Block B — Continuous state space: generators and adjoints
For a diffusion the Koopman semigroup is generated by a second-order differential operator. This block derives it, shows it is the Lebesgue adjoint of Fokker–Planck, and then builds the $\rho_0$-weighted adjoint that the paper actually uses. Interleaved question sets: Q1 on how observables are dual to densities, Q2 on the round-trip operators $\K^\dagger\K$ and $\K\K^\dagger$ and the sibling trajectory, and a closing coda (Q3) on how much of the theory survives without stationarity.
Problem 6 — Backward Kolmogorov from Itô
Goal: derive the paper's Eq. (12) and understand why it is called $L^\dagger$.
(a) Apply Itô's lemma to $g(X_t)$ for smooth $g$ and identify the drift of $dg$.
Solution
Itô: $dg=\nabla g\cdot dX+\tfrac12\,dX^\top(\nabla^2g)\,dX$. Substituting $dX=b\,dt+\sqrt D\,dW$ and using $dW_idW_j=\delta_{ij}dt$, $dt\,dW=0$, $dt^2=0$: $dX^\top\nabla^2g\,dX=\tr(\sqrt D^\top\nabla^2g\sqrt D)\,dt=\tr(D\nabla^2g)\,dt$. Hence $$dg=\Big[b\cdot\nabla g+\tfrac12\tr(D\nabla^2g)\Big]dt+\nabla g^\top\sqrt D\,dW.$$ The bracket is the drift; the second term has zero mean.
(b) Take conditional expectations to derive $\partial_tg=L^\dagger g$ with $L^\dagger g=b\cdot\nabla g+\tfrac12\nabla\!\cdot(D\nabla g)$ for $g(x,t)=\E[g_0(X_t)\mid X_0=x]$. Say where the Markov property is used.
Solution
For constant $D$, $\tfrac12\tr(D\nabla^2g)=\tfrac12\nabla\!\cdot(D\nabla g)$, so write the drift as $L^\dagger g$. Integrating (a) from $0$ to $t$ and conditioning on $X_0=x$, the stochastic integral has zero conditional expectation (it is a martingale), giving Dynkin's formula $$\E[g_0(X_t)\mid X_0=x]=g_0(x)+\E\Big[\int_0^t(L^\dagger g_0)(X_s)\,ds\,\Big|\,X_0=x\Big].$$ Differentiating in $t$: $\partial_t\,\K_tg_0(x)=\E[(L^\dagger g_0)(X_t)\mid X_0=x]=\K_t(L^\dagger g_0)(x)$.
Why this is not yet a PDE. The right-hand side applies $L^\dagger$ to the initial function $g_0$ and then propagates; it never mentions the unknown $g(\cdot,t)=\K_tg_0$. A PDE in $x$ needs $\partial_tg=L^\dagger(\K_tg_0)=L^\dagger g(\cdot,t)$, i.e. $\K_tL^\dagger=L^\dagger\K_t$.
Why they commute. $L^\dagger$ is the generator of $\K$: by Dynkin at small time, $L^\dagger f=\lim_{h\to0}(\K_hf-f)/h$ for smooth $f$, so think of $L^\dagger$ as "$\K_{dt}$". The semigroup law $\K_t\K_h=\K_h\K_t=\K_{t+h}$ (Chapman–Kolmogorov, Problem 1(d), i.e. the Markov property) then gives the commutation at once: $$\K_tL^\dagger g_0=\lim_{h\to0}\K_t\frac{\K_h-I}{h}\,g_0=\lim_{h\to0}\frac{\K_h-I}{h}\,\K_tg_0=L^\dagger\K_tg_0 .$$ Hence $\partial_tg=L^\dagger g$, $g(x,0)=g_0(x)$: the paper's Eq. (12), whose solution operator is by definition $e^{L^\dagger t}$, Eq. (13). (Because $L^\dagger$ is unbounded, $e^{L^\dagger t}$ is not a power series; the difference quotient is what makes the commutation rigorous, and the last step needs $\K_tg_0$ smooth enough for $L^\dagger$ to act on it, which holds for OU with non-degenerate $D$. Note that citing Eq. (13) itself here would be circular.)
A route that avoids the commutation step. Fix the horizon $t$ and set $M_s=g(X_s,\,t-s)$ for $0\le s\le t$. By the Markov property and time-homogeneity, $M_s=\E[g_0(X_t)\mid\mathcal F_s]$, which is a martingale (tower property). Itô applied to $g(X_s,t-s)$ in both arguments gives $$dM_s=\big(-\partial_tg+L^\dagger g\big)(X_s,t-s)\,ds+\nabla g^\top\sqrt D\,dW_s .$$ A martingale has no $ds$ term, and $X_s$ can sit anywhere, so $\partial_tg=L^\dagger g$ pointwise. Here the Markov property is used exactly once, in the identity $\E[g_0(X_t)\mid\mathcal F_s]=g(X_s,t-s)$. (Øksendal, Stochastic Differential Equations, Thm 8.1.1; Pavliotis 2014, §2.5.) The same identity by direct integration: $M_s=\int p_{t-s}(z\mid X_s)\,g_0(z)\,dz$ (definition of the transition density), and for $r<s$ the martingale property is Chapman–Kolmogorov, $\E[M_s\mid\mathcal F_r]=\int p_{s-r}(x\mid X_r)\Big[\int p_{t-s}(z\mid x)\,g_0(z)\,dz\Big]dx=\int\Big[\int p_{s-r}(x\mid X_r)\,p_{t-s}(z\mid x)\,dx\Big]g_0(z)\,dz=\int p_{t-r}(z\mid X_r)\,g_0(z)\,dz=M_r$: swap the order of integration (Fubini), then compose the two kernels (Problem 1(d)).
(c) Show by integration by parts that $L^\dagger$ is the Lebesgue adjoint of the Fokker–Planck operator $L$, assuming all boundary terms vanish. Note which term changes sign and which does not.
Solution
The one rule. For a scalar $\phi$ and a vector field $F$ on $\R^n$, the product rule is $\nabla\!\cdot(\phi F)=\nabla\phi\cdot F+\phi\,\nabla\!\cdot F$. Integrating over $\R^n$, the left side is a total divergence and equals the flux of $\phi F$ through a sphere of radius $R\to\infty$ (the boundary term). Assuming it vanishes, $$\int\phi\,(\nabla\!\cdot F)\,dx=-\int\nabla\phi\cdot F\,dx :$$ move the divergence off $F$ onto $\phi$ as a gradient, at the cost of one minus sign. In one dimension this is $\int\phi F'=-\int\phi'F$.
Drift term (one application). Take $\phi=g$, $F=bp$: $$\int g\,\big[-\nabla\!\cdot(bp)\big]\,dx=+\int\nabla g\cdot(bp)\,dx=\int(b\cdot\nabla g)\,p\,dx .$$ The minus sign built into $L$ cancels the one from the rule; the last step regroups scalars, $\nabla g\cdot(bp)=(b\cdot\nabla g)\,p$.
