Ornstein–Uhlenbeck equation, transition law, and fluctuations

On $[0,T]$, $g(s)=e^{\gamma s}$ is deterministic, continuous, and square-integrable. Thus $W_t={\int }_0^{t}g(s)d{B}_s$ is defined by [F3], and [F4] supplies a continuous version. For a partition $0=t_0<\cdots <t_n=t$, the deterministic step sums ${\sum }_{k}g(t_k)(B_{t_{k+1}}-B_{t_k})$ converge in $L^2$ to $W_t$ by the Itô isometry and uniform approximation of $g$. Discrete summation by parts rewrites them as $g(t)B_t-g(0)B_0-{\sum }_{k=0}^{n-1}{B}_{t_{k+1}}(g(t_{k+1})-g(t_k))$ up to an endpoint term tending to zero. Brownian path continuity makes the last sums converge pathwise to ${\int }_0^t{g}^{\mathrm{\prime }}(s)B_{s}ds$. The $L^2$ limit and this almost-sure limit agree in probability, so $W_t=e^{\gamma t}{B}_t-\gamma {\int }_0^t{e}^{\gamma s}{B}_{s}ds$ almost surely for each fixed $t$. Both sides have continuous versions; equality on rational $t$ holds on one probability-one event and therefore extends to all $t\in [0,T]$.

Split the stochastic integral at $s$ and use its linearity: $X_t=e^{-\gamma (t-s)}{X}_s+\sigma {\int }_s^t{e}^{-\gamma (t-u)}d{B}_u\mathrm{.}$ Step-function approximants to the last deterministic-integrand integral depend only on increments after $s$, so [A1] makes them independent of ${\mathcal{F}}_s$. Their $L^2$ limit remains independent of ${\mathcal{F}}_s$: for bounded ${\mathcal{F}}_s$-measurable $Y$ and bounded continuous $\phi$, convergence in probability plus boundedness gives convergence of $\mathbb{E}[Y\phi (\cdot )]$, preserving the factorization. By [F4] the limit is centered Gaussian; [F3] computes its variance as ${\sigma }^2{\int }_s^t{e}^{-2\gamma (t-u)}du=V(1-e^{-2\gamma (t-s)})$. This proves the conditional transition law and the Markov property.

Set $X_t=e^{-\gamma t}{X}_0+\sigma e^{-\gamma t}{W}_t$. By [F4] it is continuous and adapted. Step 1.1 rewrites it as $X_t=e^{-\gamma t}{X}_0+\sigma B_t-\sigma \gamma {\int }_0^t{e}^{-\gamma (t-s)}{B}_{s}ds\mathrm{.}$ Integrating this continuous expression in $t$, ordinary Fubini [F5] gives ${\int }_0^t{X}_{s}ds=X_0(1-e^{-\gamma t})\mathrm{/}\gamma +\sigma {\int }_0^t{B}_{s}ds-\sigma {\int }_0^t{B}_u(1-e^{-\gamma (t-u)})du\mathrm{.}$ Substitution into $X_0-\gamma {\int }_0^t{X}_{s}ds+\sigma B_t$ cancels the two $\int B_{s}ds$ terms and leaves exactly the displayed expression for $X_t$. Thus [F1] holds.

Taking expectations in the explicit solution, the stochastic integral has mean zero by [F4], giving $\mathbb{E}{X}_t=e^{-\gamma t}\mathbb{E}{X}_0$. The integral is independent of $X_0$ and centered, so its variance adds to $e^{-2\gamma t}\mathrm{Var}(X_0)$; [F3] evaluates the added variance as $V(1-e^{-2\gamma t})$. The independent Gaussian term tends in law to $N(0,V)$ because its variance tends to $V$ and $e^{-\gamma t}{X}_0\to 0$ in $L^2$; for any bounded continuous test, restrict to a compact interval, use uniform continuity and the second-moment tail bound, then take the interval radius to infinity.

Write $r=e^{-\gamma t}$ and $\pi (x)=(2\pi V{)}^{-1\mathrm{/}2}{e}^{-x^2\mathrm{/}(2V)}$. The transition density is $K_t(x,y)=[2\pi V(1-r^2){]}^{-1\mathrm{/}2}\exp \text{\hspace{0.17em}⁣}(-\frac{(y-rx{)}^2}{2V(1-r^2)})\mathrm{.}$ Multiplying by $\pi (x)$ gives a symmetric prefactor and exponent proportional to $-[x^2+y^2-2rxy]\mathrm{/}[2V(1-r^2)]$, unchanged when $x$ and $y$ are exchanged. Hence $\pi (x)K_t(x,y)=\pi (y)K_t(y,x)$, the detailed-balance identity.

