Gaussian smoothing densities, scores, and endpoint limits

Fix $0\le s<t\le T$ and put $H(r)=g(r)1_{(s,t]}(r)$. By [F4], choose elementary predictable approximants $H_n(r,\omega )={\sum }_k{\xi }_{n,k}(\omega )1_{(r_{n,k},r_{n,k+1}]}(r)$ converging to $H$ in $L^2(dr\otimes P)$. On each interval define the deterministic coefficient ${\bar{\xi }}_{n,k}=E[{\xi }_{n,k}]$ and let ${\bar{H}}_n={\sum }_k{\bar{\xi }}_{n,k}{1}_{(r_{n,k},r_{n,k+1}]}$. Boundedness of the elementary coefficients makes these expectations finite. For every $r$ in the kth interval, nonnegativity of $\mathrm{Var}({\xi }_{n,k}-H(r))$ from [F6] gives $\mathrm{\mid }{\bar{\xi }}_{n,k}-H(r){\mathrm{\mid }}^2\le E\mathrm{\mid }{\xi }_{n,k}-H(r){\mathrm{\mid }}^2$. Integrating and applying [F7], with Tonelli [F8] to exchange the time integral and expectation of this nonnegative integrand, yields $\mathrm{\parallel }{\bar{H}}_n-H{\mathrm{\parallel }}_{L^2([0,T])}^2\le E{\int }_0^T\mathrm{\mid }{H}_n-H{\mathrm{\mid }}^{2}dr\to 0$. Set $H_n^0={\bar{H}}_n{1}_{(s,t]}$; it remains a deterministic step integrand and its error cannot increase, because $H=0$ outside $(s,t]$.

Write $H_n^0={\sum }_j{a}_{n,j}{1}_{(u_{n,j},u_{n,j+1}]}$ on a partition of $[s,t]$. By [F3], its vector integral is $Z_n={\sum }_j{a}_{n,j}(W_{u_{n,j+1}}-W_{u_{n,j}})$. For $\theta \in {\mathbb{R}}^d$, independence and the Gaussian increment laws in [F1] give $E{e}^{i\theta \cdot Z_n}={\prod }_j\exp (-a_{n,j}^2(u_{n,j+1}-u_{n,j})\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}2)=\exp (-\mathrm{\mid }\theta {\mathrm{\mid }}^2{\int }_s^t(H_n^0{)}^{2}dr\mathrm{/}2)$. The same factorization holds with any indicator $1_A$, $A\in {\mathcal{F}}_s$, inserted: each increment is independent of the sigma-algebra at its left endpoint, which contains $A$ and all preceding increments, so repeated factorization gives $E[1_A{e}^{i\theta \cdot Z_n}]=P(A)E{e}^{i\theta \cdot Z_n}$.

Let $Z=M_t-M_s$. By [F5] applied to the deterministic step integrands, $E\mathrm{\mid }{Z}_n-Z{\mathrm{\mid }}^2=d{\int }_s^t\mathrm{\mid }{H}_n^0(r)-g(r){\mathrm{\mid }}^{2}dr\to 0$. Since $\mathrm{\mid }{e}^{i\theta \cdot z}-e^{i\theta \cdot w}\mathrm{\mid }\le \mathrm{\mid }\theta \mathrm{\mid }\mathrm{\mid }z-w\mathrm{\mid }$, nonnegativity of the variance of $\mathrm{\mid }{Z}_n-Z\mathrm{\mid }$ in [F6] gives convergence of both the ordinary and $1_A$-weighted characteristic functions. Also ${\int }_s^t(H_n^0{)}^{2}dr\to {\int }_s^{t}g(r{)}^{2}dr=v(t)-v(s)$ by the square-integrable norm convergence in step 1.1 and the norm-continuity result [F7]. Passing to the limit in step 1.2 therefore gives $E{e}^{i\theta \cdot Z}=\exp (-\mathrm{\mid }\theta {\mathrm{\mid }}^2(v(t)-v(s))\mathrm{/}2)$ and $E[1_A{e}^{i\theta \cdot Z}]=P(A)\exp (-\mathrm{\mid }\theta {\mathrm{\mid }}^2(v(t)-v(s))\mathrm{/}2)$ for every $\theta$ and every $A\in {\mathcal{F}}_s$.

The multivariate-normal convention [F9] identifies the Gaussian law from its one-dimensional normal projections, and [F10] gives the characteristic function in step 2.1 for those projections. For any $A\in {\mathcal{F}}_s$ with $P(A)>0$, the probability measure $B\mapsto P(Z\in B,A)\mathrm{/}P(A)$ has the same characteristic function by the second identity, so uniqueness in [F10] identifies it with that same Gaussian law. If $P(A)=0$, the joint probability is zero. Thus $Z$ is independent of ${\mathcal{F}}_s$.

