Tutorials › Stochastic Calculus › Invariant Densities for Ornstein-Uhlenbeck and Gradient Diffusions

Home  /  Tutorials  /  Stability, Invariant Measures and Long-Time Behavior

Stability, Invariant Measures and Long-Time Behavior · Tutorial 282 of 600

Invariant Densities for Ornstein-Uhlenbeck and Gradient Diffusions

Learn how drift, noise strength, and potential determine a normalizable stationary density, and how to verify the result using probability flux and the Fokker–Planck equation.

Advanced 12 min read

What You'll Learn

  • Write the stationary Fokker–Planck equation as a zero-flux condition.
  • Derive the normalized invariant Gaussian density for an Ornstein–Uhlenbeck diffusion.
  • Compute the density associated with a gradient drift and constant diffusion coefficient.
  • Check normalizability before interpreting a stationary solution as a probability density.
  • Verify candidate densities by substitution and, where possible, by checking the transition law.

From Stationary Flux to an Invariant Density

A stationary density is a probability density that does not change under the diffusion’s time evolution. The stationary Fokker–Planck equation gives a way to find candidates, but two checks matter: the candidate must integrate to one, and it must satisfy the equation with the correct boundary behavior. For a gradient drift and constant noise, these checks lead to an explicit formula involving the potential.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). All SDEs below are interpreted in the Itô sense. For each finite \(T\), an integrand \(H_s\) used in an Itô integral is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). In the SDEs here the diffusion coefficient is constant, so the integrand is that constant and satisfies this condition. The drift describes the \(dt\) term; the diffusion coefficient describes the \(dB_t\) term.

Consider the one-dimensional diffusion \(dX_t=\mu(X_t)\,dt+\sigma\,dB_t\), where \(\sigma>0\) is constant. Its density \(p(t,x)\), when it exists and is sufficiently regular, satisfies the Fokker–Planck equation. Put \(D=\sigma^2/2\); then \(D\) is the coefficient multiplying the second spatial derivative.

$$ \partial_t p(t,x)=-\partial_x\big(\mu(x)p(t,x)\big)+D\,\partial_{xx}p(t,x) =-\partial_x J(t,x), \qquad J(t,x)=\mu(x)p(t,x)-D\,\partial_xp(t,x). $$

The quantity \(J\) is the probability flux. A stationary density \(\rho\) must satisfy \(\partial_xJ=0\), so its flux is constant in space. When the boundary conditions require zero flux, the equation to solve is \(J(x)=0\), or \(\mu(x)\rho(x)-D\rho'(x)=0\). The density still needs to be nonnegative, integrable, and normalized.

Definition: A probability measure \(\pi\) is invariant for a diffusion if, whenever \(X_0\) has law \(\pi\), \(X_t\) has law \(\pi\) for every \(t\ge0\). A normalized stationary solution of the Fokker–Planck equation is a density that solves the time-independent equation and integrates to one. Under suitable regularity and boundary conditions, such a density is invariant; solving the differential equation alone does not establish those conditions.

Gradient Drift: The Potential Formula

Suppose the drift is the negative derivative of a potential \(V\): \(\mu(x)=-V'(x)\). If \(\rho\) is positive and has zero flux, then \(D\rho'(x)=-V'(x)\rho(x)\). Dividing by \(D\rho(x)\) and integrating gives \(\log\rho(x)=-V(x)/D+\text{constant}\). Thus the candidate density is proportional to the exponential of minus the potential divided by the noise scale.

Formula: For \(dX_t=-V'(X_t)\,dt+\sigma\,dB_t\), set \(D=\sigma^2/2\). If \(Z=\int_{\mathbb{R}}e^{-V(x)/D}\,dx\) is finite and positive, the zero-flux candidate is \(\rho(x)=Z^{-1}e^{-V(x)/D}\). It satisfies \(\mu\rho-D\rho'=0\), hence the stationary Fokker–Planck equation. Normalizability is exactly the finiteness of \(Z\).

