Langevin splittings and invariant measure error

Put $S=\sqrt{2\gamma {\beta }^{-1}}{M}^{1\mathrm{/}2}$. Multiplying the O equation by the deterministic integrating factor $e^{\gamma t}$ and applying [F3] componentwise gives $d(e^{\gamma t}{P}_t)=e^{\gamma t}Sd{W}_t$. Integrating from $0$ to $h$ and multiplying by $e^{-\gamma h}$ yields [F1, F2, F3, A1, given, algebra] $P_h=e^{-\gamma h}{P}_0+{\int }_0^h{e}^{-\gamma (h-s)}Sd{W}_s\mathrm{.}$ The integrand is deterministic and square-integrable. By [F2] the integral is centered Gaussian, independent of the starting sigma algebra, with covariance ${\int }_0^h{e}^{-2\gamma (h-s)}S{S}^{T}ds=2\gamma {\beta }^{-1}M{\int }_0^h{e}^{-2\gamma (h-s)}ds={\beta }^{-1}(1-e^{-2\gamma h})M\mathrm{.}$

By [F4], the displayed covariance has square root $\sqrt{{\beta }^{-1}(1-e^{-2\gamma h})}{M}^{1\mathrm{/}2}$. A centered Gaussian vector with this covariance has the same law as that matrix times $\xi \sim N_d(0,I_d)$; [F2] gives independence from the starting sigma algebra. This proves the exact transition and conditional moments.

The power bound gives $\mathrm{\parallel }{A}^{j}G{G}^T(A^T{)}^j\mathrm{\parallel }\le C^2{\theta }^{2j}\mathrm{\parallel }G{\mathrm{\parallel }}^2$. The geometric series therefore converges entrywise and in matrix norm. Each summand is symmetric positive semidefinite because $x^T{A}^{j}G{G}^T(A^T{)}^{j}x=\mathrm{\mid }{G}^T(A^T{)}^{j}x{\mathrm{\mid }}^2\ge 0$, so $P$ is symmetric positive semidefinite. Shifting the summation index gives $AP{A}^T+G{G}^T=P$.

Let $Z\sim N_d(0,P)$ and $\xi \sim N_m(0,I_m)$ be independent; [F3] supplies their finite product probability space. By [F1]–[F2], $AZ+G\xi$ has characteristic function $u\mapsto \exp [-\frac{1}{2}{u}^T(AP{A}^T+G{G}^T)u]=\exp [-\frac{1}{2}{u}^{T}Pu]$. The covariance identity in step 1.1 and [F4] show that $AZ+G\xi$ has the same law as $Z$. Thus $N_d(0,P)$ is invariant.

Let $\nu$ be any invariant probability law and let $\phi (u)=\int e^{iu\cdot x}\nu (dx)$. Invariance and independence of a fresh innovation imply [A1, F1, F2, F3, F4, F5, given, step 1.1, step 2.1, algebra] $\phi (u)=\phi (A^{T}u)\exp \text{\hspace{0.17em}⁣}(-\frac{1}{2}{u}^{T}G{G}^{T}u)\mathrm{.}$ Iterating $n$ times gives $\phi (u)=\phi ((A^T{)}^{n}u)\exp \text{\hspace{0.17em}⁣}(-\frac{1}{2}{u}^T[{\sum }_{j=0}^{n-1}{A}^{j}G{G}^T(A^T{)}^j]u)\mathrm{.}$ The power bound gives $(A^T{)}^{n}u\to 0$, so the dominated convergence theorem [F5] in the integral defining $\phi$ gives $\phi ((A^T{)}^{n}u)\to 1$. Step 1.1 makes the matrix sum converge to $P$, whence $\phi (u)=e^{-u^{T}Pu\mathrm{/}2}$, the characteristic function of the invariant Gaussian from step 2.1. By [F4], $\nu =N_d(0,P)$. If $P^{\mathrm{\prime }}$ is another finite matrix solution of the Lyapunov equation, $D=P^{\mathrm{\prime }}-P$ satisfies $D=AD{A}^T$, hence $D=A^{n}D(A^T{)}^n\to 0$ by the power bound; therefore $P^{\mathrm{\prime }}=P$. This proves uniqueness of the invariant law and covariance solution.

