Time Reversal of Diffusions

Concept Definition

Time reversal studies the law of a diffusion when observed backwards in time, asking: if \((X_t)_{0 \le t \le T}\) is a diffusion, what SDE does the reversed process \(\widetilde{X}_t := X_{T-t}\) satisfy?

Setup. Let \((X_t)_{0 \le t \le T}\) solve the forward SDE on a filtered space \((\Omega, \mathcal{F}, (\mathcal{F}_t), \mathbb{P})\):

\[ \mathrm{d}X_t^{i} = b^{i}(t, X_t)\,\mathrm{d}t + \sigma^{i\alpha}(t, X_t)\,\mathrm{d}W_t^{\alpha}, \]

where we define the covariance matrix \(a^{ij}(t,x) := \sigma^{i\alpha}(t,x)\sigma^{j\alpha}(t,x)\) (Einstein summation on \(\alpha\)). Assume \(X_t\) admits a smooth positive density \(p(t, x)\) for each \(t \in (0, T]\).

Define the time-reversed process:

\[ \widetilde{X}_t := X_{T-t}, \qquad 0 \le t \le T. \]

Theorem (Haussmann–Pardoux, 1986; Anderson, 1982)

Under regularity conditions ensuring \(p(t,x) > 0\) and sufficient smoothness, \((\widetilde{X}_t)_{0 \le t \le T}\) solves the SDE

\[ \mathrm{d}\widetilde{X}_t^{i} = \widetilde{b}^{i}(t, \widetilde{X}_t)\,\mathrm{d}t + \sigma^{i\alpha}(T-t, \widetilde{X}_t)\,\mathrm{d}\widetilde{W}_t^{\alpha}, \]

where \((\widetilde{W}_t)\) is a Brownian motion with respect to the backward filtration \(\widetilde{\mathcal{F}}_t = \sigma(X_s : T-t \le s \le T)\), and the reversed drift is

\[ \boxed{ \widetilde{b}^{i}(t, x) = -b^{i}(T-t, x) + \frac{1}{p(T-t, x)}\,\frac{\partial}{\partial x_j}\!\bigl(a^{ij}(T-t, x)\,p(T-t, x)\bigr). } \]

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Explanation

The Score Term and Its Meaning

The reversed drift contains two parts:

\[ \widetilde{b}^{i}(t, x) = \underbrace{-b^{i}(T-t, x)}_{\text{negated forward drift}} + \underbrace{\frac{\partial a^{ij}}{\partial x_j}(T-t,x) + a^{ij}(T-t,x)\,\partial_{x_j} \log p(T-t, x)}_{\text{score correction}}. \]

The score \(\nabla \log p(t,x) = \nabla_x \log p(t,x)\) is the gradient of the log-density of \(X_t\). For constant \(a = \sigma\sigma^\top\) (spatially homogeneous diffusion), the formula simplifies to

\[ \widetilde{b}^{i}(t, x) = -b^{i}(T-t, x) + a^{ij}\,\partial_{x_j} \log p(T-t, x). \]

Intuition. When running a diffusion backward, the noise term is unchanged (the diffusion matrix is the same), but the drift must be corrected to account for the flow of probability mass encoded in \(p\). The score term steers the reversed process toward regions of higher density, counteracting the tendency of the forward noise to spread out.

The Backward Filtration

A critical point: \((\widetilde{W}_t)\) is a Brownian motion adapted to the backward filtration \(\widetilde{\mathcal{F}}_t\), not the forward one. This means \(\widetilde{X}\) is a semimartingale in the backward sense. One must distinguish:

• Forward SDE: adapted to \((\mathcal{F}_t)\), driven by \(W\).
• Reversed SDE: adapted to \((\widetilde{\mathcal{F}}_t)\), driven by a new BM \(\widetilde{W}\) whose existence is guaranteed by the Doob–Meyer decomposition of \(\widetilde{X}\) as a backward semimartingale.

Conflating forward and backward filtrations is the most common source of error in time reversal arguments.

