Proof. We first establish the identities in a finite connected subtree \(D\). A branch whose root has field \(z\) assigns that root probabilities proportional to \(e^{z}\) and \(e^{-z}\). Summing its root spin against the edge interaction gives a factor proportional to \(\cosh(z+\beta s)\) when its parent spin is \(s\). Its contribution to the parent’s half log-odds is therefore \[\frac12\log\frac{\cosh(z+\beta)}{\cosh(z-\beta)} =\operatorname{atanh}(\theta\tanh z)=\psi(z).\] The branch likelihoods multiply because the components are disjoint conditional on the parent spin. Adding their half log-odds and the own observation \(X_v(t)\) proves the finite cavity formulas. Since \(|\psi|\le\beta\), the site field differs from \(X_v(t)\) by at most \(d\beta\), and a component field by at most \(b\beta\).

We next follow a path \(v_0,\ldots,v_r\) in \(D\). With the observations fixed, sum the posterior in (3) over all spins off the path. Given the preceding path spins, the conditional mean of the spin at \(v_i\) uses only \(\sigma_{v_{i-1}}\) and the field on the side excluding that predecessor: \[\mathbb E[\sigma_{v_i}\mid \sigma_{v_0},\ldots,\sigma_{v_{i-1}}, \mathcal F_t^D] =\tanh(z+\beta\sigma_{v_{i-1}}) =p_i+q_i\sigma_{v_{i-1}}, \qquad z=h_{v_i\to v_{i-1}}^D(t).\] Here \(p_i\) is an intercept depending on the observations, and \[q_i=\frac{\tanh(z+\beta)-\tanh(z-\beta)}{2} =\theta\,\frac{1-\tanh^2z}{1-\theta^2\tanh^2z} =\theta a(z).\] Iterating these affine conditional means makes the slope of the conditional mean at \(v_r\), as a function of \(\sigma_{v_0}\), equal to \(\prod_{i=1}^r q_i\). Taking covariance with the first spin gives \[C_{v_0v_r}^D(t) =(1-(m_{v_0}^D(t))^2)\prod_{i=1}^r q_i.\] This proves the finite path formula, including the empty product at \(r=0\). The coefficient \(c_{vu}^D\) is directed even though the covariance is symmetric: its normalization uses the variance at \(v\). In fact, (4) gives \[\frac{\partial}{\partial x_u}\operatorname{atanh}M_v^D(x) =\frac{\mathop{\mathrm{Cov}}_{\pi_x^D}(\eta_v,\eta_u)} {1-M_v^D(x)^2},\] which is the path product just obtained. Thus \(c_{vu}^D\) is the finite field’s response to the observation at \(u\).

Now fix \(t\) and exhaust the tree by increasing finite balls. The conditional expectations of a fixed spin, or a product of two fixed spins, converge almost surely and in \(L^2\) to conditioning on \(\mathcal F_t^0\). This is increasing-sigma-field convergence for bounded random variables. Intersecting the balls with \(T_{v\to w}\) gives the corresponding convergence for the component posterior. The finite field bounds place each mean in the random compact interval \[[\tanh(X_v(t)-k\beta),\,\tanh(X_v(t)+k\beta)] \ \subset\ (-1,1),\qquad k=d\ \hbox{or}\ b.\] Thus the limiting means remain strictly inside \((-1,1)\), and their inverse hyperbolic tangents converge to finite fields. For a fixed path, take balls large enough to contain it. Their intersections with the finitely many directed components on the path exhaust those components. The finite cavity sums and path product therefore pass to the limit; convergence of the two-spin moments identifies the limit as the conditional covariance. Any other centered exhaustion gives the same conditional expectations, since its histories generate the same sigma-fields.

It remains to choose process versions without presupposing a continuous infinite-dimensional filter. On the canonical space of continuous observation paths, evaluate the finite field formulas on \(B_R(v)\) for a site and on \(B_R(v)\cap T_{v\to w}\) for a component, with \(R=0,1,\ldots\). Define each raw field to be the pointwise \(\limsup_{R\to\infty}\) of these finite fields. Each finite expression is a continuous adapted terminal-array function. Its upper limit is Borel and progressively measurable, for the whole observation filtration or for the specified component filtration respectively. The finite bounds make the upper limits finite and preserve (6) at every input and time. At each fixed time they agree almost surely with the conditional fields proved above. Their hyperbolic tangents and the finite products in (7) give the claimed mean and covariance versions.

These formulas use only the terminal array at the evaluated time, and in particular only the histories up to that time. Centering every ball at its indexed vertex makes the formulas commute with graph relabeling. Joint measurability and Fubini’s theorem turn the fixed-time algebraic identities into identities for almost every time, almost surely. Countability of the vertices, oriented edges, finite paths, and integer time horizons permits all these time-probability statements to hold simultaneously. Continuity and identities simultaneous at every time are not needed here; they will be obtained after the filtering argument in Section 4.

Finally, direct differentiation gives \(\psi'(z)=\theta(1-\tanh^2z)/(1-\theta^2\tanh^2z)\). Since \(0<\theta<1\), its bounds and \(0<a\le1\) follow. The covariance bound follows from the finite product and \(1-m_v^2\le1\). ◻

Proof. For a signal with marginal law \(\nu\) and posterior mean \(u\), the conditional signal law given \(\varepsilon=\eta\) has density \(1+\eta u\) with respect to \(\nu\), for \(\eta\in\{-1,1\}\). Indeed, condition the indicator of \(\{\varepsilon=\eta\}\) on the signal and use its posterior probability \((1+\eta u)/2\). Conditional independence then gives joint marginal density \(1+UV\) relative to the product of the two signal laws. The signed difference of the two conditional densities gives the combined posterior mean \((U+V)/(1+UV)\). Pushing the product signal law forward to the product of the marginal laws of \(U,V\), its second moment is \[\mathbb E_{\mathrm{prod}}\frac{(U+V)^2}{1+UV}.\] The integrand is at most \(2\), so it is legitimate to average it under the symmetry \(V\mapsto -V\). The result is \[\frac{U^2+V^2-2U^2V^2}{1-U^2V^2} =U^2+V^2- \frac{U^2V^2(2-U^2-V^2)}{1-U^2V^2},\] which proves (8). The loss is nonnegative. Under the additional bound by \(\theta\), its numerator factor \(2-U^2-V^2\) is at least \(2(1-\theta^2)\) and its denominator is at most one. The expectation of \(U^2V^2\) under the product law is \(\mathbb EU^2\,\mathbb EV^2\), proving the final assertion. ◻

Proof. Root \(T_{v\to w}\) at \(v\). It is a rooted \(b\)-ary tree; write \(|u|\) for the depth of \(u\) in it. For an integer \(n\ge0\), form the component-observable statistic \[Z_n=\sum_{|u|\le n}\theta^{|u|}X_u(t).\] The broadcast covariance is \(\mathbb E[\sigma_u\sigma_z]=\theta^{\mathop{\mathrm{dist}}(u,z)}\). Since \(b\theta^2=1\), \[ \mathbb E[\sigma_vZ_n] =t\sum_{k=0}^n b^k\theta^{2k}=t(n+1). \tag{10}\] The Brownian part of \(Z_n\) has variance \(t(n+1)\) and is independent of its spin part.

We count the latter part’s second moment. If an ordered pair \((u,z)\) has depths \(k,\ell\) and its last common ancestor has depth \(j\), its weighted spin covariance, before the factor \(t^2\), is \[\theta^{k+\ell+\mathop{\mathrm{dist}}(u,z)} =\theta^{2(k+\ell-j)} =b^{-(k+\ell-j)}.\] When \(j<\min(k,\ell)\), choose the common ancestor, the two distinct first children in their ordered roles, and then the remaining descendants. The exact count is \[b^j\,b(b-1)\,b^{k-j-1}b^{\ell-j-1} =(b-1)b^{k+\ell-j-1}.\] When \(j=\min(k,\ell)\), the shallower vertex is an ancestor of the deeper one. There are \(b^{\max(k,\ell)}\) such ordered pairs; this includes the \(b^k\) diagonal pairs when \(k=\ell=j\). Each count is at most \(b^{k+\ell-j}\). Thus the weighted contribution for each triple \((k,\ell,j)\) is at most one, and \[\begin{align*} \mathbb EZ_n^2 &\le t(n+1)+t^2 \sum_{k,\ell=0}^n\bigl(\min(k,\ell)+1\bigr)\\ &=t(n+1)+t^2\sum_{j=0}^n(n-j+1)^2 \le t(n+1)+t^2(n+1)^3. \tag{11}\end{align*}\] Because \(Z_n\) is measurable from the component observations, \(\mathbb E[\sigma_vZ_n]=\mathbb E[(\tanh Y)Z_n]\). Cauchy–Schwarz and (10)–(11) therefore give \[ M(t)\ge\frac{t(n+1)}{1+t(n+1)^2}. \tag{12}\] For \(0<t\le1\), take \(n=\lfloor t^{-1/2}\rfloor\). Then \(t^{-1/2}\le n+1\le2t^{-1/2}\), so the last bound is at least \(\sqrt t/5\). For \(t\ge1\), take \(n=0\) to obtain \(M(t)\ge t/(1+t)\ge1/2\). This proves the lower bound with \(c=1/5\).