Diffusion term (two applications). First $\phi=g$, $F=D\nabla p$: $\int g\,\nabla\!\cdot(D\nabla p)\,dx=-\int\nabla g\cdot D\nabla p\,dx$. Use symmetry of $D$ to move it onto the other gradient: $\nabla g\cdot D\nabla p=\sum_{ij}\partial_ig\,D_{ij}\,\partial_jp=(D\nabla g)\cdot\nabla p$. Then $\phi=p$, $F=D\nabla g$: $$-\int(D\nabla g)\cdot\nabla p\,dx=+\int p\,\nabla\!\cdot(D\nabla g)\,dx .$$ Two minus signs, net plus. Multiplying by $\tfrac12$ and adding the drift term, $$\int g\,(Lp)\,dx=\int(L^\dagger g)\,p\,dx,\qquad L^\dagger g=b\cdot\nabla g+\tfrac12\nabla\!\cdot(D\nabla g).$$ Since $D$ is constant, $\nabla\!\cdot(D\nabla g)=\sum_{ij}D_{ij}\partial_i\partial_jg=\tr(D\nabla^2g)$, matching the drift in (a).
Boundary terms. Each dropped term is $\lim_{R\to\infty}\oint_{|x|=R}\phi F\cdot n\,dS$. Every $F$ above contains $p$ or $\nabla p$, and $p$ is a density (Gaussian for OU), so these vanish provided $g$ and its derivatives grow at most polynomially, which covers the linear observables of Section 3.
Signs. The first-order (advective) part flips sign under the adjoint: one integration by parts, one minus. (It is anti-self-adjoint only when $\nabla\!\cdot b=0$; in general $(b\cdot\nabla)^\dagger g=-b\cdot\nabla g-(\nabla\!\cdot b)g$, and for OU $\nabla\!\cdot(Ax)=\tr A\ne0$. The extra zeroth-order term is what makes $L$ act on densities and $L^\dagger$ on observables.) The second-order (diffusive) part picks up two minus signs and keeps its sign: it is self-adjoint. Irreversibility can therefore only enter through the drift; pure diffusion is always reversible. Problem 9 makes this quantitative for the OU process.
(d) Specialise to the OU drift $b(x)=Ax$ and recover the paper's Eqs. (11)–(13).
Solution
$Lp=-\nabla\!\cdot(Axp)+\tfrac12\nabla\!\cdot(D\nabla p)$ (Eq. 11), $L^\dagger g=(Ax)\cdot\nabla g+\tfrac12\nabla\!\cdot(D\nabla g)$ (Eq. 12), $\K_\tau=e^{L^\dagger\tau}$ (Eq. 13). Expanding the drift term of $L$: $-\nabla\!\cdot(Axp)=-(\tr A)\,p-(Ax)\cdot\nabla p$, which is convenient in Problem 8(c).
In what sense are observables dual to densities? Throughout the set "duality" between observables and densities is meant in the linear-operator sense, and it comes in three levels of care. The basic object is the pairing $$\ip{f,\rho}=\int f(x)\,\rho(x)\,dx=\E_{X\sim\rho}[f(X)] :$$ it takes an (observable, density) pair to a scalar — the expected value of the measurement $f$ under the distribution $\rho$ — and it is bilinear (linear in $f$ because expectation is linear; linear in $\rho$ because mixing distributions mixes expectations) and nondegenerate ($\ip{f,\rho}=0$ for all $\rho$ forces $f=0$, and symmetrically). Nondegeneracy is what earns the name "duality pairing" rather than merely "bilinear form": each density is the linear functional $f\mapsto\E_\rho[f]$ on observables, and each observable is the linear functional $\rho\mapsto\E_\rho[f]$ on densities. One caveat about "linear in $\rho$": densities themselves form a convex set, not a vector space (nonnegative, integrate to $1$); the duality is really with their linear span, the signed measures. This matters here: the interesting singular functions on the density side are exactly such differences — mean-zero objects like $\rho_A-\rho_B$, not densities.
Level two: the duality is asymmetric. In the Banach-space sense the classical statement is $(L^1)^*=L^\infty$ — every continuous linear functional on integrable densities is integration against a bounded observable — but the converse fails: $(L^\infty)^*$ is strictly larger than $L^1$. So strictly, observables are the dual of densities, and $L^1$ is only the predual of $L^\infty$. The transfer operators inherit this asymmetry. The fundamental identity $$\ip{\K_\tau f,\rho}=\ip{f,\P_\tau\rho}$$ is one probabilistic fact — $\E[f(X_\tau)]$ with $X_0\sim\rho$ — computed two ways: condition on the start (push the observable backward), or push the density forward and integrate at the end. It says $\K_\tau$ and $\P_\tau$ are adjoint with respect to the pairing; the strictly correct reading is that $\K_\tau$ (on $L^\infty$) is the adjoint of $\P_\tau$ (on $L^1$), while calling $\P_\tau$ the adjoint of $\K_\tau$ uses the extra fact that the adjoint of $\K_\tau$ happens to preserve the nice subspace $L^1\subset(L^\infty)^*$ — which it does, because pushing forward an honest density gives an honest density.
Level three: the paper collapses the duality into one Hilbert space. A Hilbert space is self-dual — the Riesz representation theorem says every continuous linear functional is an inner product against some element. Working in $L^2(\rho_0)$, the dictionary between the two pictures is $f\leftrightarrow f\rho_0$: the observable $f$ corresponds to the signed measure $f\rho_0\,dx$, "$\rho_0$ reweighted by $f$". Under this dictionary the naive pairing becomes the inner product, $\ip{f,g\rho_0}=\int fg\,\rho_0\,dx=\ip{f,g}_{\rho_0}$, the observable/density duality becomes an identification, and the abstract adjoint becomes the computable operator of Problem 7(c): $\K^\dagger_\tau=\rho_0^{-1}\P_\tau\rho_0$ is exactly density evolution read through the dictionary. This collapse is load-bearing: an SVD needs an adjoint, an adjoint needs an inner product, and the choice of inner product decides which identification of densities with observables has been made. The $\rho_0$ weight is the choice under which reversibility becomes literal self-adjointness of $\K_\tau$ and the constant observable $\mathbf 1$ and the stationary density $\rho_0$ become the same element — the trivial singular pair of Problem 7(b).
Q1(i) Verify that $\ip{f,\rho}=\E_\rho[f]$ is bilinear and nondegenerate.
Answer
Bilinearity. In $f$: $\E_\rho[\alpha f+\beta g]=\alpha\E_\rho[f]+\beta\E_\rho[g]$ is linearity of the integral. In $\rho$: $\int f\,(\alpha\rho_1+\beta\rho_2)\,dx=\alpha\ip{f,\rho_1}+\beta\ip{f,\rho_2}$, again linearity of the integral — noting that $\alpha\rho_1+\beta\rho_2$ is in general a signed measure, so the statement lives on the linear span of the densities. Nondegeneracy. If $\ip{f,\rho}=0$ for all densities $\rho$, test against densities concentrating on a point $x$ (e.g. narrow Gaussians): continuity of $f$ gives $f(x)=0$ for every $x$. Conversely if $\E_\rho[f]=0$ for all bounded $f$, take $f=\operatorname{sign}(\rho)$ (or test against indicators) to force $\rho=0$.
