scaling the residual after the first term by $c^2$ gives $(-0.85175805,0.45680087,0.40821765)$; after subtracting both terms, the last measured convergence orders are $(2.9928,2.9921,2.9935)$. The stationary law $(0.33618234,0.31623932,0.34757835)$ annihilates both predicted coefficient vectors and the exact pooled coverage gap.

Coarsened initial information: cancellation is a hyperplane, not stationarity

The statewise expansion also resolves what happens when the initial state is not observed exactly. Let $\mathcal G_0$ be pre-origin information such that, conditional on $Y_0$, the future chain and smoothing noise are independent of $\mathcal G_0$. Write its posterior over the latent state as the row vector $\nu=(\nu_1,\ldots,\nu_n)$.

Conditional on a realization of $\mathcal G_0$ for which $\nu$ is fixed, the score CDF is the mixture of the fixed-state CDFs. If $b$ and $d$ denote respectively the $c^{-1}$ and $c^{-2}$ coefficient vectors in the preceding display, then

There is an exact dimension count behind this cancellation. Suppose the statewise expansion is known through order $k$, with coefficient columns $c_1,\ldots,c_k$ satisfying $\pi c_j=0$, and put

Moreover, the strictly positive stationary law $\pi$ is in the relative interior of this slice. Indeed, its translation space is exactly $\{v:vA_k=0\}$; sufficiently small perturbations $\pi+v$ remain in the simplex. Thus, if $b\ne0$, the first-order cancellation slice has dimension $n-2$: it is only the stationary point for two states, but contains full-support nonstationary posteriors for every $n\ge3$.

If $\mathbf1,b,d$ are linearly independent, simultaneous first- and second-order cancellation has dimension $n-3$. It is generically unique in three states but again contains nonstationary posteriors once $n\ge4$.

For $\nu=\pi$, the definition of $q_c$ supplies exact pooled cancellation at every $c$. Membership in one of these affine slices only removes the displayed asymptotic coefficients; it does not imply exact coverage at finite $c$.

The proof is just conditional total probability applied to the uniform finite-state expansion, but the distinction matters statistically. This is a model-based statement for a specified latent posterior and score, not distribution-free coverage conditional on observed features. Exact distribution-free conditional guarantees are generally impossible without restrictions; see Barber, Candès, Ramdas, and Tibshirani (2021).

For the same nonreversible three-state certificate, mixing only states 1 and 3 with

gives $\nu b=0$ but $\nu d=-0.11147983$. Exact Feynman–Kac inversion therefore gives:

After subtracting the predicted $-0.11147983/c^2$ term, the last measured convergence order is $2.9913$. The example is therefore a numerical certificate of second-order cancellation, not evidence of exact or feature-conditional validity.

A four-state example shows that even the first two coefficients can vanish away from stationarity. For

the vectors $\mathbf1,b,d$ have rank three. The stationary law and a full-support posterior on the resulting one-dimensional double-cancellation slice are

The measured mixed-gap order is $2.990920$. This verifies the predicted $O(c^{-3})$ consequence without asserting that the displayed scaled values converge to a universal third coefficient.

Smooth nonlinear functions of several path averages

Linearity of the score is not needed for the leading crossover. Let $F$ be an $n\times d$ matrix of state observables, put $\bar f=\pi F$, $\widetilde F=F-\mathbf1\bar f$, and define

Let $\psi:\mathbb R^d\to\mathbb R$ be $C^4$ on a neighborhood of the convex hull of the state values and consider the nonlinear path score

with independent Gaussian smoothing. The stationary Green–Kubo covariance rate is

If $q_c$ is the stationary pooled $(1-\alpha)$ quantile and $z=z_{1-\alpha}$, then

The first formula separates curvature-induced mean shift from the variance inflation along the gradient. The second says that only the directional derivative of the nonlinear score along the initial-state Poisson corrector enters at first order. In particular, a nonlinear score can have a smaller, zero, or reversed leading conditional error even though the underlying path average has the same Green–Kubo covariance.

For the proof, the exact semigroup identity gives $\mathbb E_iA_c-\bar f=-H_{i\cdot}/c+O(e^{-\gamma c}/c)$. Under a stationary start, the covariance is $\Gamma/c+O(c^{-2})$. Finite-state exponential mixing gives stationary centered moments of orders three and four of size $O(c^{-2})$, while the difference between fixed- and stationary-start second moments is $O(c^{-2})$.

A fourth-order Taylor expansion of the Gaussian-smoothed CDF therefore gives both displays; the stationary quantile displacement affects the fixed-start gap only at order $c^{-2}$. Since $\pi H=0$, the leading conditional coefficients cancel under stationary weighting. This is still latent-start model-conditional coverage, not distribution-free feature-conditional validity.

The exact two-state certificate uses $\psi(a)=0.4a+0.6a^2$. Here $\bar f=0$, $\Gamma=1$, and $H_+=-1/2$, so the predicted positive-start limit is $-0.4\mu\phi(z)/(2\sigma)=-0.07019933$. The exact Poisson–Beta occupation mixture gives

The post-leading residual has measured order $1.973546$. At $c=128$, the scaled pooled-threshold shift is $0.81052907$, versus the predicted curvature-plus-variance coefficient $0.80504825$. Thus the certificate is nonlinear in the occupation average and independently retains the exact stationary pooling identity.

Critical score maps: the missing second order

If $\nabla\psi(\bar f)=0$, the preceding $c^{-1}$ conditional coefficient vanishes, but this does not in general make the fixed-start error exponentially small. For indices $a,b$, define the state-vector and its Poisson solution

Here $B_i$ is the second-moment boundary covariance. With the same fixed finite chain, Gaussian smoothing, and $C^4$ score map as above, the critical-point expansion is

The two terms have different meanings. The Hessian contracts with the fixed-start boundary correction to the covariance; the third derivative contracts the initial mean boundary layer with the stationary Green–Kubo covariance.

Both coefficient vectors have stationary weighted sum zero. Thus integrating over the stationary start still cancels a conditional discrepancy that is generally nonzero state by state.

A direct moment proof keeps the approximation scope visible. The backward equations for the first three centered integral moments give

Taylor-expanding $\psi$ and then the Gaussian-smoothed CDF gives the claimed coefficient; the common stationary quantile terms cancel. This is the finite-state spectral/moment mechanism behind Edgeworth expansions for additive Markov functionals; compare Hervé and Pène (2010).

The result holds for fixed $Q,F,\mu,\sigma$ and an interior target as $c\to\infty$. No uniform remainder is claimed when the spectral gap degenerates.

An exact unequal-rate two-state certificate makes the order visible. Take jump rates $\alpha=1.7$ from $+$ to $-$ and $\beta=0.4$ in reverse, $F=(1,-1)$, $\mu=1$, $\sigma=.5$, and $\psi(a)=(a-\bar f)^2$.

Then $\bar f=-0.61904762$, $\psi'(\bar f)=0$, and the Poisson formula predicts $c^2$-scaled gaps $(-0.31908624,0.07507911)$. The exact Bessel occupation-time law gives

After subtracting the predicted $c^{-2}$ terms, the last measured convergence orders are $(3.001863,3.001862)$. The stationary weights $(0.19047619,0.80952381)$ annihilate both the predicted coefficient vector and every exact pooled gap.