Write $z=(q,p)$ and $b(q,p)=(M^{-1}p,-\mathrm{\nabla }V(q)-\gamma p)$. The bounded Hessian makes $b$ globally Lipschitz with linear growth. By [F5], the phase-space Lebesgue measure in the definition of $Z$ is ${\lambda }_d\times {\lambda }_d$ on Borel sets, and Tonelli applied to the nonnegative Borel Gibbs factor gives $Z=Z_q{Z}_p,Z_p:={\int }_{{\mathbb{R}}^d}{e}^{-\beta p^T{M}^{-1}p\mathrm{/}2}dp\mathrm{.}$ Applying [F6] with $C=\beta M^{-1}$ yields $Z_p=\frac{(2\pi {)}^{d\mathrm{/}2}}{\sqrt{\det (\beta M^{-1})}}=(2\pi \mathrm{/}\beta {)}^{d\mathrm{/}2}\sqrt{\det M},$ so $Z$ is finite and positive. To verify the momentum second moment, let ${\lambda }_{\ast }>0$ be the smallest eigenvalue of $M^{-1}$. Then $p^T{M}^{-1}p\ge {\lambda }_{\ast }\mathrm{\mid }p{\mathrm{\mid }}^2$. The function $r\mapsto r{e}^{-\beta {\lambda }_{\ast }r\mathrm{/}4}$ is bounded on $[0,\mathrm{\infty })$, so for a finite $C_{\ast }$, $\mathrm{\mid }p{\mathrm{\mid }}^2{e}^{-\beta p^T{M}^{-1}p\mathrm{/}2}\le C_{\ast }{e}^{-\beta {\lambda }_{\ast }\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}4}\mathrm{.}$ The right side has finite integral by [F6] with $C=(\beta {\lambda }_{\ast }\mathrm{/}2)I$. Tonelli now gives ${\mathbb{E}}_{\mu }\mathrm{\mid }{P}_0{\mathrm{\mid }}^2<\mathrm{\infty }$, while the stated $q$-moment assumption gives ${\mathbb{E}}_{\mu }\mathrm{\mid }{Q}_0{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus the Gibbs initial state is square-integrable. With the ${\mathcal{F}}_0$-measurability and future-increment independence in the Statement, [F1] gives a unique global strong solution. Since the coefficients are time-independent, the law of the state after duration $t$ depends only on the starting state and elapsed time; write its transition operator as $P_t\varphi (z)={\mathbb{E}}_z\varphi (Q_t,P_t)$. For $\varphi (z)=\cos (u\cdot z)$ or $\sin (u\cdot z)$, the solution map is twice differentiable in its initial state: its first derivative $J_t$ solves ${\dot{J}}_t=Db(Z_t)J_t$, $J_0=I$, hence $\mathrm{\parallel }{J}_t\mathrm{\parallel }\le e^{t\mathrm{\parallel }Db{\mathrm{\parallel }}_{\mathrm{\infty }}}$; its second derivative solves the corresponding variation equation with forcing $D^{2}b(Z_t)[J_t,J_t]$ and is bounded on every finite interval because $D^{2}b$ is bounded. Difference quotients in the integral equation and these bounds justify differentiating under expectation, so $P_t\varphi$ and its first two spatial derivatives are bounded and continuous.

Uniqueness of the strong solution and independence of future Brownian increments give $P_{t+s}=P_s{P}_t$. Apply [F2] to $g=P_t\varphi$ over a short interval $s$: $(P_{s}g-g)\mathrm{/}s\to Lg$ as $s\downarrow 0$, where [A1, F1, F2, F4, given, step 1.1] $Lg=(M^{-1}p)\cdot {\mathrm{\nabla }}_{q}g+(-\mathrm{\nabla }V-\gamma p)\cdot {\mathrm{\nabla }}_{p}g+\gamma {\beta }^{-1}\mathrm{tr}(M{D}_{pp}^{2}g)\mathrm{.}$ Continuity of $Lg$ and the finite second-moment bound in [F1] give this limit. Hence ${\mathrm{\partial }}_t{P}_t\varphi =L{P}_t\varphi$; the right side is integrable against $\mu$ because the derivatives of $P_t\varphi$ are bounded, $p$ has Gaussian tails, and the stated $q$ integrability controls $\mathrm{\nabla }V$.

