Invariant Measures and Stationarity

Concept Definition

Let \((X_t)_{t \ge 0}\) be a time-homogeneous Markov process on \(\mathbb{R}^d\) with transition semigroup \((P_t)_{t \ge 0}\) defined by

\[ P_t f(x) := \mathbb{E}^x[f(X_t)], \qquad f \text{ bounded measurable}. \]

The semigroup acts on measures from the right: for a probability measure \(\mu\),

\[ (\mu P_t)(A) := \int_{\mathbb{R}^d} P_t \mathbf{1}_A(x)\,\mu(\mathrm{d}x) = \mathbb{P}^{\mu}(X_t \in A). \]

Definition: Invariant Measure

A probability measure \(\pi\) on \(\mathbb{R}^d\) is invariant for \((P_t)\) if

\[ \pi P_t = \pi \qquad \text{for all } t \ge 0, \]

equivalently, for all bounded measurable \(f\) and all \(t \ge 0\):

\[ \int_{\mathbb{R}^d} P_t f(x)\,\pi(\mathrm{d}x) = \int_{\mathbb{R}^d} f(x)\,\pi(\mathrm{d}x). \]

In other words: if \(X_0 \sim \pi\) then \(X_t \sim \pi\) for all \(t \ge 0\).

Definition: Stationarity

A process \((X_t)\) is stationary if its finite-dimensional distributions are invariant under time shifts: for all \(h \ge 0\) and all \(0 \le t_1 < \cdots < t_k\),

\[ (X_{t_1 + h}, \dots, X_{t_k + h}) \stackrel{d}{=} (X_{t_1}, \dots, X_{t_k}). \]

Starting a Markov process from an invariant measure \(\pi\) yields a stationary process.

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Explanation

Generator Characterisation

Let \(\mathcal{L}\) be the infinitesimal generator of \((P_t)\). If \(\pi\) is invariant, differentiating \(\int P_t f\,\mathrm{d}\pi = \int f\,\mathrm{d}\pi\) at \(t = 0\) gives

\[ \int_{\mathbb{R}^d} (\mathcal{L}f)(x)\,\pi(\mathrm{d}x) = 0 \qquad \text{for all } f \in \mathrm{Dom}(\mathcal{L}). \]

The interchange of differentiation and integration is justified by dominated convergence when \(f \in \mathrm{Dom}(\mathcal{L})\) (so \(\frac{P_t f - f}{t} \to \mathcal{L}f\) in \(L^\infty\) and is uniformly bounded). In terms of the formal \(L^2\)-adjoint \(\mathcal{L}^*\), this is \(\mathcal{L}^* \pi = 0\).

Fokker–Planck Equation for Stationary Densities

For a diffusion with drift \(b\) and covariance matrix \(a = \sigma\sigma^\top\), if \(\pi\) has a density (also denoted \(\pi\)), stationarity requires \(\mathcal{L}^*\pi = 0\):

\[ 0 = -\frac{\partial}{\partial x_i}\!\bigl(b^{i}(x)\,\pi(x)\bigr) + \frac{1}{2}\,\frac{\partial^2}{\partial x_i \partial x_j}\!\bigl(a^{ij}(x)\,\pi(x)\bigr). \]

This is the stationary Fokker–Planck equation. It is an elliptic PDE for the unknown density \(\pi\).

Existence of Invariant Measures

An invariant measure need not exist for every diffusion. Sufficient conditions include:

• Lyapunov / Foster criterion: there exist a function \(V \ge 1\), a compact set \(C\), and constants \(\alpha > 0\), \(K < \infty\) such that \(\mathcal{L}V(x) \le -\alpha V(x) + K\,\mathbf{1}_C(x)\). This implies tightness of the time-averages \(\frac{1}{T}\int_0^T P_t(x, \cdot)\,\mathrm{d}t\) and, by Prokhorov, existence of an invariant measure.
• Gradient diffusion: if \(\int e^{-V(x)}\,\mathrm{d}x < \infty\), the gradient diffusion has invariant density \(\pi \propto e^{-V}\) (see Example below).

