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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. Define the path average and a noisy upper nonconformity score by

where $Z$ is independent of the chain and $\sigma>0$. 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 $\bar F(q_c;c)=1-\alpha$. 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 $1-\alpha$.

The crossover

The conditional mean already exposes the changed rate:

For long horizons it falls as $1/(2c)$, not exponentially. 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 $e^{-2c}$.

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.

Any finite irreducible chain

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

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

Because $\pi Q^\#=0$, the coefficient vector has stationary weighted sum zero, and the weighted coverage gaps cancel exactly at every $c$ by the definition of $q_c$. With unequal two-state rates the leading errors are equal and opposite only when the rates are equal, so an exact marginal guarantee can conceal conditional errors of different sizes.

This is a fixed-$Q$, long-horizon statement. It does not claim a remainder uniform over chains whose spectral gap degenerates.

Overlapping calibration windows

There is a second, distinct memory effect when the threshold is estimated. Start the stationary chain at time $-\infty$ and form length-$D$ calibration windows every $h$ units, $A_j=\frac1D\int_{jh}^{jh+D}Y_t\,dt$, with covariance $\Gamma(\tau)=\operatorname{Cov}(A_0,A_\tau)$. 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 variance inflation and effective size are

With $D=1$, $h=0.1$, $N=200$, and $m=1/\varepsilon=128$, the certificate gives $\mathcal I_N=9.8734$ and $N_{\rm eff}=20.26$. Thus tenfold overlapping sampling supplies about one independent path average per window length, not ten.

The empirical threshold under dependence

Let $Q_k$ be the $k$th order statistic of $N$ calibration scores. For iid scores the next score has marginal coverage $k/(N+1)$, but the coverage conditional on the realized calibration sample is random:

Thus exchangeability gives exact coverage only after averaging over calibration samples. For $N=k=9$ the mean is $0.9$, yet $\Pr(C_{9,9}<0.8)=0.8^9=0.134217728$.

Overlap alone already breaks iid rank coverage. For sliding length-$L$ block averages of the sampled chain with adjacent correlation $\rho$, an exact lifted recursion on the block state gives the coverage of the $k$th order statistic. With $N=k=9$, $L=4$, $\mu=1$, and $\sigma=0.5$, independent latent samples ($\rho=0$) give exact coverage $0.88580278$, not $0.9$; at $\rho=0.8$ it is $0.86318921$.

Thinning restores the rank value up to an explicit budget. Let $(X_t)$ be a stationary process with absolute-regularity coefficient $\beta_X$, and let the scores come from nonoverlapping length-$L$ blocks with starts $s_0<\cdots<s_N$, mixing gaps $h_i=s_{i+1}-s_i-L+1$, and a common continuous marginal. Then

This is a marginal guarantee. For the binary chain $\beta_X(h)=|\rho|^h/2$; in the $N=k=9$, $L=4$ example at $\rho=0.8$, the exact coverages for strides $d=4,12,20,28$ are $0.88738519$, $0.89873625$, $0.89980327$, and $0.89996739$.

The training-conditional law transfers as well. With $\eta_{\rm cal}$ the total-variation distance of the calibration panel from iid, $\eta_{\rm test}$ the cost of replacing the test score by an independent draw, and $C_P$ the coverage conditional on the panel,

Therefore vanishing dependence budgets recover the nondegenerate iid beta law, not the point mass at its mean. Simply replacing $N$ by an “effective sample size” in the iid beta law has no justification.

A predictor with slow internal state

Let the predictor's internal state be the exponential filter $dX_t=-\kappa(X_t-Y_t)dt$, and put $m=1/\varepsilon$. Conditional on observing $X_0=x$,

For $m\gg\kappa$, the endpoint regime forgets like $e^{-2mt}$ while the predictor state forgets like $e^{-\kappa t}$. 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 it does not claim

The coverage expansions are population, model-conditional results indexed by the initial latent state, not distribution-free feature-conditional conformal validity. The dependence bounds are generally conservative and marginal or training-conditional; they do not give exact coverage for a general score. None of these results is a distribution-free pointwise conditional conformal guarantee.

Calibration-label reuse is not covered: a memorizing learner fitted on the calibration labels gives $\Pr(S_{N+1}\le Q_k)=0$ for every $1\le k\le N$, and fast mixing cannot repair within-panel label leakage.

The full note

The full note also covers the exact occupation-time law, unequal switching rates, the second-order coefficient, coarsened initial information, nonlinear and critical score maps, regime-aware thresholds, sharp iid calibration design and two-endpoint tail allocation, optimal spacing, slack-free training-conditional bounds, atoms and randomized ties, and fitted predictors. The sequential coupling step uses the absolute-regularity construction of Berbee (1979).

Full note: path-conformal-full.html. Certificate: verify_path_crossover.py.