Also $c\{q_c-0.5z_{.9}\}=0.58859718$ at $c=512$, versus the predicted curvature shift $\mu\Gamma=0.58740957$. These are latent-start model-conditional errors, not distribution-free conditional or feature-conditional conformal coverage.

Regime-aware and locally weighted thresholds

Given posterior $p$, the model-conditional CDF is $F_p=pF_++(1-p)F_-$. Let $q_p(c)$ solve $F_p(q_p;c)=1-\alpha$. The implicit-function theorem gives the exact sensitivity

At long horizons this reduces to

Thus the fully regime-aware thresholds differ from the pooled threshold by $\pm\mu/(2c)$ to first order. A posterior-weighted threshold interpolates continuously: it attains population coverage conditional on the stated information model, while widening the interval in the high-score regime and narrowing it in the other. This is a model-based conditional statement, not distribution-free conditional conformal validity.

A numerical benchmark

Take $\mu=1$, $\sigma=0.5$, and target coverage $0.9$. The table uses the exact pooled quantile. The last column is the positive-start error of the corresponding terminal score $\mu Y_\Delta+\sigma Z$.

At $c=4$, endpoint memory is essentially gone but the path-dependent coverage error is still 3.17 percentage points. The certificate independently simulates exponential holding times and recovers the exact coverages.

Overlapping calibration windows

There is a second, distinct memory effect when the threshold is estimated. Start the stationary chain at time $-\infty$, write $m=1/\varepsilon$, and form length-$D$ calibration windows every $h$ units,

Since $\operatorname{Cov}(Y_s,Y_t)=e^{-\lambda|t-s|}$ with $\lambda=2m$, direct integration gives the exact covariance $\Gamma(\tau)=\operatorname{Cov}(A_0,A_\tau)$:

The two branches agree at $\tau=D$. In the fast limit, $\Gamma(0)\sim\varepsilon/D$, but the correlation does not vanish for overlapping windows:

Fast switching shrinks every window average, yet two densely sampled averages share essentially the same path segment. For the sample mean of $N$ windows, the exact linear-statistic variance inflation and effective size are

With $D=1$, $h=0.1$, $N=200$, and $m=128$, the certificate gives $\mathcal I_N=9.8734$ and $N_{\rm eff}=20.26$; the triangular fast-limit values are $9.8350$ and $20.34$. Thus tenfold overlapping sampling supplies about one independent path average per window length, not ten.

What this implies for an empirical quantile

Let each calibration score be $S_j=\mu A_j+\sigma Z_j$ with independent Gaussian $Z_j$, and let $q_\mu$ be the stationary population $p$-quantile. For $I_j=\mathbf 1\{S_j\le q_\mu\}$, symmetry gives $q_\mu=\sigma z_p+O(\mu^2)$, and a Taylor expansion gives, for every nonzero lag,

Consequently,

The remainder statement is for fixed $N$. This is a local variance calculation for the empirical CDF, not a distribution-free coverage theorem.

It isolates why naive calibration counts overstate precision. Thinning to $h\ge D$ removes geometric overlap; the remaining dependence then falls as $e^{-2m(h-D)}$, matching the role of thinning in Markov conformal methods.

An exact fixed-$N$ terminal-threshold benchmark

The random order statistic can be analyzed exactly in a terminal-emission benchmark. Sample the stationary telegraph chain at equal spacings and write $Y_j\in\{-1,+1\}$, with adjacent correlation $\rho=e^{-2mh}$.

Conditional on $Y_j$, let the score $S_j$ have a continuous density, independently across times. Use $S_0,\ldots,S_{N-1}$ for calibration, let $Q_k$ be their $k$th order statistic, and test on the immediately following $S_N$.

Let $F$ be the stationary mixture CDF, put $U=F(S)$, and define the posterior contrast and its integral

Because the stationary state is balanced, $R(0)=R(1)=0$. Write $b_{n,j}(u)={n\choose j}u^j(1-u)^{n-j}$, with $b_{n,j}=0$ outside $0\le j\le n$. Then, for fixed $N$ and $1\le k\le N$,

The proof is a finite path-likelihood expansion. For a state path $y_0,\ldots,y_N$,

The derivative at $\rho=0$ is therefore the sum over adjacent pairs. There are $N-1$ calibration–calibration edges and one final calibration–test edge. Conditional on the test quantile $u$, the remaining counts are binomial.

Integrating the two possible pair types gives $A_{CC}$ and $A_{CT}$. Since the path likelihood is a polynomial in $\rho$, the remainder is $O_N(\rho^2)$ without an asymptotic mixing argument.

This is an exact marginal coverage expansion. At $\rho=0$ the scores are iid and the usual rank value $k/(N+1)$ returns.

Dependence breaks that finite-sample identity even though every score has the same marginal distribution. It does not imply or claim distribution-free conditional coverage.

For Gaussian emissions $S_j=Y_j+0.5Z_j$, $N=9$, and $k=9$, the iid coverage is exactly $0.9$. The coefficient is $(N-1)A_{CC}+A_{CT}=-0.0108539431$; after subtracting it, the measured residual rate is $2.01937$. Exact Markov-binomial propagation gives coverage $0.87995908$ at $\rho=0.8$, versus $0.88055333$ in 300,000 direct simulations.

What iid marginal coverage does not say

Even at $\rho=0$, the value $k/(N+1)$ is not the coverage conditional on the realized calibration sample. First let $Q_k$ be the $k$th order statistic of $N$ iid scores with continuous CDF $F$. Once those calibration scores are fixed, an independent test score has conditional coverage

The proof is immediate but important: the probability integral transforms $F(S_i)$ are iid uniform, and $F(Q_k)$ is their $k$th order statistic. Continuity is not essential.

For an arbitrary score law, attach independent $V_i\sim\operatorname{Unif}(0,1)$ and order $(S_i,V_i)$ lexicographically. Writing

the randomized distributional transform $H(S_i,V_i)$ is exactly uniform, including when $S_i$ is discrete. If $\widetilde Q_k$ is the $k$th ordered pair, then conditional on the realized augmented calibration panel,

Thus the following identities hold either for continuous scores or for arbitrary scores with randomized lexicographic ties. This is the distributional transform studied by Rüschendorf (2009):

Thus exchangeability gives exact coverage only after averaging over calibration samples. For $N=k=9$, $C_{9,9}\sim\operatorname{Beta}(9,1)$: its mean is $0.9$, its standard deviation is $0.09045$, and its 5%, 10%, and 50% quantiles are $0.71687$, $0.77426$, and $0.92587$. In particular,

This is an exact training-conditional statement about the random threshold, not a failure of marginal conformal validity and not feature-conditional validity. It also gives the iid reference law before considering overlap or serial dependence. In the atomic case the conditioning includes the realized auxiliary tie variables.

