Pathwise relative-entropy error of the Langevin Euler interpolation

The finite Euler recursion and $\mathrm{\mid }g(x)\mathrm{\mid }\le \mathrm{\mid }g(0)\mathrm{\mid }+L\mathrm{\mid }x\mathrm{\mid }$ give $E\mathrm{\mid }{X}_{t_k}{\mathrm{\mid }}^2<\mathrm{\infty }$ by induction, using the finite Brownian second moments. Its interpolation has $E{\sup }_{t\le T}\mathrm{\mid }{X}_t{\mathrm{\mid }}^2<\mathrm{\infty }$: on each of the finitely many intervals its supremum is bounded by $\mathrm{\mid }{X}_{t_k}\mathrm{\mid }+{\gamma }_{k+1}\mathrm{\mid }g(X_{t_k})\mathrm{\mid }+\sqrt{2}{\sup }_{t_k\le t\le t_{k+1}}\mathrm{\mid }{W}_t-W_{t_k}\mathrm{\mid }$, whose square is integrable by [F2]. Thus every stated Euler gradient moment is finite. The diffusion law exists by [F2]. Apply [F1] on the Euler probability space with $b=-g$, $a_t=-g(X_{\eta (t)})$, $c_t=-g(X_t)$ and ${\theta }_t=\frac{g(X_{\eta (t)})-g(X_t)}{\sqrt{2}},Z_T=\exp ({\int }_0^T{\theta }_t\cdot d{W}_t-\frac{1}{2}{\int }_0^T\mathrm{\mid }{\theta }_t{\mathrm{\mid }}^{2}dt)\mathrm{.}$ It is a true positive density, and under $Q=Z_{T}P$, $X$ has diffusion path law $\mathsf{D}$.

For $h=t-t_k$, the increment representation and [F3] imply $E\mathrm{\mid }{X}_t-X_{t_k}{\mathrm{\mid }}^2=h^{2}E\mathrm{\mid }g(X_{t_k}){\mathrm{\mid }}^2+2dh\mathrm{.}$ The mixed term vanishes because $g(X_{t_k})$ is ${\mathcal{F}}_{t_k}$-measurable and square integrable, while $W_t-W_{t_k}$ is centered and independent of that sigma-algebra. Lipschitz continuity and Tonelli now give $E{\int }_0^T\mathrm{\mid }{\theta }_t{\mathrm{\mid }}^{2}dt\le \frac{L^2}{2}{\sum }_{k=0}^{N-1}\{\frac{{\gamma }_{k+1}^3}{3}E\mathrm{\mid }g(X_{t_k}){\mathrm{\mid }}^2+d{\gamma }_{k+1}^2\}<\mathrm{\infty }\mathrm{.}$ Indeed integrate $h^2$ and $2dh$ on $[0,{\gamma }_{k+1}]$. The entropy identity of [F1] yields $D(P\mathrm{\parallel }Q)=\frac{1}{2}E\int \mathrm{\mid }{\theta }_t{\mathrm{\mid }}^{2}dt$.

The exact path-law identity, rather than just its upper bound, follows because $Z_T$ is a measurable function of the Euler path. In fact $W_t=\frac{X_t-X_0+{\int }_0^{t}g(X_{\eta (s)})ds}{\sqrt{2}}$ is a Borel functional of that continuous path. On deterministic refining partitions containing all mesh points, the left-endpoint step approximations to $\theta$ converge at almost every time on every continuous path: on an open Euler interval $\theta$ is continuous and at its left endpoint it equals zero. Their squared errors are bounded by a fixed constant times $1+{\sup }_{t\le T}\mathrm{\mid }{X}_t{\mathrm{\mid }}^2$, integrable by step 1.1. Dominated convergence and the Itô isometry therefore make the corresponding stochastic sums converge to $\int \theta dW$ in $L^2(P)$. Each sum is a Borel path functional, since both $\theta$ and the recovered Brownian increments are such functionals. Choose a subsequence whose squared $L^2$ errors are at most $2^{-3j}$; Markov's inequality bounds the probability of error exceeding $2^{-j}$ by $2^{-j}$, so the union bound on the tails shows almost sure convergence. Define its Borel limit arbitrarily on the Borel nonconvergence set. Also $\int \mathrm{\mid }\theta {\mathrm{\mid }}^{2}dt$ is a Borel path functional, by ordinary interval Riemann approximation. Hence $Z_T=z(X)$ almost surely for some positive Borel function $z$ (reset its value to one where the constructed exponential is not finite and positive).

For every Borel path set $C$, $\mathsf{D}(C)=Q(X\in C)=E_P[Z_T{1}_{\{X\in C\}}]={\int }_{C}zd\mathsf{E}$. Since $z>0$ almost everywhere, the two path laws are equivalent and $d\mathsf{D}\mathrm{/}d\mathsf{E}=z$. The logarithm is integrable under $\mathsf{E}$ because its pullback equals $\int \theta dW-\frac{1}{2}\int \mathrm{\mid }\theta {\mathrm{\mid }}^{2}dt$, with the first term square integrable and the second integrable by step 2.1. Thus $D(\mathsf{E}\mathrm{\parallel }\mathsf{D})=E_P[-\log Z_T]=\frac{1}{2}E{\int }_0^T\mathrm{\mid }{\theta }_t{\mathrm{\mid }}^{2}dt=\frac{1}{4}E{\int }_0^T\mathrm{\mid }g(X_t)-g(X_{\eta (t)}){\mathrm{\mid }}^{2}dt\mathrm{.}$ Combining with step 2.1 proves the asserted upper bound.

Terminal evaluation is a measurable statistic on continuous path space. Data processing in [F1] therefore gives $D(\mathcal{L}(X_T)\mathrm{\parallel }\mathcal{L}(Y_T))\le D(\mathsf{E}\mathrm{\parallel }\mathsf{D})$. If $L=0$, $g$ is constant, Euler interpolation already solves the diffusion and both entropies and the displayed bound are zero. At $t=0$ the path laws have the same initial law; the trivial zero-horizon case likewise has entropy zero. All positive step lengths, including a single interval, were covered without a minimum-step restriction. AC has exactly the inherited uses stated in [A1].