Connection to Doob's \(h\)-Transform

In the constant-\(a\) case, the score correction \(a\,\nabla\log p\) can be understood via a Doob \(h\)-transform: weighting the path measure by \(h(t,x) = p(t,x)\) (the density itself) produces a new drift that guides the process toward the support of \(p\). Time reversal is essentially a Doob \(h\)-transform with \(h = p\).

Connection to Score-Based Generative Models

The score term \(\nabla\log p(t,x)\) is precisely what modern score-based diffusion models (e.g. DDPM, score matching) learn from data. In the practically common case of constant \(a = \sigma\sigma^\top\) (spatially homogeneous noise), the reversed SDE takes the explicit form

\[ \mathrm{d}\widetilde{X}_t = \bigl(-b(T-t,\widetilde{X}_t) + a\,\nabla\log p(T-t,\widetilde{X}_t)\bigr)\mathrm{d}t + \sigma\,\mathrm{d}\widetilde{W}_t. \]

For spatially varying \(\sigma\), the full formula from the theorem above applies, with the additional divergence term \(\partial_{x_j} a^{ij}\). The practical approach: train a neural network to approximate \(\nabla\log p(t,x)\) at all noise levels \(t\), then run the reversed SDE from pure noise to generate new samples. The Haussmann–Pardoux theorem is the mathematical foundation underlying this entire class of generative models.

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Diagram / Example

Reversible Stationary Case

If \(X_0 \sim \pi\) where \(\pi\) is the invariant measure and the process is reversible (detailed balance holds), then

\[ (X_t)_{0 \le t \le T} \stackrel{d}{=} (X_{T-t})_{0 \le t \le T}. \]

In this case \(p(t, x) = \pi(x)\) for all \(t\), the score is \(\nabla\log\pi(x)\), and one can verify that \(\widetilde{b}^i = b^i\): the reversed drift equals the forward drift, consistent with time symmetry.

Example — Ornstein–Uhlenbeck process. Consider

\[ \mathrm{d}X_t = -\theta X_t\,\mathrm{d}t + \sigma\,\mathrm{d}W_t, \qquad \theta > 0. \]

The invariant measure is \(\pi = \mathcal{N}(0, \sigma^2/2\theta)\), so \(\nabla\log\pi(x) = -x/(\sigma^2/2\theta) = -2\theta x/\sigma^2\). Substituting into the reversed drift formula (constant \(a = \sigma^2\)):

\[ \widetilde{b}(t,x) = -(-\theta x) + \sigma^2 \cdot \left(-\frac{2\theta x}{\sigma^2}\right) = \theta x - 2\theta x = -\theta x = b(x). \checkmark \]

Time reversal produces the same OU process.

Non-Reversible Case: Drifted Brownian Motion

Let \(\mathrm{d}X_t = \mu\,\mathrm{d}t + \sigma\,\mathrm{d}W_t\) on \([0,T]\) with \(X_0 = 0\). Then \(p(t,x) = \phi(x; \mu t, \sigma^2 t)\) (Gaussian) and

\[ \partial_x \log p(t,x) = -\frac{x - \mu t}{\sigma^2 t}. \]

The reversed drift is

\[ \widetilde{b}(t, x) = -\mu + \sigma^2 \cdot \left(-\frac{x - \mu(T-t)}{\sigma^2(T-t)}\right) = -\mu - \frac{x - \mu(T-t)}{T-t}. \]

This is a Brownian bridge drift. To read the direction: \(\widetilde{X}_0 = X_T\) (the reversed process starts at the forward endpoint) and \(\widetilde{X}_T = X_0 = 0\) (it ends at the forward starting point). The drift \(-(x - \mu(T-t))/(T-t)\) pulls \(\widetilde{X}\) toward \(0\) as \(t \to T\), confirming it is a bridge from \(X_T\) back to \(0\), with the original drift negated. Time reversal converts free Brownian motion (with drift) into a conditioned process pinned at both endpoints.

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Proof / Derivation

We sketch the formal derivation of the reversed drift formula in the constant-\(a\) case.

