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Forward Kolmogorov duality gives reversed martingales

Statement

Let $X$ be a continuous Markov process on $[0, T]$ in $R^{d}$ with positive transition densities $q$ and positive $C^{1, 2}$ marginal densities $p$ . Suppose its marginal and transition identities define the Bayes-reversed kernels. Write $L_{r} f = b_{r, i} \partial_{i} f + \frac{1}{2} a_{r, i j} \partial_{i j} f$ , with Borel coefficients and symmetric $a$ , and $L_{r}^{*} z = - \partial_{i} (b_{r, i} z) + \frac{1}{2} \partial_{i} \partial_{j} (a_{r, i j} z)$ . Assume $\partial_{r} p_{r} = L_{r}^{*} p_{r}$ classically. Put $A_{i j} (r, x) = p_{r} (x) a_{r, i j} (x)$ , suppose its first state derivatives exist, and define $C_{i} = - p_{r} b_{r, i} + \sum_{j} \partial_{j} A_{i j}$ . Impose the following forward analytic hypotheses. For every compact $K \subset R^{d}$ , $\int_{0}^{T} ⁣ \int_{K} [\sum_{i} ∣ C_{i} ∣ + \sum_{i, j} p_{r} ∣ a_{r, i j} ∣] d x d r < \infty .$ For every $t \in (0, T]$ and $f \in C_{c}^{\infty} (R^{d})$ , the actual forward conditional expectation $g_{t, f} (r, x) = \int q (r, x; t, y) f (y) d y, 0 < r < t,$ has a $C^{1, 2}$ version on $(0, t) \times R^{d}$ satisfying the backward Kolmogorov equation $\partial_{r} g_{t, f} + L_{r} g_{t, f} = 0$ . For every $ϕ \in C_{c}^{\infty} (R^{d})$ , assume the spatial adjoint identity $\int p_{r} ϕ L_{r} g_{t, f} d x = \int g_{t, f} L_{r}^{*} (p_{r} ϕ) d x$ holds for almost every $r \in (0, t)$ ; the integrals are finite. These are assumptions on the original forward transition family, not on a reversed process.

Define $D_{r} ϕ = p_{r}^{- 1} {L_{r}^{*} (p_{r} ϕ) - ϕ L_{r}^{*} p_{r}}$ , using the classical product rule. Then $Y_{u} = X_{T - u}$ , with its natural reversed filtration $G_{u} = σ (Y_{v} : 0 \leq v \leq u)$ , has the Bayes-reversed kernels and, for each such $ϕ$ , $ϕ (Y_{u}) - ϕ (Y_{0}) - \int_{0}^{u} D_{T - v} ϕ (Y_{v}) d v$ is a true martingale. In particular $D_{T - u}$ is its extended martingale generator on compact smooth tests. No pointwise small-time transition-quotient limit is asserted.

Statement references

Anderson, Section 5, printed pp. 322–324: uses both forward and backward Kolmogorov equations. The finite-time martingale identity under the explicit hypotheses above is proved below.

Facts & Assumptions

Given: the forward process, densities, operators and analytic hypotheses in the Statement.

[F1]

The marginal and Chapman–Kolmogorov identities make the Bayes kernels probability kernels; Bayes' formula identifies them with the reversed conditional laws. Bayes-reversed transition kernels for a diffusion

[F2]

Fubini applies to an absolutely integrable time-state integrand, and dominated convergence applies to random variables bounded by an integrable random variable. Fubini's theorem for L^1 functions on a sigma-finite product Dominated convergence

[F3]

Given compact $K$ in an open set $O \subset R^{d}$ , there is a smooth compactly supported function in $O$ , between zero and one, equal to one near $K$ . Test function cutoffs and euclidean localization

[F4]

A lambda-system containing a generating pi-system contains the generated sigma-algebra. Dynkin's pi-lambda theorem

Proof

Given: all hypotheses and notation in the Statement.