For the generator $L$ in step 2.1, the Gibbs density $\rho =Z^{-1}{e}^{-\beta H}$ has derivatives ${\mathrm{\nabla }}_q\rho =-\beta \mathrm{\nabla }V\rho$ and ${\mathrm{\nabla }}_p\rho =-\beta M^{-1}p\rho$. Its formal adjoint is [given, step 2.1, algebra] $L^{\ast }\rho =-{\mathrm{\nabla }}_q\cdot (M^{-1}p\rho )-{\mathrm{\nabla }}_p\cdot ((-\mathrm{\nabla }V-\gamma p)\rho )+\gamma {\beta }^{-1}\mathrm{tr}(M{D}_{pp}^2\rho )\mathrm{.}$ The $q$-transport term equals $\beta (M^{-1}p\cdot \mathrm{\nabla }V)\rho$ and the force part of the $p$-divergence equals $-\beta (\mathrm{\nabla }V\cdot M^{-1}p)\rho$, so they cancel. The friction part is $\gamma d\rho -\gamma \beta (p^T{M}^{-1}p)\rho$. Direct differentiation gives $\mathrm{tr}(M{D}_{pp}^2\rho )=({\beta }^2{p}^T{M}^{-1}p-\beta d)\rho$, so diffusion cancels friction and $L^{\ast }\rho =0$.

Define $s(x)=0$ for $x\le 0$ and $s(x)=e^{-1\mathrm{/}x}$ for $x>0$, then $\eta (r)=s(2-r)\mathrm{/}(s(2-r)+s(r-1))$. This smooth function equals one for $r\le 1$, zero for $r\ge 2$, and has bounded first and second derivatives; every right derivative of $s$ tends to zero at the origin. The product cutoffs ${\chi }_R(q,p)=\eta (\mathrm{\mid }q\mathrm{\mid }\mathrm{/}R)\eta (\mathrm{\mid }p\mathrm{\mid }\mathrm{/}R)$ are smooth because each factor is constant near its origin; they equal one when $\mathrm{\mid }q\mathrm{\mid },\mathrm{\mid }p\mathrm{\mid }\le R$, vanish when either norm is at least $2R$, and satisfy $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$, $\mathrm{\mid }{D}^2{\chi }_R\mathrm{\mid }\le C\mathrm{/}{R}^2$. For $g=P_t\varphi$, multiply the divergence terms by ${\chi }_R$ and integrate by parts coordinatewise using [F7]; product-measure integration is justified by [F5]. The $q$-cutoff boundary term is bounded by $C\mathrm{\mid }p\mathrm{\mid }\mathrm{/}R$ times the Gibbs density; the $p$-cutoff terms are bounded by $C(1+\mathrm{\mid }\mathrm{\nabla }V(q)\mathrm{\mid }+\mathrm{\mid }p\mathrm{\mid })\mathrm{/}R$ times that density, with second-derivative cutoff terms bounded by $C\mathrm{/}{R}^2$. These bounds are integrable by Gaussian $p$ tails and the hypotheses on $q$ and $\mathrm{\nabla }V$. Dominated convergence and step 3.1 therefore give $\int Lgd\mu =\int g{L}^{\ast }\rho dqdp=0$. By step 2.1, $t\mapsto \int P_t\varphi d\mu$ has derivative zero and is constant.

Apply step 4.1 to $\varphi (z)=\cos (u\cdot z)$ and $\sin (u\cdot z)$ for arbitrary $u\in {\mathbb{R}}^{2d}$. The resulting scalar characteristic functions of projections of $\mu P_t$ and $\mu$ agree; [F3] and the Cramér–Wold device identify the phase-space laws. Thus $\mu P_t=\mu$ for every $t\ge 0$. AC is inherited through [A1].

