Ornstein-Uhlenbeck Densities from the Forward Equation

From Mean Reversion to a Forward Density

The Ornstein–Uhlenbeck process is a diffusion whose drift pulls the state toward a level \(\theta\), while Brownian noise spreads its distribution. Its forward equation describes how the probability density changes with time. For a Gaussian initial law, the equation admits a Gaussian solution at every positive time: the mean relaxes toward \(\theta\), and the variance changes through the competing effects of mean reversion and noise.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Let \(\kappa>0\) and \(\sigma>0\), and consider the SDE \(dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t\). Here \(\kappa(\theta-X_t)\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. Suppose \(X_0\) has law \(N(m_0,v_0)\), with \(v_0>0\), and is independent of the Brownian motion. For each finite \(T\), the stochastic integrals below have deterministic, progressively measurable integrands \(H_s\) satisfying \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); thus they belong to the Ito-integrand class \(\mathcal{L}^2\).

For a density \(p(t,x)\), the one-dimensional forward equation for drift \(\mu(x)\) and constant diffusion coefficient \(\sigma\) is \(\partial_t p=-\partial_x(\mu p)+\tfrac12\sigma^2\partial_{xx}p\). Substituting \(\mu(x)=\kappa(\theta-x)\) gives \(\partial_t p=\kappa\partial_x((x-\theta)p)+\tfrac12\sigma^2\partial_{xx}p\). We seek a nonnegative density on the whole real line with total mass one and initial density \(N(m_0,v_0)\). The Gaussian decay at spatial infinity makes the displayed derivatives and integrations by parts well-defined.

Deriving the Evolving Mean and Variance

Set \(p(t,x)=p_{m(t),v(t)}(x)\), where \(v(t)>0\). To determine which functions \(m\) and \(v\) make this ansatz solve the forward equation, put \(y=x-m(t)\) and \(D=\tfrac12\sigma^2\). Direct differentiation gives

For the right-hand side of the forward equation, use \(x-\theta=y+(m-\theta)\). Dividing that side by \(p\) gives

Both sides are polynomials in \(y\). Equating the coefficients of \(y\) and \(y^2\) yields \(m'=-\kappa(m-\theta)\) and \(v'=-2\kappa v+2D=-2\kappa v+\sigma^2\). The constant coefficients agree as well: the variance equation implies \(-v'/(2v)=\kappa-D/v\), exactly the constant term on the right. Thus these two ordinary differential equations are necessary and sufficient for this Gaussian ansatz to solve the PDE.

Proof. The equation \(m'=-\kappa(m-\theta)\) with \(m(0)=m_0\) becomes \(\frac{d}{dt}(e^{\kappa t}(m-\theta))=0\), so \(m(t)=\theta+(m_0-\theta)e^{-\kappa t}\). For the variance equation, multiplying \(v'+2\kappa v=\sigma^2\) by \(e^{2\kappa t}\) gives \(\frac{d}{dt}(e^{2\kappa t}v)=\sigma^2e^{2\kappa t}\). Integrating from \(0\) to \(t\) and using \(v(0)=v_0\) yields the stated formula. Since \(v_0e^{-2\kappa t}>0\) and the second term is nonnegative for \(t\ge0\), \(v(t)>0\). The coefficient calculation above then shows that \(p_{m(t),v(t)}\) satisfies the PDE. At \(t=0\) it is precisely the prescribed initial density, and its Gaussian form integrates to one. \(\square\)

This substitution constructs a classical Gaussian solution. To identify it as the law of the SDE, we now derive the explicit stochastic solution independently; this also checks the mean and variance without relying on the PDE calculation.

Checking the Density from the Stochastic Solution

Proof. Define \(Y_t=e^{\kappa t}(X_t-\theta)\). Ito’s formula applied to \(f(t,x)=e^{\kappa t}(x-\theta)\) has derivatives \(f_t=\kappa e^{\kappa t}(x-\theta)\), \(f_x=e^{\kappa t}\), and \(f_{xx}=0\). The full Ito correction is \(\tfrac12 f_{xx}\sigma^2\,dt=0\). Consequently, \(dY_t=e^{\kappa t}\,dX_t+\kappa e^{\kappa t}(X_t-\theta)\,dt=\sigma e^{\kappa t}\,dB_t\); the drift terms cancel. Integrating and multiplying by \(e^{-\kappa t}\) gives the asserted formula.

For each fixed \(t\), the integrand \(s\mapsto e^{-\kappa(t-s)}\) on \([0,t]\) is deterministic and belongs to \(\mathcal{L}^2\), because its squared integral is finite. Its Ito integral is centered Gaussian with variance given by the Ito isometry: \(\int_0^t e^{-2\kappa(t-s)}\,ds=(1-e^{-2\kappa t})/(2\kappa)\). The Gaussian assertion follows, for example, by approximating this deterministic integrand by step functions: each step-function integral is a linear combination of independent Gaussian Brownian increments, and the approximations converge in \(L^2\). The initial variable is independent of the Brownian motion, so the two random terms in the solution are independent. Therefore the mean is \(\theta+e^{-\kappa t}(m_0-\theta)\), and the variance is \(e^{-2\kappa t}v_0+\sigma^2(1-e^{-2\kappa t})/(2\kappa)\), as claimed. \(\square\)

The PDE substitution and the stochastic representation agree on both parameters. Hence the Gaussian solution constructed above is the density of \(X_t\). The verification uses the explicit SDE solution rather than assuming uniqueness of a PDE solution.

Transition Densities from a Fixed Starting Point

A deterministic starting point is a degenerate Gaussian initial law, so its time-zero law is a point mass rather than an ordinary density. At any positive time, however, the Brownian noise gives a positive variance when \(\sigma>0\). The same explicit solution then supplies the transition density from the fixed starting state.

Common Pitfall: The Noise Term Has a Time-Dependent Weight

A frequent error is to use \(\sigma^2t\) as the variance contributed by the Brownian noise. That is the variance of \(\sigma B_t\), but the Ornstein–Uhlenbeck solution weights a shock at time \(s\) by \(e^{-\kappa(t-s)}\). The correct contribution is \(\sigma^2\int_0^t e^{-2\kappa(t-s)}\,ds=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t})\). For example, in the fixed-start calculation above, replacing the integral by \(\sigma^2t\) would give \(2\ln2\), not \(3/4\). The correction is to integrate the squared weight: \(2(1-e^{-2\ln2})/2=3/4\).

The mean and variance equations are also useful checks on a proposed density. The mean must move toward \(\theta\) at rate \(\kappa\), while the variance must satisfy \(v'=-2\kappa v+\sigma^2\). A Gaussian formula that violates either equation cannot solve this forward equation. These checks concern the transient law at each time; they do not require solving a stationary or zero-current equation.