Conditioning only on the original tied scores averages over those variables and need not have the beta law. The certificate evaluates the transform interval by interval for the asymmetric atomic law $(0.17,0.46,0.37)$ and recovers the beta CDF on a grid to roundoff for interior and extreme order statistics. The distinction and high-probability training-conditional goals are studied more generally by Bian and Barber (2023).

The sharp iid calibration design

The same beta law gives an exact finite-sample design rule. Let $p=1-\alpha$ be the desired conditional coverage and $1-\delta$ the desired probability, over calibration samples, of attaining it. Since $C_{N,k}=U_{(k)}$ for iid uniforms,

Therefore the least conservative finite order statistic is

This choice is sharp: any smaller order statistic fails the requested confidence. A finite choice exists if and only if the sample maximum works, namely $p^N\le\delta$. Thus the smallest feasible calibration size is

If this condition fails, only an infinite threshold can give the requested distribution-free guarantee. The price of the finite PAC guarantee is visible in its marginal coverage, $k_{\rm PAC}/(N+1)$, which is generally above $p$.

At $p=0.9$, the smallest calibration sizes are $22$, $29$, and $44$ for confidence levels $0.90$, $0.95$, and $0.99$, respectively; at each minimum the threshold must be the sample maximum. With $N=100$, the corresponding orders are $95$, $96$, and $97$. By contrast, the conventional marginal choice $k=91$ has marginal coverage $91/101=0.90099$ but only

This is the classical one-sided distribution-free tolerance-limit calculation, not a new conformal phenomenon; see Wilks (1941). Its role here is to make the marginal/training-conditional distinction numerically exact before adding dependence.

Two finite endpoints: the exact beta-spacing law

A different exact law applies when both ends of the score interval are calibration order statistics. For iid continuous scores, let $Q_\ell$ and $Q_u$ have ranks $1\le\ell<u\le N$, and condition on the calibration panel. The next-score coverage of $[Q_\ell,Q_u]$ is

This follows from the Dirichlet law of the $N+1$ uniform spacings. In particular,

The law depends only on the rank width $r$, not on the location $\ell$. Location still controls how the excluded probability is split between the two tails and can materially affect interval length. The exact training-conditional confidence is

Consequently the smallest admissible rank width is the same binomial-tail inversion as above with $r$ in place of $k$. There is, however, a stricter feasibility constraint: two observed endpoints permit only $r\le N-1$. The sample range is best, so a finite two-endpoint interval exists if and only if

At $p=0.9$, the smallest panel sizes are $38$, $46$, and $64$ for confidence levels $0.90$, $0.95$, and $0.99$; the only feasible width at each minimum is respectively $37$, $45$, and $63$. With $N=100$, the required widths are $95$, $96$, and $97$. The conventional marginal width $r=91$ still has confidence only

For example, the range of $N=9$ observations has $C_{9,1,9}\sim\operatorname{Beta}(8,2)$, mean $0.8$, and standard deviation $0.12060$. The accompanying certificate checks its lower tail by independent quadrature and simulation and verifies every design number above. Randomized lexicographic ties give the same result for arbitrary score laws when the conditioning includes both endpoints' auxiliary tie variables.

Exact allocation of the two omitted tails

The beta-spacing law deliberately forgets where the omitted mass lies. That information is also available exactly. Put

Then the full spacing law is

Thus endpoint placement does not change the beta law of $C$, but it does change the joint law of the lower and upper errors. For desired caps $a,b\ge0$ with $a+b\le1$, the exact confidence over random calibration panels that both conditional tails meet their caps is

The count interpretation is immediate: at least $\ell$ of the $N$ probability-integral transforms must fall in $[0,a]$, and at least $N-u+1$ must fall in $[1-b,1]$. Consequently the sample range $(\ell,u)=(1,N)$ maximizes this simultaneous confidence among intervals with two observed endpoints. Its confidence reduces by inclusion–exclusion to

Therefore some finite-endpoint interval meets confidence $1-\delta$ if and only if

For symmetric five-percent tail caps, the smallest panel sizes are $58$, $72$, and $104$ at confidence levels $0.90$, $0.95$, and $0.99$; the achieved sample-range confidences are respectively $0.90012479$, $0.95071897$, and $0.99037281$. The certificate evaluates the multinomial sum for $(N,\ell,u,a,b)=(20,3,17,0.20,0.15)$ and agrees independently with a two-dimensional Dirichlet-density integral and simulation.

This is a theorem for a two-order-statistic score interval. It should not be confused with the usual absolute-residual split-conformal interval, which is generated by one upper score threshold and therefore uses the one-sided law.

Nor is this feature-conditional coverage: it is high-probability coverage conditional on the realized calibration panel. The calculation is another classical distribution-free tolerance-interval consequence of the uniform-spacing argument in Wilks (1941).

Dependent transfer of the full tail-allocation law

The scalar covered-mass transfer extends to the complete three-component allocation. Let $X$ be the calibration panel with law $\mu$, let $\nu=F^{\otimes N}$, and let $P_x$ be the conditional test-score law given $X=x$. Define

For the two calibration ranks $\ell<u$, put

If $D\sim\operatorname{Dirichlet}(\ell,u-\ell,N+1-u)$, then, for the $\ell^1$ transport cost on the probability simplex,

Indeed, for a fixed panel the three cells form a measurable partition, so their probability-vector distance is at most twice the total variation between $P_x$ and $F$. Integrating gives the test term.

A maximal coupling of $\mu$ and $\nu$ fails with probability $\eta_{\rm cal}$, and the $\ell^1$ diameter of the simplex is two, giving the calibration term. The factor two is sharp at this partition step.

This vector result yields an explicit dependent two-tail PAC certificate. For any $0<\gamma<\min(a,b)$, write $\Psi_{N,\ell,u}(a,b)$ for the exact iid multinomial confidence above. Then

The proof couples $V_P$ to $D$: if their $\ell^1$ distance is at most $\gamma$ and the reference tail masses lie $\gamma$ inside both caps, both dependent caps hold; Markov's inequality controls the complementary coupling event. The slack is one-dimensional and should be optimized, not fixed by convention.

For the $N=100$ sample range with two five-percent caps and $\eta_{\rm cal}+\eta_{\rm test}=10^{-4}$, the iid confidence is $0.988185503$. The optimized slack is $0.00829018248$, giving certified dependent confidence $0.947808420$. A separate exact contamination benchmark has $\eta_{\rm test}=0.072$ and mean three-cell $\ell^1$ displacement $0.144$, attaining the factor two in the conditional partition inequality.

This remains a PAC statement over the random calibration panel. It controls the joint law of two training-conditional tail errors; it is neither pointwise validity for every panel nor coverage conditional on a test feature or latent state. For separated path blocks, Berbee coupling supplies $\eta_{\rm cal}$ from the calibration gaps and $\eta_{\rm test}$ from the final test gap exactly as in the scalar transfer below.

An exact lifted recursion for overlapping and thinned path scores

The random threshold can also be treated exactly when the score itself is a sliding path functional. Let $Y_j\in\{-1,+1\}$ be the same stationary chain and define consecutive length-$L$ blocks and noisy scores

with independent standard Gaussian $Z_j$. Adjacent scores share $L-1$ latent states.