By [F11], $M$ has an adapted continuous version. For any finite increasing list $0=t_0<\cdots <t_n$, the increments through time $t_{j-1}$ are ${\mathcal{F}}_{t_{j-1}}$-measurable. Step 3.1 makes the next increment independent of that sigma-algebra; induction on $j$ therefore factors the joint probabilities of all increment Borel sets into the product of their marginal probabilities. This proves mutual independence over the list.

For $b(t,x)=0$ and $\sigma (t,x)=g(t)I_d$, the state-Lipschitz constant is zero and $\mathrm{\parallel }\sigma (t,x){\mathrm{\parallel }}_F=\sqrt{d}g(t)\le \sqrt{d}{g}_{\max }$, so the linear-growth bound in [F2] holds with $K=\sqrt{d}{g}_{\max }$. Since $\mathrm{\mid }{X}_0\mathrm{\mid }\le R$ almost surely, $E\mathrm{\mid }{X}_0{\mathrm{\mid }}^2\le R^2$. The complete usual filtration, Brownian independence, and initial-value hypotheses are exactly those of [F2]; [F3] therefore gives the unique strong solution used in the definition.

Since $g(r)\ge g_{\min }>0$, $v(t)>0$ for $t>0$ and $v(t)-v(s)\ge g_{\min }^{}(t-s)>0$ for $s<t$. The one-dimensional density $(2\pi a{)}^{-1\mathrm{/}2}{e}^{-u^2\mathrm{/}(2a)}$ is obtained from the standard normal density in [F8] by the $C^1$ scaling $u\mapsto \sqrt{a}u$ in [F10], so it is positive and has integral one by [F9]. The product formula for ${\varphi }_a$ is therefore positive; Tonelli [F11] gives ${\int }_{{\mathbb{R}}^d}{\varphi }_a(z)dz={\prod }_{j=1}^{d}1=1$.

The bound ${\varphi }_{v(t)}(x-z)\le (2\pi v(t){)}^{-d\mathrm{/}2}$ and $\mu ({\mathbb{R}}^d)=1$ show $p_t(x)<\mathrm{\infty }$. Since ${\varphi }_{v(t)}(x-z)>0$ for every $z$ and $\mu$ has mass one, $p_t(x)>0$. Tonelli and translation change of variables give ${\int }_{{\mathbb{R}}^d}{p}_t(x)dx={\int }_{{\mathbb{R}}^d}({\int }_{{\mathbb{R}}^d}{\varphi }_{v(t)}(x-z)dx)\mu (dz)={\int }_{{\mathbb{R}}^d}1\mu (dz)=1.$ The same argument proves that $q(s,x;t,\cdot )$ is positive, finite, and normalized for every $s<t$ and $x$.

Write $Z_t=X_t-X_0={\int }_0^{t}g(r)d{W}_r$. By [F4], $Z_t$ is independent of $X_0$, which is measurable at time zero, and has law $N_d(0,v(t)I_d)$. By [F12] this law is the product of $d$ scalar normal laws of variance $v(t)$; [F8]–[F11] show that this product law has density ${\varphi }_{v(t)}$. The independent-random-elements product law [F6] and the continuous addition map give, for every Borel $A$, $P(X_t\in A)={\int }_{{\mathbb{R}}^d}{\int }_{{\mathbb{R}}^d}{\mathbf{1}}_A(z+w){\varphi }_{v(t)}(w)dw\mu (dz)={\int }_A({\int }_{{\mathbb{R}}^d}{\varphi }_{v(t)}(x-z)\mu (dz))dx,$ where Tonelli and translation were used in the last equality. Thus $p_t$ is the law density.

Fix $0\le s<t\le T$. The vector $Z_{s,t}=X_t-X_s$ is independent of ${\mathcal{F}}_s$ by [F4], and $X_s$ is ${\mathcal{F}}_s$-measurable by the strong-solution conclusion of [F3]. Thus $X_s$ and $Z_{s,t}$ are independent random elements. The same product-law calculation as in step 2.1, now with $Z_{s,t}$ of law $N_d(0,(v(t)-v(s))I_d)$, gives for Borel $A,B$ $P(X_s\in A,X_t\in B)={\int }_A({\int }_B{\varphi }_{v(t)-v(s)}(y-x)dy){\mu }_s(dx)\mathrm{.}$ This is the asserted transition-density identity.