Q1(ii) In which direction is the Banach duality between $L^1$ and $L^\infty$ exact, and what extra fact lets one nevertheless call $\P_\tau$ "the adjoint of $\K_\tau$"?
Answer
$(L^1)^*=L^\infty$ is exact: every continuous linear functional on $L^1$ is $\rho\mapsto\int f\rho$ for a unique $f\in L^\infty$. The converse fails — $(L^\infty)^*\supsetneq L^1$. Hence the clean statement is that $\K_\tau$ is the Banach adjoint of $\P_\tau$. To reverse the roles one must check that the adjoint of $\K_\tau$, a priori an operator on the big space $(L^\infty)^*$, maps the subspace $L^1$ into itself; it does, because the adjoint acts on a density by pushing it forward through the transition kernel, and the pushforward of an integrable density is an integrable density (Problem 1(a)). With that check, $\P_\tau$ is the restriction of $\K_\tau^*$ to $L^1$ — the pre-adjoint.
Q1(iii) Under the dictionary $f\leftrightarrow f\rho_0$, what do the naive pairing and the pair $(\mathbf 1,\rho_0)$ become?
Answer
The pairing of an observable $f$ with the signed measure $g\rho_0$ becomes the $L^2(\rho_0)$ inner product: $\ip{f,g\rho_0}=\int fg\,\rho_0=\ip{f,g}_{\rho_0}$. The dictionary sends the stationary density $\rho_0=\mathbf 1\cdot\rho_0$ to the constant observable $\mathbf 1$, so the two fixed points $\K_\tau\mathbf 1=\mathbf 1$ (conservation of expectations of constants) and $\P_\tau\rho_0=\rho_0$ (stationarity) merge into a single element of $L^2(\rho_0)$ — the trivial singular pair with $\lambda_0=1$. Everything downstream (the SVD of Problem 7, the coherent pairs of Block A, the CCA of Block D) is phrased inside this one self-dual space.
Problem 7 — Contraction, the trivial pair, and the $\rho_0$-adjoint
Goal: pin down why all singular values are $\le1$, and derive the paper's Eq. (16) for continuous state space.
(a) Show $\K_\tau\mathbf 1=\mathbf 1$ and, using Jensen's inequality and stationarity, $\|\K_\tau f\|_{\rho_0}\le\|f\|_{\rho_0}$.
Solution
$\K_\tau\mathbf 1=\mathbf 1$ is Problem 1(b). Jensen for the conditional expectation: $(\K_\tau f)(x)^2=\big(\E[f(X_\tau)\mid X_0=x]\big)^2\le\E[f(X_\tau)^2\mid X_0=x]=\K_\tau(f^2)(x)$. Then $$\|\K_\tau f\|_{\rho_0}^2=\int\rho_0\,(\K_\tau f)^2\le\int\rho_0\,\K_\tau(f^2)=\int(\P_\tau\rho_0)\,f^2=\int\rho_0f^2=\|f\|_{\rho_0}^2,$$ using duality (Problem 1c) and then stationarity $\P_\tau\rho_0=\rho_0$. So $\K_\tau$ is a contraction on $L^2(\rho_0)$. (This is where stationarity is essential: for a non-stationary process one needs $\mu$ and $\nu$ and the contraction goes from $L^2(\nu)$ to $L^2(\mu)$.)
(b) Conclude that every singular value of $\K_\tau$ on $L^2(\rho_0)$ is $\le1$, that $u_0=v_0=\mathbf 1$ is a singular pair with $\lambda_0=1$, and that for stable dynamics "eigenvalue closest to one" and "second largest" coincide.
Solution
The largest singular value of an operator is its operator norm, which for a contraction is $\le1$; hence all $\lambda_i\le1$. $\K_\tau\mathbf 1=\mathbf 1$ and $\K^\dagger_\tau\mathbf 1=\mathbf 1$ (the reversed process also conserves probability, or directly: $\ip{\K^\dagger_\tau\mathbf 1,g}=\ip{\mathbf 1,\K_\tau g}=\int\rho_0\K_\tau g=\int\rho_0g=\ip{\mathbf 1,g}$ for all $g$), so $(\mathbf 1,\mathbf 1)$ is a singular pair with $\lambda=1$ — the paper's trivial pair, Eq. (7). All other singular functions are orthogonal to $\mathbf 1$ in $L^2(\rho_0)$, i.e. have zero mean under $\rho_0$; that is the "soft balance" condition noted in Block A.
Since $1$ is the maximum possible, the non-trivial singular value closest to $1$ is simply the second largest, $\lambda_1$. The paper's more careful phrase "closest to one" is needed only for its extension to unstable dynamics, where $\rho_0$ is not a probability density, the contraction argument fails, and singular values above $1$ appear (Problem 12).
(c) Show that the $\rho_0$-adjoint of $\K_\tau$ is $\K^\dagger_\tau=\rho_0^{-1}\,\P_\tau\,\rho_0$ (multiply by $\rho_0$, push forward, divide by $\rho_0$) — the paper's Eq. (16) — and that at the level of generators the $\rho_0$-adjoint of $L^\dagger$ is $\rho_0^{-1}L\rho_0$.
Solution
Start from Lebesgue duality and insert $\rho_0$: $$\ip{g,\K_\tau f}_{\rho_0}=\int(g\rho_0)\,(\K_\tau f)\,dx=\int\P_\tau(g\rho_0)\,f\,dx=\int\rho_0\Big[\rho_0^{-1}\P_\tau(\rho_0g)\Big]f\,dx=\ip{\rho_0^{-1}\P_\tau\rho_0\,g,\;f}_{\rho_0}.$$ So $\K^\dagger_\tau=\diag(\rho_0)^{-1}e^{L\tau}\diag(\rho_0)$, Eq. (16): the Perron–Frobenius operator conjugated by the stationary density. Differentiating at $\tau=0$, the generator of $\K^\dagger_\tau$ is $\rho_0^{-1}L(\rho_0\,\cdot)$: the $\rho_0$-adjoint of $L^\dagger$ is the Fokker–Planck operator conjugated by $\rho_0$. The Lebesgue adjoint and the $\rho_0$-adjoint differ exactly by this conjugation.
(d) Show directly from Bayes' rule that $\K^\dagger_\tau f(y)=\E[f(X_t)\mid X_{t+\tau}=y]$, and reconcile with (c).