For $t>0$ let $\mu =e^{-\gamma t}x$, $v=V(1-e^{-2\gamma t})$, and $p(t,x,y)$ be the normal density with mean $\mu$ and variance $v$. Direct differentiation gives ${\mathrm{\partial }}_{t}p=p(\frac{{\mu }^{\mathrm{\prime }}(y-\mu )}{v}+\frac{v^{\mathrm{\prime }}}{2}[\frac{(y-\mu {)}^2}{v^2}-\frac{1}{v}]),{\mu }^{\mathrm{\prime }}=-\gamma \mu ,v^{\mathrm{\prime }}={\sigma }^2-2\gamma v\mathrm{.}$ Also direct $y$-differentiation gives $\gamma {\mathrm{\partial }}_y(yp)+({\sigma }^2\mathrm{/}2){\mathrm{\partial }}_{yy}p$ equal to the same expression after writing $y=\mu +(y-\mu )$. Thus ${\mathrm{\partial }}_{t}p=\gamma {\mathrm{\partial }}_y(yp)+({\sigma }^2\mathrm{/}2){\mathrm{\partial }}_{yy}p$. The Gaussian formula has total integral one and converges to point mass at $x$ as $t\downarrow 0$; this is the stated positive-time Fokker–Planck solution.

If $X$ and $\tilde{X}$ are two continuous solutions with the same data, their difference $z$ satisfies $z(t)=-\gamma {\int }_0^{t}z(s)ds$. Let $D(t)={\int }_0^t\mathrm{\mid }z(s)\mathrm{\mid }ds$. Then $D(0)=0$, $D\ge 0$, and $D^{\mathrm{\prime }}(t)=\mathrm{\mid }z(t)\mathrm{\mid }\le \gamma D(t)$ by the equation. The derivative of $e^{-\gamma t}D(t)$ is nonpositive, so this nonnegative function is at most its initial value zero. Hence $D=0$ and continuity gives $z=0$ at every time.

If $X_0$ is centered normal with variance $V$, step 1.2 gives centered Gaussian finite-dimensional laws, since each stochastic integral over a disjoint increment interval is an independent Gaussian and $X_0$ is independent of future increments. Its one-time variance is $V$. For $s\le t$, the split formula in 2.1 and independence of the later increment yield $\mathbb{E}[X_s{X}_t]=e^{-\gamma (t-s)}\mathbb{E}[X_s^2]=V{e}^{-\gamma (t-s)}$. Thus covariance depends only on $\mathrm{\mid }t-s\mathrm{\mid }$. A centered Gaussian finite-dimensional law is determined by its covariance matrix; shifting every time by the same amount preserves that matrix and proves stationarity.

For $R>0$, the derivative and componentwise FTC in [F6] give ${\int }_0^R{e}^{-(\gamma -i\omega )t}dt=(1-e^{-(\gamma -i\omega )R})\mathrm{/}(\gamma -i\omega )$. The modulus formula gives $\mathrm{\mid }{e}^{-(\gamma -i\omega )R}\mathrm{\mid }=e^{-\gamma R}\to 0$ and $\mathrm{\mid }{e}^{-(\gamma -i\omega )t}\mathrm{\mid }=e^{-\gamma t}$, whose integral is finite. Thus the stationary covariance is integrable. By evenness, $S(\omega )=\frac{V}{\pi }\mathrm{Re}{\int }_0^{\mathrm{\infty }}{e}^{-(\gamma -i\omega )t}dt=\frac{V}{\pi }\mathrm{Re}\frac{1}{\gamma -i\omega }=\frac{V\gamma }{\pi ({\gamma }^2+{\omega }^2)}\mathrm{.}$ The integral converges absolutely because $\gamma >0$.

For the stationary time average, Fubini applies to the random product because $\mathbb{E}\mathrm{\mid }{X}_s{X}_t\mathrm{\mid }\le (\mathbb{E}{X}_s^2\mathbb{E}{X}_t^2{)}^{1\mathrm{/}2}=V$, so its absolute integral over the finite square and probability space is at most $V{T}^2$. Changing variables in the two triangular halves gives $\mathrm{Var}\text{\hspace{0.17em}⁣}\left(T^{-1}{\int }_0^T{X}_{t}dt\right)=\frac{2V}{T^2}{\int }_0^T(T-s)e^{-\gamma s}ds\le \frac{2V}{\gamma T}\mathrm{.}$ The bound tends to zero, so these averages converge to zero in $L^2$ and in probability. It asserts no almost-sure theorem for arbitrary observables.