The correct Markov state is therefore the whole block $W_j$, not the endpoint $Y_j$. On $\mathcal W=\{-1,+1\}^L$, its stationary law and transition kernel are

For a fixed prospective test score $x$, put $p_w(x)=\Phi((x-\mu\bar w)/\sigma)$ and let

Initialize $M_0(0,w;x)=\varpi_\rho(w)$. Conditioning first on the calibration emission and then shifting the block gives the exact recursion

Because the score distribution is continuous, the exact marginal coverage of the $k$th calibration order statistic is then

This identity follows only from the Markov property and conditional independence of the Gaussian noises; it is not a mixing or large-$N$ approximation. Its cost is $O(Nk2^L)$ per quadrature point because every block has only two successors. When $L=1$ it reduces to the terminal-emission dynamic program above.

Overlap alone already breaks iid rank coverage. With $N=k=9$, $L=4$, $\mu=1$, and $\sigma=0.5$, independent latent samples ($\rho=0$) give exact coverage $0.88580278$, not $0.9$; 300,000 direct simulations give $0.88494667$.

At $\rho=0.8$, overlap and latent persistence combine to give $0.86318921$, versus $0.86350000$ in simulation. Thus setting the endpoint mixing coefficient to zero does not remove dependence created by shared path segments.

What thinning restores

Now sample window starts every $d$ latent steps:

The exact recursion above remains valid after replacing $P_\rho$ by $P_\rho^d$. This gives a precise distinction between removing overlap and removing all dependence.

If $d\ge L$, the closest states in consecutive windows are separated by $g=d-L+1$ transitions. For every current block $w$,

Indeed, conditional on the last state of $w$, the first state of the next window has bias $\rho^g$; conditional on that first state, the rest of the window has exactly its stationary Markov law. Total variation is therefore just the Bernoulli first-state distance. Coupling the $N$ successive sampled block transitions to $N$ independent stationary blocks gives the finite-sample marginal bound

In particular, at $\rho=0$, nonoverlapping windows are iid and recover the exact rank value $k/(N+1)$—still a marginal exchangeability statement, not conditional conformal validity. At nonzero $\rho$, $d=L$ removes shared observations but adjacent blocks remain dependent; an additional gap makes the error decay geometrically.

For the same $N=k=9$ and $L=4$ example, exact coverage at $\rho=0$ is $0.90000000$ for every $d\ge4$. At $\rho=0.8$, the exact coverages for $d=4,12,20,28$ are respectively $0.88738519$, $0.89873625$, $0.89980327$, and $0.89996739$.

This separates the overlap correction from the remaining Markov correction. The certificate evaluates $P_\rho^d$ directly and independently simulates the disjoint $\rho=0$ case.

Absolute regularity beyond the binary chain

The same finite-sample conclusion has a distribution-free dependence bound. Let $(X_t)_{t\in\mathbb Z}$ be a stationary process on a standard Borel space, put $B_j=(X_{jd},\ldots,X_{jd+L-1})$ with $d\ge L$, and let $S_j=s(B_j,U_j)$, where the $U_j$ are iid, independent of $X$, and the common score distribution is continuous. To remove normalization ambiguity, define absolute regularity by total variation:

If $Q_k^{(d)}$ is the $k$th order statistic of $S_0,\ldots,S_{N-1}$, then

To prove it, apply Berbee's maximal coupling successively to the $N+1$ blocks. The end of one block and start of the next are $h=d-L+1$ steps apart, so the coupled block vector differs from iid stationary blocks with probability at most $N\beta_X(h)$. Total variation then bounds the rank event, while the iid coupled scores have exact coverage $k/(N+1)$.

This is a marginal guarantee. It neither conditions on the observed calibration sample nor supplies conditional conformal validity.

For a stationary finite Markov chain with transition matrix $P$ and invariant law $\pi$, the coefficient is explicit:

The binary formula above is the special case $\beta_X(h)=|\rho|^h/2$. An independent three-state, nonreversible certificate gives $\beta_X(4)=0.05331706$ and $\beta_X(12)=0.00028149$; the corresponding exact rank coverages are $0.89938377$ and $0.90000727$. It also enumerates five sampled states and verifies their joint-law distance from iid draws is bounded by $4\beta_X(h)$.

Irregularly spaced nonoverlapping windows

The equal-stride theorem has a sharper nonuniform form. Let $B_i=(X_{s_i},\ldots,X_{s_i+L-1})$ for deterministic starts $s_0<\cdots<s_N$ satisfying $s_{i+1}\ge s_i+L$, and define the actual mixing gaps

Let $S_i=s(B_i,U_i)$ have a common continuous marginal and let $Q_k$ be the $k$th order statistic of the first $N$ scores. Successively applying Berbee coupling with the coefficient belonging to each particular gap gives

The proof is the same telescoping maximal coupling as above, except that the probability of a mismatch at step $i$ is at most $\beta_X(h_i)$. A union bound controls the probability that any of the $N+1$ coupled blocks differs from its independent stationary copy. Thus the theorem uses the full deterministic grid; no minimum-gap or average-gap reduction is needed.

For $N=k=9$, four-point windows, $\rho=0.8$, and block-start strides $(8,12,16,20,24,28,32,36,40)$, the mixing gaps are $(5,9,13,17,21,25,29,33,37)$. The exact lifted-block recursion gives coverage $0.89874791$, while the nonuniform coupling budget is $0.27741672$. The cruder worst-gap expression $N\beta_X(5)=1.47456$ is vacuous before capping at one.

Using the average gap is not a valid shortcut. In the binary chain, $h\mapsto\beta_X(h)=0.8^h/2$ is convex.

For the same nine gaps, $N\beta_X(21)=0.04150517$, but exact enumeration of the ten sampled terminal states gives joint total variation $0.16384000$ from iid draws. Hence the average-gap surrogate is smaller than the quantity it purports to bound; the gapwise sum $0.27741672$ remains valid.

Optimal allocation of a fixed spacing budget

The preceding warning concerns evaluating an already unequal grid. It also has a positive design counterpart. Suppose a usable envelope is

and every one of the $N$ gaps must be at least $h_0$. Write $x_i=h_i-h_0\ge0$ and fix the total excess spacing $\sum_i x_i=H$. The uncapped marginal-coupling envelope is

Strict convexity and Jensen's inequality give the exact design solution

attained uniquely by equal gaps in continuous time. For integer gaps, the minimizers differ by at most one.

This minimizes the certified exponential envelope, not necessarily the unknown exact coverage error. It does not license replacing unequal observed gaps by their average: Jensen gives the opposite inequality for such a substitution.

The training-conditional bound has unequal gap roles. For a fixed slack $\gamma$, its dependence penalty is bounded by

More generally, for positive weights $w_i$, the exact convex solution is the logarithmic water fill

For $w_0=\cdots=w_{N-2}=1$ and $w_{N-1}=1/\gamma$, if $H\ge\lambda^{-1}\log(1/\gamma)$ all gaps are active and

Below that threshold, all excess spacing goes to the final test gap. Thus equal spacing is optimal for the marginal rank-event envelope, while the PAC transfer deliberately leaves a larger last gap because test decoupling is multiplied by $1/\gamma$.