For the upper bound, condition on the component root spin. Its own observation history and the \(b\) child-component histories are conditionally independent. The own signal has posterior mean \(\tanh X_v(t)\), by the one-spin version of (3), and \[q_0(t):=\mathbb E\tanh^2X_v(t)\le\mathbb EX_v(t)^2=t+t^2.\] One child’s entire observation history has posterior mean \(\theta\tanh Y\) about the root spin: conditional on the child spin, the root has mean \(\theta\) times that spin. Its second moment is \(\theta^2M(t)\), and its magnitude is at most \(\theta\). These statements concern full infinite component histories; their conditional independence follows from the broadcast construction and the independent Brownian noises, first for cylinder events and then for the sigma-fields they generate.

There are at least two child signals because \(b\ge2\). Combine two of them first. Lemma 3 subtracts at least \(2(1-\theta^2)\theta^4M(t)^2\). Add the other \(b-2\) child signals and the own signal by subadditivity. The combined signal is the full component history, and hence \[M(t)\le b\theta^2M(t) -2(1-\theta^2)\theta^4M(t)^2+q_0(t).\] The linear terms cancel at \(b\theta^2=1\). Since \(q_0(t)\le2t\) for \(t\le1\), this yields \(M(t)\le C\sqrt t\). At a whole vertex there are \(d\) child components, so subadditivity alone gives \[\mathbb Em_v(t)^2\le q_0(t)+d\theta^2M(t)\le C\sqrt t \qquad(0<t\le1).\] Both second moments are at most one for \(t\ge1\), completing (9). ◻

Proof. Let \(\mathcal G=\mathcal F_t^{T_{v\to w}}\). Since \(\sigma_v\) is a uniform sign and \(\mathbb E[\sigma_v\mid\mathcal G]=\tanh Y\), conditioning the indicator of \(\{\sigma_v=\eta\}\) on \(\mathcal G\) proves the first identity. The component observations are independent of \(\sigma_w\) conditional on \(\sigma_v\), and \(\mathbb E[\sigma_w\mid\sigma_v]=\theta\sigma_v\). It follows that \(\mathbb E[\sigma_w\mid\mathcal G]=\theta\tanh Y\); the same indicator calculation proves the second identity. The multiplying densities are bounded, so the formulas apply to every marginally integrable \(f(Y)\). If \(f\) is even, its product with \(\tanh Y\) is odd. Symmetry of the marginal law of \(Y\) makes that correction vanish. ◻

Proof. Set \(G=\psi(Y)\) and condition on the attaching spin \(\sigma_w=\eta\). We have \(|G|\le\beta\). The second identity in (13), symmetry, and Cauchy–Schwarz give \[\mathbb E[G^2\mid\sigma_w=\eta]=\mathbb EG^2,\qquad |\mathbb E[G\mid\sigma_w=\eta]|^2 =\theta^2|\mathbb E[G\tanh Y]|^2 \le\theta^2\mathbb EG^2.\] Moreover \(|\psi(y)|\ge\theta|\tanh y|\), since \(|\operatorname{atanh}u|\ge|u|\). Consequently \[ \mathop{\mathrm{Var}}(G\mid\sigma_w=\eta) \ge(1-\theta^2)\mathbb EG^2 \ge(1-\theta^2)\theta^2M(t). \tag{16}\] The conditional variance is the useful quantity here: unlike an uncentered second moment, it cannot be eliminated by adding a shift.

For any random variable \(G\) supported on \([-\beta,\beta]\), there is a constant \(c_\beta>0\) such that \[ \mathbb E\tanh^2(z+G)\ge c_\beta\mathop{\mathrm{Var}}(G) \qquad\text{for every }z\in\mathbb R. \tag{17}\] To see this, first suppose \(|z|\le\beta+1\). Then \(|z+G|\le2\beta+1\). The ratio \(\tanh^2 u/u^2\), extended continuously to be one at \(u=0\), has a positive minimum on that compact interval. Hence \(\mathbb E\tanh^2(z+G)\ge c_\beta\mathbb E(z+G)^2\ge c_\beta\mathop{\mathrm{Var}}(G)\). If \(|z|>\beta+1\), then \(|z+G|>1\), so the left side is at least \(\tanh^2(1)\); use \(\mathop{\mathrm{Var}}(G)\le\beta^2\) and decrease \(c_\beta\) if needed. This proves (17) for the conditional laws in (16) as well.

For \(u=\tanh^2h\), the elementary bounds on \(a\) give \[a(h)^2\le a(h) =1-\frac{(1-\theta^2)u}{1-\theta^2u} \le1-(1-\theta^2)u.\] Apply this with \(h=z+G\). Equations (17) and (16), followed by \(M(t)\ge\omega(t)/5\), prove (14), for example with \[c_0=\min\left\{\frac12,\, \frac{c_\beta(1-\theta^2)^2\theta^2}{5}\right\}.\] The same estimate may be used with a random shift after conditioning on data that fix the shift and are independent of the selected branch given its attaching spin.

For the path estimate, choose at each target \(v_i\), using only the path geometry, one neighbor \(s_i\) that is not on the path. There are \(d-2\ge1\) choices at an internal target and \(d-1\) at the last target. The components \(R_i=T_{s_i\to v_i}\) are disjoint and infinite. In particular, when \(d=3\) an internal target still has one full infinite side component.

Let \[\mathcal H=\sigma\bigl( \sigma_{v_0},\ldots,\sigma_{v_r},\ X_u(s):u\notin\textstyle\bigcup_iR_i,\ 0\le s\le t\bigr).\] Conditional on \(\mathcal H\), the selected component observation arrays are independent, and the law in \(R_i\) depends only on its attaching spin \(\sigma_{v_i}\). This follows directly by rooting the broadcast at \(v_0\): the multiplier on each attaching edge, the multipliers inside its selected component, and its Brownian motions form one independent group. The outside observations use none of these groups once the path spins are specified. The assertion for the full observation arrays follows from the corresponding assertion on finite cylinder events.

At this fixed time, the cavity identity gives, almost surely, \[h_{v_i\to v_{i-1}} =\psi(h_{s_i\to v_i})+ \left(X_{v_i} +\sum_{\substack{u\sim v_i\\u\ne v_{i-1},\,s_i}} \psi(h_{u\to v_i})\right).\] When all other selected arrays are held fixed, the parenthesized quantity is a fixed shift independent of the array in \(R_i\) conditional on \(\sigma_{v_i}\). Integrate \(R_1\) first. Every factor at a later target \(v_j\), \(j>1\), uses its component on the side excluding \(v_{j-1}\), and hence does not use \(R_1\). Thus (14) removes the first squared factor at a cost at most \(1-c_0\omega(t)\). Repeating in the order \(R_2,\ldots,R_r\) gives \[\mathbb E\left[\prod_{i=1}^r a(h_{v_i\to v_{i-1}})^2 \,\middle|\,\mathcal H\right] \le (1-c_0\omega(t))^r \le e^{-c_0\omega(t)r}.\] Taking a conditional expectation given only the path spins proves (15). Reversing the deterministic path gives the stated reverse product. Finally average over its spins. ◻

Proof. There are \(db^{r-1}\) vertices at distance \(r\ge1\) from \(v\). If \(b\theta^2<1\), the pointwise bound \(a\le1\) gives \[A_v(t)\le1+\sum_{r\ge1}db^{r-1}\theta^{2r} =1+\frac{d\theta^2}{1-b\theta^2}.\] At equality, \(db^{r-1}\theta^{2r}=d/b\). Tonelli’s theorem and Lemma 6 give, for \(t>0\), \[\mathbb EA_v(t) \le1+\frac db\sum_{r\ge1}e^{-c_0\omega(t)r} =1+\frac{d/b}{e^{c_0\omega(t)}-1} \le1+\frac{d}{bc_0\omega(t)} \le\frac{C}{\omega(t)}.\] Here \(e^x-1\ge x\) and \(\omega(t)\le1\). Since \[\int_0^T\frac{\,\mathrm dt}{\omega(t)} = \begin{cases} 2\sqrt T,&0\le T\le1,\\ T+1,&T\ge1, \end{cases}\] another application of Tonelli proves the integrated bound. Finally \(\sum_u C_{vu}^2=(1-m_v^2)^2A_v\le A_v\) by (7). ◻

Proof. The preceding conditional-increment and covariation calculation works with the observations restricted to \(D\), and proves the Brownian assertion. The finite posterior formula (3) gives \(m_v^D(t)=M_v^D(X_D(t))\) for a smooth function \(M_v^D\). It is a bounded continuous martingale in the raw filtration, since it is a version of \(\mathbb E[\sigma_v\mid\mathcal F_t^D]\). It is continuous in \(L^2\), so Lemma 9 also makes it a martingale in the augmented finite observation filtration.

