Ergodic Averages and Long-Run Expectations

From Invariant Distributions to Time Averages

An invariant distribution describes the law of a diffusion when its distribution is unchanged by the passage of time. An ergodic time average makes a stronger, pathwise statement: along almost every trajectory, the fraction or average accumulated over a long interval approaches an expectation under that invariant distribution. The distinction matters. Invariance alone does not guarantee that a trajectory samples the full invariant distribution, and convergence of time averages requires an ergodicity assumption as well as integrability of the observable.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. For a diffusion with SDE $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$, the drift is $\mu$ and the diffusion coefficient is $\sigma$. The Brownian integral is the Itô integral: on each finite interval its integrand is adapted, progressively measurable, and in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The results below concern a nonexplosive, time-homogeneous diffusion with a well-defined Markov transition law.

Definition: A probability measure $\pi$ is invariant for $X$ if $X_0\sim\pi$ implies $X_t\sim\pi$ for every $t\ge0$. The stationary process obtained with this initialization is ergodic if every event invariant under all time shifts has probability zero or one. For a measurable observable $f$, its continuous-time average is $\frac1T\int_0^T f(X_t)\,dt$, whenever the integral exists.

Ergodicity is a property of the stationary process, not merely the existence of an invariant density. For example, a process that remains forever in whichever one of two invariant regions it starts in can have invariant probability measures without being ergodic under a mixture of those measures.

The Continuous-Time Ergodic Theorem

Theorem (Birkhoff ergodic theorem for a stationary diffusion): Suppose $X$ is a time-homogeneous diffusion initialized with its invariant probability measure $\pi$, and its stationary process is ergodic. If $f$ is measurable and $\int |f(x)|\,\pi(dx)<\infty$, then $$ \frac1T\int_0^T f(X_t)\,dt\longrightarrow \int f(x)\,\pi(dx) \quad\text{almost surely as }T\to\infty. $$ This is the continuous-time Birkhoff ergodic theorem, a standard result in ergodic theory; its general proof is beyond this tutorial. The integrability condition ensures that the invariant expectation is finite and the time integral is defined almost surely on finite intervals.

The theorem is stated for stationary initialization. It does not, by itself, prove convergence from every fixed starting point. Such a conclusion can follow from additional mixing or recurrence results, but those require their own hypotheses. For bounded $f$, the displayed averages are bounded by $\sup_x|f(x)|$, so almost-sure convergence also gives convergence of their expectations by dominated convergence. For an unbounded integrable $f$, the theorem guarantees almost-sure convergence but does not, without further conditions, guarantee convergence of expectations of the averages.

For a stationary Markov process, ergodicity can also be expressed through shift-invariant events: such events have probability zero or one. In applications, a convenient way to establish the theorem's hypothesis is to verify that the stationary process is mixing; mixing implies ergodicity. The Ornstein–Uhlenbeck examples below use the standard fact that a stationary Gaussian process whose covariance tends to zero at large time separation is mixing and hence ergodic.

Ornstein–Uhlenbeck Stationarity and a Coupling Argument

Consider the Ornstein–Uhlenbeck SDE $$ dX_t=-\kappa(X_t-m)\,dt+\sigma\,dB_t,\qquad \kappa>0,\quad \sigma>0. $$ Its diffusion coefficient is constant, and the drift pulls the state toward $m$. The stochastic integral in its solution has integrand $\sigma e^{-\kappa(t-s)}$, which is adapted and progressively measurable and satisfies $$ \mathbb{E}\int_0^t \sigma^2e^{-2\kappa(t-s)}\,ds =\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t})<\infty. $$ Thus it belongs to $\mathcal{L}^2$ on each finite interval.

For completeness, the explicit solution follows by applying Itô's formula to $g(t,x)=e^{\kappa t}(x-m)$. Its derivatives are $g_t=\kappa e^{\kappa t}(x-m)$, $g_x=e^{\kappa t}$, and $g_{xx}=0$; the Itô correction is $\tfrac12g_{xx}\sigma^2\,dt=0$. Substitution of the SDE gives $d\,g(t,X_t)=\sigma e^{\kappa t}\,dB_t$. Integrating and rearranging yields $$ X_t=m+e^{-\kappa t}(X_0-m)+\sigma\int_0^t e^{-\kappa(t-s)}\,dB_s. $$ The integral is centered Gaussian with variance $\sigma^2\int_0^t e^{-2\kappa(t-s)}ds=\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa t})$. Consequently, if $X_0$ is independent of future Brownian increments and has distribution $N(m,\sigma^2/(2\kappa))$, then $X_t$ has that same distribution. This is the invariant Gaussian law.

