Ornstein-Uhlenbeck Transition Laws and Conditional Moments

The Ornstein-Uhlenbeck Transition Problem

Mean reversion combines a deterministic pull toward a level with continuous random fluctuations. Its transition law describes both effects over a finite time interval: the conditional mean moves toward the mean-reversion level, while the conditional variance accumulates from the Brownian noise. The integrating-factor calculation gives these quantities explicitly, and the generator provides an independent check.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\), with the usual conditions on the filtration and with Brownian increments independent of the past. Fix constants \(\kappa>0\), \(m\in\mathbb{R}\), and \(\sigma\in\mathbb{R}\). For \(0\le s\le t\le T\), consider the Ito SDE \(dX_u=\kappa(m-X_u)\,du+\sigma\,dB_u\), with \(X_s\) square-integrable and \(\mathcal{F}_s\)-measurable. The drift is \(\kappa(m-x)\), and the diffusion coefficient is the constant \(\sigma\). Its coefficients are globally Lipschitz and satisfy linear growth, since their absolute values are bounded by a constant times \(1+|x|\).

All stochastic integrals below are Ito integrals. On \([s,T]\), their integrands are progressively measurable and belong to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_s^T H_u^2\,du<\infty\). The deterministic integrands used here satisfy this condition. The diffusion coefficient is constant, so the Ito and Stratonovich forms of this particular SDE coincide; no Stratonovich correction arises. In Ito multiplication, \(dB_u\,dB_u=du\), \(dB_u\,du=0\), and \(du\,du=0\).

Integrating Factor and Explicit Solution

Proof. Apply Ito's formula to \(f(u,x)=e^{\kappa u}(x-m)\). Its derivatives are \(f_u=\kappa e^{\kappa u}(x-m)\), \(f_x=e^{\kappa u}\), and \(f_{xx}=0\); thus the Ito correction is explicitly \(\tfrac12 f_{xx}\sigma^2\,du=0\). Substituting the drift and diffusion gives \(d(e^{\kappa u}(X_u-m))=\kappa e^{\kappa u}(X_u-m)\,du+e^{\kappa u}[\kappa(m-X_u)\,du+\sigma\,dB_u]=\sigma e^{\kappa u}\,dB_u\). Integrating from \(s\) to \(t\) and rearranging proves the displayed solution. The integral \(Z_t=\int_s^t e^{\kappa u}\,dB_u\) is a martingale: its integrand is deterministic and square-integrable on \([s,T]\), so the Ito integral is integrable, and for \(s\le r\le t\), the integral over \([r,t]\) has conditional mean zero given \(\mathcal{F}_r\); hence \(\mathbb{E}[Z_t\mid\mathcal{F}_r]=Z_r\). If \(X\) and \(\widetilde X\) are two solutions with the same initial value and Brownian motion, their difference \(D\) satisfies \(dD_u=-\kappa D_u\,du\), \(D_s=0\). Multiplying by \(e^{\kappa u}\) gives \(d(e^{\kappa u}D_u)=0\), so \(D_u=0\) for all \(u\), almost surely. This proves pathwise uniqueness. The explicit formula also shows that the law of the solution is determined by the law of \(X_s\) and the Brownian input; thus for a fixed initial law it determines uniqueness in law. The solution is strong because the formula uses the given Brownian motion and initial value, rather than merely specifying a distribution on some probability space. \(\square\)

For fixed \(t\), the integrand \(u\mapsto e^{-\kappa(t-u)}\) is deterministic on \([s,t]\). The corresponding integral is centered, independent of \(\mathcal{F}_s\), Gaussian, and has variance given by the Ito isometry. In detail, \(\int_s^t e^{-2\kappa(t-u)}\,du=(1-e^{-2\kappa(t-s)})/(2\kappa)\). These facts give the stated conditional law. They do not say that the convolution, viewed as a process with changing terminal time, is a martingale.