Apply finite-dimensional Itô’s formula to \(M_v^D(X_D)\) using \(\,\mathrm dX_u=m_u^D\,\mathrm dt+\,\mathrm dW_u^D\). By (4), its continuous local martingale part is \(\sum_{u\in D}\int C_{vu}^D\,\mathrm dW_u^D\). The remaining part is continuous and has finite variation locally. Since \(m_v^D\) is itself a martingale, that part is also a local martingale and must vanish. The covariance coefficients are bounded, so the finite Brownian sum is square integrable on every finite horizon. The free prior has mean zero. This proves the first identity in (22), and orthogonality of the Brownian coordinates and Itô’s isometry prove the second. ◻

Proof. Choose \(\varepsilon_\delta>0\) with \(2\varepsilon_\delta/\sqrt{2\pi\delta}\le1/2\). A Gaussian random variable \(Z\) of variance \(s\ge\delta\), with any mean, has \(\mathbb P(|Z|\le\varepsilon_\delta)\le1/2\) by its density bound. Since \[a(z)^2\le a(z) \le 1-(1-\theta^2)\tanh^2z,\] it follows that \[\mathbb E[a(Z)^2]\le1-\kappa_\delta,\qquad \kappa_\delta=\frac{1-\theta^2}{2} \tanh^2\varepsilon_\delta\in(0,1).\] The estimate is uniform in the mean.

Let \(v=v_0,\ldots,v_r=u\) be the path in \(D\). In the squared covariance formula, discard the initial factor \((1-(m_v^D)^2)^2\le1\). At the fixed time \(s\) the remaining factors depend only on terminal observations. Conditional on all spins and all terminal Brownian coordinates other than \(B_{v_1}(s)\), the cavity recursion has the form \[h_{v_1\to v_0}^D(s)=B_{v_1}(s)+z,\] where the shift \(z\) is fixed: all branch fields in this recursion exclude \(v_1\). The later fields \(h_{v_i\to v_{i-1}}^D\), \(i>1\), also exclude \(v_1\). Conditional on the spins, the terminal Brownian coordinates are independent Gaussians of variance \(s\). Integrating \(B_{v_1}(s)\) first therefore removes the first squared factor at a cost at most \(1-\kappa_\delta\), without changing the later ones. Integrate \(B_{v_2}(s),\ldots,B_{v_r}(s)\) in that order. Iteration proves (23), even conditional on the spins. At a leaf the shift is simply its planted spin drift; the argument still applies. No infinite side branch is used here.

For fixed \(v,u,s\), exhaust the tree by finite connected subtrees containing the path. Bounded conditional-expectation convergence from Lemma 2 gives \(C_{vu}^D(s)\to C_{vu}(s)\) in \(L^2\). Passing to the limit proves the whole-tree inequality. ◻

Proof. For \(R\) large enough to contain the path from \(v\) to \(u\), the finite coefficient \(C_{vu}^{D_R}\) is a bounded continuous function of the terminal observations in \(D_R\), and is therefore a continuous raw adapted process. It converges to \(C_{vu}\) in the required space by (24). For the response coefficient use \[c_{vu}^{D_R}=\theta^{\mathop{\mathrm{dist}}(v,u)} \prod_{i=1}^{\mathop{\mathrm{dist}}(v,u)} a(h_{v_i\to v_{i-1}}^{D_R}).\] The component convergence in Lemma 2 gives fixed-time convergence to \(c_{vu}\); the common bound \(|c_{vu}^{D_R}|\le\theta^{\mathop{\mathrm{dist}}(v,u)}\le1\) again gives convergence in the stated \(L^2\) space. This finite coefficient is also a continuous raw adapted function of finitely many terminal observations.

For either finite coefficient, sample its values at the left endpoints of a deterministic mesh. As the mesh tends to zero, continuity and the common bound imply convergence in \(L^2(\,\mathrm ds\,\,\mathrm d\mathbb P)\) by dominated convergence. Choose a diagonal sequence of radii and meshes. These are the asserted elementary processes. A subsequence converges \(\,\mathrm ds\,\,\mathrm d\mathbb P\)-almost everywhere; its pointwise limit where finite, with value zero elsewhere, is predictable because each elementary process is predictable. This representative agrees with the original coefficient in the required product-measure sense. ◻

Proof. We first prove that the future of \(W\) is not exposed by either observation past. Fix a deterministic \(t\ge0\), write \(W_{\le t}\) for the restriction of \(W\) to \([0,t]\), and define its future increment array by \[Z^t_v(s)=W_v(t+s)-W_v(t),\qquad s\ge0.\] In the original law \(\mathsf Q\), \(Z^t\) is independent of the raw observation past \(\mathcal F_t^0\). Indeed, the Brownian property in Proposition 8 gives the assertion for every finite set of sites and future times. Rational times generate the Borel sigma field of a continuous path; a monotone-class argument over the countably many sites and rational times gives the assertion for the whole increment array. Causality also makes \(W_{\le t}\) \(\mathcal F_t^0\)-measurable.

It follows that, for every bounded Borel function \(F\) of the observation past, \[ \mathbb E_{\mathsf Q}[F(X_{\le t})\mid W] =\mathbb E_{\mathsf Q}[F(X_{\le t})\mid W_{\le t}] \quad\text{almost surely}. \tag{31}\] To verify the identity, denote the right side by \(J(W_{\le t})\) and test against \(u(W_{\le t})v(Z^t)\), with bounded Borel \(u,v\). Independence of \(Z^t\) from \(\mathcal F_t^0\) factors \(\mathbb Ev(Z^t)\) out of the expectation containing \(F u\). It does the same for \(J u\), and the remaining expectations agree by the definition of \(J\). These products generate \(\sigma(W)\): its past together with \(Z^t\) reconstructs every path value of \(W\). This proves (31).

Conditional independence in (30) gives, for bounded past functions \(F,G\), \[\mathbb E_{\widehat{\mathsf Q}} [F(X^1_{\le t})G(X^2_{\le t})\mid W] =\mathbb E_{\mathsf Q}[F(X_{\le t})\mid W] \mathbb E_{\mathsf Q}[G(X_{\le t})\mid W].\] Both factors on the right depend only on \(W_{\le t}\) by (31). Product Wiener measure makes \(Z^t\) independent of \(W_{\le t}\), and hence \[\begin{align*} &\mathbb E_{\widehat{\mathsf Q}} [u(W_{\le t})F(X^1_{\le t})G(X^2_{\le t})v(Z^t)]\\ &\qquad=\mathbb E_{\widehat{\mathsf Q}} [u(W_{\le t})F(X^1_{\le t})G(X^2_{\le t})] \mathbb E_{\mathsf w}[v(Z^t)]. \end{align*}\] Another monotone-class argument shows that \(Z^t\) is independent of the full sigma field \(\mathcal H_t^0\). Thus every finite subvector of \(W\) is Brownian in the raw joined filtration.

Each \(W_v\) is a continuous square-integrable martingale in the raw joined filtration and is \(L^2\)-continuous because its marginal law is Wiener measure. Lemma 9 therefore makes it a martingale in \((\mathcal H_t)\). The quadratic covariations, determined by the Brownian paths, remain \([W_u,W_v]_t=\mathbf 1_{\{u=v\}}t\). Lévy’s martingale characterization applied to each finite subvector proves the claimed Brownian property in \((\mathcal H_t)\).

We next check invariance. The selected field functions commute with graph relabeling, so the innovation map does too. Since the observation law is invariant, \(\mathsf Q\) is invariant under \((X,W)\mapsto(gX,gW)\). For one fixed automorphism \(g\), uniqueness of regular conditional probabilities now gives \[ \mathsf K(gw,\cdot)=g_*\mathsf K(w,\cdot) \quad\text{for $\mathsf w$-almost every }w. \tag{32}\] To obtain equality as probability measures, first check it for a countable determining algebra generating the Borel sets of \(\mathsf X\) and intersect the resulting full-measure sets. Applying this identity to both factors in (30) proves invariance of the joined law. This argument is made separately for each fixed \(g\).

It remains to justify the stochastic equations in the enlarged filtration. We give the approximation argument because an Itô integral cannot simply be identified across filtrations from a formal equality of pair laws. For either coefficient \(k_{vu}=c_{vu}\) or \(k_{vu}=C_{vu}\), fix \(v,u\) and a finite horizon \(T\). Lemma 12, applied to the \(X\)-marginal of \(\mathsf Q\), supplies bounded causal Borel elementary processes \[G_n(t,X)=\sum_i \gamma_{n,i}(X_{\le t_{n,i}}) \mathbf 1_{(t_{n,i},t_{n,i+1}]}(t), \qquad \int_0^T\mathbb E_{\mathsf Q}|G_n-k_{vu}|^2\,\mathrm dt\longrightarrow0,\] where each \(\gamma_{n,i}\) is a bounded Borel function of finitely many observations at time \(t_{n,i}\), viewed as a function of the indicated past.

The elementary integral against \(W_u\) is the Borel path functional \[I_n(t;X,W)=\sum_i\gamma_{n,i}(X_{\le t_{n,i}}) \bigl(W_u(t\wedge t_{n,i+1})-W_u(t\wedge t_{n,i})\bigr).\] The Brownian isometry and Doob’s maximal inequality show that these integrals converge in \(L^2(C([0,T]))\) to the original Itô integral. A deterministic subsequence converges uniformly almost surely; its limit on the Borel set of uniformly Cauchy inputs, with the zero path as default, is a Borel version of that integral as a function of \((X,W)\). The same subsequence has that pathwise limit in each joined pair, since \((X^j,W)\) has law \(\mathsf Q\). On the other hand the same isometry and maximal inequality in \((\mathcal H_t)\) show that \(I_n(\cdot;X^j,W)\) converges to the joined Itô integral of \(k_{vu}(\cdot,X^j)\). The two limits agree. This proves transport of each individual integral, including its continuous path version.