For the nine-gap binary example, $h_0=1$, $H=180$, $C=1/2$, and $\lambda=-\log0.8$. Equal gaps reduce the marginal envelope from $0.27741672$ on the displayed irregular grid to $0.04150517$. With $\gamma=0.05$, the weighted optimum uses eight calibration gaps $19.50831835$ and a final test gap $32.93345322$; its dependence penalty is $0.05789768$, versus $0.12912721$ under equal gaps with the same total span.

A sharper training-conditional transfer

A PAC statement follows by separating two kinds of dependence instead of coupling all $N+1$ scores at once. Let $P$ be the joint law of the calibration vector $X=(S_1,\ldots,S_N)$ and test score $S_{N+1}$, let $\mu=P_X$, let $F$ be their common continuous marginal, and put $\nu=F^{\otimes N}$.

This notation treats scalar continuous scores; for arbitrary scores, replace each score by its lexicographic pair, $F$ by the pair's common marginal law, and $F(Q_k)$ by the randomized transform $H(\widetilde Q_k)$ above. Define

The first distance measures non-iid calibration panels. The second keeps the actual panel law fixed and measures only the cost of replacing the test score by an independent draw. For a realized panel $x$, write

Then for every target $0<p<1$ and slack $0<\gamma<1-p$,

Here $I_z(a,b)$ is the beta CDF. To prove the bound, put $B=\{C_P<p,\ C_0\ge p+\gamma\}$ and $A_B=\{X\in B,S_{N+1}\le Q_k(X)\}$. Since $P$ and $\mu\otimes F$ have the same panel marginal,

Also $\mu\{C_0<p+\gamma\}\le\nu\{C_0<p+\gamma\}+\eta_{\rm cal}$, while under $\nu$ the variable $C_0$ has the $\operatorname{Beta}(k,N+1-k)$ law. Combining the two events proves the display.

For the thinned block construction above, with $h=d-L+1$, successive Berbee couplings and the definition of absolute regularity give

The first bound couples the $N$ calibration blocks across their $N-1$ gaps. The second decouples the test block from the whole preceding panel across the final gap. This improves the previous $N\beta+2N\beta/\gamma$ penalty without assuming a Markov score process.

For the irregular starts above the same argument keeps the two roles distinct:

The last gap therefore controls test decoupling; it should not be averaged with the calibration gaps. In the binary example with target $p=0.8$, exact enumeration gives training-conditional failure $0.13934752$.

Optimizing the displayed Berbee bound over $\gamma$ gives $0.44014194$ at $\gamma=0.00887176$. The bound is conservative, but it is finite-sample and makes no effective-sample-size substitution.

This is a genuine high-probability statement about coverage conditional on the realized calibration panel; it is not pointwise conditional validity and not feature-conditional validity. The free slack should be optimized. In particular, simply replacing $N$ by an ``effective sample size'' in the iid beta law has no justification.

The certificate makes the distinction exact using a binary Markov chain and the randomized score $S=(Y+U)/2$, whose stationary marginal is uniform but whose value reveals the state. With $N=k=9$, target $p=0.8$, latent correlation $0.8$, and strides $28,40,48$, the exact training-conditional failure probabilities are $0.13428285$, $0.13422219$, and $0.13421848$; the iid value is $0.8^9=0.134217728$. Optimizing the separated Berbee bound gives $0.22299480$, $0.15511260$, and $0.14256883$.

Using the exact calibration-panel distances gives $0.21737533$, $0.15472630$, and $0.14250402$; the exact test-decoupling distance equals $\beta$ in this chain. The conservatism is visible, but so is convergence to the exact iid training-conditional law.

The whole training-conditional coverage law

The same two TV budgets control more than one lower tail. They control the complete random law of coverage conditional on the calibration panel.

With $C_P,C_0,\mu,\nu,\eta_{\rm cal},\eta_{\rm test}$ as above, let $B\sim\operatorname{Beta}(k,N+1-k)$. Then, for the ordinary distance on $[0,1]$,

The proof is a two-coupling argument. Disintegrating the joint and decoupled laws over their common panel marginal gives

Next take a maximal coupling $X\sim\mu$ and $X^0\sim\nu$. On the event $X=X^0$, the two values of $C_0$ coincide; since $0\le C_0\le1$,

Under $\nu$, $C_0(X^0)$ has the beta order-statistic law. The triangle inequality for $W_1$ proves the result. Equivalently, the duality of Kantorovich and Rubinstein (1958) says that every 1-Lipschitz diagnostic $\varphi:[0,1]\to\mathbb R$ obeys

For irregular nonoverlapping blocks, the explicit version is

Here the first $N-1$ terms couple the calibration panel and the final term decouples the test score. Thus the formula preserves the distinction between those two roles even though their coefficients are both one for this distributional metric.

This also identifies what fast mixing cannot do at fixed calibration size. Put $s=N+1$, $m=k/s$, and $M_{N,k}=E|B-m|$. Taking $\varphi(c)=|c-m|$ gives the direct comparison

The irreducible iid term is available in closed form:

Indeed, $d\{x^k(1-x)^{s-k}\}/dx=s(m-x)x^{k-1}(1-x)^{s-k-1}$; integrating on either side of the mean gives half the absolute deviation. Stirling's formula then gives, uniformly while $m$ stays in a compact subset of $(0,1)$,

This turns the law-level transfer into a direct finite calibration design. For target $p$, take $k_N=\lceil p(N+1)\rceil\le N$, put $m_N=k_N/(N+1)$, and let $\eta=\eta_{\rm cal}+\eta_{\rm test}$. Then

Thus the least $N$ for which the right side is at most a desired tolerance $\tau$ is an exact sufficient design under the stated TV budgets. For fixed interior $p$ and $\tau-\eta\downarrow0$, its leading size is

Therefore vanishing dependence budgets recover the nondegenerate iid beta law, not the point mass at its mean. For $N=k=9$, $m=0.9$ and $M_{9,9}=0.2(0.9)^{10}=0.06973568802$. In the randomized binary certificate, strides $28,40,48$ give exact $W_1$ distances $0.0000214995$, $0.0000014743$, and $0.0000002473$, against separated-TV budgets $0.0030847990$, $0.0002118554$, and $0.0000355421$.

The corresponding values of $E|C_P-0.9|$ are $0.0697455929$, $0.0697363574$, and $0.0697358002$, converging to the irreducible beta fluctuation. At target $p=.9$, the exact rank-rounded designs for tolerances $.05,.03,.02$ need respectively $N=29,69,149$ calibration scores when $\eta=0$; allowing a dependence budget $.005$ raises the $.03$ design to $N=89$. The certificate checks the beta identity against independent quadrature, and at total beta size $s=20480$ its largest relative error against the square-root law over $m\in\{.1,.25,.5,.75,.9\}$ is below $5\times10^{-5}$.

This is a law over random calibration panels. It does not assert that $C_P(x)$ is close to $m$ for every realized panel, and it is not feature-conditional conformal coverage. The notation $M_{N,k}$ is kept separate from the signed final-memory coefficient $D_{N,k}$ below, resolving the collision recorded in issue #92.

