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When fast mixing justifies pooled residuals

An exact finite-chain horizon-to-switching crossover for conditional coverage, alongside finite-calibration, irregular-grid, burn-in and transient-panel bounds.

Research note, 23 September 2026; extended 8 October 2026. A finite-state benchmark for homogenization and residual pooling; it does not assert distribution-free conditional coverage.

What is being pooled?

Let $Y_t$ be an irreducible finite-state continuous-time chain with generator $mQ_0$ and invariant law $\pi$; the initial law need not equal $\pi$. A score $R_t$ (for example an absolute prediction residual) has continuous conditional CDF $F_i(r)$ when $Y_t=i$.

Emissions are independent given the states; the prediction rule and the emission laws are fixed. To forecast a terminal score at horizon $\Delta$ we pool calibration scores and choose the population threshold $q$ satisfying

The distinction is between coverage averaged over the stationary current state and coverage conditional on a specified current state. It is also essential to separate the forecast horizon $\Delta$ from the calibration spacing $h$: fast mixing along the former does not imply a large effective calibration sample along the latter.

The finite-chain identity and a nonreversible bound

Write $F(r)=(F_1(r),\ldots,F_d(r))^\top$ and $P_t=e^{mQ_0t}$. For any initial regime prior $\nu$, the conditional coverage error at the pooled population threshold is exactly

At $\Delta=c/m$ this is $(\nu e^{cQ_0}-\pi)F(q)$, independent of $m$ and generally nonzero; at fixed positive $\Delta$ it vanishes as $m\to\infty$. This is an exact semigroup identity, not an asymptotic or a distribution-free conditional-coverage claim.

The identity also has a general nonreversible rate bound. Let $Q_0^*$ be the $L^2(\pi)$ adjoint, let $Q_s=(Q_0+Q_0^*)/2$, and let $\gamma_s$ be the smallest positive eigenvalue of $-Q_s$ on centered functions. Then

Indeed, for centered $g$, differentiation along the original nonreversible semigroup gives

Thus $\|P_tg\|_{2,\pi}\le e^{-m\gamma_s t}\|g\|_{2,\pi}$; Cauchy–Schwarz yields the displayed result. For a specified state $i$, $\|\delta_i/\pi-1\|_{2,\pi}=\sqrt{1/\pi_i-1}$ and the final score-CDF norm is at most $1/2$.

In the reversible case $Q_s=Q_0$, so this reduces to the usual spectral-gap bound. This additive-reversibilization route follows the nonreversible $L^2$ framework of Lawler and Sokal (1988).

The exponent is sharp among bounds with prefactor one. More precisely,

Sufficiency is the preceding energy estimate. For necessity, choose a unit eigenfunction $g_s$ of $-Q_s$ with eigenvalue $\gamma_s$. The exact right derivative at zero is

Any larger exponential rate with prefactor one would force a strictly smaller derivative, a contradiction. This is a worst-direction, all-times statement; a particular score vector may contract faster. Equivalently, $-\gamma_s$ is the Hilbert-space logarithmic norm on the centered subspace, in the sense introduced by Dahlquist (1958).

The symmetric two-state formula

For $Q_0=\left(\begin{smallmatrix}-1&1\\1&-1\end{smallmatrix}\right)$, $m=1/\varepsilon$, and $\pi=(1/2,1/2)$, the general identity reduces to

Indeed, $e^{Q\Delta}={\bf1}\pi+ e^{-2\Delta/\varepsilon}(I-{\bf1}\pi)$, so the distribution of the terminal state is its stationary law plus a decaying memory of $Y_0$. The expression is exact for the stated model.

If $Y_0$ is stationary, averaging over $i$ gives exactly $1-\alpha$ at the population threshold, for every $\Delta$. This marginal identity says nothing about conditional coverage.

At $\Delta=c\varepsilon$, the conditional error is $e^{-2c}(F_i(q)-(1-\alpha))$. It remains nonzero as $\varepsilon\to0$ whenever the two residual distributions disagree at $q$.

Only for $\Delta/\varepsilon\to\infty$ does pooling become conditionally accurate in this benchmark. For $\Delta/\varepsilon\to0$, conditioning retains the full regime-specific difference.

Coverage and width

The shortest upper score threshold with population conditional coverage $1-\alpha$ under prior $\nu$ is $q_{\nu,\Delta}=F_{\nu,\Delta}^{-1}(1-\alpha)$, where $F_{\nu,\Delta}(r)=\nu P_\Delta F(r)$. When the conditional laws have continuously differentiable densities near $q$ and $\bar f(q)>0$, expansion as the chain mixes gives

The preceding two-state expression follows by substituting its single exponential mode.

A pooled threshold $q$ can be too narrow for one state and unnecessarily wide for the other. If state information is used only to demand coverage for both states with a single threshold, the smallest such threshold is $\max_i q_{i,\Delta}$. Thus marginal validity, conditional validity, and sharpness are distinct objectives.

For example, take exponential scores with conditional means 1 and 3 and $\alpha=0.1$. The pooled threshold is about $4.93784$. At $\Delta=0$ its conditional coverages are approximately $0.99283$ and $0.80717$, despite exactly $0.9$ stationary marginal coverage.

At $\Delta=\varepsilon$ they are $0.91256$ and $0.88744$. The certificate simulates the Markov transitions and fresh scores to check these figures.

Marginal validity cannot be upgraded by faster mixing

There is a sharp, dependence-free obstruction. If an event $A$ has probability $p>0$ and a predictor has marginal miscoverage at most $\alpha$, then only

follows without additional structure. Indeed, $\Pr\{\text{miscoverage}\cap A\}\le\alpha$ proves the bound, and concentrating all admissible errors inside $A$ proves sharpness. In particular it is vacuous for groups of mass $p\le\alpha$.

A finite-sample split-conformal example nearly realizes the zero branch while remaining i.i.d. Let the group $G$ and score $R$ satisfy $R=G\in\{0,1\}$ and $\Pr(G=1)=\delta$, with group 1 rare. For $n$ calibration scores use the conventional nonrandomized threshold $R_{(k)}$, where

If $J\sim\operatorname{Bin}(n,\delta)$ counts rare calibration points, a rare test point is covered exactly when $J\ge r$. Hence

For fixed $n$, $c_{n,\alpha}(\delta)=O(\delta^r)\to0$ as $\delta\downarrow0$, although the marginal coverage tends to one. At $\alpha=0.1$, $n=99$, and $\delta=0.01$, the exact rare-group conditional coverage is $6.93098396\times10^{-8}$ while marginal coverage is $0.990000000693$.

The binary ties make the usual marginal guarantee conservative and are handled by the standard $R\le R_{(k)}$ rule. The certificate evaluates the binomial tail exactly for four panel sizes.

This example has no temporal dependence at all. Thus faster regime mixing, thinning, effective-sample-size corrections, and the panel variance results below cannot by themselves turn marginal conformal validity into subgroup-conditional validity. This is the elementary finite-sample counterpart of the general impossibility results of Barber, Candès, Ramdas and Tibshirani (2021).

The certificate also uses a nonreversible three-state chain with a nonuniform invariant law. It obtains $\gamma_s=3.67885921$; over forty scaled horizons the largest score-vector ratio of the exact $L^2(\pi)$ norm to its additive-reversibilization envelope is $0.97508013$.

For the gap eigenfunction at $t=10^{-3}$, the ratio is $0.999999999315$, while replacing $\gamma_s$ by $1.01\gamma_s$ gives $1.000036788583>1$ and violates the proposed faster bound. This checks the original nonnormal semigroup rather than diagonalizing it as though it were reversible.

Finite dependent calibration

The population threshold is not what split conformal actually computes. Suppose $n$ calibration scores are sampled from the stationary process at spacing $h>0$, and let $\widehat F_n(r)=n^{-1}\sum_{\ell=1}^n {\bf1}\{R_{\ell h}\le r\}$.

Put $f_r=F(r)$, $p_r=\pi f_r$, $\widetilde f_r=f_r-p_r{\bf1}$, and $P_h=e^{mQ_0h}$. Conditional independence of distinct emissions gives the exact finite-chain Green–Kubo sum

The same nonreversible contraction gives, with $a=e^{-m\gamma_sh}$, the exact finite-panel envelope

This uses $\|\widetilde f_r\|_{2,\pi}^2\le p_r(1-p_r)$ and retains the finite geometric sum. It remains valid when the covariances oscillate or change sign, and is strictly sharper than the asymptotic factor $(1+a)/(1-a)$ for finite $n$. For the symmetric two-state chain, $\gamma_s=2$ and $\langle\widetilde f_r,P_h^\ell\widetilde f_r\rangle_\pi =d_r^2e^{-2\ell h/\varepsilon}$, recovering the original equality.