Solution
With $X_t\sim\rho_0$ and $X_{t+\tau}\sim\rho_0$, Bayes' rule gives the backward transition density $p(x\mid y)=p_\tau(y\mid x)\rho_0(x)/\rho_0(y)$, so $$\E[f(X_t)\mid X_{t+\tau}=y]=\int f(x)\,p(x\mid y)\,dx=\int f(x)\,\frac{p_\tau(y\mid x)\rho_0(x)}{\rho_0(y)}\,dx=\frac{1}{\rho_0(y)}\int p_\tau(y\mid x)\,\big[f(x)\rho_0(x)\big]\,dx=\frac{1}{\rho_0(y)}\,\P_\tau(f\rho_0)(y).$$ In the third equality the factor $1/\rho_0(y)$ does not depend on $x$ and comes out of the integral. The last equality is the definition of $\P_\tau$ from Problem 1(a), $\P_\tau q\,(y)=\int p_\tau(y\mid x)\,q(x)\,dx$, applied to $q=f\rho_0$: $\P_\tau$ is linear and accepts any integrable function, not only normalised densities. This is (c): multiply by $\rho_0$, push forward, divide by $\rho_0$.
Checking the adjoint relation directly. On one side, $\ip{g,\K_\tau f}_{\rho_0}=\int g(x)\,\E[f(X_{t+\tau})\mid X_t=x]\,\rho_0(x)\,dx=\E[g(X_t)f(X_{t+\tau})]$. On the other, $$\ip{\K^\dagger_\tau g,f}_{\rho_0}=\int\E[g(X_t)\mid X_{t+\tau}=y]\,f(y)\,\rho_0(y)\,dy .$$ Two rules turn this into the same joint expectation. (1) Since $X_{t+\tau}\sim\rho_0$, integrating any $h(y)$ against $\rho_0(y)\,dy$ is $\E[h(X_{t+\tau})]$; with $h(y)=\E[g(X_t)\mid X_{t+\tau}=y]\,f(y)$ this gives $\E\big[\E[g(X_t)\mid X_{t+\tau}]\,f(X_{t+\tau})\big]$. (2) $f(X_{t+\tau})$ is a function of the conditioning variable, so it may be moved inside the conditional expectation ("taking out what is known"), and the tower property $\E[\E[\,\cdot\mid X_{t+\tau}]]=\E[\,\cdot\,]$ (Block 0, interlude) then collapses the nesting: $$=\E\Big[\E\big[g(X_t)f(X_{t+\tau})\mid X_{t+\tau}\big]\Big]=\E\big[g(X_t)f(X_{t+\tau})\big].$$ Same joint expectation, once conditioned on the past and once on the future, so the two definitions agree. This is the continuous version of Problem 3(b): the adjoint conditions backward on the future, and it is the Koopman operator of the time-reversed process. Reversibility, $\K^\dagger_\tau=\K_\tau$, says that forward- and backward-conditioned expectations coincide, which is detailed balance $p_\tau(y\mid x)\rho_0(x)=p_\tau(x\mid y)\rho_0(y)$.
The same check by direct integration. Insert the Bayes kernel on one side and the forward kernel on the other: $$\ip{\K^\dagger_\tau g,f}_{\rho_0}=\int\Big[\int g(x)\,\frac{p_\tau(y\mid x)\rho_0(x)}{\rho_0(y)}\,dx\Big]f(y)\,\rho_0(y)\,dy=\iint g(x)f(y)\,p_\tau(y\mid x)\rho_0(x)\,dx\,dy ,$$ bringing $f(y)\rho_0(y)$ inside the $x$-integral (it is constant in $x$) and cancelling $\rho_0(y)$ against $1/\rho_0(y)$ where $\rho_0(y)>0$. And $$\ip{g,\K_\tau f}_{\rho_0}=\int g(x)\Big[\int p_\tau(y\mid x)f(y)\,dy\Big]\rho_0(x)\,dx=\iint g(x)f(y)\,p_\tau(y\mid x)\rho_0(x)\,dx\,dy ,$$ bringing $g(x)\rho_0(x)$ inside the $y$-integral. The two double integrals are the same (Fubini: the order of integration is immaterial, and $f,g\in L^2(\rho_0)$ makes the integrand absolutely integrable by Cauchy–Schwarz), and both equal $\E[g(X_t)f(X_{t+\tau})]$ because $\rho_0(x)p_\tau(y\mid x)$ is the joint density of $(X_t,X_{t+\tau})$ (chain rule for densities: joint = marginal of the first times conditional of the second). In this version the adjoint relation is the statement that one joint density has two chain-rule factorisations, marginal-of-$x$ times forward kernel and marginal-of-$y$ times Bayes kernel; no conditional-expectation rules are needed.
The two round trips and the sibling trajectory. $\K_\tau$ relates two different time slices: its input is a function of the future state and its output a function of the present one (Block A's opening remark). An operator between two different slices need not have a usable spectrum of its own — for non-reversible dynamics the eigenvalues of $\K_\tau$ can be complex and its eigenfunctions non-orthogonal. The compositions $\K^\dagger_\tau\K_\tau$ and $\K_\tau\K^\dagger_\tau$ repair this: each starts and ends on a single slice, and each is self-adjoint (by the rule $(ST)^\dagger=T^\dagger S^\dagger$, so $(\K^\dagger\K)^\dagger=\K^\dagger\K$) and positive ($\ip{u,\K^\dagger\K u}_{\rho_0}=\|\K u\|_{\rho_0}^2\ge0$), hence has real eigenvalues in $[0,1]$ (contraction, part (a)) and orthogonal eigenfunctions. The SVD of $\K_\tau$ is nothing but the simultaneous eigendecomposition of the two round trips, matched up by $\K_\tau u_i=\lambda_iv_i$: on the future slice $\K^\dagger\K\,u_i=\lambda_i^2u_i$, on the present slice $\K\K^\dagger\,v_i=\lambda_i^2v_i$, with the same eigenvalues $\lambda_i^2$ (a general fact about $T^\dagger T$ versus $TT^\dagger$ away from the kernel). Note the direction bookkeeping: $\K^\dagger$ alone is the forward transport of observables — the Koopman operator of the time-reversed process, i.e. $\P_\tau$ read through the dictionary $f\leftrightarrow f\rho_0$ of the previous remark — so $\K^\dagger\K$ is backward-then-forward, a round trip, not a transport.
Type bookkeeping. As operators, both $\K$ and $\K^\dagger$ are maps $L^2(\rho_0)\to L^2(\rho_0)$: they eat a function on state space and return a function on state space, and functions carry no time stamp, so the composition $\K^\dagger\K$ is legal by itself. The slice talk is an interpretation added when the functions are evaluated along a trajectory, and the rule is: $\K$'s input slot is the later slice and its output the earlier one ($f$ is evaluated at the future state, $\K f$ at the present one, straight from the definition $\K f(x)=\E[f(X_{t+\tau})\mid X_t=x]$); $\K^\dagger$'s input slot is the earlier slice and its output the later. So $\K f$ — a present-slice function, "my best current estimate of a future quantity" — is exactly what $\K^\dagger$ accepts, and the argument of $\K^\dagger\K f$ lives on the same slice as the argument of $f$. Stationarity makes only the lag between the slices meaningful, so that common slice may be given any time label. Q2(v)–(vi) run this bookkeeping through the computation explicitly.
