The Ornstein-Uhlenbeck process dX = -θX dt + σ dW is solved using the integrating factor e^{θt}. What is the explicit solution?
AX(t) = X(0)e^{θt} + σ∫₀ᵗ e^{θ(t-s)} dW(s)
BX(t) = X(0)e^{-θt} + σ∫₀ᵗ e^{-θ(t-s)} dW(s)
CX(t) = X(0) + σW(t) - θ∫₀ᵗ X(s)ds
DX(t) = X(0)e^{-θt} + σW(t)
Apply Itô's formula to Y(t) = X(t)e^{θt}: dY = e^{θt}(dX + θX dt) = e^{θt}(σ dW) = σe^{θt} dW. Integrating: Y(t) = Y(0) + σ∫₀ᵗ e^{θs} dW(s), so X(t) = e^{-θt}Y(t) = X(0)e^{-θt} + σ∫₀ᵗ e^{-θ(t-s)} dW(s). The exponential decay e^{-θt} on the initial condition shows mean reversion; the integral term is a weighted average of past noise with exponentially decaying weights — recent noise matters more than distant noise.
Question 2 Multiple Choice
The stationary variance of the OU process is σ²/(2θ). If θ is doubled (faster mean reversion) while σ stays constant, the stationary variance:
ADoubles — faster mean reversion increases fluctuations
BHalves — faster mean reversion pulls the process back more quickly, reducing the spread
CStays the same — variance depends only on σ
DQuadruples — variance is proportional to θ²
The stationary variance σ²/(2θ) is inversely proportional to θ. Stronger mean reversion (larger θ) gives the process less time to wander before being pulled back, reducing the equilibrium variance. This matches physical intuition: a stiffer spring (larger θ) produces smaller oscillations for the same noise level (σ). The balance between noise injection (σ) and restoring force (θ) determines the equilibrium spread.
Question 3 Short Answer
Explain why the Ornstein-Uhlenbeck process is the unique stationary Gaussian Markov process (up to affine transformation).
Think about your answer, then reveal below.
Model answer: A stationary Gaussian Markov process must have an exponentially decaying autocorrelation R(τ) = ce^{-θ|τ|} — Markov demands that the autocorrelation satisfy R(s+t) = R(s)R(t)/R(0) (the Chapman-Kolmogorov condition for Gaussians), and the only continuous solution is exponential. A Gaussian process is fully determined by its mean and covariance, so the stationary distribution and the exponential autocorrelation uniquely determine the process as the OU process. No other diffusion has all three properties simultaneously: Gaussian marginals, the Markov property, and stationarity.
Brownian motion is Gaussian and Markov but not stationary (its variance grows with time). A stationary Gaussian process with non-exponential autocorrelation (like a squared-exponential kernel) loses the Markov property. The OU process sits at the unique intersection of these three properties, making it the natural continuous-time analogue of an AR(1) process.