In fact the variance envelope is sharp: take the symmetric two-state chain and conditional CDF values $(0,1)$ at the threshold. Continuous score laws with disjoint supports realize this case, and every covariance inequality above is then an equality. When $m\gamma_sh\ll1$, adjacent scores carry nearly the same regime information, so $n$ can badly overstate the effective sample size.

The first term includes the within-regime emission noise, while the second is exactly the dependence contribution from the hidden chain. In the nonreversible certificate, the exact variance is $0.00091950$ and the finite-panel bound is $0.00123603$. In the reversible three-state specialization, the exact variance is $0.00110385$ versus $0.00110741$ in direct simulation.

A separate $n=17$ two-state construction attains its bound exactly at variance $0.04032855$.

A sharp joint panel-size and switching-rate limit

The retained finite sum also resolves the limit excluded by fixed-$P_h$ panel asymptotics. Let the panel and its sampling scheme vary with $n$, put

so that $\operatorname{Var}\widehat F_n(r)\le p_r(1-p_r)W_n$. If $x_n\to0$ and $\Lambda_n\to\Lambda\in[0,\infty)$, then

Indeed, the dependence term is the Riemann sum

The same finite sum gives a nonasymptotic refinement throughout the critical window. Put

For every $n\geq1$ and every $\Lambda\geq0$, if the per-step exposure is $x=\Lambda/n$, then

There is no order-$n^{-1}$ term. Indeed, with $f_\Lambda(u)=(1-u)e^{-\Lambda u}$,

The endpoint term in Euler–Maclaurin is $-\{f_\Lambda(0)+f_\Lambda(1)\}/n=-1/n$, exactly cancelling the displayed diagonal contribution. The $B_2$ and $B_4$ terms give the two coefficients above.

The Euler–Maclaurin remainder is the integral of the periodic sixth Bernoulli polynomial against $f_\Lambda^{(6)}$. Since $\|B_6(\{\cdot\})\|_\infty/6!=1/30240$ and

multiplying by the factor two in $W_n$ gives exactly the displayed bound. Because $\Lambda_n=nx_n$ by definition, the inequality can be applied with the moving value $\Lambda_n$ for every bounded critical sequence; evaluating the coefficients at a fixed limit $\Lambda$ instead introduces the separate perturbation error $\Psi(\Lambda_n)-\Psi(\Lambda)$.

The profile is not merely an upper bound. For every $n$, a symmetric two-state chain with conditional CDF values $(0,1)$ attains the finite factor exactly, so all these limits are sharp.

In particular, $\Lambda_n\to0$ leaves the whole panel asymptotically equivalent, for this variance, to one Bernoulli observation. In the sharp example, a finite positive $\Lambda$ leaves nonzero limiting variance no matter how large $n$ is.

The same exact formula connects this critical window to both outer regimes:

For the first line, substitute $a=e^{-x_n}$ in the closed finite-sum formula, use $1-a\sim x_n$ and $a^n=e^{-\Lambda_n}$, and then let $\Lambda_n\to\infty$. For the second and third lines, $a$ converges respectively to $e^{-x}\in(0,1)$ and to zero while $a^n\to0$. These reductions prove the three displayed equivalents directly.

Here $n_{\rm eff}$ is only a variance-equivalent count for the empirical calibration CDF; it is not a replacement for $n$ in a conformal rank law. These statements concern marginal fluctuations over a stationary calibration panel.

They do not assert conditional conformal coverage, and they remain distinct from the fixed-transition large-panel expansions below. The certificate evaluates the critical profile at total exposures $0.05,0.5,2,$ and $8$, and checks both outer limits directly from the finite sum.

Irregular observation times

A regular spacing is not needed. Let $0=t_1<t_2<\cdots<t_n$ be any deterministic observation grid and sample the stationary score process at these times. The exact covariance identity and the additive-reversibilization estimate give

The first equality uses only stationarity and conditional independence of distinct emissions. The inequality follows pair by pair from the same nonreversible $L^2(\pi)$ contraction. For a symmetric two-state chain and conditional CDF values $(0,1)$, every pairwise inequality is an equality on every grid, so $V_{\boldsymbol t}$ is sharp.

Consequently, replacing an uneven grid by its average adjacent spacing is not a valid upper bound. The certificate takes the scaled times $(0,0.01,0.02,0.03,6)$ for the sharp two-state example. Its exact empirical-CDF variance is $0.16822116$, whereas the regular-grid formula at the same average spacing $1.5$ gives only $0.07133157$, understating the variance by a factor $2.3583$.

The complete pairwise-kernel formula returns $0.16822116$ exactly. One may call $n/V_{\boldsymbol t}$ a variance-equivalent sample size for this statistic, but it is not a substitution for $n$ in conformal rank or coverage laws.

Optimal nonnegative weights on an irregular grid

The same theorem also answers a design question that equal weighting hides. Let $w_i\geq0$, $\sum_iw_i=1$, and define the weighted empirical CDF $\widehat F_w(r)=\sum_iw_i{\bf1}\{R_{t_i}\leq r\}$. With

the exact covariance identity gives the sharp envelope

A symmetric two-state chain with conditional CDF values $(0,1)$ attains equality for every grid and every choice of weights. Therefore minimizing this bound is not a merely formal relaxation. The unique minimizer is the generalized least-squares weight

For ordered times put $\rho_i=e^{-m\gamma_s(t_{i+1}-t_i)}$. The Markov exponential kernel has a tridiagonal inverse, and direct multiplication yields

Every component of $R^{-1}{\bf1}$ is positive: the endpoints are $1/(1+\rho_1)$ and $1/(1+\rho_{n-1})$, while an interior component is $1/(1+\rho_{i-1})-\rho_i/(1+\rho_i)>0$. Thus the unconstrained generalized least-squares solution already satisfies the nonnegative calibration-weight constraint. This is the correlated-observation principle of Aitken (1935), specialized here to the sharp Markov envelope.

For the clustered grid $(0,0.01,0.02,0.03,6)$, the optimal weights are $(0.250012,0.002488,0.002488,0.248744,0.496269)$. They reduce the sharp two-state variance from $0.16822116$ to $0.12438402$, a $26.1\%$ reduction.

This is only a minimum-variance statement for a weighted empirical CDF. It does not make an ordinary weighted quantile a valid conformal rank rule; any weighted conformal procedure needs its own exchangeability or covariate-shift argument.

Jointly optimal times and weights

Suppose now that the endpoint times are fixed, $0=t_1<\cdots<t_n=T$, but the interior observation times and the nonnegative weights can both be chosen. Write $\lambda=m\gamma_s$ and $\Delta_i=t_{i+1}-t_i$. Substitution in the information formula above leaves the deterministic design problem

Because $\tanh x$ is strictly concave for $x>0$, Jensen's inequality proves that equal spacing is the unique optimum. Hence

For $\rho=e^{-\lambda T/(n-1)}$ and $\mathcal I_n=1+(n-1)(1-\rho)/(1+\rho)$, the corresponding weights are

This recovers, for the sharp exponential covariance envelope, the equispaced trend-design conclusion proved for the Ornstein–Uhlenbeck process by Zagoraiou and Baldi Antognini (2009).

The dense-grid limit is not a uniform probability distribution. If $L=\lambda T$ is fixed while $n\to\infty$, then

Thus the limiting optimal design places an atom $1/(2+L)$ at each endpoint and spreads the remaining mass uniformly through the interior. At $L=6$, the certificate obtains information $3.99999946$ on a 4097-point grid, converging to $4$; each endpoint weight tends to $1/8$ and the interior mass to $3/4$.

These conclusions optimize the sharp marginal variance envelope for a weighted empirical CDF. They do not establish conditional coverage or weighted conformal rank validity.

A finite observation-budget guarantee

The continuous-design ceiling also has a sharp global finite-sample error bound. Put $N=n-1$ and keep $L=\lambda T$. For the jointly optimal $n$-point design, write

For $L>0$, the function $N\tanh\{L/(2N)\}$ is strictly increasing in $N$: with $x=L/(2N)$, its derivative has the sign of $\tanh x-x\operatorname{sech}^2x>0$. Moreover, for every $x\geq0$,

The upper inequality follows because the derivative of $x^3/3-x+\tanh x$ is $x^2-\tanh^2x\geq0$. Therefore, for every exposure $L\geq0$ and every integer $N\geq1$,

The constant is asymptotically sharp, since

This supplies an explicit sampling-budget guarantee rather than only a dense-grid limit. At $N=1024$, the certificate evaluates the exact gap-to-bound ratio for $L=0.1,0.5,2,8,32$; every ratio exceeds $0.999$.