Uniqueness typically requires irreducibility (every open set is reachable) and a non-degeneracy condition on \(a\).

Reversibility and Detailed Balance

A stronger condition than invariance is reversibility.

Definition: Reversibility

A stationary process \((X_t)\) with invariant measure \(\pi\) is reversible (satisfies detailed balance) if for all bounded measurable \(f, g\) and all \(t \ge 0\):

\[ \int f(x)\,(P_t g)(x)\,\pi(\mathrm{d}x) = \int g(x)\,(P_t f)(x)\,\pi(\mathrm{d}x). \]

Equivalently, \(\mathcal{L}\) is self-adjoint in \(L^2(\pi)\).

Reversibility implies that the process is time-symmetric: \((X_t)_{0 \le t \le T}\) and \((X_{T-t})_{0 \le t \le T}\) have the same law whenever \(X_0 \sim \pi\). This connects directly to time reversal of diffusions; see Time Reversal of Diffusions.

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Diagram / Example

Example: Gradient Diffusion (Langevin Equation)

Consider

\[ \mathrm{d}X_t = -\nabla V(X_t)\,\mathrm{d}t + \sqrt{2}\,\mathrm{d}W_t, \]

where \(V : \mathbb{R}^d \to \mathbb{R}\) is smooth and \(Z := \int_{\mathbb{R}^d} e^{-V(x)}\,\mathrm{d}x < \infty\).

Claim. \(\pi(x) = Z^{-1} e^{-V(x)}\) is an invariant density.

Verification via Fokker–Planck. Here \(b^i = -\partial_i V\) and \(a^{ij} = 2\delta^{ij}\). The stationary equation \(\mathcal{L}^*\pi = 0\) becomes

\[ \frac{\partial}{\partial x_i}\!\bigl(\partial_i V \cdot \pi\bigr) + \Delta \pi = 0. \]

Substituting \(\pi = Z^{-1}e^{-V}\) and computing each term separately (for a fixed index \(i\), no sum):

\[ \frac{\partial}{\partial x_i}\!\bigl(\partial_i V \cdot e^{-V}\bigr) = e^{-V}\bigl(\partial_i^2 V - (\partial_i V)^2\bigr), \qquad \partial_i^2(e^{-V}) = e^{-V}\bigl((\partial_i V)^2 - \partial_i^2 V\bigr). \]

Summing over \(i\): the two contributions cancel term by term, giving \(\mathcal{L}^*\pi = 0\). \(\square\)

Reversibility. This process is reversible with respect to \(\pi\); the gradient structure ensures detailed balance.

Summary Table

Property	Condition	Implication
Invariance	\(\pi P_t = \pi\)	Law preserved in time
Stationarity	Start from \(\pi\)	Shift-invariant distributions
Generator condition	\(\int \mathcal{L}f\,\mathrm{d}\pi = 0\)	Characterises invariant \(\pi\)
Reversibility	\(\mathcal{L}\) self-adjoint in \(L^2(\pi)\)	Time-symmetric dynamics

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Proof / Derivation

We verify the generator condition \(\int \mathcal{L}f\,\mathrm{d}\pi = 0\) is equivalent to invariance under mild assumptions.

Forward direction. If \(\pi P_t = \pi\), then \(\int P_t f\,\mathrm{d}\pi = \int f\,\mathrm{d}\pi\) for all \(t \ge 0\). The left side is differentiable in \(t\); differentiating at \(t = 0\) and using the definition of the generator gives \(\int \mathcal{L}f\,\mathrm{d}\pi = 0\).