Under this stationary initialization, the covariance at lag $u\ge0$ is $\frac{\sigma^2}{2\kappa}e^{-\kappa u}$. It tends to zero as $u\to\infty$, so the stationary Gaussian process is mixing and therefore ergodic by the standard Gaussian-process result stated above. The ergodic theorem applies to every $f$ integrable under this Gaussian law.

Proposition: For the Ornstein–Uhlenbeck process with $\kappa>0$, if $f$ is Lipschitz and integrable under the invariant Gaussian law, then, for every deterministic initial value $X_0=x$, $$ \frac1T\int_0^T f(X_t)\,dt\longrightarrow \int f\,d\pi \quad\text{almost surely}. $$ Here $\pi=N(m,\sigma^2/(2\kappa))$.

Proof. On the same probability space and driven by the same Brownian motion, let $X_t^*$ be the stationary solution with $X_0^*\sim\pi$, independent of future Brownian increments. The explicit solution shows $X_t-X_t^*=e^{-\kappa t}(x-X_0^*)$. If $L$ is a Lipschitz constant for $f$, then $$ \left|\frac1T\int_0^T f(X_t)\,dt-\frac1T\int_0^T f(X_t^*)\,dt\right| \le \frac{L|x-X_0^*|}{T}\int_0^T e^{-\kappa t}\,dt \le \frac{L|x-X_0^*|}{\kappa T}. $$ Since $X_0^*$ is finite almost surely, the last expression tends to zero almost surely. The stationary ergodic theorem gives convergence of the average for $X^*$ to $\int f\,d\pi$; the difference bound transfers that limit to $X$. $\square$

This proposition covers $f(x)=x$, but not $f(x)=x^2$, which is not globally Lipschitz. For the square, use the same coupling: $$ |X_t^2-(X_t^*)^2| \le e^{-\kappa t}|x-X_0^*|(|X_t|+|X_t^*|). $$ The expectation of $\int_0^\infty e^{-\kappa t}(|X_t|+|X_t^*|)\,dt$ is finite: the explicit solution gives uniformly bounded first moments for both processes, and $\int_0^\infty e^{-\kappa t}dt=1/\kappa$. Hence that integral is finite almost surely. Its contribution to the difference of the time averages, after division by $T$, tends to zero. The stationary theorem therefore also transfers to $x^2$. This argument illustrates why a result for unbounded observables needs a suitable integrability check rather than an automatic appeal to the Lipschitz proposition.

Worked Ornstein–Uhlenbeck Averages

Worked Example: Averages of the State and Its Square

Let $dX_t=-3(X_t-2)\,dt+3\,dB_t$. Find the long-run averages of $X_t$ and $X_t^2$, first under stationary initialization and then from any deterministic initial value.

Here $\kappa=3$, $m=2$, and $\sigma=3$. The invariant variance is $\sigma^2/(2\kappa)=9/(2\cdot3)=3/2$, so the invariant law is $N(2,3/2)$. Its first moment is $2$. Its second moment is variance plus squared mean: $$ \int x^2\,\pi(dx)=\frac32+2^2=\frac32+4=\frac{11}{2}. $$ Both observables are integrable. The ergodic theorem gives time-average limits $2$ and $11/2$ almost surely from stationary initialization. The coupling argument above gives the same limits from any deterministic initial value: $x$ is Lipschitz, and the displayed square-difference estimate handles $x^2$.

Check: decomposing $X_t^2=(X_t-2)^2+4(X_t-2)+4$, the corresponding invariant expectations are $3/2$, $4\cdot0=0$, and $4$. Their sum is $3/2+0+4=11/2$, agreeing with the second-moment calculation.

Worked Example: A Fourth-Power Observable

Let $dX_t=-2X_t\,dt+2\,dB_t$, with stationary initialization. Find the long-run average of $X_t^4$.