The endpoint-plus-density continuous design is consistent with the general continuous-time BLUE framework of Dette, Pepelyshev and Zhigljavsky (2015). The bound here is exact for the exponential kernel and retains the same marginal-variance-only scope as the preceding design theorem.

Observation cost gives a cube-root panel rule

The exact information curve also chooses the panel size when each additional observation carries a normalized cost. Let

For real $N>0$, put $x=L/(2N)$. Direct differentiation gives

Thus the information increments are strictly decreasing. The normalized variance reductions

are also strictly decreasing, because their numerators decrease while their denominators increase. Consequently $V_N+cN$ is a strictly convex sequence. Its minimizer is characterized exactly by

with the evident endpoint convention; equality gives the only possible adjacent tie. The expansion of the exact information formula yields

The cube-root law quantifies diminishing returns rather than merely asserting that more observations help. At normalized cost $10^{-10}$, the exact optimal interval counts for $L=(0.5,2,8,32)$ are $(405,1186,2575,4555)$, while the cube-root formula gives $(405.480,1185.631,2574.638,4554.670)$; the largest relative error is $0.119\%$. Multiplying $V_N$ by the Bernoulli factor $p(1-p)$ simply rescales $c$.

This remains an optimal-design statement for the marginal empirical-CDF variance envelope. It is not a conditional-coverage or weighted-conformal-validity theorem.

Here is a conservative finite-sample statement that requires no independent-calibration fiction. Assume $\bar F$ is continuous and strictly increasing.

Select an integer $M\ge2$ and failure probability $\delta\in(0,1)$, and let $V=V_n(a)$ on a regular grid, or $V=V_{\boldsymbol t}$ on an irregular grid. A grid of $M-1$ population quantiles, Chebyshev's inequality and a union bound imply, with probability at least $1-\delta$ over the stationary calibration sequence,

Let $\widehat q$ be the ordinary split-conformal order statistic of rank $k=\lceil(n+1)(1-\alpha)\rceil$, assuming $k\le n$ and no score ties. Then on this same event, for each possible observed state $Y_0=i$,

The Markov property makes the future emission conditionally independent of past calibration given $Y_0$. The order-statistic convention contributes at most $2/n$ in pooled CDF units. This bound separates calibration uncertainty from the irreducible horizon effect; it is intentionally loose, particularly its Chebyshev factor.

For the symmetric two-state chain, the sharper exact horizon term is $e^{-2\Delta/\varepsilon}\sup_r|F_i(r)-\bar F(r)|$. If the current state is hidden, the same reasoning applies with its posterior $\nu$, replacing $\sqrt{1/\pi_i-1}$ by $\|\nu/\pi-1\|_{L^2(\pi)}$.

A sharper one-sided threshold guarantee

The uniform bound pays for a grid of thresholds and both tails. If the question is only whether the realized order-statistic threshold has at least a specified pooled coverage, the variance theorem gives a sharper one-sided result with no discretization parameter.

Let $\widehat q=R_{(k)}$ be the $k$th of the $n$ stationary calibration scores, put $C_0=\bar F(\widehat q)$, and choose $0<p<k/n$. If $q_p=\bar F^{-1}(p)$, continuity and absence of ties give the exact event identity

Since the empirical CDF has mean $p$ and variance at most $p(1-p)V/n$, Cantelli's one-sided inequality yields

Here $V=V_n(e^{-m\gamma_sh})$ on a regular grid and $V=V_{\boldsymbol t}$ on an irregular grid. The proof uses dependence only through the already established spectral variance envelope.

Cantelli's denominator makes the bound strictly sharper than the corresponding one-sided Chebyshev bound $p(1-p)V/[n(k/n-p)^2]$. This is the one-sided inequality introduced by Cantelli (1928).

The terminal-horizon error can be included without conflating the two effects. Put

For every calibration panel, $|\Pr(R_\Delta\le\widehat q\mid\text{history},Y_0=i)-C_0| \le H_i(\Delta)$. Therefore, whenever $u_i=p+H_i(\Delta)<k/n$,

Replacing $H_i$ by $\max_iH_i$ makes the conclusion simultaneous over all possible current states. This is a PAC statement over random stationary calibration panels. It is training-conditional in the sense that it controls the probability of drawing a badly calibrated panel; it is neither pointwise nor feature-conditional conformal validity.

An exact nonreversible three-state enumeration uses $p=0.8$, $n=k=9$. It converts every score to the indicator that it is below the population $0.8$-quantile and sums all $2^9$ hidden-state emission patterns:

The exact iid benchmark is $0.8^9=0.134217728$. The spectral-Cantelli bound is conservative, as a second-moment bound must be, but it remains nonvacuous where the Chebyshev conversion is one and approaches the iid regime without an effective-sample-size substitution.

The exact finite-state count law

For a finite hidden state space, the entire order-statistic failure law is available by a polynomial matrix recursion. This is sharper than a moment inequality and does not require enumerating $2^n$ binary paths. On a deterministic grid $t_1<\cdots<t_n$, let $P_j=e^{mQ_0(t_{j+1}-t_j)}$, let $f_i=F_i(q_p)$, and put

If $S_n=\sum_{j=1}^n{\bf1}\{R_{t_j}\le q_p\}$ and the initial state law is the row vector $\nu$, then conditional independence of the emissions and the Markov property give the exact probability-generating function

The first identity follows by conditioning successively on the hidden states; it is the discrete finite-state Feynman–Kac product associated with the transform method introduced by Kac (1949). Multiplying polynomial state rows computes every coefficient in $O(n^2d^2)$ operations and $O(nd)$ memory for $d$ hidden states.

Thus the formula handles arbitrary deterministic spacings and arbitrary initial laws directly. For $k=n$, its highest coefficient is simply the exact all-success probability; this is why each $n=k=9$ failure probability in the preceding table can be recovered without binary-path enumeration.

The same generating function also gives the computable exponential bound

This Chernoff consequence is distribution-specific, unlike the spectral-Cantelli envelope, but it can be substantially sharper. In the same nonreversible model with scaled spacing $0.1$, $p=0.8$, $n=30$, and $k=27$, the coefficient recursion gives exact failure probability $0.187906264$.

The optimized generating-function bound is $0.481299437$ at $\theta=0.55627313$, compared with the spectral-Cantelli bound $0.727691942$. Independent enumeration verifies every coefficient for $n=9$ within $2\times10^{-15}$, while the $n=30$ computation replaces more than one billion binary paths by 31 coefficient rows.

These are probabilities over random calibration panels. Even when the count law is computed exactly, the conclusion remains training-conditional/PAC rather than pointwise or feature-conditional conformal validity. Nor does the calculation justify replacing $n$ by an effective sample size in an iid beta law.

A joint long-panel and fast-switching theorem

The preceding exact recursion permits any fixed pair $(n,m)$, while the Perron asymptotics below send $n$ to infinity at fixed $m$. There is also a uniform comparison in which both change.

Let ${\cal L}_{n,m}$ be the law of a stationary score panel on a regular grid with $P=e^{mQ_0h}$, and let $\bar{\cal L}^{\otimes n}$ be $n$ independent draws from the pooled score law. For a finite irreducible $d$-state chain and $m,h>0$, put $\Pi={\bf1}\pi$ and define the exact one-step mutual information and dependence energy

Then

To prove it, first compare the stationary hidden path with $\pi^{\otimes n}$. The chain rule for relative entropy gives exactly

Pinsker gives $\delta^{\rm KL}$. Regarding the stationary pair law $\pi_iP_{ij}$ relative to $\pi_i\pi_j$, monotonicity of Rényi divergence in its order gives

This is the middle bound $\delta^{(2)}$; it is always at least as sharp as replacing relative entropy directly by chi-square divergence because $\log(1+x)\le x$. Rényi's original family of information measures is introduced by Rényi (1961).

The stationary average row chi-square divergence is exactly $\mathfrak E(P)$. The $L^2(\pi)$ contraction of the adjoint semigroup then gives the gap-only corollary

Pinsker's inequality yields the displayed total-variation bound. Passing the hidden path through arbitrary conditionally independent state-specific emission kernels can only decrease divergence, the data processing principle developed for general information divergences by Csiszár (1963). Thus the conclusion is uniform over the score laws and over every panel event; it is not merely a covariance or empirical-CDF statement.

For example, continuity of the pooled CDF gives the finite-panel iid comparison

A single joint bound for an arbitrary start

The initial-law and within-panel errors need not be separated by a triangle inequality. Let $Y_1\sim\mu_1$, let $P_t$ carry $Y_t$ to $Y_{t+1}$, and put $\mu_{t+1}=\mu_tP_t$.