Backward direction. Suppose \(\int \mathcal{L}f\,\mathrm{d}\pi = 0\) for all \(f \in \mathrm{Dom}(\mathcal{L})\). Set \(h(t) := \int P_t f\,\mathrm{d}\pi\). Then

\[ h'(t) = \int \mathcal{L}(P_t f)\,\mathrm{d}\pi = 0, \]

where the first equality uses \(\frac{\mathrm{d}}{\mathrm{d}t}P_t f = \mathcal{L} P_t f\) (the Kolmogorov forward equation for the semigroup), and the second applies the hypothesis with test function \(P_t f \in \mathrm{Dom}(\mathcal{L})\). Hence \(h(t) = h(0) = \int f\,\mathrm{d}\pi\) for all \(t \ge 0\), confirming \(\pi P_t = \pi\). \(\square\)

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What to Remember

• An invariant measure \(\pi\) satisfies \(\pi P_t = \pi\): the law is preserved under the dynamics.
• Starting from \(\pi\) produces a stationary process.
• Stationary densities solve the Fokker–Planck equation \(\mathcal{L}^*\pi = 0\).
• Existence follows from Lyapunov criteria; uniqueness from irreducibility and non-degeneracy.
• Reversibility (detailed balance) is stronger than invariance and implies time-symmetry of the process.

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Exercises

Exercise 1. Consider the one-dimensional Ornstein–Uhlenbeck process \(\mathrm{d}X_t = -\theta X_t\,\mathrm{d}t + \sigma\,\mathrm{d}W_t\) with \(\theta > 0\). Using the stationary Fokker–Planck equation \(\mathcal{L}^*\pi = 0\), show that the invariant density is Gaussian with mean \(0\) and variance \(\sigma^2/(2\theta)\).

Solution to Exercise 1

The OU process has \(b(x) = -\theta x\) and \(\sigma(x) = \sigma\) (constant), so \(a(x) = \sigma^2\). The generator is

\[ \mathcal{L}f(x) = -\theta x\,f'(x) + \frac{\sigma^2}{2}\,f''(x) \]

The stationary Fokker–Planck equation \(\mathcal{L}^*\pi = 0\) in one dimension is

\[ -\frac{\mathrm{d}}{\mathrm{d}x}\bigl(b(x)\,\pi(x)\bigr) + \frac{1}{2}\frac{\mathrm{d}^2}{\mathrm{d}x^2}\bigl(\sigma^2\,\pi(x)\bigr) = 0 \]

Since \(\sigma^2\) is constant, this becomes

\[ \frac{\mathrm{d}}{\mathrm{d}x}\bigl(\theta x\,\pi(x)\bigr) + \frac{\sigma^2}{2}\,\pi''(x) = 0 \]

Integrating once (the integration constant must be zero for \(\pi\) to be integrable):

\[ \theta x\,\pi(x) + \frac{\sigma^2}{2}\,\pi'(x) = 0 \]

This gives

\[ \frac{\pi'(x)}{\pi(x)} = -\frac{2\theta x}{\sigma^2} \]

Integrating:

\[ \log\pi(x) = -\frac{\theta x^2}{\sigma^2} + C_0 \]

so \(\pi(x) \propto \exp(-\theta x^2/\sigma^2)\). Normalizing:

\[ \pi(x) = \sqrt{\frac{\theta}{\pi\sigma^2}}\,\exp\!\left(-\frac{\theta x^2}{\sigma^2}\right) \]

This is a Gaussian density \(\mathcal{N}(0, \sigma^2/(2\theta))\) with mean \(0\) and variance \(\sigma^2/(2\theta)\).

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Exercise 2. Let \(V(x) = \frac{1}{4}x^4 - \frac{1}{2}x^2\) (a double-well potential) on \(\mathbb{R}\). Consider the gradient diffusion \(\mathrm{d}X_t = -V'(X_t)\,\mathrm{d}t + \sqrt{2}\,\mathrm{d}W_t\). Write down the invariant density \(\pi(x)\) (up to a normalising constant). Sketch \(\pi(x)\) and identify the locations of its maxima. Is this process reversible?

Solution to Exercise 2

With \(V(x) = \frac{1}{4}x^4 - \frac{1}{2}x^2\) and the gradient diffusion \(\mathrm{d}X_t = -V'(X_t)\,\mathrm{d}t + \sqrt{2}\,\mathrm{d}W_t\), the invariant density is

\[ \pi(x) \propto e^{-V(x)} = \exp\!\left(-\frac{1}{4}x^4 + \frac{1}{2}x^2\right) \]

The maxima of \(\pi\) occur where \(V(x)\) is minimized. Setting \(V'(x) = x^3 - x = x(x^2 - 1) = 0\) gives critical points \(x = 0, \pm 1\). We have \(V''(x) = 3x^2 - 1\), so \(V''(0) = -1 < 0\) (local maximum of \(V\), hence local minimum of \(\pi\)) and \(V''(\pm 1) = 2 > 0\) (local minima of \(V\), hence local maxima of \(\pi\)).