Why calibration mixing alone is insufficient

The final test gap is independently necessary, not just an artifact of the bound. Keep the randomized state-revealing score $S=(Y+U)/2$ and a stationary symmetric binary chain with one-step correlation $a\in(0,1)$.

Separate the $N$ calibration observations by a stride $d$ but leave the last calibration-to-test gap at one step. As $d\to\infty$, the calibration-panel law converges in total variation to $N$ iid uniforms, while the test score retains correlation $a$ with the state revealed by the last calibration score.

The limiting law is explicit for every order statistic, not only the maximum. Let $Q_k$ be the $k$th calibration order statistic, let $y\in\{0,1\}$ be the state revealed by the last calibration score, and let $j$ count lower-half scores among the first $N-1$ observations. Put

and write $I_x(r,s)$ for the regularized beta CDF, extended as zero for $x\le0$ and one for $x\ge1$. Conditional on $(j,y)$, either $2Q_k\sim\operatorname{Beta}(k,z+1-k)$ when $z\ge k$, or $2Q_k-1\sim\operatorname{Beta}(k-z,N+1-k)$ when $z<k$. Hence the exact limiting training-conditional failure probability at target $p$ is

At $a=0$ this mixture collapses to the iid law $I_p(k,N+1-k)$. For fixed $a>0$ it generally does not.

The marginal effect has a particularly simple closed form. If $B\sim\operatorname{Binomial}(N+1,1/2)$, then

Indeed, conditionally on the iid threshold $Q_k=q$, the last score has lower-minus-upper-half mean

The test CDF is $C_a=Q_k+a(1-2Y_N)\min(Q_k,1-Q_k)$. Integrating the two pieces against the $\operatorname{Beta}(k,N+1-k)$ density and integrating by parts gives the binomial-tail coefficient above.

In particular $D_{N,N+1-k}=-D_{N,k}$, so the central order statistic has no marginal shift when $N$ is odd, although its training-conditional distribution can still change. Thus convergence of the calibration panel to iid does not by itself restore either the iid training-conditional law or, in general, the iid mean $k/(N+1)$.

The coefficient also has an explicit finite-rate description. Put $n=N+1$ and $I(\theta)=\theta\log(2\theta)+(1-\theta)\log(2(1-\theta))$. Directly separating the two binomial tails in the preceding display and applying the Chernoff bound gives, for $k>n/2$,

while for $k<n/2$,

Consequently, if $k/n\to\tau\ne1/2$, then $ND_{N,k}\to-\tau$ below the median and $ND_{N,k}\to1-\tau$ above it. For the usual upper conformal order with $k/n\to1-\alpha>1/2$, fixed final-memory correlation $a$ therefore raises mean coverage by $a\alpha/N+o(N^{-1})$, with an exponentially small remainder whenever the rank fraction stays a fixed distance from $1/2$. The median transition has a different, Gaussian scale: if $x_n=(2k-n)/\sqrt n\to x$, the binomial central limit theorem yields

This convergence also has a fully explicit finite-$n$ error. Set $x_n=(2k-n)/\sqrt n$ and let $C_{\rm BE}=0.4748$, the iid Berry–Esseen constant available from Shevtsova's bound. Applying that inequality to the standardized binomial CDF at the two adjacent lattice points $x_n-2/\sqrt n$ and $x_n$, and bounding the intervening normal mass by $2/\sqrt{2\pi n}$, gives

In particular, uniformly over $|x_n|\le K$, the right side is at most

The certificate checks the two finite-$N$ Chernoff inequalities for every order at $N=2,\ldots,200$ and checks the Berry–Esseen envelope over the same complete rank grid, as well as the central crossover at $n=32,128,512$. These are marginal-mean statements in the fixed-test-memory limit; they neither remove the changed training-conditional tail nor assert feature-conditional coverage.

For the sample maximum $M=Q_N$, the limiting coverage conditional on the realized iid calibration panel is exactly

The three cases follow by conditioning the next state on the half of the unit interval containing $S_N$. Thus iid convergence of the calibration panel does not force convergence to the iid training-conditional law unless the last-gap correlation also vanishes.

At $N=k=9$, target $p=0.8$, and $a=0.8$, direct integration over the three panel regions gives

whereas the iid beta failure probability is $0.8^9=0.134217728$. In the exact Markov certificate, increasing the eight within-panel strides from $4$ to $64$ drives panel total variation from $0.451236768$ to $6.87\times10^{-7}$, while the failure probability converges from $0.158846744$ to $0.152243886$. The test-decoupling distance remains $a/2=0.4$.

This is a training-conditional failure probability averaged over random calibration panels; it is neither pointwise nor feature-conditional coverage. It proves that $\eta_{\rm cal}$ alone cannot justify the iid beta law, even when it tends to zero.

The all-order certificate makes the distinction quantitative for $N=9$, $p=0.8$, and $a=0.8$:

The finite mixture agrees with exact Markov-path enumeration for all nine order statistics to $4.98\times10^{-14}$ once the within-panel stride is 128. The closed marginal coefficient agrees with the independent component-mean mixture for every $2\le N\le20$ and every $k$ to $2.50\times10^{-16}$, while its reflection identity holds to machine zero. This exact calculation complements the marginal mixing and thinning guarantees of Zheng and Proutiere (2024): it isolates the separate final-gap memory that a calibration-only mixing statement cannot remove.

Unequal stationary masses

The half-interval split is not essential. Let the lower latent state have stationary mass $r\in(0,1)$ and emit a uniform score on $[0,r]$; let the upper state emit uniformly on $[r,1]$.

The stationary score is again exactly uniform. If $L_N\in\{0,1\}$ indicates that the last calibration score is in the lower interval and the retained test-gap eigenvalue is $a\in[0,1)$, then

Let $j$ be the number of lower scores among the first $N-1$ observations and put $z=j+L_N$. Conditional on $(j,L_N)$, the same order-statistic argument gives the exact failure probability

At $a=0$ this again collapses to the iid beta law. The marginal mean has the exact affine form

To prove the coefficient, condition on the total lower count $Z=z$. Exchangeability makes the chance that the last observation is lower equal to $z/N$, while the relevant rescaled beta order statistic has mean $d_k(z)$.

This also gives the reflection identity $D_{N,N+1-k}(1-r)=-D_{N,k}(r)$. The symmetric formula above is the special case $r=1/2$.

The outer large-panel limit changes with the partition imbalance. If $k/(N+1)\to\tau\ne r$, then

Indeed, the binomial covariance identity, with $W\sim\operatorname{Binomial}(N-1,r)$, is

Off the transition $\tau=r$, binomial concentration selects one smooth branch of $d_k$ with exponentially high probability; its first difference gives the displayed limits. This is an unequal-rate final-gap effect, not a change to the iid calibration baseline.