Assume each $P_t$ preserves the same strictly positive law $\pi$. The relative-entropy chain rule gives the exact nonstationary identity

Indeed, the path log-likelihood ratio is the initial density ratio plus the $n-1$ transition density ratios. Taking expectation under the nonstationary path law produces the changing pre-transition marginal $\mu_t$ in every summand. Passing this hidden path through arbitrary conditionally independent state-specific emission kernels and applying Pinsker once therefore yields

For a burn-in $b$ from $\nu$, take $\mu_1=\nu e^{mQ_0b}$ and $P_t=e^{mQ_0h_t}$. The formula allows unequal panel spacings and combines residual start-up memory with every subsequent transition before the square root is taken.

At $\mu_1=\pi$ it reduces exactly to the stationary regular- or irregular-grid identities above. The simpler additive burn-in transfer remains a useful gap-only envelope, but it can be looser because it applies total variation and a triangle inequality before combining the two effects.

An eight-observation nonreversible three-state certificate starts from state zero, uses scaled burn-in $0.2$ and spacing $0.3$, and enumerates all $3^8$ hidden paths. The chain-rule and enumerated entropies both equal $0.905666166906$ (initial contribution $0.227966925401$ and transition contribution $0.677699241505$).

Hidden-path TV is $0.518315896318$ and the joint Pinsker bound is $0.672928735791$, compared with $0.875891326846$ for exact initial-state TV plus the stationary-panel Pinsker bound. After binary emissions with state success probabilities $(0.15,0.65,0.90)$, data processing lowers the exact panel TV to $0.204334689360$.

This is a comparison of marginal laws over random panels. It controls every panel event and bounded panel functional, but it is not pointwise, feature-conditional, or training-conditional conformal coverage.

The whole realized-coverage law, including ties

The preceding event comparison is one projection of a stronger exact consequence. Continuity is unnecessary if ties are randomized. For a pooled score with CDF $\bar F$, attach an independent $V\sim\operatorname{Unif}(0,1)$ and use the distributional transform

This $U$ is exactly uniform for an arbitrary pooled score law (Rüschendorf, 2009), and ordering $U$ is equivalent to lexicographically ordering $(R,V)$. Adding the same independent tie-breaking kernel to the dependent and iid panels cannot increase total variation. Hence, if $C_0^\circ=U_{(k)}$,

Thus, writing $I_x(k,n+1-k)$ for the beta CDF,

More generally, every measurable $\phi:[0,1]\to[0,1]$ obeys

For example, for every integer $r\ge1$,

where the overline denotes a rising factorial. In particular the randomized pooled coverage mean differs from $k/(n+1)$ by at most the same panel-law error. This is a distributional finite-sample result; it does not substitute an effective sample size into a beta law.

Several future predictions: the beta-binomial benchmark

A deployed threshold is normally reused. Take $n$ calibration scores, let $C_0^\circ=U_{(k)}$, and append $L$ future randomized transforms. Define

Under the iid pooled comparator, conditional on $B=C_0^\circ\sim\operatorname{Beta}(k,n+1-k)$, the $Z_\ell$ are iid Bernoulli$(B)$. Therefore their unconditional law is exchangeable, not independent, and

This is the beta-binomial law. In particular, with $p=k/(n+1)$,

Thus replacing the $L$ indicators by independent Bernoulli$(p)$ variables is wrong even before time-series dependence is introduced. The shared finite-sample threshold creates the positive covariance.

Now apply the score-path comparison to the complete string of $n+L$ calibration and future scores before taking order statistics or indicators. Data processing gives

The same bound controls the full indicator vector and every event or bounded diagnostic built from it. On irregular grids or from a nonstationary start, replace the right side by the corresponding exact joint path-entropy bound for all $n+L$ observations.

This remains a marginal comparison over random calibration and future sequences. It does not make the realized indicators conditionally independent and does not assert pointwise or feature-conditional coverage.

The iid universal finite-batch law is derived by Marques F. (2023) via exchangeability and Pólya's urn. The contribution here is its quantitative transfer to fast hidden Markov scores through the joint total-variation theorem.

For the exact two-state certificate, state zero emits uniformly on $(0,1/2)$ and state one uniformly on $(1/2,1)$, so the pooled law is uniform and the hidden path is recoverable from every score. With $n=7$, $k=6$, $L=5$, and one-step correlation $0.2$, enumeration of all $2^{12}$ paths plus the conditional beta-binomial calculation gives count-law distance $0.038520368837$, below the joint score-law bound $0.332784201141$. Even in the iid comparator, its distance from the naive independent-binomial law is $0.128876657197$; the exact count variance is $1.354166666667$, versus $0.9375$ under independence.

Several coverage levels: the Dirichlet--multinomial benchmark

Monitoring nested conformal levels creates a second shared-threshold dependence. Choose ranks $0=k_0<k_1<\cdots<k_q<k_{q+1}=n+1$, and let the $q$ order statistics split $[0,1]$ into $q+1$ cells. Their random cell probabilities $S=(S_1,\ldots,S_{q+1})$ obey

If $N_j$ is the number of $L$ future randomized scores falling in cell $j$, then $N\mid S\sim\operatorname{Multinomial}(L,S)$, so

This is the Dirichlet--multinomial law. With $p_j=\alpha_j/(n+1)$, its covariance matrix is

The factor $(n+1+L)/(n+2)$ is the exact inflation relative to a multinomial with fixed probabilities. Thus neither different future observations nor different monitored levels may be treated as independent merely because the underlying scores are iid.

Apply the same joint score-path comparison to all $n+L$ observations before forming every order statistic and cell count. Data processing gives the multilevel transfer theorem

This controls the joint law of all nested future-coverage counts and every bounded simultaneous diagnostic. It remains marginal over the random calibration and future sequence; it is not simultaneous feature-conditional coverage.

The Dirichlet law of uniform spacings is classical; see Pyke (1965). The hidden-Markov contribution is the quantitative transfer bound.

In the same exact two-state certificate, take $n=7$, ranks $(3,6)$, and $L=5$. Enumeration of all $2^{12}$ hidden paths gives joint-count distance $0.060077182905$ from the Dirichlet--multinomial law, below $0.332784201141$. The iid Dirichlet--multinomial comparator is itself $0.165035363400$ from the fixed-probability multinomial, and its covariance inflation is $13/9=1.444444444444$.

If ties are instead retained deterministically, let $C_0=\bar F(R_{(k)})$ and define $H_{\bar F}(c)=\Pr_{R\sim\bar F}\{\bar F(R)\le c\}$. Under the iid pooled panel, the exact atomic comparator is

Consequently the dependent panel obeys

Because $H_{\bar F}(c)\le c$, deterministic tie retention is conservative in the iid comparator and also gives the simpler bound $\Pr_\pi\{C_0\le c\}\le I_c(k,n+1-k)+\delta^{\rm KL}_{n,m}$. The exact $H_{\bar F}$ formula separates that discreteness effect from calibration dependence and can be much sharper than this envelope.

The terminal-state relaxation bound converts this marginal law into a transparent state-specific PAC statement without claiming conditional conformal validity. For a specified current state $i$, define the randomized realized future coverage $D_i^\circ(\widehat q)$, using an independent tie uniform for the future score as well as the calibration scores. Since $|D_i^\circ(\widehat q)-C_0^\circ|\le H_i(\Delta)$ for every panel, putting $u_i=\min\{1,p+H_i(\Delta)\}$ gives

The probability is over random stationary calibration panels, while $i$ and $\Delta$ are fixed. It is therefore a PAC guarantee for a state-specific future coverage functional, not pointwise or feature-conditional coverage.

If the panel follows an initial law $\nu$ after burn-in $b$, one may replace the right-side panel error by the joint $\delta^{\rm KL}_{\mu,\boldsymbol P}$ above. A simpler gap-only envelope is

An exact nonreversible three-state certificate with $n=12$, $k=10$, and scaled spacing $mh=1$ obtains $\delta^{\rm KL}_{n,m}=0.044220298466$. Across 97 pooled-CDF levels, the largest exact beta-CDF discrepancy is $0.000384664522$; the exact first three moment discrepancies are respectively $7.319\times10^{-5}$, $8.519\times10^{-5}$, and $7.268\times10^{-5}$. The count recursion evaluates each dependent CDF value without Monte Carlo, and numerical tail integration checks the moment consequences independently.