Here is a precise route from this calculation to invariance. Assume the SDE has a conservative Markov transition semigroup \(P_t\), and assume its coefficients and a candidate density \(\rho\) are regular enough that, for every smooth compactly supported test function \(f\), differentiation under the integral and integration by parts are valid for \(P_t f\). In particular, assume the boundary terms vanish. These analytic conditions hold in the examples below. The infinitesimal generator is \(\mathcal{L}g=\mu g'+Dg''\).

Proposition: Under the regularity and boundary assumptions just stated, if \(\rho\) is a probability density and \(-(\mu\rho)'+D\rho''=0\), then the probability measure with density \(\rho\) is invariant.

Proof. Fix a smooth compactly supported \(f\), and define \(F(t)=\int_{\mathbb{R}}P_t f(x)\rho(x)\,dx\). The backward equation for the transition semigroup gives \(\partial_tP_t f=\mathcal{L}P_t f\). By the assumed differentiation and boundary conditions,

$$ F'(t)=\int_{\mathbb{R}}\big(\mu(x)(P_tf)'(x)+D(P_tf)''(x)\big)\rho(x)\,dx =\int_{\mathbb{R}}P_tf(x)\big(-(\mu\rho)'(x)+D\rho''(x)\big)\,dx=0. $$

Therefore \(F(t)=F(0)=\int f(x)\rho(x)\,dx\) for every \(t\ge0\). Equality of integrals against all smooth compactly supported test functions determines probability measures, so the law at time \(t\) equals the initial law with density \(\rho\). This is invariance. \(\square\)

Worked Densities and Checks

Worked Example: Ornstein–Uhlenbeck Density

Consider \(dX_t=-\lambda(X_t-m)\,dt+\sigma\,dB_t\), with \(\lambda>0\) and \(\sigma>0\). Find and verify its invariant density.

Write the drift as \(-V'(x)\), where \(V(x)=\frac{\lambda}{2}(x-m)^2\). Since \(D=\sigma^2/2\), the zero-flux formula gives \(\rho(x)\) proportional to \(\exp[-\lambda(x-m)^2/\sigma^2]\). The Gaussian integral is \(\int_{\mathbb{R}}e^{-a(x-m)^2}\,dx=\sqrt{\pi/a}\) for \(a>0\). Taking \(a=\lambda/\sigma^2\) gives the normalizing constant:

$$ \rho(x)=\sqrt{\frac{\lambda}{\pi\sigma^2}}\, \exp\!\left(-\frac{\lambda(x-m)^2}{\sigma^2}\right). $$

To verify by substitution, differentiate the density: \(\rho'(x)=-\frac{2\lambda(x-m)}{\sigma^2}\rho(x)\). The flux is therefore

$$ J(x)=-\lambda(x-m)\rho(x)-\frac{\sigma^2}{2}\left(-\frac{2\lambda(x-m)}{\sigma^2}\rho(x)\right)=0. $$

So \(-J'(x)=0\), as required by the stationary Fokker–Planck equation. In this case invariance can also be checked directly. Starting at \(X_0=x\), the solution is \(X_t=m+e^{-\lambda t}(x-m)+\sigma\int_0^t e^{-\lambda(t-s)}\,dB_s\). The deterministic integrand in this Itô integral is progressively measurable and has finite squared integral on every finite time interval. If \(X_0\) is independent of the future Brownian increments and has the displayed Gaussian law, then \(X_t\) is Gaussian with mean \(m\) and variance \(e^{-2\lambda t}\frac{\sigma^2}{2\lambda}+\sigma^2\int_0^t e^{-2\lambda(t-s)}\,ds =e^{-2\lambda t}\frac{\sigma^2}{2\lambda}+\frac{\sigma^2}{2\lambda}(1-e^{-2\lambda t}) =\frac{\sigma^2}{2\lambda}\). This reproduces the same density, an independent check of invariance.

Worked Example: A Shifted Quadratic Potential

Take \(V(x)=(x-2)^2\) and \(D=1\), so the SDE is \(dX_t=-2(X_t-2)\,dt+\sqrt{2}\,dB_t\). Compute and verify the normalized zero-flux density.