Conditional Moments from the Generator

For a twice differentiable function \(f\), the generator of this diffusion is \(\mathcal{L}f(x)=\kappa(m-x)f'(x)+\tfrac12\sigma^2f''(x)\). Define \(h=t-s\), \(a=e^{-\kappa h}\), and \(q(h,x)=m+a(x-m)\). The transition formula gives the conditional first and second moments: \(\mathbb{E}[X_t\mid\mathcal{F}_s]=q(h,X_s)\) and \(\mathbb{E}[X_t^2\mid\mathcal{F}_s]=q(h,X_s)^2+\frac{\sigma^2}{2\kappa}(1-a^2)\). The second identity uses \(\mathbb{E}[Y^2]=(\mathbb{E}Y)^2+\operatorname{Var}(Y)\), applied to the conditional Gaussian distribution.

Proof. Since \(\partial_xq=a\) and \(\partial_{xx}q=0\), we have \(\mathcal{L}_xq=\kappa(m-x)a\). Also \(\partial_hq=-\kappa a(x-m)=\kappa(m-x)a\), proving the first identity. For the second, \(\partial_x(q^2)=2qa\), \(\partial_{xx}(q^2)=2a^2\), and \(v\) is constant in \(x\). Consequently \(\mathcal{L}_x(q^2+v)=2\kappa(m-x)qa+\sigma^2a^2\). On the other hand, \(\partial_h(q^2)=2q[-\kappa a(x-m)]=2\kappa(m-x)qa\), and \(\partial_hv=\sigma^2a^2\), by differentiating \(a=e^{-\kappa h}\). Adding these derivatives gives the same expression as the generator. The identities hold also when \(\sigma=0\), since then \(v=0\) and its derivative vanishes. \(\square\)

This is an independent moment check: the time evolution of each formula agrees with the generator applied to its spatial function. The calculation can also be connected directly to Ito's formula. For \(f(x)=x\), the derivatives are \(f'=1\), \(f''=0\), and the correction \(\tfrac12\sigma^2f''\,du\) is zero. For \(g(x)=x^2\), the derivatives are \(g'=2x\), \(g''=2\), and the correction is \(\tfrac12\sigma^2(2)\,du=\sigma^2\,du\). Thus Ito gives \(dX_u=\kappa(m-X_u)\,du+\sigma\,dB_u\) and \(d(X_u^2)=[2\kappa X_u(m-X_u)+\sigma^2]\,du+2\sigma X_u\,dB_u\). On finite intervals the explicit solution has finite second moments, so the integrands \(1\) and \(X_u\) in these stochastic integrals are square-integrable. Their Ito integrals are true martingales: they are integrable and have conditional mean-zero increments by the defining properties of the Ito integral. Taking conditional expectations therefore yields the generator moment equations, consistent with the formulas above.

Gaussian Initial Values and Stationarity

A conditional transition law is not the same statement as a stationary distribution. A stationary distribution is an initial law that is preserved over time. The transition formula lets us check preservation directly when the initial state is Gaussian and independent of future Brownian increments.

Common Pitfall: A Terminal-Time-Dependent Integral

Write the stochastic convolution as \(J_t=\int_s^t e^{-\kappa(t-u)}\,dB_u\). It is incorrect to call \((J_t)_{t\ge s}\) a martingale merely because each fixed-endpoint Ito integral has mean zero. For \(s\le r\le q\), splitting at \(r\) gives \(J_q=e^{-\kappa(q-r)}J_r+\int_r^q e^{-\kappa(q-u)}\,dB_u\). The second term has conditional mean zero given \(\mathcal{F}_r\), so \(\mathbb{E}[J_q\mid\mathcal{F}_r]=e^{-\kappa(q-r)}J_r\), which is generally not \(J_r\). For instance, if \(\kappa=1\) and \(q-r=\ln2\), the conditional expectation is \(J_r/2\), not \(J_r\). In contrast, \(Z_t=\int_s^t e^{\kappa u}\,dB_u\) is a martingale, as verified in the integrating-factor proof: its integrand does not depend on the terminal time. The distinction matters because a centered random variable at each time need not be a martingale.

Another frequent error is omitting the factor \(1/(2\kappa)\) in the transition variance. The Ito isometry requires integrating the squared kernel, not just evaluating its endpoint: \(\sigma^2\int_s^t e^{-2\kappa(t-u)}\,du=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa(t-s)})\). This variance is zero at \(t=s\), is nonnegative for every \(t\ge s\), and tends to \(\sigma^2/(2\kappa)\) as \(t-s\) grows. These checks help catch both a missing denominator and an incorrect exponent.