The missing critical window has a Gaussian interpolation. If

then the same covariance identity and the binomial central limit theorem give

There is also a finite-rate form. For fixed $r\in(0,1)$ and $|x_N|\le K$,

Indeed, away from the single crossing increment, $N\{d_k(W+1)-d_k(W)\}$ tends to $1/(1-r)$ below the rank boundary and $-1/r$ above it. The crossing atom has probability $O(N^{-1/2})$, so it vanishes after the displayed scaling. Weighting the two branch limits by $\Pr(W<k)\to\Phi(x)$ gives the formula.

It joins the one-sided limits continuously: $-(1-r)$ as $x\to-\infty$ and $r$ as $x\to+\infty$. At $r=1/2$ it reduces to the symmetric crossover $\Phi(x)-1/2$. For the rate, on the binomial central region the two rational increments differ from their branch constants by $O_{r,K}(N^{-1/2}+|W-(N-1)r|/N)$, whose expectation is $O(N^{-1/2})$.

Chernoff controls the complementary tails, the crossing mass is $O(N^{-1/2})$, and the Berry–Esseen inequality controls the remaining binomial CDF error at the same order.

For $N=9$, $p=0.8$, $a=0.8$, and $r=0.3$, the exact values are:

A separate unequal-transition Markov-path enumeration agrees with the mixture for every order statistic within $1.05\times10^{-13}$ after within-panel stride 128. Across $2\le N\le20$, four unequal masses, and every rank, the component mean agrees with the displayed marginal coefficient within $1.17\times10^{-15}$; the reflection discrepancy is $7.68\times10^{-17}$. At $N=4096$, six critical-window checks with $r\in\{0.3,0.7\}$ and $x\in\{-1,0,1\}$ are within $0.019394$ of the Gaussian limit; over $N=256,1024,4096$, the largest $\sqrt N$-scaled error is $1.241183$.

These failure probabilities average a training-conditional event over random panels, whereas the last column is marginal coverage. Neither is feature-conditional validity.

A coefficient-only slack-free rearrangement bound

The slack can in fact be removed without knowing the actual panel order-statistic law. Let $U_0\sim\operatorname{Beta}(k,N+1-k)$, put $W_0=(U_0-p)_+$, and write $Q_0$ for the lower quantile function of $W_0$. Then the two separated total-variation budgets alone imply

Take a maximal coupling $(X,X_0)$ of the actual panel law $\mu$ and the iid law $\nu$, so $P(X\ne X_0)=\eta_{\rm cal}$. On the failure set $A=\{C_P<p\}$, disintegration gives

The matched part of $A$ has mass at least $(\mu(A)-\eta_{\rm cal})_+$. Among all subprobabilities of the iid panel law with that mass, the lower rearrangement inequality says that the smallest possible integral of $(C_0-p)_+$ is the corresponding lower-quantile integral.

This proves the display. The rearrangement step is the probability-space form of the inequality of Hardy and Littlewood (1930).

If $G(u)=I_u(k,N+1-k)$ and the test budget is below $E[(U_0-p)_+]$, let $q\ge p$ solve

then the bound is $\min\{1,\eta_{\rm cal}+G(q)\}$. It is strictly sharper than optimizing the free-slack theorem at small test budgets. If $g(p)$ is the beta density at the target, the excess above the iid failure probability plus $\eta_{\rm cal}$ is

whereas the optimized slack bound has leading excess $2\sqrt{g(p)\eta_{\rm test}}+O(\eta_{\rm test})$. For $N=k=9$ and $p=.8$, the certificate obtains the constants $1.737788959$ and $2.457605000$, versus the exact limits $1.737785625$ and $2.457600000$.

Using only Berbee's coefficient bounds at strides $28,40,48$, the new slack-free bounds are $0.19910268$, $0.14913574$, and $0.14014678$, improving the optimized slack bounds $0.22299480$, $0.15511260$, and $0.14256883$. Replacing the coefficient budgets by the exact two TV distances improves them further to $0.19348321$, $0.14874945$, and $0.14008197$.

A sharper slack-free bound when the panel law is known

The free slack is unnecessary if the distribution of the random independent-test coverage under the actual panel law can be computed. Let $U=C_0(X)$ under $\mu$, put $W=(U-p)_+$, and write $Q_W(s)=\inf\{w:\Pr(W\le w)\ge s\}$ for its quantile function. Disintegration of total variation, using the common panel marginal of $P$ and $\mu\otimes F$, gives

On the failure set $A=\{C_P<p\}$, one has $W\le(C_0-C_P)_+$. The lower rearrangement inequality therefore yields

This is the strongest bound obtainable from the distribution of $U$ and this $L^1$ deficit budget: panels with $U$ closest to, but above, $p$ are the cheapest ones on which to lose coverage. If $U$ has continuous CDF $G$ and the budget is not already large enough to make the bound one, let $q\ge p$ solve

then the bound is simply $G(q)$. This actual-law refinement improves the coefficient-only rearrangement because it uses the dependent order-statistic distribution rather than merely a TV neighborhood of the iid beta law.

For the binary-chain certificate, that panel law is an exact finite mixture of affine beta order statistics. At strides $28,40,48$, the rearrangement bounds are $0.19149163$, $0.14861292$, and $0.14005907$, improving the corresponding exact-TV slack bounds $0.21737533$, $0.15472630$, and $0.14250402$. They remain above the exact failures $0.13428285$, $0.13422219$, and $0.13421848$.

Atoms and randomized ties

The continuity condition can be removed, but tie handling then matters. For arbitrary stationary scores, including discrete scores, attach iid $V_j\sim\operatorname{Unif}(0,1)$ independent of everything else and order the pairs $\widetilde S_j=(S_j,V_j)$ lexicographically. If $\widetilde Q_k$ is the $k$th calibration pair, the same coupling proof gives

The iid coupled pairs have no ties, so exchangeability makes the test rank exactly uniform. This is the randomized tie-breaking used in exact conformal validity; see Vovk, Gammerman, and Shafer (2005).

More strongly, the randomized-transform argument above gives the exact $\operatorname{Beta}(k,N+1-k)$ law for coverage conditional on the realized augmented iid calibration panel. Consequently the sharp iid PAC design and both separated-TV training-conditional transfers above apply verbatim to arbitrary score distributions after augmenting the panel and test score with their tie variables. This is distinct from deterministic tie retention and from conditioning only on the unaugmented scores.

If ties are instead retained and the event is $S_N\le Q_k$, the dependence bound remains valid but its iid center is generally not $k/(N+1)$. Writing $F_-(s)=\Pr(S<s)$, the exact iid center is

The first formula conditions on the test score: inclusion occurs exactly when fewer than $k$ calibration scores are strictly smaller. The inequality follows by comparing with independent uniform tie-breaking. Thus atoms create deterministic conservatism even under independence; they must not be misread as a dependence effect or as conditional validity.

For the nonreversible three-state chain, collapse the first two states to score zero and assign score one to the third. With $N=k=9$, deterministic iid coverage is $0.98322654$, whereas randomized iid coverage is exactly $0.9$.

At strides $1,4,12$, randomized dependent coverage is respectively $0.88703565$, $0.89984374$, and $0.90000932$; both the deterministic and randomized versions satisfy their corresponding $N\beta_X$ bounds. The certificate evaluates the finite-state recursion and the iid atomic formula independently.