The canonical density factorizes over $q$ and $p$. By [F4] its normalizer is the product of the two one-dimensional Gaussian integrals, and [F5] gives $Z_q=\sqrt{2\pi \mathrm{/}(\beta k)}$ and $Z_p=\sqrt{2\pi m\mathrm{/}\beta }$. The normalized factors are centered. For $c>0$, [F6] on $[-R,R]$ gives ${\int }_{-R}^R{x}^2{e}^{-c{x}^2\mathrm{/}2}dx=\frac{1}{c}({\int }_{-R}^R{e}^{-c{x}^2\mathrm{/}2}dx-2R{e}^{-c{R}^2\mathrm{/}2})\mathrm{.}$ Letting $R\to \mathrm{\infty }$ and using $R{e}^{-c{R}^2\mathrm{/}2}\to 0$ gives ${\int }_{\mathbb{R}}{x}^2{e}^{-c{x}^2\mathrm{/}2}dx=c^{-1}{\int }_{\mathbb{R}}{e}^{-c{x}^2\mathrm{/}2}dx$; oddness gives zero mean. Therefore the canonical variances are $1\mathrm{/}(\beta k)$ and $m\mathrm{/}\beta$, with zero cross-covariance. These are the canonical values used below.

Set $v=p\mathrm{/}m$, $s=h\mathrm{/}2$, and ${\sigma }^2={\beta }^{-1}(1-a^2)\mathrm{/}m$. Applying [F2] in sequence and using $F(q)=-kq$ gives [F1, F2, F3, given, algebra] $q^{\mathrm{\prime }}=(1-r(1+a))q+s(1+a)v+s\sigma \xi ,$ $v^{\mathrm{\prime }}=-s{\omega }^2(1+a)(1-r)q+(a-r(1+a))v+(1-r)\sigma \xi ,$ where $\xi \sim N(0,1)$. Thus $z^{\mathrm{\prime }}=(q^{\mathrm{\prime }},v^{\mathrm{\prime }}{)}^T=Az+g\xi$, with $A=\left(\begin{array}{cc}1-r(1+a)& s(1+a)\\ -s{\omega }^2(1+a)(1-r)& a-r(1+a)\end{array}\right),g=\sigma \left(\begin{array}{c}s\\ 1-r\end{array}\right)\mathrm{.}$

Direct expansion gives $\det A=a$ and $\mathrm{tr}A=(1+a)(1-2r)$. The characteristic polynomial is ${\lambda }^2-(1+a)(1-2r)\lambda +a$. Since $0<a<1$, for $0<r<1$ its values at $1$ and $-1$ satisfy [F1, step 1.2, algebra] $1-\mathrm{tr}A+\det A=2r(1+a)>0,1+\mathrm{tr}A+\det A=2(1+a)(1-r)>0.$ If its roots are nonreal, they are conjugate with modulus $\sqrt{a}<1$. If they are real, their product $a>0$ makes them have the same sign. For two positive roots, the first displayed product is positive and their product is below one, so neither can be at least one. For two negative roots, the second displayed product is positive and their product is below one, so neither can be at most $-1$. Hence both roots have modulus less than one. At $r=1$, the polynomial has root $-1$. At $r>1$, its value at $-1$ is negative while it tends to positive infinity as $\lambda \to -\mathrm{\infty }$, so it has a root ${\lambda }_u<-1$; its other root is $a\mathrm{/}{\lambda }_u$ and has modulus less than one. Therefore $\rho (A)<1$ exactly when $0<r<1$.

For $0<r<1$, the characteristic roots have modulus less than one by step 2.1. If they differ, Cayley–Hamilton writes $A^n$ as a linear combination of their $n$th powers. If they coincide at $\lambda$, Cayley–Hamilton gives $(A-\lambda I{)}^2=0$, so $A^n={\lambda }^{n}I+n{\lambda }^{n-1}(A-\lambda I)$; for $\mathrm{\mid }\lambda \mathrm{\mid }<1$ this is bounded by $C{\theta }^n$ for any fixed $\mathrm{\mid }\lambda \mathrm{\mid }<\theta <1$. Thus the power-decay hypothesis of [L1] holds. Its discrete Lyapunov covariance in $(q,v)$ coordinates is [F1, F3, L1, step 1.2, step 2.1, algebra] $P=\frac{1}{\beta m}\left(\begin{array}{cc}{\omega }^{-2}& 0\\ 0& 1-r\end{array}\right)\mathrm{.}$ Indeed, substituting $A,g$ from step 1.2 gives $AP{A}^T+g{g}^T=P$: the off-diagonal identity reduces to $s(1-r)[(1+a)(a-1)+(1-a^2)]=0$; the first diagonal identity reduces, after multiplication by $\beta m{\omega }^2$, to $[1-r(1+a){]}^2+r(1+a)[2-r(1+a)]=1;$ and the second diagonal identity, after division by $1-r$ and multiplication by $\beta m$, reduces to $[a-r(1+a){]}^2+r(1+a{)}^2(1-r)+(1-a^2)(1-r)=1.$ Both displayed scalar identities follow by expansion and $r=s^2{\omega }^2$. Thus [L1] gives the unique invariant Gaussian with this covariance. Transforming back by $p=mv$ gives $\mathrm{Var}(Q)=1\mathrm{/}(\beta k)$, $\mathrm{Var}(P)=m(1-r)\mathrm{/}\beta$, and zero covariance.