For either coefficient family the integrated squared row is finite: this is (18) for \(c\), and follows for \(C\) also from \(C_{vu}=(1-m_v^2)c_{vu}\). If \(F\subset F'\) are finite vertex sets, the joined integral tails satisfy \[\mathbb E_{\widehat{\mathsf Q}}\sup_{t\le T} \left|\sum_{u\in F'\setminus F}\int_0^t k_{vu}(s,X^j)\,\mathrm dW_u(s)\right|^2 \le4\sum_{u\in F'\setminus F}\mathbb E_{\mathsf Q} \int_0^T k_{vu}(s,X)^2\,\mathrm ds.\] The right side tends to zero along an exhaustion. The drift at any fixed time is a Borel observation functional, obtained by time integration; its absolute integral is finite almost surely because \(|m_v|\le1\) and \(\mathbb E\int_0^T A_v\,\mathrm dt<\infty\). Assigning zero to exceptional nonfinite integral values supplies a Borel scalar version if needed. Transport the original equations with finite partial Brownian sums at rational times and let the partial sums increase: their residuals converge in probability under \(\mathsf Q\), hence under each identical joined pair law, while their joined integral limits have just been identified. The equations therefore hold at all rational times in the joining. Continuity of the fields and Brownian sums, and absolute continuity of the drift integrals, extend them to all times. Taking countable intersections over vertices and integer horizons finishes the transport assertion. ◻

Proof. Subtract the two field equations in the common filtration. The martingale part is \(M_o=\sum_u\int\Delta c_{ou}\,\mathrm dW_u\), and, outside the \(\,\mathrm dt\,\,\mathrm d\widehat{\mathsf Q}\)-null set where a row diverges, \[\sum_u(\Delta c_{ou})^2\le2(A_o+A'_o).\] The integrated expectation of the right side is finite. Thus the sum defining \(M_o\) converges in maximal \(L^2\), and its bracket is \[\langle M_o\rangle_t =\int_0^t\sum_u(\Delta c_{ou}(s))^2\,\mathrm ds.\] This follows first for finite sums from the zero cross-covariations of distinct Brownian coordinates, then for the limit by the \(L^1\) convergence of the integrated squared row tails. Itô’s formula for \((\Delta h_o)^2\) gives \[\begin{align*} (\Delta h_o(t))^2={}&2\int_0^t\Delta h_o(s)\,\mathrm dM_o(s) +\int_0^t\sum_u(\Delta c_{ou}(s))^2\,\mathrm ds\\ &+2\int_0^t\Delta h_o(s) (m_o(s)A_o(s)-m'_o(s)A'_o(s))\,\mathrm ds. \end{align*}\] The first integral is a square-integrable martingale: by (34), its expected bracket is at most \(8H_T^2\mathbb E\int_0^T(A_o+A'_o)\,\mathrm ds\). The absolute values of the other two integrands have expectations bounded respectively by \(2\mathbb E(A_o+A'_o)\) and \(2H_T\mathbb E(A_o+A'_o)\), which are integrable in time. Taking expectations therefore gives an integral of an \(L^1([0,T])\) function. This proves absolute continuity and its almost-everywhere derivative (35). ◻

Proof. Since the products in (40) are at most one, \(L_r\le V\). Cauchy–Schwarz in the sum \(Z_r\) gives \[Q_r\le2\mathbb EZ_r^2 \le2r\sum_{i=1}^r\mathbb ED_{e_i}^2=2r^2S.\] Use this with (39) in the noise and drift bounds. The contributions are bounded by \(V\) times sums of \(r^2\), \(1\), and \(r\), respectively, against \[n_r\theta^{2r}=\frac db(b\theta^2)^r.\] All three series converge when \(b\theta^2<1\). Equation (35) therefore gives \(V'\le C V\) almost everywhere on \([0,T]\), for a finite constant depending on the fixed parameters. Absolute continuity and \(V(0)=0\) imply \(V=0\) there by Gronwall’s inequality. The horizon \(T\) was arbitrary. ◻

Proof. We first prove a correlation bound for an assignment \(f\) that is centered and measurable, at vertex \(v\), from the entire observation paths at the vertices in a fixed ball \(B_R(v)\). Square integrability is enough; write \(\|f\|^2=\mathbb E_\nu f_o^2\). Write \(\mathbb E_{\mathrm{pl}}\) for expectation in the planted spin-and-Brownian construction, whose observation marginal is \(\nu\).

Suppose that \(\mathop{\mathrm{dist}}(u,v)=k>2R\), and let \(a\) and \(c\) be the respective endpoints of the path joining \(B_R(u)\) to \(B_R(v)\). Then \(\mathop{\mathrm{dist}}(a,c)=k-2R\). In the planted broadcast construction, conditional on \((\sigma_a,\sigma_c)\), the observation arrays in these two balls are independent. The conditional law in the first ball depends only on \(\sigma_a\), and the law in the second only on \(\sigma_c\). To see this, root the broadcast at \(a\). Once the endpoint spins are fixed, the edge multipliers inside the two balls are taken from disjoint branches off the connecting path. They are independent of that path and of one another, and the Brownian paths at the two disjoint sets of vertices are independent as well. The assertion for products of square-integrable functions follows from its bounded version by truncation.

The endpoint spins are uniform signs. Since \(f_u,f_v\) are centered, their conditional expectations have the form \[\mathbb E_{\mathrm{pl}}[f_u\mid\sigma_a]=\alpha_u\sigma_a,\qquad \mathbb E_{\mathrm{pl}}[f_v\mid\sigma_c]=\alpha_v\sigma_c.\] \(L^2\)-contraction of conditional expectation gives \(|\alpha_u|,|\alpha_v|\le\|f\|\). The broadcast correlation along the connecting path is \(\mathbb E_{\mathrm{pl}}[\sigma_a\sigma_c]=\theta^{k-2R}\). It follows that \[|\mathbb E_\nu[f_uf_v]| =|\alpha_u\alpha_v|\theta^{k-2R} \le\|f\|^2 b^R b^{-k/2}.\] For \(k\le2R\) the same bound follows from Cauchy–Schwarz, since \(b^R b^{-k/2}\ge1\). We have therefore proved \[ |\mathbb E_\nu[f_uf_v]| \le C_f b^{-\mathop{\mathrm{dist}}(u,v)/2} \quad\text{for all \(u,v\)},\qquad C_f=b^R\|f\|^2. \tag{46}\] This calculation uses the planted spins to describe the law, while the assignment itself remains a function of the observations alone.

We now turn spatial decay into an operator bound. Let \(p_n(o,v)\) be the \(n\)-step transition probabilities of simple random walk on the tree and put \(q(v)=b^{-\mathop{\mathrm{dist}}(o,v)/2}\). For the deterministic neighbor average \(P\) on functions of vertices, \[(Pq)(v)=\rho q(v)\quad(v\ne o),\qquad (Pq)(o)=b^{-1/2}\le\rho q(o).\] The first identity uses one neighbor closer to \(o\) and \(b\) farther away. Positivity of \(P\) gives \(P^nq\le\rho^nq\) by induction, and hence \[\sum_vp_n(o,v)q(v)\le\rho^n.\] Only finite-step random-walk averages are used here; \(q\) need not belong to \(\ell^2(\mathcal V)\). Self-adjointness of \(\mathsf A_\nu\), its finite-walk expansion, and (46) now give, for every \(n\ge0\), \[ \|\mathsf A_\nu^nf\|^2 =\langle f,\mathsf A_\nu^{2n}f\rangle =\sum_vp_{2n}(o,v)\mathbb E_\nu[f_of_v] \le C_f\rho^{2n}. \tag{47}\]

Let \(\mu_f\) be the spectral measure of \(f\) for the self-adjoint restriction of \(\mathsf A_\nu\) to centered assignments. For \(\varepsilon>0\), (47) implies \[\mu_f\{|\lambda|\ge\rho+\varepsilon\} \le C_f\left(\frac{\rho}{\rho+\varepsilon}\right)^{2n}.\] Letting \(n\to\infty\) proves that \(\mu_f\) is supported on \([-\rho,\rho]\).

It remains to pass from local assignments to all centered ones. For \(f\in\mathcal H_{\mathrm v}(\nu)\) centered, let \[f_v^{(R)} =\mathbb E_\nu\!\left[f_v\ \middle|\ \sigma\{X_z(s):z\in B_R(v),\ s\ge0\}\right].\] These are centered square-integrable local assignments. To verify their equivariance, pull the conditional expectation at \(\gamma v\) back by a fixed automorphism \(\gamma\). Invariance of \(\nu\) pulls its conditioning sigma-field back to that for \(B_R(v)\), and equivariance pulls its integrand back to \(f_v\). Uniqueness of conditional expectation gives the required identity almost surely for that \(\gamma,v\). As \(R\to\infty\), the conditioning sigma-fields generate the complete observation array, so \(f_o^{(R)}\to f_o\) in \(L^2\). This proves density of the local assignments just used. Equally, truncating and recentering \(f\) first gives a bounded local dense family. The range of the corresponding spectral projection \(\mathbf 1_{[-\rho,\rho]}(\mathsf A_\nu)\) on the centered space is closed and contains every such local assignment. It is consequently the entire centered space, proving the lemma. Finally, \(\rho<1\) follows from \(d=b+1\) and \((b+1)^2-4b=(b-1)^2>0\). ◻

Proof. The recurrence deletes only the contribution from \(p\) at \(y\) and propagates that change forward; every field from a side child is unchanged. All these fields use components contained in \(\mathcal T_y\), so the claimed locality follows by induction. The modified fields are used as functions of the descendant observations; no posterior interpretation for a truncated model is needed.

