The Invariant Density of the Ornstein-Uhlenbeck Process

Stationarity for Mean-Reverting Diffusions

The Ornstein-Uhlenbeck process is the basic diffusion whose drift pulls the state toward a fixed level. Its invariant density describes the distribution that remains unchanged under the transition dynamics, and it is also the long-time limit of the transition densities. These are related claims, but they require separate checks: solving a stationary differential equation identifies a candidate density, while the transition law verifies that the candidate is preserved by the process.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Let \(\kappa>0\), \(\alpha\in\mathbb{R}\), and \(\eta>0\), and consider the Ornstein-Uhlenbeck SDE \(dX_t=-\kappa(X_t-\alpha)\,dt+\eta\,dB_t\). Here \(-\kappa(X_t-\alpha)\) is the drift and \(\eta\) is the diffusion coefficient. For each finite \(T\), the stochastic integrals below have deterministic, hence progressively measurable, integrands in \(\mathcal{L}^2\): their squared integrals have finite expectation on \([0,T]\).

For this process the forward, or Fokker-Planck, equation for a density \(p(t,x)\) is \(\partial_t p=\kappa\,\partial_x((x-\alpha)p)+\tfrac12\eta^2\partial_{xx}p\). A stationary density is a time-independent density solving the resulting stationary equation. We first solve that equation and then verify the candidate directly from the transition law.

Solving the Stationary Fokker-Planck Equation

Proof. The stationary equation is \(\kappa((x-\alpha)p)' + \tfrac12\eta^2p''=0\). Define the probability current \(J(x)=-\kappa(x-\alpha)p(x)-\tfrac12\eta^2p'(x)\). The stationary equation says \(J'(x)=0\), so \(J(x)=j\) for some constant \(j\). Put \(a=2\kappa/\eta^2>0\). Rearranging the current equation gives \(p'(x)+a(x-\alpha)p(x)=-2j/\eta^2\). Multiplication by the integrating factor \(\exp(a(x-\alpha)^2/2)\), followed by integration from \(\alpha\) to \(x\), yields \(p(x)=\exp(-a(x-\alpha)^2/2)\left(C-\frac{2j}{\eta^2}\int_\alpha^x\exp(a(y-\alpha)^2/2)\,dy\right)\) for a constant \(C\). If \(j>0\), the integral tends to \(+\infty\) as \(x\to+\infty\); the expression in parentheses is eventually negative, contradicting \(p\ge0\). If \(j<0\), the integral tends to \(-\infty\) as \(x\to-\infty\), and the expression in parentheses is again eventually negative. Thus \(j=0\). The density must therefore be \(C\exp(-a(x-\alpha)^2/2)\). The Gaussian integral makes its total mass one precisely when \(C=\sqrt{a/(2\pi)}=\sqrt{\kappa/(\pi\eta^2)}\). This is the stated normal density, and normalization also fixes the only remaining constant, proving uniqueness. \(\square\)

The zero-current conclusion here follows from nonnegativity on the whole real line; an arbitrary nonzero constant current cannot produce a nonnegative integrable density on \(\mathbb{R}\). The normalization uses the standard Gaussian integral. In particular, the factor \(1/2\) in the diffusion term matters: it makes the variance \(\eta^2/(2\kappa)\), not \(\eta^2/\kappa\).

Transition Densities and the Long-Time Limit

The stationary equation alone does not establish invariance. To check the dynamics, solve the SDE explicitly. Applying Ito's formula to \(f(t,x)=e^{\kappa t}(x-\alpha)\), whose derivatives are \(f_t=\kappa e^{\kappa t}(x-\alpha)\), \(f_x=e^{\kappa t}\), and \(f_{xx}=0\), gives \(d(e^{\kappa t}(X_t-\alpha))=\eta e^{\kappa t}dB_t\). Integrating and rearranging gives \(X_t=\alpha+(X_0-\alpha)e^{-\kappa t}+\eta\int_0^t e^{-\kappa(t-s)}\,dB_s\). The integrand \(H_s=\eta e^{-\kappa(t-s)}\), for fixed \(t\), is progressively measurable, and \(\mathbb{E}\int_0^tH_s^2\,ds=\eta^2\int_0^te^{-2\kappa(t-s)}\,ds<\infty\), so this Ito integral is defined in \(\mathcal{L}^2\).