Step 1: Forward Fokker–Planck. The density \(p(t,x)\) satisfies

\[ \partial_t p = -\partial_{x_i}(b^i p) + \frac{1}{2} a^{ij}\,\partial_{x_i x_j} p. \]

Step 2: Reversed process as backward semimartingale. By the theory of backward stochastic calculus (see Pardoux 1986), \(\widetilde{X}_t = X_{T-t}\) is a semimartingale under \(\widetilde{\mathcal{F}}_t\) admitting a decomposition

\[ \mathrm{d}\widetilde{X}_t = \widetilde{b}(t, \widetilde{X}_t)\,\mathrm{d}t + \sigma\,\mathrm{d}\widetilde{W}_t. \]

Step 3: Identify the drift. We work in the constant-\(a\) case (the general case requires one extra step noted below). The reversed density \(\widetilde{p}(t,x) = p(T-t,x)\) must satisfy the forward Fokker–Planck equation for \(\widetilde{X}\):

\[ \partial_t \widetilde{p} = -\partial_{x_i}(\widetilde{b}^i \widetilde{p}) + \frac{1}{2}a^{ij}\,\partial_{x_i x_j}\widetilde{p}. \]

Substituting \(\widetilde{p}(t,x) = p(T-t,x)\) so \(\partial_t \widetilde{p} = -\partial_t p(T-t,x)\), and using the forward Fokker–Planck for \(p\) gives (after rearrangement):

\[ \partial_{x_i}(\widetilde{b}^i \widetilde{p}) = \partial_{x_i}(b^i \widetilde{p}) - a^{ij}\,\partial_{x_i x_j}\widetilde{p}. \]

Dividing by \(\widetilde{p} > 0\) and expanding \(\partial_{x_j}(a^{ij}\widetilde{p})/\widetilde{p} = \partial_{x_j}a^{ij} + a^{ij}\partial_{x_j}\log\widetilde{p}\) (using constant \(a\), so \(\partial_{x_j}a^{ij} = 0\)) yields \(\widetilde{b}^i = -b^i + a^{ij}\partial_{x_j}\log p(T-t,x)\), consistent with the boxed formula. For spatially varying \(a\), the divergence term \(\partial_{x_j}a^{ij}\) is non-zero and the same calculation produces the full formula \(\widetilde{b}^i = -b^i + \frac{1}{p}\partial_{x_j}(a^{ij}p)\) stated in the theorem. \(\square\)

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What to Remember

• Time reversal introduces a score correction \(a\,\nabla\log p(t,x)\) to the negated forward drift.
• The reversed process is a semimartingale with respect to the backward filtration; \(\widetilde{W}\) is a new BM, not \(-W\).
• Under detailed balance (reversibility), the reversed drift equals the forward drift and \(\widetilde{X} \stackrel{d}{=} X\).
• The score term connects time reversal to score-based generative models (diffusion models in ML), where learning \(\nabla\log p\) enables sampling by running the SDE in reverse.

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Exercises

Exercise 1. Let \(\mathrm{d}X_t = \mu\,\mathrm{d}t + \sigma\,\mathrm{d}W_t\) on \([0, T]\) with \(X_0 = x_0\), where \(\mu, \sigma\) are constants. The density of \(X_t\) is \(p(t, x) = \phi(x;\, x_0 + \mu t,\, \sigma^2 t)\) (Gaussian). Compute the score \(\partial_x \log p(t, x)\) and substitute into the reversed drift formula to derive the SDE for \(\widetilde{X}_t = X_{T-t}\). Show that \(\widetilde{X}\) has a Brownian bridge-type drift.

Solution to Exercise 1

With \(X_0 = x_0\), the density of \(X_t\) is Gaussian: \(p(t, x) = \phi(x;\, x_0 + \mu t,\, \sigma^2 t)\), where \(\phi(x; m, v) = (2\pi v)^{-1/2}\exp(-(x-m)^2/(2v))\).