The cavity identity gives \(H_0-\widetilde H_0=\psi(h_{p\to y})\). At each subsequent step, only the predecessor message differs. Since \(|\psi|\le\beta\) and \(|\psi'|\le\theta\), induction yields \[|H_i-\widetilde H_i|\le\beta\theta^i\qquad(0\le i<L).\] The bound \(|(\log a)'|\le2\) from (38) now gives \[\left|\log \frac{\prod_{i=0}^{L-1}a(H_i)^2} {\prod_{i=0}^{L-1}a(\widetilde H_i)^2}\right| \le4\sum_{i=0}^{L-1}|H_i-\widetilde H_i| \le4\beta\sum_{i=0}^{L-1}\theta^i \le\frac{4\beta}{1-\theta}.\] This proves (60). In particular, obtaining component locality costs a multiplicative factor at most \(C_{\mathrm{cut}}\), independent of the length and time.

The outward inequality in (61) is the conditional path estimate of Lemma 6, applied to \(w_0,\ldots,w_L\). We prove the modified inward inequality by integrating intact side components from the far end back toward the cut.

At each target \(w_i\), \(0\le i<L\), choose one child off the path, using only the path geometry. There are \(b-1=d-2\ge1\) choices, including at \(w_0\) after deleting its parent and excluding \(w_1\). The selected child components are disjoint and infinite; this also covers \(d=3\) and \(L=1\).

Condition on the true path spins. The selected component observation histories are independent of each other and of the remaining observation histories, and their laws depend only on the attaching path spins. Indeed, root the broadcast at \(w_0\) and group each selected component’s attaching-edge transition, internal transitions, and Brownian noises. These groups are mutually independent; once the path spins are specified, the observations outside the selected components use none of them. This factorization passes from finite cylinder events to the sigma-fields of the full observation histories. Thus fixing the other observation histories leaves each selected component with exactly the conditional law used in (58).

Integrate the selected components in the order \(w_{L-1},w_{L-2},\ldots,w_0\). In (59), \(\widetilde H_i\) is its selected side message plus a shift fixed by the other observations. Every field still present after its factor is removed has index less than \(i\) and does not use that message. Equation (58) therefore removes each squared factor at a cost at most \(1-c_0\omega(t)\). The resulting bound is \((1-c_0\omega(t))^L\le e^{-c_0\omega(t)L}\), as required. ◻

Proof. Root the tree at \(z\) and put \(k=\lfloor\ell/2\rfloor\) and \(L=\ell-k\). The threshold ensures \(k\ge1\). Let \(\mathcal Y_k\) be the vertices at depth \(k\) whose path from \(z\) obeys the restriction \(F\). There are \[ N:=|\mathcal Y_k|=(d-|F|)b^{k-1}\ge b^k \tag{63}\] such vertices, and each has exactly \(b^L\) descendant continuations of length \(L\). Since every squared factor lies in \((0,1]\), deleting the first \(k\) factors increases either product. The average of the remaining products is consequently an equal-weight average of \(N\) continuation averages.

For \(y\in\mathcal Y_k\), let \(p(y)\) be its parent and \(\mathcal T_y=T_{y\to p(y)}\). Let \(Z_y^+\) be the average of \[\prod_{i=1}^{L}a(h_{w_i\to w_{i-1}})^2\] over its \(b^L\) continuations \(y=w_0,\ldots,w_L\). Let \(Z_y^-\) be the average of \(\prod_{i=0}^{L-1}a(\widetilde H_i)^2\) for the same continuations, using (59) separately for each one. Both variables lie in \([0,1]\) and are functions only of the observation histories in \(\mathcal T_y\). Deleting the first \(k\) factors and applying (60) gives \[ G^\varepsilon_\ell(z;F) \le\frac{C_{\mathrm{cut}}}{N} \sum_{y\in\mathcal Y_k}Z_y^\varepsilon, \qquad \varepsilon\in\{+,-\}. \tag{64}\] For \(+\) the multiplier can be replaced by one.

Averaging (61) first over the continuation spins conditional on \(\sigma_y\), and then over the continuation choices, yields \[ \mathbb E[Z_y^\varepsilon\mid\sigma_y] \le e^{-c_0\omega(t)L},\qquad\varepsilon\in\{+,-\}. \tag{65}\] Now condition on the sigma-field of true cut-level spins \[\mathcal H_k=\sigma(\sigma_y:y\in\mathcal Y_k).\] The observations in the disjoint components \(\mathcal T_y\) are independent under this conditioning, with each component law depending only on its own root spin. Indeed, after specifying the cut-level spins, all broadcast transitions and Brownian noises strictly below the cut, together with the Brownian noise at each cut vertex, can be sampled independently across these components. The same statement holds when \(F\) forbids one first neighbor, since that simply omits one collection of components. The cut-level spins themselves need not be independent. Thus the \(Z_y^\varepsilon\) are conditionally independent variables in \([0,1]\), and their conditional means given \(\mathcal H_k\) satisfy (65).

Write \(\tau_\ell=e^{-\eta\omega(t)\ell}\). Since \(L\ge\ell/2\), \(\omega(t)\ell\ge C_0\), and \(\eta\le c_0/4\), the chosen constants give \[\mathbb E[Z_y^\varepsilon\mid\mathcal H_k] \le e^{-c_0\omega(t)\ell/2}\le\tfrac12\tau_\ell.\] Apply Hoeffding’s bounded-sum inequality (Hoeffding 1963, Theorem 2) to each conditional product law. Because the distance from the conditional mean of the average to \(\tau_\ell\) is at least \(\tau_\ell/2\), it gives \[\begin{align*} \mathbb P\left(\frac1N\sum_yZ_y^\varepsilon>\tau_\ell \,\middle|\,\mathcal H_k\right) &\le\exp\left(-2N(\tau_\ell/2)^2\right)\\ &=\exp\left(-\tfrac12N e^{-2\eta\omega(t)\ell}\right) \le\exp\left(-\frac{b^{\ell/4}}{2b}\right). \end{align*}\] For the last inequality, \(k\ge\ell/2-1\), \(\omega(t)\le1\), and \(2\eta\le(\log b)/4\) imply \[N e^{-2\eta\omega(t)\ell} \ge b^{\ell/2-1}e^{-(\log b)\ell/4} =b^{\ell/4}/b.\] The conditional bound holds almost surely. Taking expectations and using (64) proves (62). ◻

Proof. Fix \(T\) and put \[H_T=2T+2d\beta,\qquad D_{\max,T}=\frac{2H_T}{1-\theta}.\] The site bound (34), the component comparison (37), and \(|(\log a)'|\le2\) imply \[ |\Delta h_v|\le H_T,\qquad |\Delta h_{j\to i}| \le\frac{H_T+\theta H_T}{1-\theta^2} =\frac{H_T}{1-\theta},\qquad |D_e|\le D_{\max,T}. \tag{69}\] Dropping \(P_r^2,(P'_r)^2\le1\) and applying Proposition 16 gives the first bound in (68); dropping \(P_r^2\) gives the second.

For the longer paths, set \[p=\lfloor r/2\rfloor,\qquad q=r-p,\qquad U=\sum_{i=1}^pD_{(v_{i-1},v_i)},\qquad Z=\sum_{i=p+1}^rD_{(v_{i-1},v_i)}.\] Since \((U+Z)^2\le2U^2+2Z^2\), \[ Q_r\le2\mathbb E[(P_r^2+(P'_r)^2)U^2] +2\mathbb E[(P_r^2+(P'_r)^2)Z^2]. \tag{70}\] Figure [fig:half-path-averages] displays the two path averages that control these terms.

First fix the prefix \((v_0,\ldots,v_p)\) and vary all length-\(q\) extensions from \(v_p\) that avoid \(v_{p-1}\) on their first step. The pointwise stabilizer of the prefix is transitive on these extensions: map the successive descendant branches of one extension to those of the other, and extend independently to the remaining regular rooted subtrees. The invariance of the joined law consequently makes the expectation of the product for each complete path, multiplied by \(U^2\), the same for every extension. In a fixed joined observation configuration, \(U\) also denotes the same quantity in every term of this finite average: its indexed directed edges stay fixed. This statement places no locality restriction on their fields. Dropping the prefix factors from the product gives \[ \mathbb E[P_r^2U^2] \le\mathbb E[U^2G_q^+(v_p;\{v_{p-1}\})]. \tag{71}\]

For the other term, fix the suffix \((v_p,\ldots,v_r)\) and vary a reversed prefix \(w_0=v_p,w_1,\ldots,w_p\) with \(w_1\ne v_{p+1}\). The resulting complete directed path is \[(w_p,w_{p-1},\ldots,w_1,v_p,v_{p+1},\ldots,v_r).\] The pointwise stabilizer of the suffix is transitive on these choices; it may move the original starting vertex \(v_0\), which invariance under the full automorphism group permits. The suffix sum \(Z\) has the same indexed edges in every term. In the direction of the complete path, its prefix product is exactly \[\prod_{i=1}^p a(h_{w_{i-1}\to w_i})^2,\] the inward product along the reversed geometric prefix. Dropping the suffix factors therefore yields \[ \mathbb E[P_r^2Z^2] \le\mathbb E[Z^2G_p^-(v_p;\{v_{p+1}\})]. \tag{72}\] Each path average is finite. Thus its use of equivariance requires only a finite intersection of almost-sure identities for fixed automorphisms, consistent with the theorem’s convention.