Therefore \(\pi(x)\) has two maxima at \(x = \pm 1\) and a local minimum at \(x = 0\). The density is bimodal, reflecting the double-well structure of \(V\).

Reversibility: Yes, this process is reversible. It is a gradient diffusion with constant diffusion matrix (\(a = 2I\)), which automatically satisfies detailed balance with respect to \(\pi(x) \propto e^{-V(x)}\). The generator

\[ \mathcal{L}f = -V'(x)f'(x) + f''(x) \]

is self-adjoint in \(L^2(\pi)\): for \(f, g \in C_c^\infty(\mathbb{R})\),

\[ \int (\mathcal{L}f)\,g\,\pi\,\mathrm{d}x = -\int f'\,g'\,\pi\,\mathrm{d}x = \int f\,(\mathcal{L}g)\,\pi\,\mathrm{d}x \]

which follows by integration by parts.

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Exercise 3. Construct a Lyapunov function \(V(x) = 1 + |x|^2\) for the diffusion \(\mathrm{d}X_t = -X_t\,\mathrm{d}t + \mathrm{d}W_t\) in \(\mathbb{R}^d\). Verify the Foster–Lyapunov criterion \(\mathcal{L}V(x) \le -\alpha V(x) + K\,\mathbf{1}_C(x)\) by computing \(\mathcal{L}V\), and find explicit constants \(\alpha > 0\), \(K < \infty\), and a compact set \(C\).

Solution to Exercise 3

The diffusion is \(\mathrm{d}X_t = -X_t\,\mathrm{d}t + \mathrm{d}W_t\) in \(\mathbb{R}^d\), with \(b(x) = -x\) and \(a = I_d\). The Lyapunov function is \(V(x) = 1 + |x|^2\).

Computing \(\mathcal{L}V\):

\[ \partial_i V = 2x^i, \qquad \partial_i\partial_j V = 2\delta^{ij} \]

\[ \mathcal{L}V(x) = b^i(x)\,\partial_i V(x) + \frac{1}{2}a^{ij}\,\partial_i\partial_j V(x) = (-x^i)(2x^i) + \frac{1}{2}\delta^{ij}(2\delta^{ij}) \]

\[ = -2|x|^2 + d \]

We want \(\mathcal{L}V(x) \le -\alpha V(x) + K\,\mathbf{1}_C(x)\). Since \(V(x) = 1 + |x|^2\):

\[ -2|x|^2 + d \le -\alpha(1 + |x|^2) + K\,\mathbf{1}_C(x) \]

Choose \(\alpha = 1\). Then we need

\[ -2|x|^2 + d \le -(1 + |x|^2) + K\,\mathbf{1}_C(x) \]

\[ -|x|^2 + d + 1 \le K\,\mathbf{1}_C(x) \]

For \(|x|^2 \ge d + 1\), the left side is \(\le 0\) and the inequality holds without the indicator. Choose \(C = \{x : |x|^2 \le d + 1\}\) (a compact ball) and \(K = d + 1\). Then for \(x \in C\): \(-|x|^2 + d + 1 \le d + 1 = K\), so the inequality holds.

Therefore the Foster–Lyapunov criterion is satisfied with \(\alpha = 1\), \(K = d + 1\), and \(C = \{x : |x| \le \sqrt{d+1}\}\).

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Exercise 4. Consider the two-dimensional diffusion

\[ \mathrm{d}X_t^1 = -X_t^1\,\mathrm{d}t + X_t^2\,\mathrm{d}t + \mathrm{d}W_t^1, \qquad \mathrm{d}X_t^2 = -X_t^1\,\mathrm{d}t - X_t^2\,\mathrm{d}t + \mathrm{d}W_t^2. \]

Find the invariant measure (hint: try a Gaussian ansatz). Is this process reversible? Justify your answer by checking whether \(\mathcal{L}\) is self-adjoint in \(L^2(\pi)\).