A separate three-atom score law gives a maximum randomized-transform beta-CDF discrepancy of $0.000419796730$ on 99 levels and a maximum deterministic atomic-comparator discrepancy of $0.000420433415$. At $c=0.8$, $H_{\bar F}(c)=0.649319371728$: the dependent lower tail is $0.149730544921$, the exact iid atomic comparator is $0.150150978336$, and the looser continuous-beta envelope is $0.558345748480$.

The Hilbert--Schmidt expression is intrinsic: it is the sum of the squared singular values of $P-\Pi$ on centered $L^2(\pi)$. If $Q_0$ is reversible with relaxation rates $0=\lambda_1<\lambda_2\le\cdots\le\lambda_d$, the identity becomes

Thus iid calibration-panel rank laws become uniformly accurate under the intrinsic sufficient condition $(n_m-1)\mathfrak E(P_m)\to0$. The simpler condition $n_me^{-2m\gamma_sh}\to0$ remains sufficient, but can waste a factor $d-1$ by treating every mode as if it relaxed at the gap. Equivalently, the reversible effective rank $r_{\rm eff}=e^{2m\lambda_2h}\mathfrak E(P)$ lies between $1$ and $d-1$ and counts the modes still visible at the panel spacing.

This remains a marginal law over calibration panels; it does not turn ordinary pooled conformal prediction into pointwise, training-conditional, or feature-conditional coverage. From an initial law $\nu$ followed by a burn-in $b$, the exact joint entropy bound $\delta^{\rm KL}_{\mu,\boldsymbol P}$ can be used directly. If only the stationary Hilbert–Schmidt or gap envelope is desired, the simpler triangle bound adds $\tfrac12\|\nu/\pi-1\|_{2,\pi}e^{-m\gamma_sb}$.

A ten-state reversible certificate makes the improvement concrete. Two five-state clusters mix rapidly internally but communicate through one slow contrast. At scaled time $10$,

so $r_{\rm eff}=1$ to numerical precision. For $n=20$, the exact Hilbert--Schmidt TV bound is $0.417131357540$, whereas the gap-only bound is the vacuous value $1$. The certificate also checks the weighted row-$\chi^2$, Frobenius, singular-value, and reversible eigenvalue formulas independently.

The Rényi step matters away from the weak-dependence limit. For the same ten-state chain at scaled time $0.2$ and a two-observation panel, the exact-KL, Rényi-2, and direct Hilbert--Schmidt bounds are respectively $0.560996559452$, $0.601934420847$, and $0.729386305243$. The middle quantity is a rigorous transform of the same computable dependence energy, not a new mixing assumption.

Irregular panels: the whole spacing vector matters

The comparison is not restricted to a regular grid. Let the successive spacings be $h_1,\ldots,h_{n-1}$ and put $P_j=e^{mQ_0h_j}$. Stationarity is preserved across every unequal transition, so the relative-entropy chain rule gives the exact identity

Applying data processing, Pinsker, and the same rowwise chi-square comparison separately at every gap yields, for arbitrary conditionally independent state-specific emission kernels,

If $Q_0$ is reversible with relaxation rates $\lambda_r$, the exact energy in the first bound is

Thus an irregular triangular array is asymptotically iid whenever this summed modal energy tends to zero. The whole spacing vector is essential: Jensen's inequality gives $\sum_j e^{-2m\gamma_sh_j}\ge (n-1)e^{-2m\gamma_s\bar h}$. Replacing unequal gaps by their mean would therefore make the displayed envelope smaller, not safer; a few short gaps can dominate it.

The nonreversible three-state certificate uses scaled spacings $(0.35,0.50,0.75,1.10,1.70)$. Full enumeration of all $3^6$ hidden paths gives KL divergence $0.088542841901$ and TV distance $0.179564893897$. The exact-KL, summed Rényi-2, Hilbert--Schmidt, and gap-only bounds are respectively $0.210407749264$, $0.282956573231$, $0.289820479269$, and $0.325134088385$.

The information scale remains sharp with unequal gaps. In the symmetric two-state chain set $a_j=e^{-2mh_j}$. The path-to-sign bijection now gives

where the $Z_j$ are independent fair signs under the iid reference. Hence path TV is exactly heterogeneous Bernoulli-product TV. Its Hellinger affinity with the iid law is also exact:

Writing $\psi(a)=-\log[(\sqrt{1+a}+\sqrt{1-a})/2]$, the ratio $\psi(a)/a^2$ extends to a strictly positive continuous function on $[0,1]$, with endpoint values $1/8$ and $(\log2)/2$. Consequently, for any triangular array of gaps, without an infinitesimal-gap assumption,

This finite-product calculation is the Bernoulli specialization of the affinity criterion in Kakutani's (1948) equivalence--singularity theorem. The interior critical value needs more information than the sum of squares.

If $\max_j|a_j|\to0$ and $\sum_j a_j^2\to c$, the triangular-array log-likelihood is locally asymptotically normal in the sense of Le Cam (1960), and the TV distance tends to $2\Phi(\sqrt c/2)-1$. Without the maximum condition, the same $c$ need not give that limit: at $c=1/4$, one persistent correlation $a=1/2$ has TV $0.25$, while $65{,}536$ equal correlations with the same total squared correlation give $0.19741205$, approaching the diffuse LAN limit $0.19741265$.

There is nevertheless a complete mixed-profile limit. Suppose a fixed number $q$ of correlations converge to $b_1,\ldots,b_q$, while among the remaining correlations the maximum tends to zero and the sum of squares tends to $c$.

For independent fair signs $Z_1,\ldots,Z_q$ and an independent $G\sim N(0,1)$, put $B=\prod_{\ell=1}^q(1+b_\ell Z_\ell)$. Then

The proof factors the likelihood into persistent and diffuse parts, applies the Bernoulli triangular-array LAN limit only to the latter, and uses $\mathbb E_0L^2=\prod_j(1+a_j^2)$ for uniform integrability. Thus the entire finite-information profile, not merely its squared norm, is the invariant. With persistent correlations $0.5$ and $0.3$ and diffuse information $c=0.25$, the exact TV at $16{,}384$ diffuse signs is $0.312896499$, versus the product Bernoulli/lognormal limit $0.312895750$.

With equally many correlations $a$ and $2a$ and total squared correlation tending to one, the exact certificate reaches $0.38297545$ at $a=2^{-6}$, versus the limit $0.38292492$. This is still a marginal comparison of random panel laws, not conditional conformal coverage.

The scale cannot be improved in general even on a regular grid. For the stationary symmetric two-state chain, encode the states by $X_j\in\{-1,1\}$ and put $a=e^{-2mh}$. The transition signs $Z_j=X_jX_{j+1}$ are independent with $\Pr(Z_j=1)=(1+a)/2$; under an iid stationary panel they are independent fair signs.

The path total variation is therefore exactly the distance between $n-1$ copies of $\operatorname{Bernoulli}((1+a)/2)$ and fair Bernoulli variables. As $a\downarrow0$, it tends to zero if $(n-1)a^2\to0$, to one if $(n-1)a^2\to\infty$, and, if $(n-1)a^2\to c\in(0,\infty)$, to

Continuous emissions on disjoint state-specific supports reveal the state and preserve this distance exactly, so the example is a genuine continuous-score counterexample outside the subcritical scale. The certificate checks the exact Bernoulli-product formula by full path enumeration and obtains $0.38292359$ at $a=2^{-7}$, $n-1=a^{-2}$, against the critical limit $0.38292492$. For the nonreversible three-state chain and a 200-score panel, the spectral bounds at scaled spacings $mh=1,1.5,2$ are respectively $0.35621999$, $0.05660622$, and $0.00899518$.

Overlapping emissions attenuate the sharp constant

Disjoint supports prove that the hidden-path scale is sharp uniformly over all score laws, but they are the most informative possible emissions. The sharp constant for a fixed overlapping score channel is smaller and can be identified.

Keep the stationary symmetric hidden signs $X_j$, with transition correlation $a$, and let $K_+$ and $K_-$ be the two score laws. With $ar K=(K_++K_-)/2$, suppose

The number $\tau$ measures how much of the hidden sign one score reveals: it is zero for identical emissions and one for mutually singular emissions. If ${\cal R}_{n,a}$ denotes the emitted-panel law and ${\cal R}_{n,0}=\bar K^{\otimes n}$ its iid pooled counterpart, direct differentiation at hidden independence gives the observed score and Fisher information

The edge summands are one-dependent under the iid law, but distinct summands are uncorrelated because $\mathbb E_{\bar K}g=0$. Thus the hidden-path information $(n-1)$ is multiplied by $\tau^2$ after the score channel; data processing can change the critical constant, not merely make a finite bound smaller.

There is also an exact finite-panel identity. Take two independent copies of the hidden chain and put $U_j=X_jX'_j$.