Suppose \(r\ge2\ell_*(t)\). Then \(p,q\ge\ell_*(t)\). Apply (66) to the right side of (71) with \(\zeta=U^2\). The unconditional path variance bound and (69) give \(\mathbb EU^2\le C_{\mathrm{var}}pS\) and \(\|U^2\|_\infty\le p^2D_{\max,T}^2\). Hence \[ \mathbb E[P_r^2U^2] \le C_{\mathrm{cut}}C_{\mathrm{var}} e^{-\eta\omega(t)q}pS+D_{\max,T}^2p^2\varepsilon_q. \tag{73}\] Applying the same bound to (72), now with \(\zeta=Z^2\), gives \[ \mathbb E[P_r^2Z^2] \le C_{\mathrm{cut}}C_{\mathrm{var}} e^{-\eta\omega(t)p}qS+D_{\max,T}^2q^2\varepsilon_p. \tag{74}\] These applications use the probability, under the joined law, of the corresponding event determined by one observation marginal. They do not require a conditional variance bound or independence of that event from the discrepancy. The identical reasoning applies to \((P'_r)^2\) using the second marginal’s path averages.

For \(r\ge2\), both \(p\) and \(q\) are at least \(r/3\). Inserting the four estimates into (70), and using \(p+q=r\) and \(p^2+q^2\le r^2\), proves the explicit inequality \[ Q_r\le4C_{\mathrm{cut}}C_{\mathrm{var}} e^{-\eta\omega(t)r/3}rS +4D_{\max,T}^2r^2 \exp\left(-\frac{b^{r/12}}{2b}\right). \tag{75}\]

Finally fix \(v_0\) and average the entire directed path over \(\mathcal P_r(v_0;\varnothing)\). The stabilizer of \(v_0\) is transitive on these paths, and the vertex quantity \(\Delta h_{v_0}\) is fixed in every term. Invariance gives the exact identity \[L_r=\mathbb E[(\Delta h_{v_0})^2G_r^+(v_0;\varnothing)].\] For \(r\ge\ell_*(t)\), use (66) with \(\zeta=(\Delta h_{v_0})^2\), whose mean is \(V\) and whose essential supremum is at most \(H_T^2\). This yields \[ L_r\le C_{\mathrm{cut}}e^{-\eta\omega(t)r}V +H_T^2\exp\left(-\frac{b^{r/4}}{2b}\right). \tag{76}\] Choosing one sufficiently small positive \(c\) in the two explicit bounds gives (67). Finally, \(\ell_*(t)\le(C_0+1)/\omega(t)\) since \(\omega(t)\le1\), which proves the stated sufficient length condition. ◻

Proof. We begin with the second moment. On \(\{|X_v(t)|\le1\}\) the cavity bound gives \(|Y|\le1+b\beta\). Comparison on this compact interval therefore gives \(Y^2\le C\tanh^2Y\). On its complement we use \(Y^2\le2X_v(t)^2+2b^2\beta^2\). For \(t\le1/2\), the event \(|t\sigma_v+\sqrt t\,G|>1\) requires \(|G|>1/(2\sqrt t)\). Integration of the normal density, decreasing the positive exponent constant if necessary, yields \[\mathbb E\bigl[(1+X_v(t)^2)\mathbf 1_{\{|X_v(t)|>1\}}\bigr] \le C e^{-c/t}.\] Combining the two events with \(M(t)\le C\sqrt t\) proves that, after fixing some \(t_0\le1/2\), \[ A_2(t):=\mathbb E|Y|^2\le C\sqrt t, \qquad A_1(t):=\mathbb E|Y|\le C^{1/2}t^{1/4} \quad(0<t\le t_0). \tag{79}\] The second inequality is Cauchy–Schwarz. The cavity bound also shows that \(A_p(t):=\mathbb E|Y|^p\) is finite for every integer \(p\ge1\).

We next explain why the recursion can be expanded using independent terms without changing these absolute moments. For an integrable function \(f\), the spin-tilt identities from the posterior calculation are \[\begin{align*} \mathbb E[f(Y)\mid\sigma_v=\eta] &=\mathbb E[f(Y)(1+\eta\tanh Y)],\\ \mathbb E[f(Y)\mid\sigma_w=\eta] &=\mathbb E[f(Y)(1+\eta\theta\tanh Y)], \qquad \eta\in\{-1,1\}. \end{align*}\] These are the identities (13) from Lemma 5. Because \(Y\) has a symmetric marginal law, the term containing \(\tanh Y\) integrates to zero against every even function of \(Y\). In particular conditioning on either spin leaves each \(A_p(t)\) unchanged.

Condition now on \(\sigma_v=+1\) in the cavity recursion at \(v\). The own observation and the \(b\) child-component observations are independent under this conditional law: their spin transitions and their planted noises lie in disjoint branches given the root spin. The own observation has law \(t+\sqrt t\,G\). A child field is conditioned through its deleted-neighbor spin, so the even-function consequence of the second tilt gives, for every integer \(r\ge1\), \[ \mathbb E\bigl[|\psi(h_{u\to v}(t))|^r\mid\sigma_v=+1\bigr] =\mathbb E|\psi(Y)|^r\le\theta^r A_r(t). \tag{80}\] Here \(\psi(0)=0\) and \(0<\psi'(z)\le\theta\) imply \(|\psi(z)|\le\theta|z|\).

There is an absolute constant \(K_0\) such that \[ \mathbb E|t+\sqrt t\,G|^r\le e^{K_0r^2}t^{r/4} \qquad(r\ge1\text{ an integer},\ 0<t\le1). \tag{81}\] Indeed, \(|t+\sqrt t\,G|\le\sqrt t(1+|G|)\) in this time range, and \(\mathbb E|G|^r\le(C\sqrt r)^r\). The inequality \((x+y)^r\le2^{r-1}(x^r+y^r)\) for nonnegative \(x,y\) bounds the remaining constant by \(e^{K_0r^2}\), uniformly over integer \(r\). We also used \(t^{r/2}\le t^{r/4}\).

Choose \(K\ge2K_0\) large enough that (77) holds for \(p=1,2\), using (79). Suppose it holds for all positive integers below \(p\), where \(p\ge3\). The triangle inequality in the cavity recursion, followed by the multinomial expansion and the conditional independence just established, bounds \(A_p(t)\) by the expectation of the \(p\)th power of the sum of the absolute values of the own term and the \(b\) child terms. The terms with exponent \(p\) on a single child contribute at most \(b\theta^p A_p(t)\). The term supported on the own observation is at most \(e^{Kp^2/2}t^{p/4}\) by (81).

Every other exponent vector \((r_0,r_1,\ldots,r_b)\) has sum \(p\) and at least two positive entries. Hence every positive entry is below \(p\), and the induction hypothesis applies to its child factors. Using (81) for its own factor, if present, and discarding powers of \(\theta\le1\), its product of moments is at most \[e^{K\sum_{i=0}^b r_i^2}t^{p/4}.\] For such a vector, \(\sum_i r_i^2\le(p-1)^2+1=p^2-2(p-1)\): merging positive entries until only two remain, then moving units to the larger one, can only increase the sum of squares. The sum of all multinomial coefficients in \(b+1\) slots is \((b+1)^p\). Thus \[ (1-b\theta^p)A_p(t) \le e^{Kp^2}t^{p/4} \left[e^{-Kp^2/2}+(b+1)^p e^{-2K(p-1)}\right]. \tag{82}\] The coefficient on the left is positive uniformly over \(p\ge3\): \[1-b\theta^p=1-b^{1-p/2}\ge\kappa_b:=1-b^{-1/2}>0.\] Enlarge the same constant \(K\) once more so that \[2K>\log(b+1),\qquad e^{-9K/2}+(b+1)^3e^{-4K}\le\kappa_b.\] Both terms in brackets in (82) decrease as the integer \(p\) increases from \(3\). Their sum is therefore at most \(\kappa_b\le1-b\theta^p\) for every such \(p\). Absorption proves (77) with this single \(K\), for every integer order and all \(t\le t_0\).