For $r>1$, step 2.1 gives a real left eigenvector $\mathrm{\ell }$ for ${\lambda }_u<-1$. The innovation vector excites that mode: direct multiplication gives $\det [g,Ag]=-2{\sigma }^{2}s(1-r)\mathrm{\ne }0$, so ${\mathrm{\ell }}^{T}g\mathrm{\ne }0$ (otherwise $g$ would lie in the other right eigenspace and $g,Ag$ would be dependent). If an invariant probability existed, its projection $X={\mathrm{\ell }}^{T}z$ would satisfy $X^{\mathrm{\prime }}={\lambda }_{u}X+\zeta$ with independent nondegenerate centered normal $\zeta ={\mathrm{\ell }}^{T}g\xi$. Its characteristic function would obey $\phi (t)=\phi ({\lambda }_{u}t)e^{-\mathrm{Var}(\zeta )t^2\mathrm{/}2}$. Iterating and using $\mathrm{\mid }\phi \mathrm{\mid }\le 1$ forces $\phi (t)=0$ for every $t\mathrm{\ne }0$. For any probability law of a finite real random variable $X$, [F7] and dominated convergence applied to $e^{itX}\to 1$ with majorant $1$ give $\phi (t)\to 1$ as $t\to 0$, contradicting this forced vanishing. Hence no invariant probability exists for $r>1$. Since $r=h^{2}k\mathrm{/}(4m)$, the threshold is $h=2\sqrt{m\mathrm{/}k}$. The boundary $r=1$ has an eigenvalue $-1$, so [L1] does not give a unique invariant law there. This proves the stated stable, boundary, and unstable cases without making any nonharmonic invariant-law claim.

First suppose $\mathbb{E}\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2<\mathrm{\infty }$. Shift time to $t$ and use $W_{t+u}-W_t$ with filtration ${\mathcal{F}}_{t+u}$. Brownian motion relative to this filtration has future increments independent of ${\mathcal{F}}_t$, so the initial state has the independence required in [F5]. The drift $b$ is globally Lipschitz with linear growth and $G$ is constant; [F5] therefore supplies the exact solution and finite second moments used in the following estimates. We first establish the estimates under this extra integrability condition, then remove it.

Set $s=h\mathrm{/}2$, $a=e^{-\gamma h}$, $\mathrm{\Delta }W=W_{t+h}-W_t$, and $\eta ={\int }_0^h{e}^{-\gamma (h-r)}Sd{W}_{t+r}$. Writing the BAOAB stages from [F2] gives [F1, F2, F3, F4, given, algebra] $q^{\mathrm{\prime }}=q+s(1+a)M^{-1}p+s^2(1+a)M^{-1}F(q)+s{M}^{-1}\eta ,p^{\mathrm{\prime }}=ap+asF(q)+\eta +sF(q^{\mathrm{\prime }})\mathrm{.}$ Subtracting $z+hb(z)+G\mathrm{\Delta }W$ yields $R_q^B=s(a-1)M^{-1}p+s^2(1+a)M^{-1}F(q)+s{M}^{-1}\eta ,$ $R_p^B=(a-1+\gamma h)p+s((a-1)F(q)+F(q^{\mathrm{\prime }})-F(q))+\eta -S\mathrm{\Delta }W\mathrm{.}$ Since $\mathrm{\mid }a-1\mathrm{\mid }\le Ch$ and $\mathrm{\mid }a-1+\gamma h\mathrm{\mid }\le C{h}^2$, the deterministic terms have size at most $C{h}^2(1+\mathrm{\mid }z\mathrm{\mid })$. The displayed formula and linear growth give $\mathbb{E}[\mathrm{\mid }{q}^{\mathrm{\prime }}-q\mathrm{\mid }\mid {\mathcal{F}}_t]\le Ch(1+\mathrm{\mid }z\mathrm{\mid })$ and $\mathbb{E}[\mathrm{\mid }{q}^{\mathrm{\prime }}-q{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^2(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$. Lipschitz continuity of $F$ and [F4] therefore give $\mathrm{\mid }\mathbb{E}[R^B\mid {\mathcal{F}}_t]\mathrm{\mid }\le C{h}^2(1+\mathrm{\mid }z\mathrm{\mid })$ and $\mathbb{E}[\mathrm{\mid }{R}^B{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^3(1+\mathrm{\mid }z{\mathrm{\mid }}^2)$.

