The Ornstein-Uhlenbeck Transition Density

Mean Reversion and the Transition Law

The Ornstein–Uhlenbeck process is a diffusion whose drift pulls its state toward a fixed level. Its transition law describes how much of the starting value remains after a given time and how randomness accumulates along the way. Both features can be derived directly from the SDE by an integrating factor.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix parameters \(\kappa>0\), \(m\in\mathbb{R}\), and \(\sigma>0\), and a starting time \(s\ge0\). For \(t\ge s\), consider the Ornstein–Uhlenbeck SDE with fixed starting value \(X_s=x\):

Here \(\kappa\) is the rate of mean reversion, \(m\) is the long-run level, and \(\sigma\) is the diffusion coefficient. The drift at state \(z\) is \(\kappa(m-z)\); the diffusion coefficient is the constant \(\sigma\). The transition density at time \(t>s\) is the density, with respect to Lebesgue measure in \(y\), of the conditional law of \(X_t\) given \(X_s=x\).

Solving the SDE

Use the integrating factor \(e^{\kappa t}\). The stochastic integral that appears has integrand \(H_u=e^{\kappa u}\), which is deterministic, hence adapted and progressively measurable. On every finite interval \([s,T]\), it belongs to \(\mathcal{L}^2\), since \(\mathbb{E}\int_s^T H_u^2\,du=\int_s^T e^{2\kappa u}\,du<\infty\). Thus the Ito integral below is well-defined.

To justify the integrating-factor calculation, apply Ito’s formula to \(f(t,z)=e^{\kappa t}z\). Its derivatives are \(f_t(t,z)=\kappa e^{\kappa t}z\), \(f_z(t,z)=e^{\kappa t}\), and \(f_{zz}(t,z)=0\). Since the diffusion coefficient of \(X\) is \(\sigma\), its quadratic variation differential is \((dX_t)^2=\sigma^2dt\): the drift-times-drift and drift-times-diffusion terms vanish by \(dt\,dt=0\) and \(dt\,dB_t=0\), while \(dB_t\,dB_t=dt\). Ito’s formula, including its correction term, gives

Integrating from \(s\) to \(t\), and evaluating the ordinary integral, yields the explicit solution

The integral here has deterministic integrand \(u\mapsto e^{-\kappa(t-u)}\), which is progressively measurable and square-integrable on \([s,t]\), because its expected squared-integrand integral is the finite number \(\int_s^t e^{-2\kappa(t-u)}\,du\). This formula gives a strong solution: it is constructed using the given Brownian motion, and its value at time \(t\) is measurable with respect to \(\mathcal{F}_t\).

Proof. The Ito calculation above shows that the displayed process satisfies the SDE and the initial condition. If \(X\) and \(\widetilde X\) are two solutions with the same Brownian motion and initial value, their difference \(D_t=X_t-\widetilde X_t\) has no stochastic term and satisfies \(dD_t=-\kappa D_t\,dt\), with \(D_s=0\). Multiplying by \(e^{\kappa t}\), or integrating this ordinary differential equation, gives \(e^{\kappa t}D_t=e^{\kappa s}D_s=0\) for all \(t\ge s\). Hence \(X_t=\widetilde X_t\) for all \(t\ge s\), almost surely. \(\square\)

Conditional Mean, Variance, and Density

For fixed \(s<t\), the random part of the solution is a deterministic Brownian integral. Such an integral is centered Gaussian, with variance given by the Ito isometry. It is independent of \(\mathcal{F}_s\): for step-function integrands this follows from independence of Brownian increments after \(s\) from \(\mathcal{F}_s\), and general deterministic square-integrable integrands are limits in \(L^2\) of step-function integrands. The resulting conditional law is therefore Gaussian.

Proof. Set \(h=t-s>0\). In the explicit solution, the nonrandom terms combine as \(e^{-\kappa h}x+m(1-e^{-\kappa h})=m+(x-m)e^{-\kappa h}\), which is the conditional mean because the stochastic integral has conditional mean zero. Its conditional variance is its ordinary variance, since it is independent of \(\mathcal{F}_s\). By the Ito isometry,

This variance is strictly positive because \(\sigma>0\), \(\kappa>0\), and \(t-s>0\). A Gaussian random variable with this mean and positive variance has the stated Lebesgue density. \(\square\)

The same calculation gives a transition kernel for every starting state \(x\). If \(X_s\) is instead random and \(\mathcal{F}_s\)-measurable, the conditional law given \(\mathcal{F}_s\) is the Gaussian law with mean \(m+(X_s-m)e^{-\kappa(t-s)}\) and the same variance \(V_{s,t}\). The formula is a statement about conditional laws, not a claim that \(X_t\) is independent of its starting state.

Limiting Cases and a Common Error

At zero elapsed time, \(t=s\), the solution is exactly \(x\). Its conditional law is a point mass at \(x\), not a Lebesgue density; the displayed density formula is for \(t>s\). If \(\sigma=0\), the solution is deterministic and again has no nondegenerate Lebesgue density. The assumptions \(\kappa>0\) and \(\sigma>0\) ensure that for every positive elapsed time the variance in the theorem is positive. As \(t-s\) grows, the mean approaches \(m\), while the variance approaches \(\sigma^2/(2\kappa)\); this is the limiting Gaussian scale, not the variance at every finite time.

Common pitfall: using \(\sigma^2(t-s)\) for the variance. That is the variance of \(\sigma(B_t-B_s)\), where Brownian increments are not damped. In the Ornstein–Uhlenbeck solution, an increment from time \(u\) to \(u+du\) is weighted by \(e^{-\kappa(t-u)}\), so its contribution to the variance is weighted by the square of that factor. For example, when \(\kappa=2\), \(\sigma=2\), and \(h=(\ln2)/2\), the incorrect expression gives \(\sigma^2h=2\ln2\), whereas the correct variance is \(\tfrac34\). The correct value follows by integrating the squared weight: \(\sigma^2\int_s^t e^{-2\kappa(t-u)}du\).

It is useful to keep the roles of the two coefficients distinct. Mean reversion changes the conditional mean through the factor \(e^{-\kappa(t-s)}\) and damps the effect of earlier noise; the diffusion coefficient sets the scale of the accumulated randomness through \(\sigma^2\). The transition density records both effects in one Gaussian kernel.