For deterministic \(X_0=x\), the integral is normal with mean zero and variance \(\eta^2\int_0^te^{-2\kappa(t-s)}\,ds\). Consequently, for \(t>0\), the transition law from \(x\) is normal with mean \(m_t(x)=\alpha+(x-\alpha)e^{-\kappa t}\) and variance \(v_t=\frac{\eta^2}{2\kappa}(1-e^{-2\kappa t})\). The formula shows that \(m_t(x)\to\alpha\) and \(v_t\to\eta^2/(2\kappa)\) as \(t\to\infty\).

Proof. The transition density is \(p_t(x,y)=\frac{1}{\sqrt{2\pi v_t}}\exp(-(y-m_t(x))^2/(2v_t))\). Since \(m_t(x)\to\alpha\) and \(v_t\to v_\infty=\eta^2/(2\kappa)>0\), direct substitution gives \(p_t(x,y)\to p_\infty(y)\) for each \(y\). To obtain \(L^1\) convergence, restrict to \(t\ge1\). Then \(v_t\) lies between the positive number \(v_1\) and \(v_\infty\), and \(m_t(x)\) is bounded. These bounds imply that there are constants \(A,c>0\), independent of \(t\ge1\), such that \(p_t(x,y)\le A e^{-cy^2}\): use \((y-m)^2\ge y^2/2-m^2\) for bounded \(m\). The same type of bound holds for \(p_\infty(y)\). The dominated convergence theorem applied to \(|p_t(x,y)-p_\infty(y)|\) now gives convergence of its integral to zero.

For invariance, let \(X_0\) have density \(p_\infty\), independently of \(B\). In the explicit solution, \((X_0-\alpha)e^{-\kappa t}\) is a centered normal variable with variance \(v_\infty e^{-2\kappa t}\). It is independent of the stochastic integral, which is centered normal with variance \(v_t=v_\infty(1-e^{-2\kappa t})\). Their sum is therefore centered normal with variance \(v_\infty e^{-2\kappa t}+v_t=v_\infty\). Thus \(X_t\) has mean \(\alpha\) and variance \(v_\infty\), and has density \(p_\infty\) for every \(t\). This proves invariance as well as the long-time density limit. \(\square\)

Invariant Moments and a Generator Check

Because the invariant density is normal, its mean and variance are \(\alpha\) and \(\eta^2/(2\kappa)\). Its second moment is \(\alpha^2+\eta^2/(2\kappa)\). These values can also be checked from the generator \(\mathcal{L}f(x)=-\kappa(x-\alpha)f'(x)+\tfrac12\eta^2f''(x)\). For \(f(x)=x\), \(\mathcal{L}f=-\kappa(x-\alpha)\); for \(f(x)=x^2\), \(\mathcal{L}f=-2\kappa x(x-\alpha)+\eta^2\). Under stationarity, the expectation of each of these generator expressions is zero. The second-moment calculation is legitimate here because the invariant Gaussian has finite second moment.

Common Pitfall: Losing the Diffusion Factor

A frequent error is to omit the \(\tfrac12\) in the forward equation's second-derivative term. For example, the incorrect equation \(\kappa((x-\alpha)p)'+\eta^2p''=0\) would give a Gaussian proportional to \(\exp(-\kappa(x-\alpha)^2/(2\eta^2))\), with variance \(\eta^2/\kappa\). But the SDE's noise contribution has variance \(\eta^2\int_0^t e^{-2\kappa(t-s)}ds\), whose limit is \(\eta^2/(2\kappa)\). The stationary-equation answer must agree with this independent transition calculation. Keeping both checks in view catches the missing factor of two.