For $0\le s<r<t\le T$, put $a=v(r)-v(s)>0$ and $b=v(t)-v(r)>0$. In one coordinate, completing the square gives $\frac{(z-x{)}^2}{a}+\frac{(y-z{)}^2}{b}=\frac{a+b}{ab}{(z-\frac{bx+ay}{a+b})}^2+\frac{(y-x{)}^2}{a+b}\mathrm{.}$ Applying this identity in each of the $d$ coordinates and integrating the centered Gaussian factor using [F9]–[F11] yields $\int {\varphi }_a(z-x){\varphi }_b(y-z)dz={\varphi }_{a+b}(y-x)$. Since $a+b=v(t)-v(s)$, this is exactly the claimed Chapman–Kolmogorov identity.

For $a>0$, direct differentiation gives ${\mathrm{\partial }}_a{\varphi }_a(z)=\frac{1}{2}{\mathrm{\Delta }}_z{\varphi }_a(z)$. Every finite spatial derivative of ${\varphi }_a(z)$ is a polynomial in $z$ and $a^{-1}$ times ${\varphi }_a(z)$. Fix $ϵ>0$ and compact sets $K\subset {\mathbb{R}}^d$ and $J\subset [ϵ,T]$. Since $a=v(t)$ is bounded above and away from zero on $J$, and $z$ lies in the compact support of $\mu$, each such derivative is uniformly bounded for $x\in K$, $t\in J$, and $z\in \mathrm{supp}\mu$. Difference quotients are bounded by the same type of constant on a slightly larger compact set, by the one-variable fundamental theorem of calculus. Dominated convergence [F4] therefore permits differentiation under the finite measure $\mu$ to every spatial order and once in time. The chain rule and $v^{\mathrm{\prime }}(t)=g(t{)}^2$ give ${\mathrm{\partial }}_t{p}_t(x)=\int \frac{g(t{)}^2}{2}{\mathrm{\Delta }}_x{\varphi }_{v(t)}(x-z)\mu (dz)=\frac{g(t{)}^2}{2}{\mathrm{\Delta }}_x{p}_t(x)\mathrm{.}$ Positivity follows from the positive kernel and the nonzero probability measure.

The SDE setup in [F2] and well-posedness theorem [F3] give a common continuous-path event. For bounded continuous $f$, $X_t\to X_s$ almost surely as $t\to s$ on that event, and [F4] applied to the bounded variables $f(X_t)$ gives $\mathbb{E}f(X_t)\to \mathbb{E}f(X_s)$. By [F9], these expectations are $\int fd{\mu }_t$; also ${\mu }_0=\mu$.

Fix $t\in (0,T]$ and $f\in C_c^{\mathrm{\infty }}({\mathbb{R}}^d)$. For $0\le s<t$ set $h(s,x)={\int }_{{\mathbb{R}}^d}f(x+z){\varphi }_{v(t)-v(s)}(z)dz,h(t,x)=f(x)\mathrm{.}$ For $a=v(t)-v(s)>0$, write $P_{a}f(x)=\int f(x+z){\varphi }_a(z)dz$. Differentiating the explicit Gaussian and integrating by parts twice, with boundary terms zero because $f$ is compactly supported, gives ${\mathrm{\partial }}_a{P}_{a}f=\frac{1}{2}{P}_a\mathrm{\Delta }f$. Thus ${\mathrm{\partial }}_{s}h+\frac{1}{2}g(s{)}^2{\mathrm{\Delta }}_{x}h=0$ for $s<t$. Spatial derivatives may be moved onto $f$, so $h$, $\mathrm{\nabla }h$, and $\mathrm{\Delta }h$ are bounded by the corresponding sup norms of $f$, $\mathrm{\nabla }f$, and $\mathrm{\Delta }f$. For any such derivative $q$ and $\eta >0$, ${\sup }_x\mathrm{\mid }{P}_{a}q(x)-q(x)\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\nabla }q{\mathrm{\parallel }}_{\mathrm{\infty }}\eta +2\mathrm{\parallel }q{\mathrm{\parallel }}_{\mathrm{\infty }}{\int }_{\mathrm{\mid }z\mathrm{\mid }>\eta }{\varphi }_a(z)dz\mathrm{.}$ The tail equals ${\int }_{\mathrm{\mid }u\mathrm{\mid }>\eta \mathrm{/}\sqrt{a}}{\varphi }_1(u)du$ and tends to zero as $a\downarrow 0$ because ${\varphi }_1$ is integrable. Letting first $a\downarrow 0$ and then $\eta \downarrow 0$ proves uniform convergence of $P_{a}q$ to $q$, including $q=\mathrm{\Delta }f$; hence the backward equation and bounds extend continuously to $s=t$.