The score is

\[ \partial_x \log p(t, x) = -\frac{x - (x_0 + \mu t)}{\sigma^2 t} \]

The reversed drift formula (constant \(a = \sigma^2\)) gives

\[ \widetilde{b}(t, x) = -b(T-t, x) + a\,\partial_x \log p(T-t, x) \]

Since \(b = \mu\) (constant drift):

\[ \widetilde{b}(t, x) = -\mu + \sigma^2 \cdot \left(-\frac{x - (x_0 + \mu(T-t))}{\sigma^2(T-t)}\right) = -\mu - \frac{x - x_0 - \mu(T-t)}{T-t} \]

Simplifying:

\[ \widetilde{b}(t, x) = -\mu - \frac{x - x_0}{T-t} + \mu = -\frac{x - x_0}{T-t} + \frac{-\mu(T-t) + \mu(T-t)}{T-t} \]

Wait — let us be more careful:

\[ \widetilde{b}(t, x) = -\mu - \frac{x - x_0 - \mu(T-t)}{T-t} = -\mu - \frac{x - x_0}{T-t} + \mu = -\frac{x - x_0}{T-t} \]

The reversed SDE is

\[ \mathrm{d}\widetilde{X}_t = -\frac{\widetilde{X}_t - x_0}{T - t}\,\mathrm{d}t + \sigma\,\mathrm{d}\widetilde{W}_t \]

This is a Brownian bridge drift: \(\widetilde{X}_0 = X_T \sim \mathcal{N}(x_0 + \mu T, \sigma^2 T)\) and the drift pulls \(\widetilde{X}_t\) toward \(x_0\) as \(t \to T\), so \(\widetilde{X}_T = x_0 = X_0\). The process is a Brownian bridge from the forward endpoint back to the forward starting point.

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Exercise 2. Consider the Ornstein–Uhlenbeck process \(\mathrm{d}X_t = -\theta X_t\,\mathrm{d}t + \sigma\,\mathrm{d}W_t\) with \(\theta > 0\), started from the invariant distribution \(X_0 \sim \mathcal{N}(0, \sigma^2/(2\theta))\). Show that the reversed drift \(\widetilde{b}(t, x)\) equals the forward drift \(b(x) = -\theta x\), confirming reversibility. What property of the OU process makes this work?

Solution to Exercise 2

The OU process has \(b(x) = -\theta x\), \(a = \sigma^2\). Started from the invariant distribution \(X_0 \sim \pi = \mathcal{N}(0, \sigma^2/(2\theta))\), the process is stationary, so \(p(t, x) = \pi(x)\) for all \(t\).

The score is

\[ \partial_x \log \pi(x) = \partial_x \left(-\frac{\theta x^2}{\sigma^2}\right) = -\frac{2\theta x}{\sigma^2} \]

The reversed drift is

\[ \widetilde{b}(t, x) = -b(T-t, x) + a\,\partial_x \log p(T-t, x) = -(-\theta x) + \sigma^2 \cdot \left(-\frac{2\theta x}{\sigma^2}\right) = \theta x - 2\theta x = -\theta x \]

Therefore \(\widetilde{b}(t, x) = -\theta x = b(x)\), confirming that the reversed drift equals the forward drift.

What makes this work: The OU process is a gradient diffusion — the drift \(b(x) = -\theta x = -\nabla V(x)\) with \(V(x) = \frac{\theta}{2}x^2\) and constant diffusion \(\sigma\). Gradient diffusions with constant noise are reversible (satisfy detailed balance) with respect to their invariant measure \(\pi \propto e^{-V}\). Reversibility means \((X_t)_{0 \le t \le T} \stackrel{d}{=} (X_{T-t})_{0 \le t \le T}\) when started from \(\pi\), which forces \(\widetilde{b} = b\). The key property is the self-adjointness of \(\mathcal{L}\) in \(L^2(\pi)\).

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Exercise 3. Let \(\mathrm{d}X_t = b(X_t)\,\mathrm{d}t + \sigma\,\mathrm{d}W_t\) with constant \(\sigma > 0\) and a smooth density \(p(t, x) > 0\). Starting from the forward Fokker–Planck equation

\[ \partial_t p = -\partial_x(b\,p) + \frac{\sigma^2}{2}\,\partial_x^2 p, \]

derive the reversed drift formula \(\widetilde{b}(t, x) = -b(T-t, x) + \sigma^2\,\partial_x \log p(T-t, x)\) by requiring that the reversed density \(\widetilde{p}(t, x) = p(T-t, x)\) satisfies its own Fokker–Planck equation.