Then $U$ is a fair two-state chain with transition correlation $a^2$, and integrating the two emission likelihoods gives the vertex weight $1+\tau U_j$. Hence, with

This gives the nonasymptotic bound $d_{\rm TV}\le\tfrac12\sqrt{\chi^2}$. The Perron root of $T_bD_\tau$ is

Consequently, if $(n-1)a_n^2\to c$, the likelihood second moment tends to $e^{c\tau^2}$. Under standard dominated HMM regularity, for example $|g|\le1-\eta$, stationary hidden-Markov LAN (Bickel and Ritov, 1996) and the explicit information above give

The exact second-moment formula supplies the uniform integrability needed to pass from likelihood convergence to total variation. At $\tau=1$ this recovers the hidden-path limit; at $\tau=0$ the two panel laws coincide for every $n$ and $a$. For a binary symmetric score channel with fidelity $0.8$, $\tau=0.64$.

At $c=1$, exact enumeration at $n=18$ gives TV $0.252356800$, approaching the emitted limit $0.251031669$, whereas the recoverable hidden path has limit $0.382924923$. This is still a marginal comparison of random score panels, not conditional conformal coverage.

Sharp joint burn-in and panel limit

The additive burn-in term is not merely a proof artifact. In the same symmetric two-state model, let the initial sign imbalance before burn-in be $\eta$, put

and observe $n_m$ states after the burn-in. The path-to-sign map is a bijection, and under the iid stationary reference law $(X_0,Z_1,\ldots,Z_{n_m-1})$ are independent fair signs.

Under the Markov law they are still independent, but have means $(\delta_m,a_m,\ldots,a_m)$. Consequently the likelihood ratio is exactly

Suppose $a_m\to0$, $\delta_m\to\delta\in[-1,1]$, and $(n_m-1)a_m^2\to c\in[0,\infty)$. The log of the product is locally asymptotically normal, in the sense introduced by Le Cam (1960), and uniform integrability follows from its bounded second moment. Hence

This phase surface is available in closed form. For $a\ge0$ define

When $a>0$ and $c>0$, splitting at the unique likelihood crossing gives

The continuous boundary values are $\mathcal D(a,0)=|a-1|$ and $\mathcal D(0,c)=1$. The formula also exposes an exact two-source comparison:

The lower bounds follow by marginalizing to the initial sign or to the transition experiment; the upper bound is the product total-variation inequality. Moreover $\Psi$ is nondecreasing in both $|\delta|$ and $c$. For $|\delta|$, a binary symmetric channel degrades any larger initial bias to a smaller one while preserving the fair reference.

For $c$, write the lognormal factor as the exponential Brownian martingale $e^{B_c-c/2}$ and apply conditional Jensen to the convex function $|x-1|$. Thus the two memory sources cannot cancel, although their total-variation effects need not add.

This contains both boundary cases: $\Psi(\delta,0)=|\delta|/2$ is the surviving initial-state memory, while $\Psi(0,c)=2\Phi(\sqrt c/2)-1$ is the stationary panel limit above. Therefore iid panel convergence holds in this sharp example if and only if

A short within-panel spacing cannot compensate for insufficient burn-in, nor can burn-in remove accumulated dependence among a growing number of calibration scores. With $\delta=0.6$ and $c=1$, the exact finite-panel distance at $a=2^{-7}$ is $0.451714292$, converging to $\Psi(0.6,1)=0.451711767$. Full path enumeration independently verifies the finite factorization.

Continuous disjoint-support emissions again preserve the hidden path and make the result a score-panel statement. Independent quadrature checks the Gaussian-CDF phase formula over 16 bias/information points to $4.2\times10^{-12}$. For the same benchmark, the two marginal lower bounds are $0.3$ and $0.382924923$, while the additive upper bound is $0.682924923$, locating the exact value strictly between the unavoidable marginal information and the triangle bound.

It remains a comparison of marginal random-panel laws, not a training-conditional or feature-conditional coverage guarantee.

Long regular panels: the exact exponential rate

The generating matrix also resolves the large-panel limit, but only when the spacing is fixed. Let $P=e^{mQ_0h}$ for $h>0$, assume the finite CTMC is irreducible, and suppose $0<f_i<1$ in every state.

Then $P$ and the tilted matrix below are strictly positive. Put

where $\rho$ is the Perron root. Since

Perron–Frobenius theory gives, for every initial state law $\nu$,

with $p=\pi f=\Lambda'(0)$. Positivity makes the Perron root simple and analytic for every real $\theta$, so the differentiability hypothesis in the Gärtner–Ellis theorem is automatic. This is the finite-state specialization of the primary large-deviation results of Gärtner (1977) and Ellis (1984).

For $x>p$, the optimizing tilt is positive, and the finite-panel Chernoff exponent converges to $I(x)$. In the iid specialization, $\Lambda(\theta)=\log(1-p+pe^\theta)$ and $I$ reduces to the Bernoulli relative entropy.

For the nonreversible certificate at scaled spacing $0.1$, $p=0.8$ and $x=0.9$, the Perron calculation gives

The exact coefficients converge toward the predicted logarithmic rate; the remaining finite-$n$ difference is resolved by the sharp prefactor below. This theorem is for a fixed regular-grid transition matrix. The earlier polynomial recursion remains exact on irregular grids, but no single-matrix Perron rate is claimed there.

The sharp lattice tail prefactor

The boundary amplitude also identifies the leading finite-panel tail, rather than only its logarithmic rate. Let $r_\theta,\ell_\theta$ be right and left Perron eigenvectors normalized by $\ell_\theta^\top r_\theta=1$, define $A_\nu(\theta)=\{\nu D(e^\theta)r_\theta\} \{\ell_\theta^\top{\bf1}\}$, and put $B_\nu(\theta)=A_\nu(\theta)e^{-\Lambda(\theta)}$.

For integer thresholds $k_n$ such that $x_n=k_n/n$ stays in a compact subset of $(p,1)$, let $\theta_n>0$ solve $\Lambda'(\theta_n)=x_n$, set $v_n=\Lambda''(\theta_n)$, and write $I(x_n)=\theta_nx_n-\Lambda(\theta_n)$. Then

The proof is a finite-dimensional lattice saddle point. Fourier inversion of the exact matrix MGF gives the central Gaussian factor. Because every $0<f_i<1$ and $P$ is positive, the tilted Fourier matrix has spectral radius strictly below its real Perron root away from frequency zero, so the remaining integral is negligible.

Successive lattice masses above $k_n$ have limiting ratio $e^{-\theta_n}$; summing them supplies the factor $(1-e^{-\theta_n})^{-1}$. Finally, the replacement of $(n-1)\Lambda$ by $n\Lambda$ explains why the initial-law factor is $B_\nu=A_\nu e^{-\Lambda}$ rather than $A_\nu$.

This is the finite-state Markov-additive counterpart of the sharp lattice asymptotics of Bahadur and Ranga Rao (1960); the Markov-additive large-deviation framework is due to Ney and Nummelin (1987). For the benchmark $x_n=0.9$, $v_n=0.122606564$, $B_\pi(\theta_n)=0.984845741$, and the $n$-free prefactor is $2.66849646$. The table's fourth column shows the exact tail divided by this sharp approximation, converging to one.

A pure exponential approximation omits both the $n^{-1/2}$ factor and the initial-law amplitude. The theorem remains a fixed-$P$ result; it neither covers irregular grids nor asserts a joint limit in panel size and switching rate.

The first two relative saddle-point corrections

The same local Fourier calculation supplies the next two terms, including the derivatives of the initial-law and lattice amplitude. Write

With every quantity below evaluated at $\theta_n$, uniformly while $x_n=k_n/n$ remains in a compact subset of $(p,1)$,

Writing $H_j=H_\nu^{(j)}(\theta_n)$ and $c_j=\Lambda^{(j)}(\theta_n)$, the second coefficient is

Indeed, after the saddle scaling $t=s/\sqrt n$, expand the analytic Fourier amplitude as $H_\nu+iH_\nu't-H_\nu''t^2/2$ and the Perron exponent as $-vt^2/2-ic_3t^3/6+c_4t^4/24$. Integrating the even terms against the centered Gaussian gives the four displayed contributions; the noncentral arc is exponentially smaller by the same Perron separation used above.

Retaining the even terms through order $n^{-2}$ gives $C_2$; analyticity makes the expansion uniform on compact tilt intervals. This is the finite-state Markov-additive version of the asymptotic saddle-point expansion introduced by Daniels (1954).