At $t=0$ the weak identity has equal endpoint terms and a zero right-hand integral. For $t>0$, fix an admissible test $\psi$ and $0<\delta <t$. On the common compact spatial support, $p_s$ and ${\mathrm{\partial }}_s{p}_s$ are continuous and bounded for $s\in [\delta ,t]$ by step 1.1. Differentiation under the space integral is therefore justified by [F4]. Using step 1.1 and integrating by parts twice in each coordinate, with no boundary terms because $\psi$ has compact spatial support, gives $\frac{d}{ds}\int \psi (s,x)p_s(x)dx=\int ({\mathrm{\partial }}_s\psi (s,x)+\frac{g(s{)}^2}{2}{\mathrm{\Delta }}_x\psi (s,x))p_s(x)dx\mathrm{.}$ Integrating from $\delta$ to $t$ gives the weak identity with lower endpoint $\int \psi (\delta ,x){\mu }_{\delta }(dx)$. Its difference from $\int \psi (0,x)\mu (dx)$ tends to zero: the term ${\sup }_x\mathrm{\mid }\psi (\delta ,x)-\psi (0,x)\mathrm{\mid }$ tends to zero by the bounded continuous time derivative on the common compact support, and $\int \psi (0,\cdot )d{\mu }_{\delta }\to \int \psi (0,\cdot )d\mu$ by step 1.2. The omitted right-hand integral over $[0,\delta ]$ tends to zero because each ${\mu }_s$ has mass one and the derivatives of $\psi$ and $g$ are bounded on the common compact support. Letting $\delta \downarrow 0$ proves the weak identity, so ${\mu }_t$ is a weak forward solution.

Let $({\nu }_s{)}_{0\le s\le T}$ be any weak forward solution with initial law $\mu$. Choose a smooth cutoff ${\chi }_L$ equal to one on $\mathrm{\mid }x\mathrm{\mid }\le L$ and zero on $\mathrm{\mid }x\mathrm{\mid }\ge 2L$, with $\mathrm{\mid }\mathrm{\nabla }{\chi }_L\mathrm{\mid }\le C\mathrm{/}L$ and $\mathrm{\mid }\mathrm{\Delta }{\chi }_L\mathrm{\mid }\le C\mathrm{/}{L}^2$. For completeness, take ${\chi }_L(x)=\eta (\mathrm{\mid }x{\mathrm{\mid }}^2\mathrm{/}{L}^2)$, where $\rho (u)=0$ for $u\le 0$, $\rho (u)=e^{-1\mathrm{/}u}$ for $u>0$, and $\eta (u)=\rho (4-u)\mathrm{/}(\rho (4-u)+\rho (u-1))$. When $t<T$, extend $h$ to $t<s\le T$ by $h(s,x)=f(x)-\frac{1}{2}(s-t)g(t{)}^2\mathrm{\Delta }f(x)$. Its value and spatial derivatives through order two match at $t$, and both one-sided time derivatives equal $-\frac{1}{2}g(t{)}^2\mathrm{\Delta }f$ by step 1.3. Thus ${\psi }_L(s,x)={\chi }_L(x)h(s,x)$ is an admissible test on all of $[0,T]$. On $[0,t]$, step 1.3 gives ${\mathrm{\partial }}_s{\psi }_L+\frac{g(s{)}^2}{2}\mathrm{\Delta }{\psi }_L=\frac{g(s{)}^2}{2}(2\mathrm{\nabla }{\chi }_L\cdot \mathrm{\nabla }h+h\mathrm{\Delta }{\chi }_L),$ whose absolute value is at most $C_t(L^{-1}\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_{\mathrm{\infty }}+L^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }})$. Since each ${\nu }_s$ is a probability, the time-integrated error tends to zero as $L\to \mathrm{\infty }$. For all sufficiently large $L$, ${\chi }_L=1$ on the support of $f$ and on the support of $\mu$. Letting $L\to \mathrm{\infty }$ in the weak identity therefore gives $\int fd{\nu }_t=\int h(0,x)\mu (dx)\mathrm{.}$ The same cutoff argument applied to the weak solution ${\mu }_t$ from step 2.1 gives $\int fd{\mu }_t=\int h(0,x)\mu (dx)$. Thus ${\nu }_t$ and ${\mu }_t$ agree on every compactly supported smooth real function.