Solution to Exercise 4

The drift is \(b(x) = (-x^1 + x^2,\, -x^1 - x^2)^\top\) and \(a = I_2\). Try a Gaussian invariant measure \(\pi \sim \mathcal{N}(0, \Sigma)\) with \(\Sigma = \text{diag}(\sigma_1^2, \sigma_2^2)\) or more generally a symmetric positive definite matrix.

The drift can be written \(b(x) = Bx\) where

\[ B = \begin{pmatrix} -1 & 1 \\ -1 & -1 \end{pmatrix} \]

For a linear SDE \(\mathrm{d}X = BX\,\mathrm{d}t + \mathrm{d}W\) with constant \(a = I\), the invariant covariance \(\Sigma\) solves the Lyapunov equation

\[ B\Sigma + \Sigma B^\top + I = 0 \]

Writing \(\Sigma = \begin{pmatrix} a & c \\ c & b \end{pmatrix}\) and \(B^\top = \begin{pmatrix} -1 & -1 \\ 1 & -1 \end{pmatrix}\):

\[ B\Sigma + \Sigma B^\top = \begin{pmatrix} -2a+2c & -2c+a+b \\ a+b-2c & 2c-2b \end{pmatrix} + \text{correction} \]

Computing directly:

\[ B\Sigma = \begin{pmatrix} -a+c & -c+b \\ -a-c & -c-b \end{pmatrix}, \quad \Sigma B^\top = \begin{pmatrix} -a-c & a-c \\ -c-b & c-b \end{pmatrix} \]

\[ B\Sigma + \Sigma B^\top = \begin{pmatrix} -2a & a-2c+b \\ a-2c+b & -2b \end{pmatrix} = -I = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \]

From the diagonal entries: \(-2a = -1\) gives \(a = 1/2\), \(-2b = -1\) gives \(b = 1/2\). From the off-diagonal: \(a - 2c + b = 0\) gives \(c = (a+b)/2 = 1/2\).

Wait — checking: \(c = 1/2\) but then \(\Sigma = \frac{1}{2}\begin{pmatrix}1&1\\1&1\end{pmatrix}\) which is singular. Let me recompute.

Actually, recomputing \(\Sigma B^\top\) with \(B^\top = \begin{pmatrix}-1&-1\\1&-1\end{pmatrix}\):

\[ \Sigma B^\top = \begin{pmatrix}a&c\\c&b\end{pmatrix}\begin{pmatrix}-1&-1\\1&-1\end{pmatrix} = \begin{pmatrix}-a+c & -a-c \\ -c+b & -c-b\end{pmatrix} \]

\[ B\Sigma + \Sigma B^\top = \begin{pmatrix}-2a+2c & -a-2c+b \\ a+b-2c & 2c-2b\end{pmatrix} + \begin{pmatrix}0&0\\0&0\end{pmatrix} \]

Wait, let me redo this carefully:

\[ B\Sigma + \Sigma B^\top = \begin{pmatrix}-a+c-a+c & -c+b-a-c \\ -a-c-c+b & -c-b+c-b \end{pmatrix} \]

Hmm, let me just add element-by-element. \((B\Sigma)_{11} = -a+c\), \((\Sigma B^\top)_{11} = -a+c\), sum \(= -2a+2c\). Setting \(-2a+2c = -1\) gives \(a - c = 1/2\). \((B\Sigma+\Sigma B^\top)_{22} = (-c-b)+(-c-b) = -2c-2b\). Setting \(-2b-2c=-1\) gives \(b+c=1/2\). Off-diagonal \((1,2)\): \((-c+b)+(-a-c) = -a+b-2c = 0\). So \(b = a+2c\).