Probabilistically, each round trip asks a question about a sibling trajectory: a second realisation that shares one state with yours and evolves with independent noise otherwise. $\K^\dagger\K\,\mathbf 1_B(y)$ is the probability that a sibling sharing your past is in $B$, given that you are at $y$ (Q2(i)); $\K\K^\dagger\mathbf 1_A(x)$ is the probability that a sibling sharing your future started in $A$, given that you are at $x$ (Q2(iii)). So $\K^\dagger\K$ interrogates the future about its past — its subdominant eigenfunction $u_1$ marks the retrospective set with the most common past — and $\K\K^\dagger$ interrogates the present about its future — $v_1$ marks the predictive set with the most common future. By the law-of-total-variance identity of Block A's opening remark, $1-\lambda_1^2$ is exactly the sibling disagreement: the fraction of the variance of $u_1(X_{t+\tau})$ that the shared state fails to fix. Mind one Bayes flip when putting this into words: the round trip computes $P(\text{sibling in }B\mid\text{same past as }y)$, not $P(\text{same past}\mid\text{sibling in }B)$. And because sibling pairs never occur inside a single realisation — each visited state comes with exactly one future — these are precisely the quantities a lone time series cannot estimate directly; Problem 15(e) turns that observation into the statement that a neuron cannot check its own residual.
In Q2, $X'$ denotes a sibling of $X$: $X'_t=X_t$, independent noise afterwards.
Q2(i) Show $\K^\dagger_\tau\K_\tau\mathbf 1_B(y)=P(X'_{t+\tau}\in B\mid X_{t+\tau}=y)$, naming the rule at each step.
Answer
By Problem 7(d), $\K^\dagger_\tau g(y)=\E[g(X_t)\mid X_{t+\tau}=y]$; with $g=\K_\tau\mathbf 1_B$, $$\K^\dagger\K\,\mathbf 1_B(y)=\E\big[P(X_{t+\tau}\in B\mid X_t)\,\big|\,X_{t+\tau}=y\big].$$ Now two rules. Conditional independence given the shared state: conditionally on $X_t$, the sibling's future $X'_{t+\tau}$ is independent of your own $X_{t+\tau}$ (they use independent noise from the same start), and it has the same transition law, so $P(X_{t+\tau}\in B\mid X_t)=P(X'_{t+\tau}\in B\mid X_t)=P(X'_{t+\tau}\in B\mid X_t,X_{t+\tau})$ — adding $X_{t+\tau}$ to the conditioning changes nothing for $X'$. Tower property for the nested conditioning $\E[\,\E[\cdot\mid X_t,X_{t+\tau}]\,\mid X_{t+\tau}=y]=\E[\cdot\mid X_{t+\tau}=y]$: $$\K^\dagger\K\,\mathbf 1_B(y)=\E\big[P(X'_{t+\tau}\in B\mid X_t,X_{t+\tau})\,\big|\,X_{t+\tau}=y\big]=P(X'_{t+\tau}\in B\mid X_{t+\tau}=y).$$ Among all trajectories that share $y$'s history, this is the fraction now in $B$.
By direct integration. Use the kernels of Q2(v): $\K\mathbf 1_B(x)=\int p_\tau(y'\mid x)\mathbf 1_B(y')\,dy'=\int_Bp_\tau(y'\mid x)\,dy'$ (definition of $\K$; the indicator restricts the domain of integration), then $\K^\dagger$ with the Bayes kernel $\tilde p_\tau(x\mid y)=p_\tau(y\mid x)\rho_0(x)/\rho_0(y)$ of Problem 7(d): $$\K^\dagger\K\mathbf 1_B(y)=\int\tilde p_\tau(x\mid y)\Big[\int_Bp_\tau(y'\mid x)\,dy'\Big]dx=\int_B\Big[\int p_\tau(y'\mid x)\,\tilde p_\tau(x\mid y)\,dx\Big]dy'=\int_Bq_\tau(y'\mid y)\,dy' ,$$ swapping the order of integration (Fubini; the integrand is nonnegative) and naming the bracket $q_\tau$, the sibling kernel of Q2(v). Since $q_\tau(\cdot\mid y)$ is the density of $X'_{t+\tau}$ given $X_{t+\tau}=y$ (backward by Bayes to the shared past, forward by the dynamics), integrating it over $B$ is $P(X'_{t+\tau}\in B\mid X_{t+\tau}=y)$. The conditional-independence-plus-tower argument above is this computation with the integrals written as expectations.
Q2(ii) Show $\ip{\mathbf 1_B,\K^\dagger\K\mathbf 1_B}_{\rho_0}=\|\K\mathbf 1_B\|_{\rho_0}^2=P(X_{t+\tau}\in B,\;X'_{t+\tau}\in B)$, and that dividing by $\ip{\mathbf 1_B,\mathbf 1_B}_{\rho_0}$ gives $P(X'_{t+\tau}\in B\mid X_{t+\tau}\in B)$ — the self-coherence of $B$.
Answer
First equality: the definition of the adjoint, $\ip{\mathbf 1_B,\K^\dagger\K\mathbf 1_B}=\ip{\K\mathbf 1_B,\K\mathbf 1_B}$. Second: writing out the norm, $\|\K\mathbf 1_B\|_{\rho_0}^2=\E\big[P(X_{t+\tau}\in B\mid X_t)^2\big]$ with $X_t\sim\rho_0$; a squared conditional probability is the chance that two conditionally independent draws both succeed, $$P(B\mid X_t)^2=P(X_{t+\tau}\in B\mid X_t)\,P(X'_{t+\tau}\in B\mid X_t)=P(X_{t+\tau}\in B,\;X'_{t+\tau}\in B\mid X_t)$$ (conditional independence again), and taking $\E$ removes the conditioning (tower). For the denominator, stationarity gives $\ip{\mathbf 1_B,\mathbf 1_B}_{\rho_0}=\E[\mathbf 1_B(X_{t+\tau})]=P(X_{t+\tau}\in B)$, so the ratio is the definition of conditional probability, $P(X'_{t+\tau}\in B\mid X_{t+\tau}\in B)$: how often a trajectory's sibling agrees with it about membership in $B$. Maximising this Rayleigh quotient over mean-zero (soft) indicators is exactly the eigenproblem for $\K^\dagger\K$ restricted to $\mathbf 1^\perp$, i.e. the search for $u_1$.
By direct integration. $\|\K\mathbf 1_B\|_{\rho_0}^2=\int[\K\mathbf 1_B(x)]^2\rho_0(x)\,dx$ (definition of the norm) with $\K\mathbf 1_B(x)=\int_Bp_\tau(y\mid x)\,dy$ as in (i). The square of one integral is a double integral over two dummy variables ($(\int a)(\int b)=\iint ab$), so $$\|\K\mathbf 1_B\|^2=\int\rho_0(x)\Big[\int_Bp_\tau(y\mid x)\,dy\Big]\Big[\int_Bp_\tau(y'\mid x)\,dy'\Big]dx=\int_B\!\!\int_B\Big[\int\rho_0(x)\,p_\tau(y\mid x)\,p_\tau(y'\mid x)\,dx\Big]dy\,dy' ,$$ after swapping the order of integration. The bracket is the joint density of $(X_{t+\tau},X'_{t+\tau})$: a start $x\sim\rho_0$, then two independent runs of the dynamics from it (the product of the two transition densities given the shared start, integrated over the start). Integrating a joint density over $B\times B$ is $P(X_{t+\tau}\in B,\,X'_{t+\tau}\in B)$. The bracket is also $q_\tau(y'\mid y)\rho_0(y)$, the symmetric kernel of Q2(v).