The unnormalized density is \(e^{-(x-2)^2}\). Its integral is \(\sqrt{\pi}\), so \(\rho(x)=\pi^{-1/2}e^{-(x-2)^2}\). Differentiation gives \(\rho'(x)=-2(x-2)\rho(x)\). The drift is \(\mu(x)=-2(x-2)\), and \(D=1\), hence

$$ J(x)=\mu(x)\rho(x)-D\rho'(x) =-2(x-2)\rho(x)-\big(-2(x-2)\rho(x)\big)=0. $$

The normalization also identifies a normal distribution with mean \(2\) and variance \(1/2\): the exponent is \(-(x-2)^2=-\frac{(x-2)^2}{2(1/2)}\), and its Gaussian prefactor is \(1/\sqrt{2\pi(1/2)}=1/\sqrt{\pi}\). Thus both the normalization and the flux calculation agree with the potential formula.

Worked Example: A Logarithmic Potential and a Heavy-Tailed Density

Let \(V(x)=2\log(1+x^2)\) and \(D=1\). Then \(\mu(x)=-V'(x)=-4x/(1+x^2)\), and the SDE has diffusion coefficient \(\sqrt{2}\). Find the normalized zero-flux density.

The unnormalized density is \(e^{-V(x)}=(1+x^2)^{-2}\). To normalize, use the standard beta-integral identity \(\int_{\mathbb{R}}(1+x^2)^{-q}\,dx=\sqrt{\pi}\,\Gamma(q-\frac12)/\Gamma(q)\), valid for \(q>\frac12\). For \(q=2\), \(\Gamma(2)=1\) and \(\Gamma(3/2)=\sqrt{\pi}/2\), so the integral is \(\pi/2\). Therefore

$$ \rho(x)=\frac{2}{\pi(1+x^2)^2}. $$

Differentiate: \(\rho'(x)=-4x(1+x^2)^{-1}\rho(x)\). Substitution into the flux gives \(J(x)=-4x(1+x^2)^{-1}\rho(x)-\rho'(x)=0\), so the stationary equation holds. The same calculation also displays why normalizability must be checked: for the family \(V(x)=\kappa\log(1+x^2)/2\), the candidate has tail proportional to \(|x|^{-\kappa/D}\). Its integral is finite precisely when \(\kappa/D>1\), or \(\kappa>D\). Below or at that threshold, it is not a probability density.

Common Mistake: Losing the Noise Factor

A frequent error is to write the gradient density as proportional to \(e^{-V(x)}\) regardless of the diffusion coefficient. That omits the factor \(D=\sigma^2/2\). For example, with \(V(x)=a x^2/2\), the drift is \(-ax\). If \(\sigma=2\), then \(D=2\), and the correct exponent is \(-V/D=-ax^2/4\), not \(-ax^2/2\). The correct variance is \(\sigma^2/(2a)=2/a\); the incorrect exponent would imply variance \(1/a\). The stationary equation itself detects the error: the zero-flux condition requires \(\rho'/\rho=\mu/D\), not \(\mu\).

Also distinguish a stationary Fokker–Planck solution from an invariant probability law. A formal positive solution may fail to be integrable, or a nonzero constant flux may be incompatible with the intended boundary behavior. For a gradient drift with zero flux, the computation gives a candidate; normalization and the stated regularity and boundary conditions complete the argument for invariance.

Check Your Understanding

Use the zero-flux equation and check both normalization and boundary conditions where relevant.

  1. For \(dX_t=-3(X_t+1)\,dt+2\,dB_t\), compute the mean, variance, and normalized invariant density.
  2. For \(V(x)=c x^2/2\) and constant diffusion coefficient \(\sigma>0\), derive the density and state the conditions on \(c\) for it to be normalizable.
  3. For a proposed density \(\rho\), explain why \(\partial_xJ=0\) does not by itself imply \(J=0\).
  4. For \(V(x)=\kappa\log(1+x^2)/2\), use the tail behavior to find the condition on \(\kappa\) and \(D\) for normalizability.
  5. Verify directly that the Ornstein–Uhlenbeck variance \(\sigma^2/(2\lambda)\) is unchanged by its transition formula.

Previous

Next

Back to Tutorials