From \(a-c=1/2\) and \(b+c=1/2\): \(b = 1/2-c\) and \(a = 1/2+c\). From \(b=a+2c\): \(1/2-c = 1/2+c+2c\), so \(-c = 3c\), giving \(c = 0\). Then \(a = b = 1/2\).

So \(\Sigma = \frac{1}{2}I\) and the invariant measure is \(\pi = \mathcal{N}(0, \frac{1}{2}I)\).

Reversibility: The process is reversible if and only if \(B\Sigma = \Sigma B^\top\) (the drift is self-adjoint in \(L^2(\pi)\)). We have \(B\Sigma = \frac{1}{2}B\) and \(\Sigma B^\top = \frac{1}{2}B^\top\), so reversibility requires \(B = B^\top\). But

\[ B = \begin{pmatrix}-1&1\\-1&-1\end{pmatrix} \ne \begin{pmatrix}-1&-1\\1&-1\end{pmatrix} = B^\top \]

So the process is not reversible. The antisymmetric part of \(B\) (the rotation component \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\)) generates a probability current that circulates around the origin, breaking detailed balance.

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Exercise 5. Prove that reversibility (detailed balance) implies invariance. That is, show that if for all bounded measurable \(f, g\) and all \(t \ge 0\),

\[ \int f(x)\,(P_t g)(x)\,\pi(\mathrm{d}x) = \int g(x)\,(P_t f)(x)\,\pi(\mathrm{d}x), \]

then \(\pi P_t = \pi\).

Solution to Exercise 5

Assume detailed balance holds: for all bounded measurable \(f, g\) and \(t \ge 0\),

\[ \int f(x)\,(P_t g)(x)\,\pi(\mathrm{d}x) = \int g(x)\,(P_t f)(x)\,\pi(\mathrm{d}x) \]

To show \(\pi P_t = \pi\), we must show \(\int P_t f\,\mathrm{d}\pi = \int f\,\mathrm{d}\pi\) for all bounded measurable \(f\).

Set \(g \equiv 1\) in the detailed balance condition. Then \(P_t g = P_t 1 = 1\) (since \(P_t\) is a Markov semigroup, \(P_t 1 = 1\)). The left side becomes

\[ \int f(x) \cdot 1\,\pi(\mathrm{d}x) = \int f\,\mathrm{d}\pi \]

The right side becomes

\[ \int 1 \cdot (P_t f)(x)\,\pi(\mathrm{d}x) = \int P_t f\,\mathrm{d}\pi \]

Therefore \(\int f\,\mathrm{d}\pi = \int P_t f\,\mathrm{d}\pi\) for all bounded measurable \(f\), which is exactly \(\pi P_t = \pi\). \(\square\)

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Exercise 6. Give an example of a diffusion that has an invariant measure but is not reversible. (Hint: consider adding a non-gradient drift component to a gradient diffusion.) Verify invariance directly and explain why detailed balance fails.

Solution to Exercise 6

Consider the two-dimensional diffusion

\[ \mathrm{d}X_t = b(X_t)\,\mathrm{d}t + \sqrt{2}\,\mathrm{d}W_t \]

with drift \(b(x) = -\nabla V(x) + \gamma(x)\), where \(V(x) = \frac{1}{2}|x|^2\) and \(\gamma(x) = (-x^2, x^1)^\top\) is a divergence-free rotational field (\(\nabla \cdot \gamma = 0\)).

Invariance: The stationary Fokker–Planck equation is \(\mathcal{L}^*\pi = 0\):

\[ -\nabla \cdot (b\,\pi) + \Delta\pi = 0 \]

Try \(\pi(x) \propto e^{-V(x)} = e^{-|x|^2/2}\). Then \(\nabla\pi = -(\nabla V)\pi\) and:

\[ -\nabla\cdot(b\,\pi) + \Delta\pi = -\nabla\cdot\bigl((-\nabla V + \gamma)\pi\bigr) + \Delta\pi \]

The gradient part: \(-\nabla\cdot(-\nabla V\,\pi) + \Delta\pi = \nabla\cdot(\nabla V\,\pi) + \Delta\pi = 0\) (this is the same cancellation as for the pure gradient diffusion).