To obtain the tail, put \(L=\log(1/t)\) and choose the integer \(p=\lfloor L/(16K)\rfloor\). If \(L\ge32K\), then \(L/(32K)\le p\le L/(16K)\). Markov’s inequality gives \[\mathbb P(|Y|>t^{1/8}) \le e^{Kp^2-pL/8} \le e^{-pL/16} \le e^{-L^2/(512K)}.\] Take \(t_{\mathrm{tail}}\le\min(t_0,e^{-32K})\). Both \(|Y|^p\) and the indicator of the tail event are even functions of \(Y\), so the spin-tilt identities prove all the conditional assertions as well. ◻

Proof. Write \(Y=h_{j\to i}(t)\) and \(Y'=h'_{j\to i}(t)\) for the edge used to define \(S(t)\). The two cavity equations on this edge and \(|\psi'|\le\theta\) give the earlier comparison \[|Y-Y'|\le\frac{|\Delta h_j|+\theta|\Delta h_i|}{1-\theta^2}.\] Invariance and Cauchy–Schwarz consequently give \[ \mathbb E|Y-Y'|^2\le\frac{V(t)}{(1-\theta)^2}, \qquad |Y-Y'|\le\frac{H_T}{1-\theta}. \tag{85}\] Also \[|(\log a)'(z)| =\frac{2(1-\theta^2)|\tanh z|}{1-\theta^2\tanh^2z} \le2|\tanh z|\le2.\] The global bound gives \(S(t)\le4V(t)/(1-\theta)^2\) and \(|D_e(t)|\le2H_T/(1-\theta)\).

On the event \(|Y|,|Y'|\le t^{1/8}\) the segment between them also lies in that interval, and \(|(\log a)'(z)|\le2t^{1/8}\) there. The mean-value theorem and (85) therefore give \[\mathbb E[D_e(t)^2;\ |Y|,|Y'|\le t^{1/8}] \le\frac{4}{(1-\theta)^2}t^{1/4}V(t).\] For the complementary event the global square-integrable bound gives \(CV(t)\), while the deterministic bound on \(D_e\) and a union bound using (78) in the two marginals give \(C_Te^{-c\log^2(1/t)}\). Thus \[ \mathbb E[D_e(t)^2;\ \max(|Y|,|Y'|)>t^{1/8}] \le\min\{CV(t),C_Te^{-c\log^2(1/t)}\}. \tag{86}\] The argument uses only the two marginal probabilities; it requires no independence between the two fields.

Set \(u=V(t)>0\) and \(J=\log^2(1/t)\ge1\). If \(u\le e^{-cJ/2}\), then \(I(t)\ge cJ/2\), so the first member of the minimum in (86) is at most \(C u I(t)/J\). If \(u>e^{-cJ/2}\), the second member divided by \(u\) is at most \(C_Te^{-cJ/2}\). Since \(J e^{-cJ/2}\) is bounded for \(J\ge1\) and \(I(t)\ge1\), this is at most \(C_T I(t)/J\). Adding the small-field contribution proves (84). ◻

Proof. For \(r\le R\), the estimates \(Q_r\le CrS\) and \(L_r\le V\) give the respective bounds \[CS\sum_{r\le R}r\le CSR^2, \qquad C\sqrt S\sum_{r\le R}\sqrt r\le C\sqrt S R^{3/2}, \qquad \sum_{r\le R}L_r\le VR.\] For the ordinary exponential terms at \(r>R\), use \[ \sum_{r\ge1}r^p e^{-c\omega r} \le C_{p,c}\omega^{-p-1} \quad\left(p\in\left\{0,\tfrac12,1\right\},\ 0<\omega\le1\right). \tag{90}\] For example, compare the sum with \(\int_0^\infty(x+1)^p e^{-c\omega x}\,\mathrm dx\) and put \(y=\omega x\); since \(\omega\le1\), the resulting integral is bounded by \(\omega^{-p-1}\int_0^\infty(y+1)^p e^{-cy}\,\mathrm dy\). This gives \(CS\omega^{-2}\) for the first sum and \(CV\omega^{-1}\) for the third. For the middle sum use \(\sqrt{x+y}\le\sqrt x+\sqrt y\); the ordinary exponential contribution is bounded by \(C\sqrt S\sum_{r\ge1}r^{1/2}e^{-c\omega r/2} \le C\sqrt S\omega^{-3/2}\).

For the exceptional terms, (88) bounds their three contributions by constants times \[V^2\sum_{r\ge1}r^2e^{-(\alpha/2)b^{\gamma r}},\qquad V\sum_{r\ge1}r e^{-(\alpha/4)b^{\gamma r}},\qquad V^2\sum_{r\ge1}e^{-(\alpha/2)b^{\gamma r}},\] respectively. Each series is finite because its exponential exponent grows exponentially in \(r\). In the middle series the square root of the \(V^2\) factor is \(V\), which is why the drift sum has this order. Finally \(\omega^{-1}+\log I\ge1\), so the ceiling in (87) gives \(R\le(C_1+1)(\omega^{-1}+\log I)\). Bounding each power of this sum by a constant times the sum of its corresponding powers proves (89). ◻

Proof. Work on the fixed horizon \([0,T]\) used above. Inserting the noise and drift bounds (42) and (43) into the variance identity (35), and using \(n_r\theta^{2r}=db^{r-1}b^{-r}=d/b\) for \(r\ge1\), gives \[ V'(t)\le C\left( V(t)+\sum_{r\ge1}Q_r(t)+\sum_{r\ge1}L_r(t) +\sqrt{V(t)}\sum_{r\ge1}\sqrt{Q_r(t)}\right) \tag{91}\] for almost every \(t\). The sums in Lemma 23 are finite at the positive times with \(V(t)>0\) where we now use this inequality.

Insert those three bounds into (91). Use \(S\le CV\), and hence \(\sqrt{VS}\le C V\), for all terms except \(S\omega^{-2}\). The bounds \(V\le H_T^2\) imply \(V^{3/2}\le H_TV\) and \(V^2\le H_T^2V\). Finally \[(\log I)^p\le C_p I\qquad(I\ge1,\ 0<p\le2),\] since \(x^p e^{-x}\) is bounded on \(x\ge0\). As \(\omega^{-1}\le \omega^{-3/2}\), all the terms just treated are at most \[C_TV\bigl(I+\omega^{-3/2}\bigr).\] This identifies the worst drift contribution explicitly: \(\sqrt{VS}\,\omega^{-3/2}\le CV\omega^{-3/2}\).

For \(0<t\le\min\{t_*,T\}\), we have \(\omega=\sqrt t\). Lemma 22 gives the remaining estimate \[S\omega^{-2} \le C_TV\left(t^{-3/4} +\frac{I}{t\log^2(1/t)}\right).\] Together with \(I\ge1\), the preceding bounds prove \[ V'(t)\le k_T(t)V(t) \left(1+\log\frac{K_T}{V(t)}\right) \quad\text{for almost every $t\in(0,T]$ with $V(t)>0$}, \tag{92}\] where, after increasing \(C_T\), we may take \[ k_T(t)= \begin{cases} C_T\left(1+t^{-3/4} +\dfrac{1}{t\log^2(1/t)}\right),&0<t\le\min\{t_*,T\},\\[2mm] C_T,&t_*<t\le T. \end{cases} \tag{93}\] For the second branch, \(\omega(t)\ge\sqrt{t_*}>0\) because \(t_*\le1\); there we use \(S\le CV\) also for \(S\omega^{-2}\). Thus that branch is bounded on the whole later interval. If \(T\le t_*\) there is no later interval. The small-time expression is used only for \(t\le t_*\le e^{-1}\), and it is integrable there: \[\int_0^{t_*}t^{-3/4}\,\mathrm dt=4t_*^{1/4},\qquad \int_0^{t_*}\frac{\mathrm dt}{t\log^2(1/t)} =\frac{1}{\log(1/t_*)}.\] In particular \(k_T\in L^1(0,T)\).

We finish with an Osgood-type comparison (Osgood 1898), proving the needed absolutely continuous form directly. Suppose that \(V\) is positive somewhere on \([0,T]\). Choose such a time \(t\), and let \(a<t\) be the left endpoint of its component in the positivity set of \(V\). Continuity and \(V(0)=0\) give \(V(a)=0\). On any closed interval \([u,t]\) with \(a<u<t\), continuity bounds \(V\) away from zero. The function \(z\mapsto\log(1+\log(K_T/z))\) has bounded derivative on the range of \(V\) on that interval. Its composition with the absolutely continuous function \(V\) is therefore absolutely continuous, and the chain rule and (92) give \[(\log I)'=-\frac{V'}{VI}\ge-k_T\quad\text{almost everywhere on }[u,t].\] Integrating yields \[\log I(u)\le\log I(t)+\int_u^t k_T(s)\,\mathrm ds \le\log I(t)+\int_a^t k_T(s)\,\mathrm ds<\infty.\] The last bound is independent of \(u\). But \(V(u)\to V(a)=0\) as \(u\downarrow a\), so \(I(u)\to\infty\) and \(\log I(u)\to\infty\), a contradiction. Hence \(V\) vanishes on \([0,T]\). Since the horizon was arbitrary, it vanishes at every time. ◻

Proof. Fix an auxiliary vertex \(o\). For each integer \(R\ge0\), choose a Borel function \(h_R:[0,1]^{B_R(o)}\to[-1,1]\) representing \[\mathbb E[\Phi(U)_o\mid U|_{B_R(o)}].\] A Borel version exists, and clipping it to \([-1,1]\) preserves this conditional expectation. Let \(G_R\) be the finite automorphism group of the rooted ball \(B_R(o)\), and average: \[q_R(y)=\frac1{|G_R|}\sum_{a\in G_R}h_R(ay).\] Every \(a\in G_R\) extends to a tree automorphism fixing \(o\): outside the ball, match the isomorphic rooted components attached to its boundary. Such an extension preserves the IID law and the ball’s label sigma-field. The assumed almost-sure equivariance therefore makes \(h_R(a(U|_{B_R(o)}))\) another version of the same conditional expectation. Thus \(q_R\) still represents that conditional expectation, and is Borel, bounded by one, and invariant under \(G_R\) on every input.

Apply this same local rule at every vertex. Explicitly, for a rooted isomorphism \(\iota:B_R(o)\to B_R(v)\), put \[A_R(x)_v=q_R\bigl((x_{\iota(u)})_{u\in B_R(o)}\bigr).\] Two choices of \(\iota\) differ by an element of \(G_R\), so the value does not depend on this choice. The maps \(A_R\) are consequently Borel and satisfy \(A_R(gx)=gA_R(x)\) for every input and automorphism. Each \(\iota\) also extends to a tree automorphism taking \(o\) to \(v\). Transporting the conditional expectation by this fixed automorphism shows that \[A_R(U)_v=\mathbb E[\Phi(U)_v\mid U|_{B_R(v)}] \quad\text{almost surely}.\] The ball sigma-fields increase to the full label sigma-field. Upward martingale convergence, together with a countable intersection over vertices and radii, gives \(A_R(U)_v\to\Phi(U)_v\) simultaneously at every vertex almost surely.

On every label input define \[\Psi(x)_v= \begin{cases} 1,&\limsup_{R\to\infty} A_R(x)_v\ge0,\\ -1,&\limsup_{R\to\infty} A_R(x)_v<0. \end{cases}\] Boundedness makes each limsup finite, and this formula defines a Borel spin map. The pointwise equivariance of every \(A_R\) passes through the limsup and the common threshold. On the simultaneous convergence event the limits belong to \(\{-1,1\}\), so \(\Psi(U)=\Phi(U)\) almost surely. ◻

Proof. First suppose \(\theta>0\). The two uniqueness propositions give \(V(t)=0\) for every deterministic time in their respective parameter ranges. Invariance permits any vertex in place of the auxiliary vertex \(o\). Intersecting the full-probability events for all vertices and nonnegative rational times gives \(h_v(q)=h'_v(q)\) simultaneously at those indices. Both field arrays are continuous, so they agree at every time and vertex. Hence their magnetizations agree, and the common innovation identity gives \[X_v(t)=W_v(t)+\int_0^t m_v(s)\,\mathrm ds =W_v(t)+\int_0^t m'_v(s)\,\mathrm ds=X'_v(t).\] Thus \(\widehat{\mathsf Q}\{X=X'\}=1\) as an equality of full path configurations. The use of rational times avoided an uncountable intersection over time.

For \(v\in\mathcal V\) and \(q\in\mathbb Q_{\ge0}\) choose the finite Borel version \[f_{v,q}(w)=\int x_v(q)\,\mathsf K(w,\,\mathrm dx)\] when the absolute integral is finite, and set it to zero otherwise. Kernel integration makes this Borel, and integrability of \(X_v(q)\) ensures that the exceptional inputs have \(\mathsf w\)-measure zero. Taking expectations in (33) and using \(X=X'\) gives \[0=\mathbb E_{\widehat{\mathsf Q}}(X_v(q)-X'_v(q))^2 =2\mathbb E_{\mathsf Q}\mathop{\mathrm{Var}}(X_v(q)\mid W).\] It follows that \(X_v(q)=f_{v,q}(W)\) almost surely. We may impose these identities simultaneously for all the countably many pairs \((v,q)\). Disintegrating them shows that, for \(\mathsf w\)-almost every \(w\), \(\mathsf K(w,\cdot)\) gives mass one to \[\{x\in\mathsf X:x_v(q)=f_{v,q}(w) \text{ for every }v,q\}.\] This set has at most one element, since continuous paths are determined by their rational-time values. Its conditional mass one also makes it nonempty. Thus \(\mathsf K(w,\cdot)\) is a point mass for almost every \(w\).

The conditional point-mass principle for measurable recovery also appears in Kurtz (Kurtz 2014, Lemma 1.3 in arXiv:1305.6747v2). We give a Borel choice of that path, including values on all other inputs. For each \(v\) and integer \(j\ge1\), let \(p_{v,j}(w)\) be the piecewise linear path whose value at \(k2^{-j}\) is \(f_{v,k2^{-j}}(w)\) for every integer \(k\ge0\). These are Borel maps to \(C([0,\infty),\mathbb R)\) with its locally uniform topology. Let \(\mathcal G\) be the set of \(w\) for which, at every vertex, these paths are Cauchy uniformly on every compact interval and \(f_{v,0}(w)=0\). This set is Borel. Indeed, the uniform Cauchy condition on \([0,N]\) can be tested by the supremum over rational times there, and the quantifiers over vertices, integers \(N\), error levels, and pairs of sequence indices are all countable. On \(\mathcal G\) the limits are continuous paths starting at zero. Define \(F_v(w)\) to be this limit for \(w\in\mathcal G\), and let \(F(w)\) be the all-zero path configuration for \(w\notin\mathcal G\). The limit map on the Cauchy set of a Polish space is Borel, so this defines a Borel map \(F:\mathsf W\to\mathsf X\).

For almost every \(w\), the rational coordinates just specified are those of the unique path configuration charged by \(\mathsf K(w,\cdot)\). Uniform continuity of each such path on compact intervals shows that its dyadic linear interpolants converge uniformly there. Consequently \(w\in\mathcal G\) and \(F(w)\) is that configuration. This proves both \(\mathsf K(w,\cdot)=\delta_{F(w)}\) almost everywhere and \(X=F(W)\) in the original law.

Fix now one automorphism \(g\). Invariance of \(\mathsf Q\) and uniqueness of conditional expectations imply \[ f_{v,q}(gw)=f_{g^{-1}v,q}(w) \quad\text{for $\mathsf w$-almost every }w \tag{94}\] for each \(v,q\). Intersecting over those countably many pairs gives all these identities simultaneously for this fixed \(g\). The dyadic interpolants then obey \(p_{v,j}(gw)=p_{g^{-1}v,j}(w)\) for every \(v,j\). On this full-measure set, \(w\in\mathcal G\) if and only if \(gw\in\mathcal G\), because its defining Cauchy conditions are the same after permuting vertices. The all-zero default is invariant as well, so \(F(gw)=gF(w)\) almost surely. The full-measure set in this argument is allowed to depend on \(g\); none is asserted for all automorphisms at once.

Define a spin map on every innovation input by \[ H_v(w)= \begin{cases} \displaystyle\lim_{n\to\infty}\frac{F_v(w)(n)}{n}, &\text{if this limit exists and belongs to $\{-1,1\}$},\\[2mm] 1,&\text{otherwise}. \end{cases} \tag{95}\] Existence of either indicated limit is a Borel condition expressed through countably many error bounds and integer indices. Thus each \(H_v\) is Borel, and so is \(H\) into the countable spin product. On the planted probability space, \[\frac{F_v(W)(n)}n=\frac{X_v(n)}n =\sigma_v+\frac{B_v(n)}n\] simultaneously for the integer indices. For every \(\varepsilon>0\), the Gaussian tail bound gives \[\sum_{n\ge1}\mathbb P(|B_v(n)|>\varepsilon n) \le\sum_{n\ge1}2e^{-\varepsilon^2n/2}<\infty.\] Borel–Cantelli, first for positive rational \(\varepsilon\) and then over the countable vertex set, gives \(B_v(n)/n\to0\) simultaneously for every vertex. Therefore \(H(W)=\sigma\) almost surely. Under product Wiener input, \(H\) has exactly the free zero-field Ising law. For the fixed \(g\) above, the identity \(F(gw)=gF(w)\) makes the sequences in (95) agree after the same permutation. Their limits and their failure cases consequently agree. The same default sign at every site gives \(H(gw)=gH(w)\) almost surely.

Finally, Wiener measure is a Borel probability measure on the Polish space \(C_0([0,\infty),\mathbb R)\). The standard Borel sampling theorem therefore supplies a Borel map \(\xi:[0,1]\to C_0([0,\infty),\mathbb R)\) which sends Lebesgue measure to Wiener measure. For independent uniform labels \((U_v)\), set \[\Phi(U)=H\bigl((\xi(U_v))_{v\in\mathcal V}\bigr).\] Applying the same \(\xi\) at every site gives product Wiener law and commutes pointwise with every vertex permutation. Thus \(\Phi\) is Borel, has law \(\mu_{d,\beta}\), and satisfies \(\Phi(gU)=g\Phi(U)\) almost surely for each fixed \(g\). Lemma 25 replaces \(\Phi\) by a Borel map with the same law that is equivariant on every label input.

When \(\theta=0\), the broadcast law is already a product of fair signs. The rule \(\Phi_v(U)=1\) for \(U_v\le1/2\) and \(\Phi_v(U)=-1\) otherwise gives that law and commutes with every permutation for every input. All choices above are for the fixed pair \((d,\beta)\). Together with the known converse in Section 1, this proves Theorem 1. ◻