Solution to Exercise 3

The forward Fokker–Planck equation for \(p(t,x)\) is

\[ \partial_t p = -\partial_x(bp) + \frac{\sigma^2}{2}\partial_x^2 p \]

The reversed density is \(\widetilde{p}(t,x) = p(T-t, x)\). This must satisfy its own Fokker–Planck equation with drift \(\widetilde{b}\):

\[ \partial_t \widetilde{p} = -\partial_x(\widetilde{b}\,\widetilde{p}) + \frac{\sigma^2}{2}\partial_x^2 \widetilde{p} \]

Since \(\partial_t \widetilde{p}(t,x) = -\partial_t p(T-t, x)\), substituting the forward Fokker–Planck:

\[ -\partial_t p(T-t,x) = \partial_x(b\,p)(T-t,x) - \frac{\sigma^2}{2}\partial_x^2 p(T-t,x) = \partial_x(b\,\widetilde{p}) - \frac{\sigma^2}{2}\partial_x^2\widetilde{p} \]

Equating the two expressions for \(\partial_t\widetilde{p}\):

\[ \partial_x(b\,\widetilde{p}) - \frac{\sigma^2}{2}\partial_x^2\widetilde{p} = -\partial_x(\widetilde{b}\,\widetilde{p}) + \frac{\sigma^2}{2}\partial_x^2\widetilde{p} \]

Rearranging:

\[ \partial_x(\widetilde{b}\,\widetilde{p}) = -\partial_x(b\,\widetilde{p}) + \sigma^2\,\partial_x^2\widetilde{p} \]

Integrating with respect to \(x\) (the constant of integration is zero for integrability):

\[ \widetilde{b}\,\widetilde{p} = -b\,\widetilde{p} + \sigma^2\,\partial_x\widetilde{p} \]

Dividing by \(\widetilde{p} > 0\):

\[ \widetilde{b}(t,x) = -b(T-t, x) + \sigma^2\,\frac{\partial_x\widetilde{p}(t,x)}{\widetilde{p}(t,x)} = -b(T-t, x) + \sigma^2\,\partial_x\log p(T-t, x) \]

This completes the derivation of the reversed drift formula. \(\square\)

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Exercise 4. Explain why the reversed Brownian motion \(\widetilde{W}_t\) is adapted to the backward filtration \(\widetilde{\mathcal{F}}_t = \sigma(X_s : T - t \le s \le T)\) and is generally not adapted to the forward filtration \((\mathcal{F}_t)\). Why does this distinction matter when writing the reversed SDE?

Solution to Exercise 4

The reversed process \(\widetilde{X}_t = X_{T-t}\) is defined by "reading the path backwards." The reversed Brownian motion \(\widetilde{W}_t\) is constructed via the Doob–Meyer decomposition of \(\widetilde{X}_t\) as a semimartingale with respect to the backward filtration.

The backward filtration is \(\widetilde{\mathcal{F}}_t = \sigma(X_s : T-t \le s \le T) = \sigma(\widetilde{X}_r : 0 \le r \le t)\). This encodes information about the future of the original process (from time \(T-t\) to \(T\)), which is the past of the reversed process.

\(\widetilde{W}_t\) is adapted to \(\widetilde{\mathcal{F}}_t\) because it is constructed from the martingale part of \(\widetilde{X}\) in the backward decomposition. However, \(\widetilde{W}_t\) is generally not adapted to the forward filtration \(\mathcal{F}_t = \sigma(X_s : 0 \le s \le t)\) because knowing the path from \(0\) to \(t\) does not determine the reversed martingale increments (which depend on the path from \(T-t\) to \(T\)).

Why this matters: The reversed SDE \(\mathrm{d}\widetilde{X}_t = \widetilde{b}\,\mathrm{d}t + \sigma\,\mathrm{d}\widetilde{W}_t\) is an Itô equation with respect to \((\widetilde{\mathcal{F}}_t)\), not \((\mathcal{F}_t)\). If one mistakenly uses the forward filtration, the stochastic integral \(\int \sigma\,\mathrm{d}\widetilde{W}\) is not well-defined (the integrand must be adapted to the filtration of the driving Brownian motion). Conflating the two filtrations leads to incorrect applications of Itô's formula and erroneous drift calculations.

