Score SDEs and reverse-time generation — Examples

Put $M_t={\int }_0^t{e}^{\gamma r}d{W}_r$, so $X_t=e^{-\gamma t}(X_0+\sqrt{2\gamma \theta }{M}_t)$. Apply [F2] to $f(t,x)=e^{-\gamma t}x$ and the continuous Itô process $X_0+\sqrt{2\gamma \theta }{M}_t$. Its drift is zero and its diffusion coefficient is $\sqrt{2\gamma \theta }{e}^{\gamma t}$; the product rule gives $d{X}_t=-\gamma X_{t}dt+\sqrt{2\gamma \theta }d{W}_t$, as asserted.

For fixed $t$, [F3] gives $M_t\sim N(0,{\int }_0^t{e}^{2\gamma r}dr)$. Equivalently the stochastic term in the Example is centered normal with variance $2\gamma \theta {\int }_0^t{e}^{-2\gamma (t-r)}dr=\theta (1-e^{-2\gamma t})$. Its deterministic-integrand Itô approximants are measurable functions of Brownian increments, and [F4] gives a Brownian-process-measurable version of their limit; that version is independent of $X_0$, and almost-sure equality preserves the joint law. The first term is $N(0,\theta e^{-2\gamma t})$, and [F5] gives $X_t\sim N(0,\theta e^{-2\gamma t}+\theta (1-e^{-2\gamma t}))=N(0,\theta )$. By [F6], its density is the one displayed in the Example.

Differentiating the density in [F6] gives ${\mathrm{\partial }}_{x}p=-(x\mathrm{/}\theta )p$ and ${\mathrm{\partial }}_{xx}p=(x^2\mathrm{/}{\theta }^2-1\mathrm{/}\theta )p$. With $b(x)=-\gamma x$ and $a=2\gamma \theta$, the Fokker–Planck right-hand side is $-{\mathrm{\partial }}_x(bp)+\frac{1}{2}{\mathrm{\partial }}_{xx}(ap)=\gamma {\mathrm{\partial }}_x(xp)+\gamma \theta {\mathrm{\partial }}_{xx}p=0$ by substitution of these derivatives. Thus $p_t=p$ is a positive smooth solution of the forward equation and [F7] gives $s(x)=-x\mathrm{/}\theta$.

The diffusion covariance $a={\sigma }^2=2\gamma \theta$ is constant, so ${\mathrm{\partial }}_{x}a=0$. Substituting $b=-\gamma x$, $a=2\gamma \theta$, and $s=-x\mathrm{/}\theta$ into the probability-flow velocity gives $v(x)=-\gamma x-\frac{1}{2}(2\gamma \theta )(-x\mathrm{/}\theta )=0$. Its global flow is ${\mathrm{\Phi }}_{t,s}(x)=x$, jointly smooth and a diffeomorphism satisfying the flow composition and both time-derivative identities since $v=0$. Step 1.2 identifies $p_t=p$ as a normalized probability density; $p$, $b$, and $a$ are smooth, and step 2.1 verifies the forward equation. Thus every hypothesis of [F8] holds, so the deterministic flow started with law $N(0,\theta )$ preserves that law at each time, including $T$.

By [A1], [F1], and [F2], for each $t>0$ the law of $W_t$ has the displayed Gaussian density. Differentiation gives ${\mathrm{\partial }}_x\log p_t(x)=-x\mathrm{/}t$, so [F3] gives the stated score. At $t=0$, [F1] and [F4] give $P_0(\{0\})=1$. Any measure with a Lebesgue density assigns mass ${\int }_{\{0\}}p(x)dx=0$ to this singleton, so $P_0$ has no Lebesgue density.