The invariant mean is $0$, and the invariant variance is $2^2/(2\cdot2)=1$. Thus the invariant law is standard normal. For a standard normal random variable $Z$, integration by parts in its density gives $\mathbb{E}[Z^4]=3\mathbb{E}[Z^2]=3$, since $\mathbb{E}[Z^2]=1$. The fourth power is integrable under the invariant Gaussian law. The stationary process is ergodic, so $$ \frac1T\int_0^T X_t^4\,dt\longrightarrow 3\quad\text{almost surely}. $$

Check: the variance used here is $4/(2\cdot2)=1$; the centered Gaussian fourth moment is $3\cdot1^2=3$. In particular, the limit is positive even though the invariant mean of $X_t$ is zero: averaging a fourth power is not the same as taking the fourth power of an average.

Worked Example: Long-Run Fraction Above the Mean

Let $dX_t=-X_t\,dt+\sqrt{2}\,dB_t$, initialized in its invariant law. What fraction of a long interval is spent strictly above zero?

Here $\kappa=1$, $m=0$, and $\sigma=\sqrt2$, so the invariant variance is $(\sqrt2)^2/(2\cdot1)=1$. Set $f(x)=\mathbf{1}_{\{x>0\}}$, using the strict inequality $x>0$. This indicator is integrable because it is bounded by $1$. The invariant distribution is symmetric about zero and has no atom at zero, so $\pi((0,\infty))=1/2$. Ergodicity gives $$ \frac1T\int_0^T \mathbf{1}_{\{X_t>0\}}\,dt\longrightarrow\frac12 \quad\text{almost surely}. $$

Check: the set $(0,\infty)$ and its reflection $(-\infty,0)$ have equal invariant probability, and together they have probability $1$ because $\pi(\{0\})=0$. Each probability is therefore $1/2$.

When the Limit Can Be Used

The theorem turns a path-dependent calculation into an invariant-distribution expectation, but the hypotheses should be checked for the observable at hand. For the stationary diffusion, verify that the invariant law is a probability measure, that its stationary process is ergodic, and that $\int|f|\,d\pi<\infty$. If the starting point is fixed rather than distributed according to $\pi$, use an additional result—such as the Ornstein–Uhlenbeck coupling proposition—rather than silently treating the process as stationary.

A common mistake is to infer that an invariant distribution alone makes every trajectory's time average equal its invariant expectation. The theorem requires ergodicity. Another is to assume that almost-sure convergence automatically implies convergence of expectations for an unbounded observable. The dominated-convergence argument applies directly to bounded observables, but an unbounded one requires an additional condition such as uniform integrability of the averages.

Question	Condition or conclusion
Which law supplies the long-run expectation?	The invariant probability measure $\pi$.
Which initialization is covered directly?	Stationary initialization, $X_0\sim\pi$.
What condition on the observable is required?	$\int \|f\|\,d\pi<\infty$.
What is the basic conclusion?	Almost-sure convergence of the time average to $\int f\,d\pi$.
Does invariance alone imply the conclusion?	No; ergodicity is also required.

Check Your Understanding

For each question, identify the invariant expectation and the hypotheses needed to use it.

For $dX_t=-4(X_t-1)\,dt+2\,dB_t$, what are the invariant mean and variance?
For that same diffusion, what is the predicted long-run average of $X_t^2$? Show the second-moment calculation.
Why does the stationary ergodic theorem not, on its own, establish convergence from every deterministic starting value?
For a stationary standard-normal Ornstein–Uhlenbeck process, what long-run fraction of time is spent in $\{x\ge0\}$, and why does using $\ge$ rather than $>$ not change the answer?
Which additional issue must be addressed before deducing convergence of expected time averages from almost-sure convergence for an unbounded observable?

Question	Condition or conclusion
Which law supplies the long-run expectation?	The invariant probability measure \(\pi\).
Which initialization is covered directly?	Stationary initialization, \(X_0\sim\pi\).
What condition on the observable is required?	\(\int \|f\|\,d\pi<\infty\).
What is the basic conclusion?	Almost-sure convergence of the time average to \(\int f\,d\pi\).
Does invariance alone imply the conclusion?	No; ergodicity is also required.