Reversed kernels, generator, and martingale problem

Let $X_t$ have continuous paths on $[0,T]$ and be Markov on ${\mathbb{R}}^d$, meaning its conditional future law given the past depends only on its current state; nonexplosive means it remains in ${\mathbb{R}}^d$ throughout this finite interval. Let it have forward generator $L_t\varphi =b_t\cdot \mathrm{\nabla }\varphi +\frac{1}{2}{a}_t:D^2\varphi ,$ where $a_t={\sigma }_t{\sigma }_t^{\mathsf{T}}$ is symmetric and $L_t^{\ast }r=-{\mathrm{\partial }}_i(b_{t,i}r)+\frac{1}{2}{\mathrm{\partial }}_i{\mathrm{\partial }}_j(a_{t,ij}r)$. Assume positive $C^{1,2}$ marginal densities $p_t$ and positive transition densities $q(s,x;t,y)$ satisfy the marginal identity and Chapman–Kolmogorov identity, and that ${\mathrm{\partial }}_t{p}_t=L_t^{\ast }{p}_t$ and ${\mathrm{\partial }}_{t}q(s,x;t,y)=L_t^{\ast ,y}q(s,x;t,y)$ hold classically. Assume for every $\varphi \in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$ that the transition-density integrals can be differentiated in time, the resulting derivatives are integrable, and integration by parts in the state variable has no boundary term.

Define $K_{u,v}$ by the Bayes-reversed kernel formula for $0\le u\le v\le T$. Then the reversed process $Y_u=X_{T-u}$ is Markov with transitions $K_{u,v}$. For compact smooth $\varphi$, define the formal adjoint candidate operator ${\tilde{L}}_u\varphi (y)=\frac{L_{T-u}^{\ast }(p_{T-u}\varphi )(y)-\varphi (y)L_{T-u}^{\ast }{p}_{T-u}(y)}{p_{T-u}(y)}\mathrm{.}$ The product rule gives the identity ${\tilde{L}}_u\varphi ={\tilde{b}}_u\cdot \mathrm{\nabla }\varphi +\frac{1}{2}{a}_{T-u}:D^2\varphi ,{\tilde{b}}_u=-b_{T-u}+\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a_{T-u}+a_{T-u}\mathrm{\nabla }\log p_{T-u},$ with $(\mathrm{\nabla }\text{\hspace{0.17em}⁣}\cdot a{)}_i={\sum }_j{\mathrm{\partial }}_j{a}_{ij}$. These Bayes-kernel and formal product identities use only the preceding hypotheses.

For the martingale conclusion, impose additionally the following forward analytic conditions. For every compact $K\subset {\mathbb{R}}^d$, with $A_{ij}(r,x)=p_r(x)a_{r,ij}(x)$ and $C_i=-p_r{b}_{r,i}+{\sum }_j{\mathrm{\partial }}_j{A}_{ij}$, ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_K[{\sum }_i\mathrm{\mid }{C}_i\mathrm{\mid }+{\sum }_{i,j}{p}_r\mathrm{\mid }{a}_{r,ij}\mathrm{\mid }]dxdr<\mathrm{\infty }\mathrm{.}$ For every $t\in (0,T]$ and $f\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$, the actual forward conditional expectation $g_{t,f}(r,x)=\int q(r,x;t,y)f(y)dy$ has a $C^{1,2}$ version on $(0,t)\times {\mathbb{R}}^d$, solves ${\mathrm{\partial }}_r{g}_{t,f}+L_r{g}_{t,f}=0$, and satisfies $\int p_r\varphi L_r{g}_{t,f}dx=\int g_{t,f}{L}_r^{\ast }(p_r\varphi )dx$ for each compact smooth $\varphi$ and almost every $r<t$. All coefficient expressions in these conditions are Borel. These are conditions on the original forward transition family.

Under these additional forward analytic conditions, ${\tilde{L}}_u$ is the extended martingale generator of $Y$: for its natural reversed filtration ${\mathcal{G}}_u=\sigma (Y_r:0\le r\le u)$, every compact-test compensated process is a martingale, hence a local martingale. The unconditional formal candidate identity alone makes no martingale or pointwise transition-quotient assertion.

Anderson, Section 5, printed pp. 322–324: gives the reverse Kolmogorov-equation approach as context; the full kernel and martingale argument is proved here.

Tang and Zhao, Section 2.2, PDF pp. 3–5: states the reverse-drift formula and explicitly gives a heuristic derivation.