\chapter{Temporal evidence on a continuous trajectory} \label{ch:temporal} \section{The inherited pairing construction} Let $\theta\in\{-1,+1\}$ be a source type. Let $F_t,G_t$ be mutually independent sign noises with means $\alpha,\beta\in[0,1]$, independent across events and independent of a sign-valued context process $Z_t$. Define \begin{equation}\label{eq:channels} A_t=\theta Z_tF_t,\qquad B_t=Z_tG_t,\qquad\kappa=\alpha\beta. \end{equation} The aligned products satisfy \begin{equation} U_t=A_tB_t=\theta F_tG_t,\qquad \E[U_t\mid\theta]=\theta\kappa. \end{equation} The context cancels exactly. Thus the products are independent even when the context process is correlated. This cancellation is a property of the declared sign model, not a universal theorem about arbitrary multimodal neural representations. \begin{proposition}[Inherited balanced-summary erasure] For an even-length record, suppose $Z$ has exactly equal numbers of positive and negative signs. The joint distribution of the two separate sums $S_A=\sum_tA_t$ and $S_B=\sum_tB_t$ is independent of $\theta$. \end{proposition} \begin{proof} Condition on a particular balanced $Z$. Each channel sum is the difference of two sums of identically distributed independent signs of equal size, so its distribution is symmetric and does not depend on the arrangement of the signs in $Z$. The channels' noises are independent, giving conditional independence of the two sums. Changing $\theta$ only reverses the first symmetric sum. Since the conditional joint law does not depend on the arrangement, averaging over balanced $Z$ preserves the conclusion. \end{proof} Equal means alone would not prove this statement. Balanced occupancy and the complete joint-law argument are essential. A stationary Markov context does not generally have exact balanced occupancy in a finite record; the 50\% ceiling from this proposition is therefore not used as a baseline claim for the new continuous-trajectory study. For the product likelihood with known $0<\kappa<1$, \begin{equation} P(U_t=u\mid\theta)=\frac{1+u\theta\kappa}{2}. \end{equation} The posterior log odds equal the prior log odds plus \begin{equation} 2\operatorname{atanh}(\kappa)\sum_tU_t. \end{equation} The signed count is sufficient for this product family. It need not be sufficient for the full sequence likelihood of $A,B$, where a model of context persistence can supply additional information. The new experiments retain full-HMM Bayes as a stronger reference. \section{A synchronization law with dependence included} Assume a stationary symmetric two-state Markov context. It can be written \begin{equation} Z_{t+1}=Z_tR_t, \end{equation} where $Z_0$ is fair and the increments $R_t$ are independent signs with mean $\rho\in[0,1]$. For a nonnegative pairing offset $\ell$, use \begin{equation}\label{eq:lagged} U_t^{(\ell)}=A_tB_{t+\ell}=\theta V_t\prod_{j=t}^{t+\ell-1}R_j, \qquad V_t=F_tG_{t+\ell}. \end{equation} The $V_t$ are independent signs of mean $\kappa$. Empty increment products are one. This representation exposes the exact dependence that endpoint-only simulations omit. \begin{theorem}[Continuous-record lag covariance]\label{thm:lag-cov} For $n$ valid pairs from one trajectory, let $S_n^{(\ell)}=\sum_{t=1}^nU_t^{(\ell)}$. Then \begin{align} \E[U_t^{(\ell)}\mid\theta]&=\theta\kappa\rho^\ell,\\ \Cov(U_t^{(\ell)},U_{t+h}^{(\ell)}\mid\theta) &=\begin{cases} \kappa^2(\rho^{2h}-\rho^{2\ell}),&1\le h<\ell,\\ 0,&h\ge\ell, \end{cases}\label{eq:lag-cov} \end{align} and \begin{equation}\label{eq:lag-var} \Var(S_n^{(\ell)}\mid\theta)=n(1-\kappa^2\rho^{2\ell})+ 2\kappa^2\sum_{h=1}^{\min(n-1,\ell-1)}(n-h)(\rho^{2h}-\rho^{2\ell}). \end{equation} For $\ell=0$ the sum is empty and the products are independent with variance $n(1-\kappa^2)$. \end{theorem} \begin{proof} Independence in \eqref{eq:lagged} gives the mean. For $0