Q2(iii) Interpret $\K\K^\dagger\mathbf 1_A(x)$ the same way.
Answer
$\K\K^\dagger\mathbf 1_A(x)=\E[\K^\dagger\mathbf 1_A(X_{t+\tau})\mid X_t=x]$ (definition of $\K$), and $\K^\dagger\mathbf 1_A(y)=P(X_t\in A\mid X_{t+\tau}=y)$ (backward conditioning, 7(d)) — "did you come from $A$." Let $X''$ be a sibling sharing the future state, $X''_{t+\tau}=X_{t+\tau}$, drawn backward with the Bayes kernel of Problem 7(d) and otherwise independent. The same conditional-independence-plus-tower argument, run with the roles of past and future exchanged, gives $$\K\K^\dagger\mathbf 1_A(x)=P(X''_t\in A\mid X_t=x):$$ among all trajectories that will end where yours will, the fraction that started in $A$. Pairing with $\mathbf 1_A$ as in (ii) gives the self-coherence of the predictive set: $P(X''_t\in A\mid X_t\in A)$.
By direct integration. $\K^\dagger\mathbf 1_A(y)=\int_A\tilde p_\tau(x'\mid y)\,dx'$ (Bayes kernel integrated over $A$), and then $$\K\K^\dagger\mathbf 1_A(x)=\int p_\tau(y\mid x)\Big[\int_A\tilde p_\tau(x'\mid y)\,dx'\Big]dy=\int_A\Big[\int\tilde p_\tau(x'\mid y)\,p_\tau(y\mid x)\,dy\Big]dx' ,$$ swapping the order of integration. The bracket is the kernel of Problem 10's setting: forward from $x$ to a future $y$, then backward by Bayes to a past $x'$, which is the density of $X''_t$ given $X_t=x$; integrating it over $A$ is $P(X''_t\in A\mid X_t=x)$.
Q2(iv) What do the two round trips become for reversible dynamics, and what does that say about the coherent pair?
Answer
Reversibility means $\K^\dagger_\tau=\K_\tau$ (Problem 7(d): detailed balance), so both round trips collapse to the same operator $\K_\tau^2$, and the singular functions coincide with each other and with the eigenfunctions of $\K_\tau$ itself: $u_i=v_i$ (up to sign) and $\lambda_i=|\text{eigenvalue}_i|$. The coherent pair then has $A=B$: the set with the most common future is also the set with the most common past, i.e. the sets are metastable — they hold their members — rather than transported from one place to another. A genuinely moving coherent set ($A\ne B$) is possible only for non-reversible dynamics, which is why the paper's saddle example (Problem 12) and the predictive/retrospective distinction of the Introduction need the non-reversible case.
Q2(v) Write $\K_\tau$ and $\K^\dagger_\tau$ as integral operators and compute $\K^\dagger\K f$ for a general observable $f$ by composing the kernels. What kernel emerges, and what are its sanity checks?
Answer
Forward kernel (definition of $\K$): $\K f(x)=\int p_\tau(y\mid x)\,f(y)\,dy$. Backward kernel (Problem 7(d), the Bayes kernel): $\K^\dagger g(y)=\int\tilde p_\tau(x\mid y)\,g(x)\,dx$ with $\tilde p_\tau(x\mid y)=p_\tau(y\mid x)\rho_0(x)/\rho_0(y)$. Substitute $g=\K f$, using a fresh dummy variable $y'$ for the inner future integral — reusing the letter $y$ is where this computation usually derails — and swap the integrals (Fubini): $$\K^\dagger\K f(y)=\int\tilde p_\tau(x\mid y)\Big[\int p_\tau(y'\mid x)f(y')\,dy'\Big]dx=\int f(y')\underbrace{\Big[\int p_\tau(y'\mid x)\,\tilde p_\tau(x\mid y)\,dx\Big]}_{=:\,q_\tau(y'\mid y)}dy'.$$ The bracket is a genuine transition kernel — nonnegative, and it integrates to $1$ in $y'$ because $p_\tau$ does and then $\tilde p_\tau$ does: go backward from $y$ by the Bayes kernel to a past state $x$, then forward from $x$ by the ordinary dynamics to $y'$. That is the sibling kernel, so $\K^\dagger\K f(y)=\E[f(X'_{t+\tau})\mid X_{t+\tau}=y]$, of which Q2(i) is the case $f=\mathbf 1_B$. Problem 10's setting states exactly this kernel (in the mirror orientation, for $\K\K^\dagger$: forward first, then Bayes-invert) and computes its Gaussian form for the OU process.
Sanity checks. $q_\tau(y'\mid y)\,\rho_0(y)=\int p_\tau(y'\mid x)\,p_\tau(y\mid x)\,\rho_0(x)\,dx$ is symmetric in $(y,y')$ — the self-adjointness of $\K^\dagger\K$ made visible at the kernel level; and $f=\mathbf 1$ gives $\K^\dagger\K\mathbf 1=\mathbf 1$, the trivial pair.
Q2(vi) In conditional-expectation form, $\K^\dagger\K f(y)=\E\big[\E[f(X_{t+\tau})\mid X_t]\,\big|\,X_{t+\tau}=y\big]$. Why does the tower property not collapse this to $f(y)$ — i.e. why is $\K^\dagger\K\ne I$? And for which dynamics is $\K^\dagger\K=I$?
Answer
The tower property (Block 0, interlude) collapses $\E[\,\E[\cdot\mid\mathcal G]\mid\mathcal H\,]$ only when $\mathcal H\subset\mathcal G$: the outer conditioning must be coarser than the inner one. Here the inner conditioning is on $X_t$ and the outer on $X_{t+\tau}$, and $\sigma(X_{t+\tau})\not\subset\sigma(X_t)$ — they are different variables — so the rule does not apply, and the naive collapse to $\E[f(X_{t+\tau})\mid X_{t+\tau}=y]=f(y)$ is illegal. The legal move is the sibling substitution of Q2(i): conditionally on $X_t$, a fresh draw $X'_{t+\tau}$ has the same law as $X_{t+\tau}$ and is independent of it, so the inner expectation equals $\E[f(X'_{t+\tau})\mid X_t,X_{t+\tau}]$, and now the outer conditioning is coarser and the tower property collapses the nesting onto the sibling.