For the benchmark, $C_1=-14.728744589$ and $C_2=901.839869479$. The last two columns divide the exact probability by the first- and second-order corrected approximations. At $n=480$ and $960$, the second correction reduces the relative discrepancies to $0.0682\%$ and $0.0094\%$, from $0.3353\%$ and $0.0899\%$ after the first correction and $2.7434\%$ and $1.4457\%$ for the leading formula.

The overshoot at small $n$ is deliberately retained: this is an asymptotic expansion, not a finite-panel error bound, and truncation is not guaranteed to improve every small panel. The statement is still for a fixed regular-grid $P$; it makes no irregular-grid or joint $n$--switching-rate claim.

The moderate-deviation bridge to the mean

The compact-interior hypothesis can be relaxed in one important direction. Keep the regular-grid transition matrix $P$ fixed, let $p=\Lambda'(0)$ and $\sigma_P^2=\Lambda''(0)>0$, and take integer thresholds $k_n$ with

If $\theta_n>0$ solves $\Lambda'(\theta_n)=p+\delta_n$, then $\theta_n\sim\delta_n/\sigma_P^2$ and $n\theta_n^2\to\infty$. For every fixed initial law $\nu$, the sharp lattice formula continues to hold:

The singular lattice factor is essential: as the saddle approaches zero, $(1-e^{-\theta_n})^{-1}\sim1/\theta_n$. After the Fourier scaling $t=s/\sqrt n$, derivatives of that amplitude cost powers of $1/\theta_n$; the first omitted even Gaussian term is therefore of relative order $(n\theta_n^2)^{-1}$. The displayed condition is exactly what makes it vanish.

Analyticity gives $B_\nu(\theta)=1+O(\theta)$, $\Lambda''(\theta)=\sigma_P^2+O(\theta)$, and $\theta=\delta/\sigma_P^2+O(\delta^2)$, yielding the second formula. Thus the initial law disappears from the leading moderate-deviation prefactor, though it remains in the exact finite-$n$ amplitude.

This is the near-mean boundary of the same Markov-additive saddle argument of Ney and Nummelin (1987) and the lattice prefactor mechanism of Bahadur and Ranga Rao (1960). It is not a central-limit theorem: when $\sqrt n\,\delta_n=O(1)$, the pole and Gaussian saddle interact and a normal-tail approximation is required instead.

In the certificate, $p=0.5$ and $k_n=\lceil np+\tfrac12n^{3/4}\rceil$. From $n=120$ to $3840$, $\delta_n$ falls from $0.15833333$ to $0.06354167$, while $n\theta_n^2$ rises from $22.2914$ to $107.7808$.

The exact-to-Perron ratio rises from $0.923553$ to $0.978915$; after also replacing the exact amplitude by the universal Gaussian prefactor, the ratio falls from $1.133454$ to $1.062593$. The theorem is fixed-$P$ and gives no irregular-grid, finite-$n$, central-zone, or joint switching-rate bound.

All $O(\cdot)$ constants above are fixed-model, fixed-initial-law constants. No uniformity is asserted as the transition matrix loses positivity or its spectral separation degenerates.

The central-limit boundary: a second-order lattice Edgeworth bridge

The complementary zone has a different uniform approximation. Assume $P$ is fixed and positive, every emission probability satisfies $0<f_i<1$.

The initial law may be any probability vector $\nu$, and may even vary with $n$. Put

For integer thresholds $k_n$, use the half-integer tail coordinate

Write $H_1(x)=x$, $H_3(x)=x^3-3x$, and $H_5(x)=x^5-10x^3+15x$. Then, uniformly whenever $x_n$ stays in a fixed compact subset of $\mathbb R$,

Both first-order terms are necessary. The skewness term comes from the cubic Taylor coefficient of the Perron eigenvalue. The boundary term is the order-one mean shift: the spectral decomposition $\mathbb E_\nu e^{\theta S_n} =B_\nu(\theta)e^{n\Lambda(\theta)}+O(\rho^n)$ implies $\mathbb E_\nu S_n=np+b_{1,\nu}+O(\rho^n)$.

Hence a positive $b_{1,\nu}$ raises the upper tail. For a stationary start, $b_{1,\pi}=0$ because the finite-panel mean is exactly $np$.

For completeness, the proof is a local Fourier expansion rather than a continuation of the singular large-deviation prefactor. At frequency $t/(\sigma\sqrt n)$, expand $\log B_\nu$ through first order and $n\Lambda$ through third order. Fourier inversion gives the Gaussian, boundary-mean and skewness terms.

Summing the lattice masses through $k_n-1$ places the discontinuity halfway between adjacent integers, producing $k_n-\tfrac12$; omitting this continuity correction leaves an order-$n^{-1/2}$ periodic term. At second order, expanding $\log B_\nu$ through quadratic order and $n\Lambda$ through fourth order and then exponentiating recombines $(\log B_\nu)''(0)$ with $\{(\log B_\nu)'(0)\}^2$ into $b_{2,\nu}=B_\nu''(0)$. This gives the $b_{2,\nu}$, $\kappa_4$, $b_{1,\nu}\kappa_3$, and $\kappa_3^2$ terms.

The midpoint Euler–Maclaurin formula contributes the separate $-H_1/(24\sigma^2)$ lattice term. Positivity of $P$ and $0<f_i<1$ give strict Fourier contraction away from zero, while the next local Taylor order makes the remainder $O_K(n^{-3/2})$. This is the finite-matrix version of the lattice Fourier expansion of Esseen (1945) and the first-order Markov-chain Edgeworth theory of Datta and McCormick (1993); arbitrary-order local lattice expansions for finite Markov chains are established by Pudlo (2010).

The remainder is uniform over the entire initial-law simplex. Indeed, $B_\nu$, $b_{1,\nu}=B_\nu'(0)$ and $b_{2,\nu}=B_\nu''(0)$ are linear in $\nu$ because the exact matrix MGF is linear in its starting law.

The exact tail and the displayed truncated expansion are therefore both affine in $\nu$. Their signed difference is affine too, so

Thus the same $O_K(n^{-3/2})$ constant works simultaneously for every start, including an adversarial sequence $\nu_n$. This conclusion uses finiteness of the regime set; it is stronger than pointwise uniformity over a preselected family of starts.

The exact-law certificate uses the nonstationary start concentrated in state zero, for which $\sigma^2=0.261098129759$, $\kappa_3=-0.350878968118$, $\kappa_4=0.328775644197$, $b_{1,\nu}=0.653945953614$, and $b_{2,\nu}=-0.497726343467$. Over target coordinates $x\in\{-1,0,1,2\}$, the maximum uncorrected Gaussian error falls from $0.062972628$ at $n=120$ to $0.022139640$ at $n=960$, with $\sqrt n$ times that error nearly constant. Including skewness but omitting the boundary-mean term still leaves error $0.016504237$ at $n=960$.

Including both first-order terms reduces the maximum errors to $0.002466128$, $0.001130799$, $0.000576090$, and $0.000271427$ for $n=120,240,480,960$; the products by $n$ remain between $0.2606$ and $0.2959$. Adding $A_{2,\nu}$ reduces the maximum errors to $4.34444\times10^{-4}$, $1.37683\times10^{-4}$, $5.32133\times10^{-5}$, $1.71588\times10^{-5}$, $6.15325\times10^{-6}$, and $2.22912\times10^{-6}$ for $n=120,240,480,960,1920,3840$. The corresponding $n^{3/2}$-scaled errors stay between $0.510$ and $0.572$.

Checking all three pure-state starts shows that this same state-zero error is the simplex-wide maximum at $n=120,480,1920$: the worst errors are $4.34444\times10^{-4}$, $5.3213\times10^{-5}$, and $6.153\times10^{-6}$, respectively.

The $O_K(n^{-3/2})$ statement is uniform only over a fixed compact set of central coordinates for this fixed regular-grid model and all initial laws. It is not an explicit finite-panel error bound, is not uniform as positivity, variance or the spectral gap degenerates, and makes no irregular-grid or joint panel-size/switching-rate claim. It concerns the calibration-count law and does not upgrade marginal split-conformal coverage to conditional coverage.

Perron curvature is exactly Green–Kubo

The tilted eigenvalue is also consistent with the finite-panel variance theorem, including its boundary term. Let $\widetilde f=f-p{\bf1}$, $\Pi={\bf1}\pi$, and

On centered vectors, $Z=\sum_{\ell\ge0}P^\ell$. Differentiating the Perron eigenvalue at zero, or equivalently summing the stationary indicator covariances, gives

The first line is the discrete Green–Kubo formula. The second and third lines are an exact finite-panel refinement: $\chi_P$ is the persistent endpoint intercept, while the final term decays geometrically. No reversibility is required.

