Questions: Ergodic Theory for Stochastic Processes

Question 1

A process is stationary if its finite-dimensional distributions are time-invariant, and ergodic if time averages equal ensemble averages. What is the logical relationship?

• A. Stationarity implies ergodicity
• B. Ergodicity implies stationarity
• C. Ergodicity requires stationarity (or at least a stationary distribution to average against), but stationarity alone does not imply ergodicity
• D. They are equivalent conditions for Markov processes

Answer: (C)

Ergodicity requires a reference stationary measure π for the ensemble average E_π[f]. So stationarity (or convergence to a stationary distribution) is a prerequisite. However, a process can be stationary without being ergodic: consider two independent OU processes with different means, chosen at random. The combined process is stationary but not ergodic — each trajectory stays near one mean forever, so time averages differ across trajectories. Ergodicity additionally requires that the process 'mixes' enough to visit its entire stationary distribution.

Question 2

For a diffusion dX = μ(X)dt + σ(X)dW on ℝ with σ(x) > 0, a sufficient condition for ergodicity is that the process is positive recurrent (returns to compact sets in finite expected time). What drives this return?

• A. The diffusion coefficient σ(x) — larger noise makes the process return faster
• B. The drift μ(x) — if μ(x) points inward strongly enough for large |x| (e.g., μ(x) ~ -θx), the process is pulled back toward the center
• C. The initial condition X(0) — ergodicity depends on starting near the center
• D. The smoothness of the sample paths — continuous paths cannot escape to infinity

Answer: (B)

Positive recurrence is driven primarily by the drift pulling the process back from large excursions. For the OU process (μ(x) = -θx), the linear restoring force ensures the process returns to any neighborhood of zero in finite expected time. If the drift is zero or points outward (μ(x) = θx for θ > 0), the process escapes to infinity and has no stationary distribution. The diffusion σ helps with accessibility (the process can reach any state) but doesn't alone ensure return. The initial condition is irrelevant for ergodicity — the process forgets it over time.

Question 3

Explain the practical significance of ergodicity for Monte Carlo estimation.

Model answer: Ergodicity justifies estimating expectations E_π[f(X)] from a single long trajectory rather than many independent samples. If X(t) is ergodic with stationary distribution π, then (1/T)∫₀ᵀ f(X(t))dt → E_π[f] a.s. In discrete simulation, (1/N)Σf(X(tₖ)) → E_π[f] as N → ∞. This is essential because generating independent samples from π may be difficult or impossible, while simulating a single trajectory of the process is straightforward. MCMC methods exploit this: run one Markov chain for a long time, and time averages converge to the target distribution's expectations.

Without ergodicity, a single trajectory might get stuck in a subset of the state space and produce biased time averages. Ergodicity guarantees this doesn't happen — the trajectory visits all regions with the correct long-run frequency. The rate of convergence (mixing time) determines how long you need to run, and this is a separate quantitative question from the qualitative guarantee of convergence.