The same point by direct integration. Composing kernels as in Q2(v), $\K^\dagger\K f(y)=\int f(y')\,q_\tau(y'\mid y)\,dy'$ with $q_\tau(y'\mid y)=\int p_\tau(y'\mid x)\,\tilde p_\tau(x\mid y)\,dx$. The naive collapse $\K^\dagger\K f=f$ would require $q_\tau(y'\mid y)=\delta(y'-y)$: that going back by the Bayes kernel and forward by the dynamics lands exactly on $y$ again. It does not. For fixed $y$ the backward step spreads over every past $x$ compatible with $y$, and the forward step from each such $x$ spreads over every future compatible with $x$, so $q_\tau(\cdot\mid y)$ has positive width whenever $p_\tau$ does. Compare the composition the tower property does license, $\K_s\K_\tau=\K_{s+\tau}$, which at the kernel level is Chapman–Kolmogorov, $\int p_s(z\mid y)\,p_\tau(y'\mid z)\,dz=p_{s+\tau}(y'\mid y)$ (Problem 1(d)): forward after forward composes into one forward kernel, whereas forward after backward does not compose into the identity. Only for a deterministic flow, $p_\tau(y\mid x)=\delta(y-\Phi_\tau(x))$, does the Bayes kernel concentrate on the single point $x=\Phi_\tau^{-1}(y)$, and then $q_\tau(y'\mid y)=\delta(y'-\Phi_\tau(\Phi_\tau^{-1}(y)))=\delta(y'-y)$.
The gap between $\K^\dagger\K$ and $I$ is exactly the conditional variance charged by the law of total variance: $\ip{f,\K^\dagger\K f}=\|\K f\|^2=\|f\|^2-\E[\operatorname{Var}(f(X_{t+\tau})\mid X_t)]$. So $\K^\dagger\K=I$ iff every conditional variance vanishes, i.e. for deterministic dynamics: the sibling is you, $q_\tau(y'\mid y)=\delta(y'-y)$, and $\K_\tau$ is unitary — the Koopman operator of a measure-preserving map. All singular values then equal $1$ and no partition is more coherent than any other; it is the noise that creates the coherence hierarchy, by degrading some future observables more than others.
Coda — does any of this survive without stationarity?
Goal: identify exactly what the stationary theory used $\rho_0$ for, and which conclusions survive when there is no $\rho_0$.
The skeleton survives — it was built for this case. Coherent sets were introduced by Froyland and collaborators for non-autonomous flows (ocean eddies tracked over a finite time window); the stationary theory of this set is the special case in which the two time slices can be identified. In general one chooses the reference $\mu$, gives the future slice its pushforward $\nu=\P\mu$, and reads the Koopman operator as a map between two different Hilbert spaces, $\K:L^2(\nu)\to L^2(\mu)$. The adjoint is backward conditioning with the Bayes kernel built from $\mu$ and $\nu$; the contraction survives, with the compatibility condition $\nu=\P\mu$ doing the work stationarity used to do (Q3(i) — Problem 7(a)'s parenthetical already says this); $(\mathbf 1,\mathbf 1)$ is still the trivial pair; and the SVD — which never needed the two spaces to be equal, only an inner product on each — still exists: $\K u_i=\lambda_iv_i$ with $v_i\in L^2(\mu)$ on the present slice and $u_i\in L^2(\nu)$ on the future one. The subdominant pair still marks the finite-time coherent pair $A=\{v_1>0\}$ at $t$, $B=\{u_1>0\}$ at $t+\tau$: mass that travels together despite the noise. The sibling round trips and their interpretation (Q2), Problem 15's variational bounds (Courant–Fischer holds for operators between different Hilbert spaces), and the Galerkin and CCA estimators of Block D all survive with slice-specific covariances; indeed the VAMP theory behind Problem 14 was formulated by Wu and Noé precisely to cover non-stationary, non-reversible processes, and CCA with slice-specific whitening is the estimator that stays correct.
What dies is the identification of the slices. (1) There is no canonical $\rho_0$: the reference $\mu$ becomes part of the problem statement, and coherent sets are coherent relative to $(t,\tau,\mu)$ — fluid dynamics takes normalised Lebesgue on the domain; for a sensory system the natural choice is the actual ensemble distribution of states at time $t$. (2) The semigroup and the single generator are gone: $\K_{t,t+\tau}$ depends on both endpoints, so Eq. (13)'s $e^{L^\dagger\tau}$ is replaced by a two-parameter propagator $U(t+\tau,t)$ with only the cocycle property $U(t_3,t_2)U(t_2,t_1)=U(t_3,t_1)$ (a time-ordered exponential of $L^\dagger(t)$); Problem 6's derivation of the instantaneous generator still holds — what is lost is exponentiating one operator. (3) Eigenvalues of $\K$ are no longer even well-typed, and reversibility cannot be stated (Q3(ii)); only the SVD survives, which vindicates this set's insistence that the SVD, not the eigendecomposition, is the fundamental object. (4) Statistically, the casualty is ergodicity: one realisation supplies one sample per time slice, so nothing is estimable from a single trajectory without extra assumptions (Q3(iii)).
What this means for a neuron. A neuron estimating lag-vector covariances from its own input stream is implicitly assuming local stationarity on its learning timescale, and real-world non-stationarity then splits by timescale. Fast "non-stationarity" is often partial observation in disguise: if the drift in the statistics is itself generated by a lawful process (a plume meandering, light changing with cloud cover), enlarging the state can restore stationarity — the same projection theme as Problems 13/15 and the olfaction discussion of the Introduction. Slow non-stationarity is what adaptation is for: re-estimate the covariances in a sliding window, i.e. re-whiten against the drifting reference $\mu_t$; luminance and contrast adaptation are exactly this, and the prediction is that adaptation should adjust not only the gain but the shape of temporal filters as the local covariances drift. Finally, Problem 12's saddle already sits outside the stationary theory: there is no normalisable stationary density, Eq. (18) is applied with the formal (indefinite) Sylvester $\Sigma$ as a fixed weight on both slices, the contraction hypotheses fail, and eigenvalues above $1$ appear with a $\tau$-dependent selection problem (12(b)). The two-measure formalism is the principled home for such transient problems: choose a genuine Gaussian reference $\mu$ at the initial time and let $\nu=\P\mu$, and all singular values return to $\le1$. Q2(iv)'s hierarchy thus gains a level: reversible stationary (metastable sets; eigenfunctions of a self-adjoint $\K$ suffice) $\subset$ non-reversible stationary (coherent pairs; SVD in one space) $\subset$ non-stationary (finite-time coherent pairs; SVD between two spaces, chosen reference, no spectra at all).
Q3(i) Let $\mu$ be any density on the present slice and $\nu=\P_{t,t+\tau}\mu$. Show that $\K:L^2(\nu)\to L^2(\mu)$ is a contraction and that $(\mathbf 1,\mathbf 1)$ is still a singular pair with $\lambda=1$. Where exactly does the stationary proof's use of $\P_\tau\rho_0=\rho_0$ get replaced?