For any fixed $t>0$, any $\theta \in {\mathbb{R}}^d$, and integer $n\ge 1$, apply step 3.1 to ${\chi }_n(x)\cos (\theta \cdot x)$ and ${\chi }_n(x)\sin (\theta \cdot x)$ and combine the resulting identities into one complex identity. Let $n\to \mathrm{\infty }$. The left side converges to the characteristic function of ${\nu }_t$ by dominated convergence [F4]. For each fixed $x$, the backward averages on the right converge by [F4] and [F10] to $e^{i\theta \cdot x}{e}^{-v(t)\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}2}$; they are bounded by one, so another application of [F4] gives the limit $e^{-v(t)\mathrm{\mid }\theta {\mathrm{\mid }}^2\mathrm{/}2}\int e^{i\theta \cdot x}\mu (dx)$. By [F1], ${\mu }_t$ is the convolution of $\mu$ with the centered Gaussian of covariance $v(t)I_d$; Fubini [F5] and [F10] show that its characteristic function is exactly this product. Hence the vector characteristic functions of ${\nu }_t$ and ${\mu }_t$ agree for every $\theta$. For each $u\in {\mathbb{R}}^d$, take $\theta =ru$ for arbitrary $r\in \mathbb{R}$; scalar uniqueness [F7] gives equality of the laws of $u\cdot x$ under the two measures, and [F8] then gives ${\nu }_t={\mu }_t$. This holds for every $t>0$; at $t=0$ both laws equal $\mu$ by the initial condition, proving uniqueness. AC is used only through the inherited Brownian/Itô interfaces and the cited Fourier-uniqueness results.

The coefficients $b(t,x)=0$ and $\sigma (t,x)=g(t)I_d$ are jointly Borel, globally Lipschitz in $x$ with constant zero, and satisfy $\mathrm{\mid }b\mathrm{\mid }+\mathrm{\parallel }\sigma {\mathrm{\parallel }}_F=\sqrt{d}g(t)\le \sqrt{d}{g}_{\max }$. The initial support gives $\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2\le R^2$, and the complete usual filtered space, adapted Brownian increments, and independence of $X_0$ are hypotheses of the statement. Thus [F3] supplies the unique continuous strong solution. By [F2], $X_{ϵ}-X_0={\int }_0^{ϵ}g(r)d{W}_r$. The characteristic-function identity in [F1] identifies this vector with $N_d(0,v(ϵ)I_d)$, whose coordinate variances sum to $dv(ϵ)$. Thus $\mathbb{E}\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2=dv(ϵ)$. Also $v(ϵ)\le g_{\max }^{}ϵ\to 0$. For every $a>0$, $\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2\ge a^2{\mathbf{1}}_{\{\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }>a\}}$; taking expectations gives $P(\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }>a)\le dv(ϵ)\mathrm{/}{a}^2\to 0$.

The support condition gives $\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2\le R^2$, and step 1.1 gives $\mathbb{E}\mathrm{\mid }{X}_{ϵ}{\mathrm{\mid }}^2\le 2R^2+2dv(T)<\mathrm{\infty }$. Hence both laws lie in the quadratic Wasserstein space. The measurable pair $(X_{ϵ},X_0)$ has a Borel law on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ by [F8]; by [F5] its marginals are ${\mu }_{ϵ}$ and $\mu$, and [F4] identifies ${\mu }_{ϵ}$ as the smoothed law from the definition. Thus it is a coupling. Its cost is $\mathbb{E}\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2=dv(ϵ)$. Taking the infimum in [F7] proves $W_2({\mu }_{ϵ},\mu )\le \sqrt{dv(ϵ)}$.

Fix bounded continuous $f$ and let $K=\bar{B}(0,R+1)$. Its restriction to compact $K$ is uniformly continuous; define ${\omega }_K(\delta )=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in K,\mathrm{\mid }x-y\mathrm{\mid }\le \delta \}$, which tends to zero as $\delta \downarrow 0$. For $0<\delta <1$, on the event $\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }\le \delta$, both vectors belong to $K$, so their $f$-values differ by at most ${\omega }_K(\delta )$. On the complementary event the difference is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 1.1 and Markov's inequality therefore give $\mid \int fd{\mu }_{ϵ}-\int fd\mu \mid \le {\omega }_K(\delta )+2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{dv(ϵ)}{{\delta }^2}\mathrm{.}$ For fixed $\delta$ the second term tends to zero as $ϵ\downarrow 0$; then letting $\delta \downarrow 0$ proves convergence for every such $f$. This is weak convergence by [F6]. The estimates establish only the endpoint statements in the theorem.