When a fitted predictor may be treated as fixed

There are two different meanings of a ``learned predictor.'' A predictor can be fitted on data separate from the calibration and test path, or it can have a dynamical internal state that continues to evolve along that path. This section addresses the first meaning. The next section addresses the second.

Let $\mathcal T$ be the sigma-field generated by an independent training sample, let $\widehat f$ and the score map $s_{\widehat f}$ be $\mathcal T$-measurable, and apply that map to the deterministic nonoverlapping blocks $B_0,\ldots,B_N$ above. Assume that, conditionally on $\mathcal T$, the common score marginal is continuous; randomized lexicographic ties give the corresponding result without continuity.

Put $\eta_B=\min\{1,\sum_{i=0}^{N-1}\beta_X(h_i)\}$. Then, almost surely in the fitted model,

After conditioning on $\mathcal T$, the fitted score map is fixed and the independent future process keeps its stationary law. Measurable transformations cannot increase absolute regularity, so the earlier gapwise Berbee proof applies verbatim. This is coverage conditional on the independent fitted model but still marginal over the calibration and test path.

It is not coverage conditional on the realized calibration panel, a test feature, or a hidden state. The training-conditional panel bounds above may likewise be applied after conditioning on $\mathcal T$.

Remote training on the same series

Independence can be weakened quantitatively. Suppose $\mathcal T\subseteq\sigma(X_t:t\le0)$ and the first future block starts at time $g\ge1$. Let $C(\mathcal T)$ be the actual probability of the split rank event conditional on the fitted model and its training history.

Let $C_0(\mathcal T)$ use the same fitted score map but an independent stationary copy of the entire future calibration/test panel. With the half-$L^1$ convention for total variation used above, disintegration and absolute regularity give

For every fitted map, the independent future panel satisfies $|C_0(\mathcal T)-k/(N+1)|\le\eta_B$. Consequently,

The second display is a PAC statement over the random training history, not deterministic conditional validity. It also keeps the roles of the gaps distinct: $\beta_X(g)$ separates training from the full future panel once, while $\eta_B$ controls dependence among future blocks.

There is no factor $N$ multiplying the training-separation term. For the stationary binary chain with correlation $0.8$, exact enumeration at $g=7$ gives joint/product distance $0.10485760=0.8^7/2$, even when the retained future panel has several additional unequal strides.

Calibration-label reuse can destroy coverage

No analogous theorem holds for an arbitrary learner fitted on the calibration labels. Let $(X_i,Y_i)$ be iid with independent continuous uniform coordinates and fit the memorizer

For the residual score $S_i=|Y_i-\widehat f(X_i)|$, every calibration score is zero. A fresh test feature differs from all calibration features almost surely, so its score is $Y_{N+1}>0$ almost surely. Hence every calibration order statistic is zero and

If the same memorizer is instead fitted on an independent training sample, calibration and test features are all unseen almost surely, their residuals equal the iid $Y_i$, and coverage returns to $k/(N+1)$. The certificate obtains empirical coverage $0.89982750$ from 400,000 honest panels with $N=k=9$ and exact zero under leakage. Thus fast mixing cannot repair within-panel label leakage.

Full conformal, leave-one-out, and other symmetric refitting schemes are separate procedures; the counterexample concerns using their data in an ordinary split threshold while pretending the fitted map is fixed. The independent-fit formulation is the classical split-conformal construction of Lei et al. (2018).

A predictor with slow internal state

The same warning can be made exact for a predictor with memory. Let its internal state be the exponential filter

This is a piecewise-deterministic Markov process, not an endpoint emission. Put $m=1/\varepsilon$ and $\nu=m/\kappa$. In stationarity,

and $\operatorname{Var}(X_t)=\kappa/(\kappa+2m)$. Thus the predictor state becomes smaller under fast switching, but it does not forget a realized state at the fast rate. Conditional on observing $X_0=x$,

When $2m=\kappa$, the continuous limit is $(1+\kappa t)e^{-\kappa t}$. For $m\gg\kappa$, the endpoint regime forgets like $e^{-2mt}$ while the predictor state forgets like $e^{-\kappa t}$.

Consider the noisy score $S_t=\mu X_t+\sigma Z$ and let $q_\mu$ be its stationary pooled $(1-\alpha)$-quantile. Symmetry gives $q_\mu=\sigma z_{1-\alpha}+O(\mu^2)$. Gaussian smoothing and the exact conditional mean yield

This is a local, model-based conditional expansion. A typical stationary $X_0$ is only $O(\sqrt{\kappa/m})$, so its amplitude also shrinks in the fast limit unless a learned model normalizes or amplifies the feature. The decay rate, however, is the predictor rate $\kappa$, not the hidden-chain rate.

The full conditional characteristic function is also computable. Given $X_0=x$, mix the two starting regimes with probabilities $(1\pm x)/2$, retain the deterministic term $e^{-\kappa t}x$, and solve the two-state Feynman–Kac system with time-dependent potential $ik\mu\kappa e^{-\kappa s}\operatorname{diag}(1,-1)$. The certificate performs this time-ordered solve and Fourier inversion, then checks it against direct piecewise-deterministic simulation.

For $m=1$, $\kappa=0.2$, $\mu=0.1$, $\sigma=0.5$, $x=0.8$, and 90% pooled coverage, at $t=5$ the endpoint memory is $e^{-10}=4.54\times10^{-5}$ but $R(5)=0.40875$. Exact conditional coverage is $0.88813$; the first-order prediction is $0.88852$.

What remains separate

The lifted recursion still controls one explicit discrete sliding-window model. The absolute-regularity theorem controls arbitrary stationary block scores through a generally conservative marginal bound and the total-variation transfer gives a correspondingly conservative training-conditional tail bound; neither gives exact coverage for a general score. Deterministic irregular spacing is covered by the gapwise theorem, and the finite-state certificate gives an exact benchmark on one such grid.

A continuous occupation integral or a nonlinear path score still requires its own augmented state or numerical approximation for exact results. An independently fitted score map is covered conditionally on its training sample, while same-series remote training pays the explicit $\beta_X(g)$ term above. Calibration-label reuse is not covered.

None of these results is a distribution-free pointwise conditional conformal guarantee.

A general dynamically updated predictor need not reduce to one exponential filter. Its full internal state must be included in $\mathcal G_0$ and in the augmented Markov dynamics. The exact linear-filter benchmark shows why terminal mixing alone can underestimate conditional coverage error by orders of magnitude.

The stationary law belongs to the classical filtered-telegraph literature; see Newman (1968) and the transition-density treatment of Pawula (1970). Related Markov conformal results analyze dependent calibration samples, temporal predictors, and blockwise hidden-state constructions: Zheng and Proutiere (2024), Barber and Pananjady (2026), and Nettasinghe et al. (2023). Chernozhukov, Wüthrich, and Zhu (2018) instead build dependence-aware permutation groups from blocks.

The calculations here diagnose what happens to the ordinary sequential order-statistic threshold when the path scores themselves overlap. The sequential coupling step uses the original absolute-regularity construction of Berbee (1979).