Answer
Run Problem 7(a) again. Jensen for the conditional expectation gives $(\K f)^2\le\K(f^2)$ pointwise, exactly as before. Then $$\|\K f\|_{L^2(\mu)}^2=\int\mu\,(\K f)^2\le\int\mu\,\K(f^2)=\int(\P\mu)\,f^2=\int\nu\,f^2=\|f\|_{L^2(\nu)}^2 .$$ The third equality is Lebesgue duality (Problem 1(c)); the fourth is where stationarity used to enter — the stationary proof used $\P\rho_0=\rho_0$, and here it is replaced by the definition $\nu=\P\mu$: the future slice must carry the pushforward of the present reference, or the pushforward step has nowhere to land. $\K\mathbf 1=\mathbf 1$ as always, $\K^\dagger\mathbf 1=\mathbf 1$ because $\K^\dagger g(y)=\E_\mu[g(X_t)\mid X_{t+\tau}=y]$ returns $1$ for $g=\mathbf 1$, and $\|\mathbf 1\|_{L^2(\mu)}=\|\mathbf 1\|_{L^2(\nu)}=1$ since both are probability densities. If instead $\nu$ is chosen independently of $\mu$, the pushforward step fails and nothing bounds the singular values — which is exactly what happens in Problem 12, where a fixed formal weight sits on both slices.
Q3(ii) Explain why the eigenvalue equation $\K f=\lambda f$ is no longer well-typed, why reversibility cannot even be stated, and why the SVD nevertheless survives.
Answer
$\K$ takes its input in $L^2(\nu)$ and returns its output in $L^2(\mu)$. The equation $\K f=\lambda f$ equates an element of $L^2(\mu)$ with an element of $L^2(\nu)$ — two different Hilbert spaces — which is meaningless without a chosen identification between the slices. Stationarity was that identification: one $\rho_0$ served both slices, so input and output lived in the same space and eigenvalues made sense. The statement of reversibility, $\K^\dagger=\K$, fails typing the same way ($\K^\dagger$ maps $L^2(\mu)\to L^2(\nu)$, the opposite direction), and with it goes TICA (Problem 14(b)) and any comparison of $u_1$ and $v_1$ "as the same function" — asking whether the predictive and retrospective sets coincide requires a way to lay one slice over the other.
The SVD asks for less: an inner product on each space separately. The adjoint of an operator between two Hilbert spaces is defined by $\ip{g,\K f}_{L^2(\mu)}=\ip{\K^\dagger g,f}_{L^2(\nu)}$, the round trips $\K^\dagger\K$ (on $L^2(\nu)$) and $\K\K^\dagger$ (on $L^2(\mu)$) are each self-adjoint on their own space, and their eigendecompositions matched by $\K u_i=\lambda_iv_i$ are the SVD, exactly as in the stationary case. Every quantity built from singular values and the two orthonormal families — coherence, the variational bounds of Problem 15, the CCA of Problem 14 — survives; every quantity built from eigenvalues or from comparing the two families does not.
Q3(iii) Why can a single realisation of a non-stationary process not estimate even the covariance $C_{00}(t)=\E[x_tx_t^\top]$? Which assumptions restore estimability, and what happens to the sibling obstruction of Problem 15(e)?
Answer
$C_{00}(t)$ is an ensemble average at the fixed time $t$, and one realisation supplies exactly one sample of $x_t$. For a stationary ergodic process the ergodic theorem converts time averages along one trajectory into ensemble averages — that licence is what every estimator in Block D silently used. Without stationarity a time average mixes samples from different distributions and estimates nothing in particular. Three assumptions restore estimability, in decreasing order of luxury:
(1) Ensembles: many independent realisations (trials), averaged across the ensemble at fixed $t$ — the setting of the fluid-dynamics literature and of trial-based neuroscience. (2) Quasi-stationarity: the statistics drift slowly compared with the mixing time, $\tau_{\rm mix}\ll T_{\rm window}\ll T_{\rm drift}$, so windowed time averages approximate slice averages and the stationary theory applies within each window with $\mu$ the local stationary density — the assumption a neuron implicitly makes, with the window set by its plasticity timescale. (3) Cyclostationarity: the statistics are periodic (day/night, breathing, sniffing), so one averages across cycles at fixed phase and $\mu$ becomes phase-indexed.
The sibling obstruction compounds rather than resolves: Problem 15(e) said a single trajectory cannot supply $\|\K f\|^2$ even with stationarity, because sibling pairs never occur in one realisation; without stationarity even the one-slice covariances join the list of quantities that need an ensemble or an assumption.
Q3(iv) Time-dependent OU: $dX=A(t)X\,dt+\sqrt{D(t)}\,dW$. Show that the slice covariances obey the differential Lyapunov equation $\dot\Sigma=A\Sigma+\Sigma A^\top+D$, and that "linear observables suffice" (Problem 11) survives slice-wise. What is the finite-dimensional shadow of the SVD?
Answer
Let $\Phi(t_2,t_1)$ be the state propagator, $\partial_{t_2}\Phi=A(t_2)\Phi$, $\Phi(t,t)=I$ (for constant $A$ it is $e^{A(t_2-t_1)}$). Variation of constants solves the SDE: $$X_{t+\tau}=\Phi(t+\tau,t)\,X_t+\eta,\qquad \eta=\int_t^{t+\tau}\Phi(t+\tau,s)\sqrt{D(s)}\,dW_s ,$$ with $\eta$ independent of $X_t$ (Itô integral over later noise). If $X_t\sim\mathcal N(0,\Sigma_t)$ then $X_{t+\tau}\sim\mathcal N(0,\Sigma_{t+\tau})$ with $$\Sigma_{t+\tau}=\Phi\,\Sigma_t\,\Phi^\top+\int_t^{t+\tau}\Phi(t+\tau,s)\,D(s)\,\Phi(t+\tau,s)^\top ds .$$ Differentiate in the upper limit, using $\partial\Phi=A\Phi$ (product rule on the first term; the integrand at $s=t+\tau$ is $D$, and the $\Phi$'s inside the integral differentiate to $A(\cdot)\Phi$): $\dot\Sigma=A\Sigma+\Sigma A^\top+D$. At constant coefficients its fixed point $\dot\Sigma=0$ is Problem 8's algebraic Lyapunov equation.
Linear observables still close: for $f(y)=w^\top y$, $\K f(x)=w^\top\Phi(t+\tau,t)\,x$ — linear again — and the backward map is Gaussian linear regression exactly as in Problem 10(a), $\E[X_t\mid X_{t+\tau}=y]=\Sigma_t\Phi^\top\Sigma_{t+\tau}^{-1}y$. So the singular functions are linear in the state, and the finite-dimensional shadow of the SVD is the SVD of the doubly-whitened propagator $$\tilde M=\Sigma_{t+\tau}^{-1/2}\,\Phi(t+\tau,t)\,\Sigma_t^{1/2},$$ whitened by $\Sigma_t$ on the input side and $\Sigma_{t+\tau}$ on the output side — Problem 10(d) with the two $\Sigma^{1/2}$'s allowed to differ, which is precisely the CCA of Problem 14 with $C_{00}$ and $C_{11}$ estimated each on its own slice. The only casualty is notation: nothing depends on $\tau$ alone any more.