Gaussian-smoothed data law and transition density

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.