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Exercise 5. Consider a two-dimensional diffusion with constant diffusion matrix \(a = I\) (the \(2 \times 2\) identity) and drift \(b(x_1, x_2) = (-x_1 + x_2,\, -x_1 - x_2)^\top\). Suppose the process is started from its invariant distribution. Is this process reversible? Compute the reversed drift \(\widetilde{b}\) and compare it to \(b\).

Solution to Exercise 5

The drift is \(b(x) = (-x^1 + x^2,\, -x^1 - x^2)^\top\), written as \(b(x) = Bx\) with \(B = \begin{pmatrix}-1&1\\-1&-1\end{pmatrix}\), and \(a = I\).

The eigenvalues of \(B\) are \(-1 \pm i\) (complex with negative real part), so \(B\) is stable and an invariant measure exists. As computed in the invariant measures chapter, the invariant covariance \(\Sigma\) solves \(B\Sigma + \Sigma B^\top + I = 0\), giving \(\Sigma = \frac{1}{2}I\).

Reversibility check: A linear diffusion \(\mathrm{d}X = BX\,\mathrm{d}t + \mathrm{d}W\) with \(a = I\) is reversible if and only if \(B\) is symmetric. Here \(B \ne B^\top\) (the off-diagonal entries \(1\) and \(-1\) differ), so the process is not reversible.

Reversed drift: With stationary \(p(t,x) = \pi(x)\) (since we start from the invariant distribution), the score is

\[ \nabla\log\pi(x) = -\Sigma^{-1}x = -2x \]

The reversed drift is

\[ \widetilde{b}(t,x) = -b(x) + a\,\nabla\log\pi(x) = -Bx + I\cdot(-2x) = (-B - 2I)x \]

Computing:

\[ -B - 2I = \begin{pmatrix}1&-1\\1&1\end{pmatrix} - \begin{pmatrix}2&0\\0&2\end{pmatrix} = \begin{pmatrix}-1&-1\\1&-1\end{pmatrix} \]

So \(\widetilde{b}(x) = B^\top x\) where \(B^\top = \begin{pmatrix}-1&-1\\1&-1\end{pmatrix}\).

Comparing: \(b(x) = Bx\) and \(\widetilde{b}(x) = B^\top x\). Since \(B \ne B^\top\), the reversed drift differs from the forward drift, confirming non-reversibility. The reversed process has the rotation component flipped (clockwise instead of counterclockwise).

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Exercise 6. In the context of score-based generative models, the forward process is often taken as \(\mathrm{d}X_t = -\frac{1}{2}\beta(t)\,X_t\,\mathrm{d}t + \sqrt{\beta(t)}\,\mathrm{d}W_t\) for a noise schedule \(\beta(t) > 0\). Write down the reversed SDE using the time-reversal formula. Identify the score term \(\nabla_x \log p(t, x)\) that must be learned. Why does the forward process eventually converge to an (approximately) standard Gaussian as \(t \to \infty\)?

Solution to Exercise 6

The forward SDE is \(\mathrm{d}X_t = -\frac{1}{2}\beta(t)X_t\,\mathrm{d}t + \sqrt{\beta(t)}\,\mathrm{d}W_t\), with time-dependent \(b(t,x) = -\frac{1}{2}\beta(t)x\), \(\sigma(t) = \sqrt{\beta(t)}\), and \(a(t) = \beta(t)\).

Since \(a\) is spatially constant, the reversed drift formula gives

\[ \widetilde{b}(t,x) = -b(T-t, x) + a(T-t)\,\nabla_x\log p(T-t, x) = \frac{1}{2}\beta(T-t)\,x + \beta(T-t)\,\nabla_x\log p(T-t, x) \]

The reversed SDE is

\[ \mathrm{d}\widetilde{X}_t = \left[\frac{1}{2}\beta(T-t)\,\widetilde{X}_t + \beta(T-t)\,\nabla_x\log p(T-t, \widetilde{X}_t)\right]\mathrm{d}t + \sqrt{\beta(T-t)}\,\mathrm{d}\widetilde{W}_t \]

The score term that must be learned is \(s_\theta(x, t) \approx \nabla_x\log p(t, x)\) — the gradient of the log-density of the forward process at time \(t\).