The rotational part: \(-\nabla\cdot(\gamma\,\pi) = -\gamma\cdot\nabla\pi - (\nabla\cdot\gamma)\pi = -\gamma\cdot(-\nabla V\,\pi) - 0 = \gamma\cdot\nabla V\,\pi\). Now \(\gamma \cdot \nabla V = (-x^2, x^1)\cdot(x^1, x^2) = -x^1 x^2 + x^1 x^2 = 0\). So the rotational drift does not affect the invariant measure.

Therefore \(\pi(x) \propto e^{-|x|^2/2}\) is invariant.

Detailed balance fails: Reversibility requires \(\mathcal{L}\) to be self-adjoint in \(L^2(\pi)\). The generator is \(\mathcal{L}f = (-\nabla V + \gamma)\cdot\nabla f + \Delta f\). The rotational component \(\gamma\cdot\nabla f\) is antisymmetric in \(L^2(\pi)\) (since \(\gamma\) is divergence-free and orthogonal to \(\nabla V\)):

\[ \int (\gamma\cdot\nabla f)\,g\,\pi\,\mathrm{d}x = -\int f\,(\gamma\cdot\nabla g)\,\pi\,\mathrm{d}x \]

This antisymmetric part breaks the self-adjointness of \(\mathcal{L}\), so detailed balance fails. Physically, the rotational drift induces a probability current circulating around the origin.

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Exercise 7. Suppose a one-dimensional diffusion \(\mathrm{d}X_t = b(X_t)\,\mathrm{d}t + \sigma(X_t)\,\mathrm{d}W_t\) on an interval \((l, r)\) has generator \(\mathcal{L}f = b\,f' + \frac{1}{2}\sigma^2 f''\). Using the stationary Fokker–Planck equation, show that any invariant density must satisfy

\[ \pi(x) = \frac{C}{\sigma^2(x)}\exp\!\left(\int^x \frac{2\,b(y)}{\sigma^2(y)}\,\mathrm{d}y\right) \]

for some normalising constant \(C > 0\). Apply this formula to recover the invariant density of the OU process from Exercise 1.

Solution to Exercise 7

The generator is \(\mathcal{L}f = bf' + \frac{1}{2}\sigma^2 f''\). The adjoint operator \(\mathcal{L}^*\) acts on densities as

\[ \mathcal{L}^*\pi = -(b\pi)' + \frac{1}{2}(\sigma^2\pi)'' \]

Setting \(\mathcal{L}^*\pi = 0\) and integrating once (with the integration constant set to zero for integrability at the boundary):

\[ -b(x)\,\pi(x) + \frac{1}{2}\bigl(\sigma^2(x)\,\pi(x)\bigr)' = 0 \]

Expanding the derivative:

\[ -b\,\pi + \frac{1}{2}(\sigma^2)'\pi + \frac{1}{2}\sigma^2\pi' = 0 \]

Dividing by \(\frac{1}{2}\sigma^2\pi\):

\[ \frac{\pi'}{\pi} = \frac{2b - (\sigma^2)'}{\sigma^2} = \frac{2b}{\sigma^2} - \frac{(\sigma^2)'}{\sigma^2} \]

Integrating:

\[ \log\pi(x) = \int^x \frac{2b(y)}{\sigma^2(y)}\,\mathrm{d}y - \log\sigma^2(x) + \text{const} \]

Exponentiating:

\[ \pi(x) = \frac{C}{\sigma^2(x)}\exp\!\left(\int^x \frac{2b(y)}{\sigma^2(y)}\,\mathrm{d}y\right) \]

Application to the OU process: With \(b(x) = -\theta x\) and \(\sigma^2 = \sigma^2\) (constant):

\[ \pi(x) = \frac{C}{\sigma^2}\exp\!\left(\int^x \frac{-2\theta y}{\sigma^2}\,\mathrm{d}y\right) = \frac{C}{\sigma^2}\exp\!\left(-\frac{\theta x^2}{\sigma^2}\right) \]

This is \(\mathcal{N}(0, \sigma^2/(2\theta))\), consistent with Exercise 1.