More generally, analyticity of the simple Perron root gives $\lim_{n\to\infty}\kappa_r(S_n)/n=\Lambda^{(r)}(0)$ for every fixed cumulant order $r$. The covariance interpretation is the discrete counterpart of the primary formulas of Green (1954) and Kubo (1957).

For the same nonreversible benchmark,

A five-point numerical derivative gives $\Lambda''(0)=0.261098129855$. The exact coefficient law has $\operatorname{Var}(S_{30})/30=0.254532972569$ and $\operatorname{Var}(S_{60})/60=0.257815515224$; the exact remaining boundary terms after slope and intercept are respectively $2.1564\times10^{-6}$ and $6.86\times10^{-12}$. This is a panel-size asymptotic at fixed $P$, distinct from the fast-switching limit obtained by changing $m$.

All Perron cumulants and near-mean tail asymmetry

The same simple-root perturbation gives every limiting cumulant rate by finite-dimensional linear algebra. Put $B=P\operatorname{diag}(f)$, $r_0={\bf1}$, $a_0=1$, and $\kappa_0=0$. For $k\ge1$, recursively define

Here $a_k=\rho^{(k)}(0)$ and $\kappa_{k,P}=\Lambda^{(k)}(0)$, so for every fixed $k$ and every initial state law,

The recursion follows by differentiating $M_\theta r(\theta)=\rho(\theta)r(\theta)$ and $\rho(\theta)=e^{\Lambda(\theta)}$; every step uses the same fundamental matrix $Z$. This is the finite-dimensional positive-matrix specialization of Kato's (1949) regular eigenvalue perturbation method.

Every finite-panel cumulant has an explicit boundary constant

The limiting slopes above do not describe the complete fixed-order asymptotics. Let $r_\theta$ and $\ell_\theta$ be analytic right and left Perron eigenvectors of $M_\theta$, normalized by $\ell_\theta^\top r_\theta=1$, and put

The scalar $A_\nu$ is independent of the reciprocal rescaling of the two eigenvectors. Positivity and the spectral gap give a complex neighborhood of zero and constants $C<\infty$, $0<q<1$ such that, uniformly there,

This holds for every fixed $k$ and every initial probability law $\nu$. It follows directly by inserting the analytic Perron projector $r_\theta\ell_\theta^\top$ into the exact matrix MGF and differentiating the uniform analytic remainder by Cauchy's formula. Thus the start-up effect is an explicit order-one constant, followed by geometric decay; it is not hidden in an unspecified $o(n)$ term.

For the nonreversible certificate, the stationary boundary constants through order four are

so the second entry exactly recovers the Green–Kubo endpoint intercept $\chi_P$. Starting instead from state zero gives

These $\beta_{k,\nu}$ are boundary cumulants, hence derivatives of $\log B_\nu=\log A_\nu-\Lambda$. They are deliberately distinct from the ordinary amplitude derivatives $b_{k,\nu}$ in the Edgeworth theorem. In particular,

For the state-zero start, this gives $b_{2,e_0}=-0.925371653715+(0.653945953614)^2 =-0.497726343467$, exactly the coefficient used in the second-order Edgeworth term.

At $n=60$, all eight constants agree with cumulants extracted independently from the exact coefficient law within $1.17\times10^{-8}$. These are fixed-$P$ boundary corrections in panel size; they are unrelated to a joint limit that changes the switching rate.

Finally, writing $v=\Lambda''(0)>0$, analytic Legendre duality gives the two-sided local expansion through fourth order

The quadratic term is the Green–Kubo Gaussian approximation; the cubic and quartic terms are the first skewness and kurtosis corrections for the dependent panel. This is a Taylor consequence of the same Gärtner–Ellis rate function above. It is distinct from the sharp lattice prefactor proved above, and is not an Edgeworth approximation to a finite-panel probability or a joint limit in $n$ and $m$.

For the benchmark, $(\kappa_{3,P},\kappa_{4,P}) =(-0.350878968118,0.328775644197)$. Richardson differentiation gives $(-0.350878967774,0.328775569108)$, while the exact coefficient law at $n=480$ gives $(-0.349342619570,0.325014708079)$. At $\delta=0.02$, the exact rate is $0.000793937413$; the quadratic, cubic and quartic errors are respectively $2.7942\times10^{-5}$, $1.6583\times10^{-6}$ and $1.0070\times10^{-7}$.

Across $\delta=0.04,0.02,0.01,0.005,0.0025$, their measured residual orders are $3.041083$, $4.042367$ and $5.047594$. In the iid specialization, the recursion agrees with cumulants extracted independently from the exact coefficient law through order eight, with maximum discrepancy $3.32\times10^{-8}$.

Arbitrary starts and burn-in

The preceding calibration confidence was stated for a stationary panel. There is an explicit transfer to any initial regime law.

Let the chain start from $\nu$, allow a burn-in $b\ge0$, and then observe the emitted panel $\mathcal P_b=(R_b,R_{b+h},\ldots,R_{b+(n-1)h})$. If $\mathcal P_\pi$ denotes the same panel started from $\pi$, then

For the first inequality, condition on the state at time $b$ and view all later states and emissions as one Markov kernel; total variation cannot increase under that kernel. This data-processing principle is the finite-state form of the contraction studied by Dobrushin (1956).

For the second inequality, the density relative to $\pi$ evolves under the adjoint semigroup. Its symmetric part is again $Q_s$, so the same energy argument gives $L^2(\pi)$ contraction at rate $\gamma_s$; Cauchy–Schwarz then converts $L^2$ to total variation.

Consequently, every stationary-panel event with probability at least $1-\delta$ holds after burn-in with probability at least $1-\delta-\zeta_b$. In particular, the uniform empirical-CDF event and the conditional-coverage inequality above remain unchanged on the good event, but their confidence becomes $1-\delta-\zeta_b$.

When the comparator is specifically the iid pooled panel, the exact nonstationary path-entropy identity above combines start-up and within-panel errors before Pinsker and can be sharper than this general transfer. To make the start-up penalty at most $\epsilon$, it is sufficient that

For a specified initial state $i$, the numerator is $\sqrt{1/\pi_i-1}$. This is a marginal probability statement over the random calibration panel. It is neither training-conditional nor feature-conditional validity, and the bound does not replace the panel size by an informal effective sample size.

The nonreversible three-state certificate enumerates the exact law of every four-score binary panel. For scaled burn-ins $mb=0,0.25,0.5,1$, the score-panel total variations are respectively $0.13175444$, $0.06439285$, $0.02503886$, and $0.00201894$; each lies below both the exact first-state total variation and the $L^2$ envelope. From state zero, the sufficient burn-in for a one-percent envelope is $mb\ge1.19131949$.

The panel also forgets the start

The total-variation transfer is deliberately uniform over every panel event. For the empirical CDF itself, later observations provide their own burn-in and give a sharper result.

Fix $r$, put $f=F(r)$, $p=\pi f$, $P=e^{mQ_0h}$, and $\mu_\ell=\nu e^{mQ_0b}P^\ell$. For $X_\ell={\bf1}\{R_{b+\ell h}\le r\}$ and $\widehat F_n(r)=n^{-1}\sum_{\ell=0}^{n-1}X_\ell$,

The identity is just the sum of the transient marginal expectations. For the bound, center $f$, move each transition semigroup onto that centered function, and apply the additive-reversibilization contraction. Since $f_i\in[0,1]$, its stationary standard deviation is at most $1/2$, giving the universal version

This prefactor and the complete factor $B_n(a)$ are sharp. For the symmetric two-state chain, a point start, and conditional success probabilities $(0,1)$, every summand has the same sign and lies in the gap eigenspace, so equality holds for every $n$, $b$, and $h$.

In particular, if $mh$ is bounded away from zero, then $B_n(a)\sim\{n(1-a)\}^{-1}$: within-panel relaxation makes the initial-law bias $O(n^{-1})$ even with no separate burn-in. This does not improve the preceding $\zeta_b$ guarantee for an arbitrary panel event; it uses the averaging structure of this particular statistic.

For an arbitrary deterministic grid $0=t_1<\cdots<t_n$, the same proof replaces $B_n(a)$ by the exact envelope

The same symmetric binary chain attains equality for every grid and burn-in. On the clustered five-point grid above with $b=0.2$, its exact start-up bias is $-0.32283968$, equal in magnitude to the displayed bound. In the nonreversible certificate on times $(0,0.07,0.31,0.9,1.8)$, the exact bias is $0.02702529$ and the score-specific bound is $0.03907180$.

The same conditioning argument gives an exact nonstationary variance, not an effective-sample-size heuristic: