Block C — The Ornstein–Uhlenbeck process

Mean: take expectations of the SDE; the noise term has zero mean, so $d\bar x/dt=A\bar x$ and $\bar x(t)=e^{At}\bar x(0)\to0$ when all eigenvalues of $A$ have negative real part. Zero mean is forced by stationarity, not just preserved.

Covariance: Itô's product rule, $d(xx^\top)=(dx)x^\top+x(dx)^\top+(dx)(dx)^\top$ with $(dx)(dx)^\top=\sqrt D\,dW\,dW^\top\sqrt D^\top=D\,dt$: $$d(xx^\top)=(Axx^\top+xx^\top A^\top+D)\,dt+\text{(zero-mean martingale terms)}.$$ Taking expectations, $\dot\Sigma=A\Sigma+\Sigma A^\top+D$. Stationarity $\dot\Sigma=0$ gives the Lyapunov equation $A\Sigma+\Sigma A^\top=-D$, Eq. (15). Read it as a balance: the drift term contracts the covariance (for stable $A$), $D$ injects variance, and $\Sigma$ is where they cancel. Scalar case: $2a\sigma^2+d=0\Rightarrow\sigma^2=d/(2|a|)$.

$A\Sigma+\Sigma A^\top=\int_0^\infty\big(Ae^{At}De^{A^\top t}+e^{At}De^{A^\top t}A^\top\big)dt=\int_0^\infty\frac{d}{dt}\big(e^{At}De^{A^\top t}\big)dt=\big[e^{At}De^{A^\top t}\big]_0^\infty=0-D=-D$, using $e^{At}\to0$ for stable $A$.

Interpretation: the stationary state is $x_t=\int_{-\infty}^te^{A(t-s)}\sqrt D\,dW_s$, a sum of all past noise kicks, each propagated forward and shrunk by $e^{A(t-s)}$; the covariance adds up their (independent) contributions. Discrete-time analogue: the Kalman-filter prediction covariance fixed point $\Sigma=\Phi\Sigma\Phi^\top+Q$ with $\Phi=e^{A\Delta t}$. Uniqueness: the Sylvester equation $A\Sigma+\Sigma A^\top=-D$ has a unique solution iff no two eigenvalues of $A$ sum to zero (Bhatia & Rosenthal 1997), which stable $A$ satisfies automatically.

$\nabla\rho_0=-\Sigma^{-1}x\,\rho_0$. Using the expansion from Problem 6(d), $$L\rho_0=-(\tr A)\rho_0-(Ax)\cdot\nabla\rho_0+\tfrac12\nabla\!\cdot(D\nabla\rho_0) =\rho_0\Big[-\tr A+x^\top A^\top\Sigma^{-1}x-\tfrac12\tr(D\Sigma^{-1})+\tfrac12x^\top\Sigma^{-1}D\Sigma^{-1}x\Big],$$ where the last two terms come from $\nabla\!\cdot(-D\Sigma^{-1}x\,\rho_0)=-\tr(D\Sigma^{-1})\rho_0+x^\top\Sigma^{-1}D\Sigma^{-1}x\,\rho_0$. Now use the Lyapunov equation to eliminate $D$: $\Sigma^{-1}D\Sigma^{-1}=-(\Sigma^{-1}A+A^\top\Sigma^{-1})$, so $\tfrac12x^\top\Sigma^{-1}D\Sigma^{-1}x=-x^\top A^\top\Sigma^{-1}x$ (the two quadratic forms are transposes of the same scalar), cancelling the second term; and $\tr(D\Sigma^{-1})=-\tr(A\Sigma\Sigma^{-1}+\Sigma A^\top\Sigma^{-1})=-2\tr A$, so $-\tfrac12\tr(D\Sigma^{-1})=\tr A$, cancelling the first. Hence $L\rho_0=0$. (Alternatively: a linear SDE with Gaussian initial condition stays Gaussian, and (a) fixes its moments.)

Componentwise, $(A\Sigma+\Sigma A^\top)_{ij}=(a_i+a_j)\Sigma_{ij}=-D_{ij}$. With $a=(-1,2)$: $\Sigma_{11}=-1/(2\cdot(-1))=\tfrac12$, $\Sigma_{22}=-1/(2\cdot2)=-\tfrac14$, $\Sigma_{12}\cdot(-1+2)=0\Rightarrow\Sigma_{12}=0$. So $\Sigma=\diag(\tfrac12,-\tfrac14)$, indefinite. Then $\rho_0\propto\exp(-x_1^2+2x_2^2)$ grows along the unstable direction and is not normalisable. The integral formula of (b) diverges for the same reason. What Eq. (14) gives at a saddle is a formal zero of $L$ (Kwon, Ao & Thouless 2005), describing the local shape of the density near the fixed point along stable directions and its growth rate along unstable ones — not a stationary distribution. Consequently $L^2(\rho_0)$, the adjoint of Problem 7, the contraction bound, and the compactness behind Eq. (7) are all unavailable as stated. This is the paper's open issue (i); its own suggested repair is to localise (window or condition on staying near the saddle), which would restore a normalisable reference measure. High that the gap exists; Medium on how much of the saddle interpretation survives a rigorous treatment.

Now $a_1+a_2=0$, so the $(1,2)$ equation reads $0\cdot\Sigma_{12}=0$: $\Sigma_{12}$ is undetermined and the Sylvester equation has a one-parameter family of solutions. This is precisely the excluded case "eigenvalues related by sign reversal" mentioned after Eq. (15). The symmetric saddle $\pm1$ is the natural first example one reaches for, and it is exactly the degenerate one; Problem 12 uses unequal rates for this reason.

$\nabla\log\rho_0=-\Sigma^{-1}x$, so $b_{\rm rev}=-Ax-D\Sigma^{-1}x$. From $D=-(A\Sigma+\Sigma A^\top)$: $D\Sigma^{-1}=-A-\Sigma A^\top\Sigma^{-1}$, hence $$b_{\rm rev}(x)=-Ax+Ax+\Sigma A^\top\Sigma^{-1}x=\Sigma A^\top\Sigma^{-1}\,x.$$ The reversed process is linear with drift matrix $A_{\rm rev}=\Sigma A^\top\Sigma^{-1}$. Intuition: naïve reversal ($-A$) would run away from the fixed point; the score term $D\nabla\log\rho_0$ pulls back toward high stationary density; for a Gaussian the two combine into a linear drift that is $A^\top$ in $\Sigma$-adapted coordinates.

Same $D$ is part of the reversal theorem. Lyapunov for the reversed drift: $A_{\rm rev}\Sigma+\Sigma A_{\rm rev}^\top=\Sigma A^\top+A\Sigma=-D$, so $\Sigma$ solves the reversed Lyapunov equation too. $A_{\rm rev}$ is similar to $A^\top$, hence has the eigenvalues of $A$; in particular it is stable, and the reversed process is a legitimate stationary OU process. The reversed process is the one whose Koopman operator is $\K^\dagger_\tau$ (Problem 7d).

$\Sigma A^\top\Sigma^{-1}=A\iff\Sigma A^\top=A\Sigma\iff(A\Sigma)^\top=A\Sigma$. For $\tilde M$: $\tilde M^\top=\Sigma^{1/2}e^{A^\top\tau}\Sigma^{-1/2}$, and $\tilde M^\top=\tilde M\iff e^{A\tau}=\Sigma e^{A^\top\tau}\Sigma^{-1}=e^{\Sigma A^\top\Sigma^{-1}\tau}=e^{A_{\rm rev}\tau}$. If $A\Sigma$ is symmetric this holds for every $\tau$; conversely, differentiating at $\tau=0$ gives $A=A_{\rm rev}$. So $$\text{reversible}\iff A_{\rm rev}=A\iff A\Sigma=\Sigma A^\top\iff\tilde M=\tilde M^\top\iff\K^\dagger_\tau=\K_\tau .$$ The antisymmetric part of $A\Sigma$ is the probability current of the stationary state; its vanishing is detailed balance (Pavliotis 2014, §4.6). When it vanishes the singular vectors of $\tilde M$ are its eigenvectors, $u_i=v_i$, and the paper's predictive/retrospective distinction collapses. The distinction is therefore a claim about broken detailed balance in the stimulus dynamics.

Write $\Sigma=\begin{pmatrix}s_{11}&s_{12}\\s_{12}&s_{22}\end{pmatrix}$. Then $A\Sigma=\begin{pmatrix}-s_{11}&-s_{12}\\cs_{11}-2s_{12}&cs_{12}-2s_{22}\end{pmatrix}$ and $A\Sigma+(A\Sigma)^\top=-I$ gives $-2s_{11}=-1$, $-3s_{12}+cs_{11}=0$, $2cs_{12}-4s_{22}=-1$, so $$s_{11}=\tfrac12,\qquad s_{12}=\tfrac c6,\qquad s_{22}=\tfrac{c^2}{12}+\tfrac14,\qquad A\Sigma=\begin{pmatrix}-\tfrac12&-\tfrac c6\\ \tfrac c6&-\tfrac12\end{pmatrix}.$$ $A\Sigma$ is symmetric iff $c=0$. So the process is reversible only when $A$ is diagonal (then $D=I$ and $A$ commute and $\Sigma$ is diagonal); any coupling $c\ne0$ makes it irreversible, with probability current $\propto c$. $A$ is lower-triangular, so its eigenvalues are $-1$ and $-2$ for every $c$, real and distinct, with real eigenvectors — the paper's standing assumption holds. This is the working example for Problems 12–13: a stable, real-spectrum OU for which $u_i\ne v_i$.

Solving the SDE from $t$ to $t+\tau$: $X_{t+\tau}=e^{A\tau}X_t+\eta$ with $\eta=\int_t^{t+\tau}e^{A(t+\tau-s)}\sqrt D\,dW_s$ independent of $X_t$, $\operatorname{Cov}(\eta)=\Sigma_\tau=\int_0^\tau e^{At}De^{A^\top t}dt$. Then $\operatorname{Cov}(X_t,X_{t+\tau})=\E[X_tX_t^\top]e^{A^\top\tau}=\Sigma e^{A^\top\tau}$. Gaussian conditioning $\E[Y\mid X=s]=C_{YX}C_{XX}^{-1}s$: $$\E[X_{t+\tau}\mid X_t=s]=e^{A\tau}\Sigma\,\Sigma^{-1}s=e^{A\tau}s,\qquad\E[X_t\mid X_{t+\tau}=r]=\Sigma e^{A^\top\tau}\,\Sigma^{-1}r,$$ the second using $\operatorname{Cov}(X_{t+\tau})=\Sigma$ by stationarity. The backward conditional is linear regression of the past on the future.

$\ip{e^{A\tau}a,b}_{\Sigma^{-1}}=a^\top e^{A^\top\tau}\Sigma^{-1}b$ and $\ip{a,\Sigma e^{A^\top\tau}\Sigma^{-1}b}_{\Sigma^{-1}}=a^\top\Sigma^{-1}\Sigma e^{A^\top\tau}\Sigma^{-1}b$ — equal. And $\Sigma e^{A^\top\tau}\Sigma^{-1}=e^{\Sigma A^\top\Sigma^{-1}\tau}=e^{A_{\rm rev}\tau}$ (conjugation commutes with the exponential). So the backward mean is the mean of the reversed OU of Problem 9, and the $\Sigma^{-1}$-adjoint is the finite-dimensional shadow of the $\rho_0$-adjoint: on linear observables $v^\top x$, $\ip{v^\top x,w^\top x}_{\rho_0}=v^\top\Sigma w$, so adjointness in $L^2(\rho_0)$ restricted to linear functions is adjointness in the $\Sigma$ (dually $\Sigma^{-1}$) inner product. $\Sigma^{-1}$ is the Fisher metric of the Gaussian family; "backward = Fisher-adjoint of forward" is a coordinate-free way to say it.

$q$ is Gaussian in $x$ (a Gaussian mixture over $y$ of Gaussians with means linear in $y$, all of fixed covariance, is Gaussian), with mean $\E_q[x\mid s]=\Sigma e^{A^\top\tau}\Sigma^{-1}\E[X_{t+\tau}\mid X_t=s]=Ms$, $M=\Sigma e^{A^\top\tau}\Sigma^{-1}e^{A\tau}$. For linear $f$, $F_\tau f(s)=\E_q[v^\top x\mid s]=v^\top Ms$. The eigen-equation $F_\tau f=\lambda^2f$ becomes $v^\top M=\lambda^2v^\top$, i.e. $M^\top v=\lambda^2v$: $$e^{A^\top\tau}\Sigma^{-1}e^{A\tau}\Sigma\;v=\lambda^2v,$$ the first equation of (18). For $B_\tau f(r)=\E\big[\E[f(X_{t+\tau})\mid X_t]\mid X_{t+\tau}=r\big]$: inner mean $e^{A\tau}X_t$, outer $e^{A\tau}\Sigma e^{A^\top\tau}\Sigma^{-1}r=:Nr$, so $B_\tau(u^\top x)=u^\top Nx$ and $N^\top u=\lambda^2u$: $$\Sigma^{-1}e^{A\tau}\Sigma e^{A^\top\tau}\;u=\lambda^2u,$$ the second equation of (18). The covariance of $q$ never entered: it would only appear when $F_\tau$ acts on quadratic and higher observables, which is Problem 11.

The second equation by direct integration. With the forward Gaussian kernel $p_\tau(y\mid x)$, whose mean is $e^{A\tau}x$ (part (a)), and the backward kernel $\tilde p_\tau(x\mid r)$, whose mean is $\Sigma e^{A^\top\tau}\Sigma^{-1}r$ (part (b)), $$B_\tau(u^\top x)(r)=\int\tilde p_\tau(x\mid r)\Big[\int p_\tau(y\mid x)\,u^\top y\,dy\Big]dx=\int\tilde p_\tau(x\mid r)\,u^\top e^{A\tau}x\,dx=u^\top e^{A\tau}\,\Sigma e^{A^\top\tau}\Sigma^{-1}r=u^\top Nr .$$ Second equality: the integral of a linear function against a probability density is the linear function of the density's mean (linearity of the integral). Third: the same rule for the outer kernel, after taking the constant row vector $u^\top e^{A\tau}$ outside the integral. The covariances of both kernels never enter because only linear functions are integrated, which is the integral form of the remark that only the means matter here.

$\tilde M^\top\tilde M=\Sigma^{1/2}e^{A^\top\tau}\Sigma^{-1}e^{A\tau}\Sigma^{1/2}$, so $M^\top=e^{A^\top\tau}\Sigma^{-1}e^{A\tau}\Sigma=\Sigma^{-1/2}(\tilde M^\top\tilde M)\Sigma^{1/2}$. Hence $M^\top v=\lambda^2v\iff\tilde M^\top\tilde M(\Sigma^{1/2}v)=\lambda^2(\Sigma^{1/2}v)$: $w=\Sigma^{1/2}v$ is a right singular vector of $\tilde M$ and $\lambda$ the corresponding singular value. Similarly $N^\top=\Sigma^{-1/2}(\tilde M\tilde M^\top)\Sigma^{1/2}$, so $w'=\Sigma^{1/2}u$ is a left singular vector of $\tilde M$. In whitened coordinates $z=\Sigma^{-1/2}x$ (identity covariance), the singular functions are simply $v_i(x)=v_i^\top x=w_i^\top z$ and $u_i(x)=w_i'^\top z$: the projections of the whitened state onto the right and left singular vectors of the whitened one-step map $\tilde M$, which is the transition matrix in $z$-coordinates ($z_{t+\tau}=\tilde Mz_t+\text{noise}$). That is the abstract's statement.

Bound: by stationarity $\Sigma=e^{A\tau}\Sigma e^{A^\top\tau}+\Sigma_\tau$ (propagate the old covariance and add the new noise), so $\tilde M\tilde M^\top=\Sigma^{-1/2}(\Sigma-\Sigma_\tau)\Sigma^{-1/2}=I-\Sigma^{-1/2}\Sigma_\tau\Sigma^{-1/2}\preceq I$. All singular values are $\le1$, matching Problem 7(b), and $\lambda_i=1$ only along directions receiving no noise over the horizon. In Problem 14 the same $\tilde M$ appears as the canonical-correlation matrix between $x_t$ and $x_{t+\tau}$ and the $\lambda_i$ as canonical correlations.

With $\xi=x/\sigma$: $g_n'=\mathrm{He}_n'/\sigma$, $g_n''=\mathrm{He}_n''/\sigma^2$, so $$L^\dagger g_n=-ax\frac{\mathrm{He}_n'}{\sigma}+\frac{\mathrm{He}_n''}{2\sigma^2}=\frac{1}{2\sigma^2}\Big[\mathrm{He}_n''-2a\sigma^2\,\xi\,\mathrm{He}_n'\Big]=\frac{1}{2\sigma^2}\Big[\mathrm{He}_n''-\xi\mathrm{He}_n'\Big]=\frac{-n\,\mathrm{He}_n}{2\sigma^2}=-na\,g_n,$$ using $2a\sigma^2=1$. Then $\K_\tau g_n=e^{L^\dagger\tau}g_n=e^{-na\tau}g_n$. The eigenfunctions are orthogonal in $L^2(\rho_0)$ (Hermite polynomials are orthogonal under the Gaussian weight), $g_0=1$ is the trivial pair, and $g_1\propto x$ is the subdominant one with eigenvalue $e^{-a\tau}$, exactly $\tilde M=e^{-a\tau}$ for the 1-D case. Degree $n$ decays $n$ times faster. Deterministic-map intuition from Problem 0(e): monomials $x^n$ were eigenfunctions of $x\mapsto e^{-a\tau}x$; noise replaces them by Hermite polynomials (the monomials' Gaussian orthogonalisation) but keeps the eigenvalues $e^{-na\tau}$.

$\K_\tau f(x)=\E[f(e^{A\tau}x+\eta)]$ with $\eta$ Gaussian independent of $x$. If $f$ is a polynomial of degree $\le k$, expanding $f(e^{A\tau}x+\eta)$ and taking the expectation over $\eta$ yields a polynomial in $x$ of degree $\le k$ (moments of $\eta$ are constants). The same holds for $\K^\dagger_\tau$, which is the Koopman operator of the reversed OU (Problem 9). So the flag of subspaces $\mathcal P_0\subset\mathcal P_1\subset\mathcal P_2\subset\cdots$ (polynomials of degree $\le k$) is invariant under $F_\tau$ and $B_\tau$; in a basis graded by degree the operators are block upper-triangular, and the spectrum is the union of the spectra of the diagonal blocks (degree-$k$ homogeneous parts modulo lower degree). The degree-$1$ block is exactly $M^\top$ (resp. $N^\top$) of Problem 10, with eigenvalues $\lambda_i^2$ = the squared singular values of $\tilde M$. In whitened coordinates the degree-$k$ block of $\K_\tau$ modulo lower degree is the symmetric tensor power $\mathrm{Sym}^k\tilde M$, so the degree-$k$ block of $F_\tau$ has eigenvalues equal to products $\lambda_{i_1}^2\cdots\lambda_{i_k}^2$. High for the invariant-flag statement; the spectrum of OU semigroups is worked out rigorously in Metafune, Pallara & Priola, J. Funct. Anal. 196 (2002). Polynomials are dense in $L^2(\rho_0)$, so nothing is missed.

For stable $A$ every $\lambda_i\le1$ (Problems 7b, 10d), so products of $k\ge2$ of them are $\le\lambda_1^{2k}\le\lambda_1^2$, with equality only in degenerate cases. The degree-$0$ block is the trivial eigenvalue $1$, the degree-$1$ block contains $\lambda_1^2$, and every higher block is smaller. Hence the non-trivial eigenfunction of $F_\tau$ with eigenvalue closest to one is $v_1^\top x$: linear. That is the paper's claim after Eq. (18).

At a saddle the argument breaks in two places. First, $\rho_0$ is not normalisable (Problem 8d), so there is no $L^2(\rho_0)$ in which to take adjoints, and $F_\tau$ is only formally defined. Second, even formally, the "singular values" of Eq. (18) straddle $1$ (Problem 12 finds $0.75$ and $1.04$ at $\tau=0.25$), so a degree-$2$ product $\lambda_1\lambda_2$ can be closer to one than either linear eigenvalue. The paper's rule "subdominant = closest to one" then does not obviously select a linear function. The Supplement (Sec. 3) states that it does not treat saddle points. Medium — a real gap, but one that the paper's proposed localisation might close.

Central differences on $[-7,7]$ with $801$ points: top eigenvalues of the discretised $L^\dagger$ are $0,-1,-2,-3,-4$ (to three decimals), and the corresponding eigenvectors have correlation $1.000$ with $\mathrm{He}_n(x/\sigma)$ on $|x|<3$. The discretisation is non-symmetric (as $L^\dagger$ is under Lebesgue measure) but has real spectrum, as it must: conjugating by $\rho_0^{1/2}$ makes it symmetric, which is the reversibility of 1-D OU.

import numpy as np
from numpy.polynomial.hermite_e import hermeval
a = 1.0; sigma = np.sqrt(1/(2*a))
N, L = 801, 7.0; x = np.linspace(-L, L, N); h = x[1]-x[0]
D1 = (np.eye(N,k=1)-np.eye(N,k=-1))/(2*h); D2 = (np.eye(N,k=1)-2*np.eye(N)+np.eye(N,k=-1))/h**2
Ldag = -a*np.diag(x)@D1 + 0.5*D2
w, V = np.linalg.eig(Ldag); idx = np.argsort(-w.real)[:5]
print(np.round(w[idx].real, 3))          # [ 0. -1. -2. -3. -4.]

$A=\begin{pmatrix}-1&1.5\\0&0.5\end{pmatrix}$, and solve_sylvester(A, A.T, -D) returns $\Sigma=\begin{pmatrix}-4&-3\\-3&-1\end{pmatrix}$ with eigenvalues $-5.85$ and $0.85$: indefinite, residual $10^{-16}$. (Eigenvalues $-1$ and $0.5$ do not sum to zero, so the solution is unique.)

Here $\ell_u$ is the left eigenvector of $A$ for the unstable eigenvalue (row of $S^{-1}$), which is orthogonal to $e_s$ by biorthogonality. All eigenvalues are real over the whole range tested (max imaginary part $0$). Note the eigenvalues: one grows like $e^{2b\tau}$ and one decays like $e^{-2a\tau}$ (their product is $\det e^{A\tau}{}^2=e^{2(b-a)\tau}$), so for large $\tau$ neither is close to one — the paper's selection rule "closest to unity" is only meaningful at short horizons.

For large $\tau$, $e^{A\tau}\approx e^{b\tau}e_u\ell_u^\top$ (dominant unstable mode; $\ell_u^\top e_u=1$). Then $$M^\top=e^{A^\top\tau}\Sigma^{-1}e^{A\tau}\Sigma\approx e^{2b\tau}(e_u^\top\Sigma^{-1}e_u)\;\ell_u\,(\ell_u^\top\Sigma),$$ a rank-one matrix $c\,\ell_u\beta^\top$ with $\beta=\Sigma\ell_u$. Its non-zero eigenvector is $\ell_u$, so $v_1\to\ell_u\perp e_s$: the predictive filter becomes orthogonal to the stable subspace, i.e. it ignores the component of the state that will decay and reads out the coefficient of the unstable mode — the only thing that determines the far future. Similarly $$N^\top=\Sigma^{-1}e^{A\tau}\Sigma e^{A^\top\tau}\approx e^{2b\tau}(\ell_u^\top\Sigma\ell_u)\;(\Sigma^{-1}e_u)\,e_u^\top,$$ whose dominant eigenvector is $\Sigma^{-1}e_u$ and whose other eigenvector, for the small eigenvalue, must lie in the null space of the rank-one part: $e_u^\top u_2=0$, so $u_2\perp e_u$. The retrospective filter ignores the unstable component — which carries no information about where the trajectory came from, since everything gets amplified along $e_u$ — and reads out the stable component, whose present value is the decayed memory of the past. Left/right eigenvector biorthogonality of $A$ is doing all the work; $\Sigma$ only fixes the finite-$\tau$ corrections. High for the 2-D mechanism; the paper flags (open issue iii) that with several unstable directions the picture is not established.

• Eq. (18) itself was derived (Problem 10) from Gaussian conditioning with $\Sigma=\operatorname{Cov}(X_t)$ and from the $\rho_0$-adjoint. With indefinite $\Sigma$, "$\E[X_t\mid X_{t+\tau}]$" under a non-normalisable $\rho_0$ has no probabilistic meaning; the matrices $M^\top$, $N^\top$ are the formal continuation of the stable-case formulas.
• The interpretation of eigenvalues as $\lambda^2\le1$, of singular functions as members of an orthonormal basis, and of sign-thresholding as approximately optimal for Eq. (4) (Problem 5a) all used the contraction on $L^2(\rho_0)$ and the balance constraint under a probability measure. None applies.
• Whitening ($\Sigma^{1/2}$) is unavailable; the reduction to an SVD of a real matrix is lost, and realness of the eigenvalues in (b) is an observed fact for this example, not a theorem.
• What survives: the limits in (c) depend only on the eigenstructure of $A$, so the qualitative claim "predictive $\perp$ stable, retrospective $\perp$ unstable" is robust. What would make the intermediate-$\tau$ objects rigorous is a normalisable reference measure concentrated near the saddle — a windowed, discounted, or committor-conditioned process — which is exactly what the paper's open issue (i) proposes. Medium that such a construction reproduces Eq. (18) up to corrections; it has not been shown.