For two starting states driven by the same $\eta$, subtract the expression for $q^{\mathrm{\prime }}$ in step 1.2. The result and the Lipschitz bound on $F$ give $\mathrm{\mid }{q}^{\mathrm{\prime }}(z)-q^{\mathrm{\prime }}(\tilde{z})\mathrm{\mid }\le C\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. Subtracting the two $R_q^B$ formulas then gives $\mathrm{\mid }{R}_q^B(z)-R_q^B(\tilde{z})\mathrm{\mid }\le C{h}^2\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. The $R_p^B$ formula gives the same bound for its linear and $(a-1)F$ terms; its last force difference is bounded by $sL(\mathrm{\mid }{q}^{\mathrm{\prime }}(z)-q^{\mathrm{\prime }}(\tilde{z})\mathrm{\mid }+\mathrm{\mid }q-\tilde{q}\mathrm{\mid })\le Ch\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$. Combining components proves $\mathrm{\mid }{R}^B(z)-R^B(\tilde{z})\mathrm{\mid }\le Ch\mathrm{\mid }z-\tilde{z}\mathrm{\mid }$.

With $b,G$ from step 1.2, the exact residual is $R^E={\int }_t^{t+h}(b(Z_r)-b(Z_t))dr$. For $0\le u\le h$, the integral equation, linear growth of $b$, Cauchy–Schwarz in time, and [F4] give, with $m(u)=\mathbb{E}[1+\mathrm{\mid }{Z}_{t+u}{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]$, [A1, F1, F4, F5, F6, step 1.2] $m(u)\le C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)+C{\int }_0^{u}m(v)dv\mathrm{.}$ Set $M(u)=C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)+C{\int }_0^{u}m(v)dv$ with the constant enlarged so $m\le M$. Then $M^{\mathrm{\prime }}(u)=Cm(u)\le CM(u)$ almost everywhere, so $(e^{-Cu}M(u){)}^{\mathrm{\prime }}\le 0$ and $m(u)\le C(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$ for $u\le 1$. Substitution in the integral equation gives $\mathbb{E}[\mathrm{\mid }{Z}_{t+u}-Z_t{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le Cu(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$, and Cauchy–Schwarz in time gives $\mathbb{E}[\mathrm{\mid }{R}^E{\mathrm{\mid }}^2\mid {\mathcal{F}}_t]\le C{h}^3(1+\mathrm{\mid }{Z}_t{\mathrm{\mid }}^2)$. For its mean, the existence of the second derivative makes $DF$ continuous, so $F\in C^1$. The ordinary chain rule along $Q$ and the linear momentum equation [F6] give $db(Z_r)=a(Z_r)dr+\widehat{G}d{W}_r$, where $a(q,p)=(M^{-1}(F(q)-\gamma p),DF(q)M^{-1}p-\gamma (F(q)-\gamma p))$ and $\widehat{G}=(M^{-1}S,-\gamma S{)}^T$. Bounded $DF$ and linear growth of $F$ give $\mathrm{\mid }a(z)\mathrm{\mid }\le C(1+\mathrm{\mid }z\mathrm{\mid })$. The constant-diffusion integral has conditional mean zero, so the preceding moment bound yields $\mathrm{\mid }\mathbb{E}[b(Z_{t+u})-b(Z_t)\mid {\mathcal{F}}_t]\mathrm{\mid }\le Cu(1+\mathrm{\mid }{Z}_t\mathrm{\mid })$; integration yields $\mathrm{\mid }\mathbb{E}[R^E\mid {\mathcal{F}}_t]\mathrm{\mid }\le C{h}^2(1+\mathrm{\mid }{Z}_t\mathrm{\mid })$.