Convergence to Gaussian: The forward SDE is a time-inhomogeneous OU process. Its solution (with \(X_0 = x_0\)) is

\[ X_t = x_0\,e^{-\frac{1}{2}\int_0^t \beta(s)\,\mathrm{d}s} + \int_0^t e^{-\frac{1}{2}\int_s^t \beta(r)\,\mathrm{d}r}\sqrt{\beta(s)}\,\mathrm{d}W_s \]

As \(t \to \infty\), the first term (the "signal") decays to zero provided \(\int_0^\infty \beta(s)\,\mathrm{d}s = \infty\). The variance of the second term (the "noise") converges to \(\int_0^\infty \beta(s)e^{-\int_s^\infty \beta(r)\,\mathrm{d}r}\,\mathrm{d}s = 1\). Therefore \(X_t \to \mathcal{N}(0, I)\) in distribution as \(t \to \infty\). The noise schedule \(\beta(t)\) controls the rate at which the data distribution is destroyed and replaced by standard Gaussian noise.

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Exercise 7. Explain the connection between time reversal and Doob's \(h\)-transform. Specifically, for a diffusion with constant \(a\) and forward drift \(b\), show that the reversed measure on path space can be obtained by an \(h\)-transform with \(h(t, x) = p(T - t, x)\), where \(p\) is the forward density. Verify that \(h\) is a space-time harmonic function for the adjoint operator.

Solution to Exercise 7

Doob's \(h\)-transform for a diffusion with generator \(\mathcal{L}\) and a positive function \(h > 0\) defines a new process with generator

\[ \mathcal{L}^h f = \frac{1}{h}\mathcal{L}(hf) - \frac{f}{h}\mathcal{L}h \]

For the space-time setting with generator \(\partial_t + \mathcal{L}\), a function \(h(t,x) > 0\) defines a new drift. In the constant-\(a\) case, \(\mathcal{L}f = b \cdot \nabla f + \frac{1}{2}\text{tr}(a\,\nabla^2 f)\), and the \(h\)-transformed process has drift

\[ b^h(t,x) = b(t,x) + a\,\frac{\nabla_x h(t,x)}{h(t,x)} = b(t,x) + a\,\nabla_x\log h(t,x) \]

Time reversal as \(h\)-transform: Set \(h(t,x) = p(T-t, x)\). The \(h\)-transformed drift starting from the negated forward drift (accounting for time reversal) is

\[ \widetilde{b}(t,x) = -b(T-t, x) + a\,\nabla_x\log h(t,x) = -b(T-t, x) + a\,\nabla_x\log p(T-t, x) \]

which is exactly the reversed drift formula.

Verification that \(h\) is space-time harmonic for the adjoint: We need to show that \((\partial_t + \mathcal{L}^*)h = 0\), where \(\mathcal{L}^*\) is the formal adjoint of \(\mathcal{L}\). For constant \(a\), \(\mathcal{L}^*g = -\nabla\cdot(bg) + \frac{1}{2}\text{tr}(a\,\nabla^2 g)\).

With \(h(t,x) = p(T-t, x)\), we have \(\partial_t h(t,x) = -\partial_t p(T-t, x)\). The forward Fokker–Planck equation states \(\partial_t p = \mathcal{L}^* p\), so

\[ \partial_t h(t,x) = -(\mathcal{L}^* p)(T-t, x) \]

Meanwhile \(\mathcal{L}^* h(t, \cdot) = (\mathcal{L}^* p)(T-t, \cdot)\) (since \(h(t,x) = p(T-t,x)\) and \(\mathcal{L}^*\) acts on \(x\)). Therefore

\[ (\partial_t + \mathcal{L}^*)h = -\mathcal{L}^* p(T-t, \cdot) + \mathcal{L}^* p(T-t, \cdot) = 0 \]

confirming that \(h(t,x) = p(T-t, x)\) is space-time harmonic for \(\partial_t + \mathcal{L}^*\). \(\square\)