By [F2], changing variables $x=\sqrt{ϵ}z$ yields $\mathbb{E}\mathrm{\mid }{W}_{ϵ}{\mathrm{\mid }}^2={\int }_{\mathbb{R}}{x}^2\frac{e^{-x^2\mathrm{/}(2ϵ)}}{\sqrt{2\pi ϵ}}dx=ϵ{\int }_{\mathbb{R}}{z}^2\frac{e^{-z^2\mathrm{/}2}}{\sqrt{2\pi }}dz=ϵ$. For the last equality, integration by parts gives $\int z^2{e}^{-z^2\mathrm{/}2}dz=\int e^{-z^2\mathrm{/}2}dz$ over $\mathbb{R}$: the boundary term $-z{e}^{-z^2\mathrm{/}2}$ tends to $0$ at both ends, and the normalized standard-normal density integrates to $1$. Hence the second moment tends to zero.

Fix any sequence ${ϵ}_n>0$ with ${ϵ}_n\to 0$. On the common continuity event from [F1], $W_{{ϵ}_n}\to W_0=0$, so for bounded continuous $f$, $f(W_{{ϵ}_n})\to f(0)$ almost surely and $\mathrm{\mid }f(W_{{ϵ}_n})\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Applying [F5] gives $\int fd{P}_{{ϵ}_n}=\mathbb{E}f(W_{{ϵ}_n})\to f(0)=\int fd{P}_0$. Since this holds for every such sequence, the limit holds as $ϵ\downarrow 0$; [F6] identifies it as $P_{ϵ}\Rightarrow P_0$. The positive-time score conclusion in step 1.1 and the absence of a score at zero prove the stated truncation consequence.

Write $K_t={\int }_0^t{e}^{B(r)\mathrm{/}2}\sqrt{\beta (r)}d{W}_r$, whose deterministic integrand is square-integrable because $B$ and $\beta$ are continuous on $[0,T]$. Since $B^{\mathrm{\prime }}(t)=\beta (t)$, apply [F2] to $f(t,x)=e^{-B(t)\mathrm{/}2}x$ and the continuous Itô process $K_t$. The product rule gives $d{X}_t=-\frac{1}{2}\beta (t)X_{t}dt+\sqrt{\beta (t)}d{W}_t$ and $X_0=0$.

By [F3], $K_t$ is centered normal with variance ${\int }_0^t{e}^{B(r)}\beta (r)dr$. Differentiating $e^{B(r)}$ gives $\frac{d}{dr}{e}^{B(r)}=e^{B(r)}\beta (r)$, so this variance is $e^{B(t)}-1$. Multiplication by $e^{-B(t)\mathrm{/}2}$ therefore gives $X_t\sim N(0,e^{-B(t)}(e^{B(t)}-1))=N(0,v_t)$. For $t>0$, positivity of $\beta$ implies $v_t>0$; [F4] supplies its displayed density.

Differentiating the density in step 1.2 gives ${\mathrm{\partial }}_x\log p_t(x)=-x\mathrm{/}{v}_t$, so [F5] yields the score for every $t>0$. At $t=0$, $X_0=0$ almost surely; its law assigns mass one to $\{0\}$ and therefore has no Lebesgue density, since every Lebesgue-density measure assigns zero mass to that singleton.

Continuity of $\beta$ on the compact interval makes $B(T)$ finite, and strict positivity gives $B(T)>0$. Hence $v_T=1-e^{-B(T)}$ lies strictly between zero and one. For any $v>0$, the density in [F4] and the substitution $x=\sqrt{v}z$ give ${\int }_{\mathbb{R}}{x}^2\frac{e^{-x^2\mathrm{/}(2v)}}{\sqrt{2\pi v}}dx=v{\int }_{\mathbb{R}}{z}^2\frac{e^{-z^2\mathrm{/}2}}{\sqrt{2\pi }}dz=v$. Indeed integration by parts yields $\int z^2{e}^{-z^2\mathrm{/}2}dz=\int e^{-z^2\mathrm{/}2}dz$ over $\mathbb{R}$, because $z{e}^{-z^2\mathrm{/}2}\to 0$ at both ends, and the normalized standard-normal density integrates to one. Thus $E_{P_T}{X}^2=v_T$ by step 1.2, whereas $E_{N(0,1)}{Z}^2=1$. If the laws were equal these finite second moments would be equal, a contradiction. Step 2.1 shows the exact positive-time score is available at $T$, and [F7] prescribes initialization from $P_T$; replacing it by $N(0,1)$ changes that initial law.