Exact OU velocity response, diffusion and Gaussian scaling

Let $V^0$ be the unforced OU solution from [F2], satisfying the integral equation in [F1]. Define the deterministic function $g(t):=\frac{1}{m}{\int }_0^t{e}^{-\gamma (t-s)}u(s)ds$. Differentiation under this finite integral gives $g(0)=0$ and $g^{\mathrm{\prime }}(t)=-\gamma g(t)+u(t)\mathrm{/}m$, hence $g(t)=-\gamma {\int }_0^{t}g(s)ds+\frac{1}{m}{\int }_0^{t}u(s)ds$. Adding this identity to the unforced equation shows that the continuous adapted process $V_t:=V_t^0+g(t)$ satisfies $V_t=V_0-\gamma {\int }_0^t{V}_{s}ds+\frac{1}{m}{\int }_0^{t}u(s)ds+\sigma B_t$, the forced equation in the Statement. If two forced solutions have the same initial state and Brownian path, their difference $z$ satisfies $z(t)=-\gamma {\int }_0^{t}z(s)ds$; with $H(t)={\int }_0^t\mathrm{\mid }z(s)\mathrm{\mid }ds$, one has $H^{\mathrm{\prime }}(t)\le \gamma H(t)$ and $H(0)=0$. Thus $(e^{-\gamma t}H(t){)}^{\mathrm{\prime }}\le 0$, so $H=0$ and $z=0$. Uniqueness makes the finite-horizon solutions agree on overlaps; gluing along integer horizons gives the solution for all $t\ge 0$ on a common probability-one event.

Since $V^0$ is centered and $g$ is deterministic, $\mathbb{E}{V}_t=g(t)$, giving the claimed mean. The parameter relation gives ${\sigma }^2\mathrm{/}(2\gamma )=k_B\mathrm{\Theta }\mathrm{/}m=v_{\mathrm{\Theta }}$, so [F2] gives stationarity, Gaussianity and the displayed covariance for the unforced solution.

For a uniform partition of $[0,t]$, the Riemann sums $S_n=(t\mathrm{/}n){\sum }_{j=0}^{n-1}{V}_{jt\mathrm{/}n}$ converge almost surely to $Y_t$ because $V$ has continuous paths. Each $S_n$ is centered Gaussian under $u=0$. By the covariance in step 2.1, its variance is a Riemann sum converging to ${\int }_0^t{\int }_0^t{v}_{\mathrm{\Theta }}{e}^{-\gamma \mathrm{\mid }s-r\mathrm{\mid }}dsdr=2v_{\mathrm{\Theta }}{\int }_0^t(t-a)e^{-\gamma a}da=2v_{\mathrm{\Theta }}(\frac{t}{\gamma }-\frac{1-e^{-\gamma t}}{{\gamma }^2})$. Hence the Gaussian characteristic functions of $S_n$ converge to that of the centered normal with this variance. For each real $\xi$, almost-sure convergence and $\mathrm{\mid }{e}^{i\xi S_n}\mathrm{\mid }=1$ let bounded convergence give $\mathbb{E}{e}^{i\xi S_n}\to \mathbb{E}{e}^{i\xi Y_t}$. Therefore $Y_t$ has the stated Gaussian law and variance. Adding the deterministic $g$ to velocity changes only the mean of its integral, not this variance.

Dividing the variance formula in step 3.1 by $2t$ and letting $t\to \mathrm{\infty }$ gives $D=v_{\mathrm{\Theta }}\mathrm{/}\gamma$. The centered displacement has Gaussian variance $2Dt-2v_{\mathrm{\Theta }}(1-e^{-\gamma t})\mathrm{/}{\gamma }^2$; after division by $t$ this tends to $2D$, proving the distributional limit. For constant $u_0$, step 1.1 gives $\mathbb{E}{V}_t=u_0(1-e^{-\gamma t})\mathrm{/}(m\gamma )$, so $\chi =1\mathrm{/}(m\gamma )$ and $D=(k_B\mathrm{\Theta }\mathrm{/}m)\mathrm{/}\gamma =\chi k_B\mathrm{\Theta }$.