1.1givenF1F2algebra

The adjoint product rule and symmetry give $p_{r} D_{r} ϕ = (- p_{r} b_{r, i} + \partial_{j} A_{i j}) \partial_{i} ϕ + \frac{1}{2} A_{i j} \partial_{i j} ϕ .$ If $K$ contains the support of $ϕ$ , boundedness of its derivatives and the first displayed integrability hypothesis imply $\int_{0}^{T} E ∣ D_{r} ϕ (X_{r}) ∣ d r = \int_{0}^{T} ⁣ \int_{K} p_{r} ∣ D_{r} ϕ ∣ d x d r < \infty .$ Thus every random time integral below is integrable. Bayes' formula gives the reversed Markov kernels and their composition by [F1].

2.1givenF2step 1.1algebra

Fix $0 < s < t \leq T$ and $f \in C_{c}^{\infty}$ , and write $g = g_{t, f}$ . On every compact subinterval of $(0, t)$ , the product $p_{r} (x) ϕ (x) g (r, x)$ has a continuous time derivative and compact state support. The ordinary fundamental theorem of calculus and [F2] therefore differentiate its state integral. The forward marginal equation, backward equation and assumed spatial adjoint identity yield $\frac{d}{d r} \int p_{r} ϕ g_{r} d x = \int ϕ g_{r} L_{r}^{*} p_{r} d x - \int p_{r} ϕ L_{r} g_{r} d x = - \int p_{r} g_{r} D_{r} ϕ d x .$ This last function is integrable on $(0, t)$ because $∣ g_{r} ∣ \leq ∥ f ∥_{\infty}$ and step 1.1 holds. Also $\int p_{r} ϕ g_{r} d x = E [ϕ (X_{r}) f (X_{t})]$ . Integrate between interior times and then approach $s$ and $t$ ; continuity of $X$ and boundedness of $ϕ, f$ justify the endpoint limits by [F2]. The same argument lets $s ↓ 0$ . Hence, for $0 \leq s < t \leq T$ , $E ⁣ [f (X_{t}) {ϕ (X_{s}) - ϕ (X_{t}) - \int_{s}^{t} D_{r} ϕ (X_{r}) d r}] = 0.$ Fubini in the last term is justified by step 1.1 and $∥ f ∥_{\infty}$ .

3.1F2F3F4step 2.1

Denote the integrable expression in braces by $Z$ . For an open $O$ , put $K_{n} = {x : ∣ x ∣ \leq n, dist (x, O^{c}) \geq 1 / n}$ , taking distance to the empty set as infinity. Each $K_{n}$ is compact in $O$ , and every $x \in O$ belongs to all sufficiently late $K_{n}$ . Take the cutoffs $f_{n}$ by the explicit rational-ball construction in [F3] (using the least code of a finite rational cover at each $n$ ), supported in $O$ , with $0 \leq f_{n} \leq 1$ and $f_{n} = 1$ on $K_{n}$ . Then $f_{n} \to 1_{O}$ pointwise. Step 2.1 and dominated convergence with bound $∣ Z ∣$ give $E [Z 1_{{X_{t} \in O}}] = 0$ . This includes $O = R^{d}$ . The Borel sets $B$ for which $E [Z 1_{{X_{t} \in B}}] = 0$ form a lambda-system, since $Z$ is integrable and its total expectation is zero. Open sets are a generating pi-system, so [F4] proves $E [Z ∣ X_{t}] = 0$ .

4.1F1F2step 1.1step 3.1∎

Set $s = T - v$ and $t = T - u$ for $0 \leq u < v \leq T$ . Step 3.1 gives the zero conditional expectation of the compensated increment between $u$ and $v$ given $Y_{u}$ . The reversed Markov property from step 1.1 makes the same conditional expectation given $G_{u}$ zero: first this holds for bounded functions of finitely many future reversed coordinates by the Markov factorization, then bounded monotone approximation gives it for the future reversed path segment, and [F2] and step 1.1 permit its integrable time integral. Thus the displayed compensated process is a true $G$ -martingale. The operator is its extended martingale generator by this identity.

Depends on

Used by

Reversed kernels, generator, and martingale problem Lemma

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

B. D. O. Anderson, Reverse-time diffusion equation models, Section 5: forward and backward Kolmogorov equations (standard reference, not scraped)