For dX = kappa(theta - X) dt + sigma dW with kappa = 2 per year, theta = 0.05, sigma = 0.10 and X_0 = 0.09, compute E[X_1]. Give the solution of the SDE you used.

For dX = kappa(theta - X) dt + sigma dW with kappa = 2 per year, theta = 0.05, sigma = 0.10 and X_0 = 0.09, compute E[X_1]. Give the solution of the SDE you used.

Approach: Multiply by the integrating factor e^{kappa t}, integrate, and take expectations so that the stochastic integral drops out and only the deterministic decay remains.

0.0554. Using the integrating factor e^{kappa t}, d(e^{kappa t} X_t) = kappa theta e^{kappa t} dt + sigma e^{kappa t} dW_t, and integrating gives the Ornstein-Uhlenbeck solution X_t = theta + (X_0 - theta) e^{-kappa t} + sigma ∫_0^t e^{-kappa(t-s)} dW_s. The stochastic integral has mean zero, so mean reversion leaves E[X_t] = theta + (X_0 - theta) e^{-kappa t}, an exponential decay of the initial gap toward the long run level. Numerically E[X_1] = 0.05 + 0.04 e^{-2} = 0.05 + 0.04(0.13534) = 0.0554. Just over 86% of the initial 4 point gap has been closed after one year, which is what kappa = 2 means in practice.

Follow-up: What is Var(X_1) for the same parameters, and how many years until the standard deviation is within 1% of its stationary value?

Key concepts: Ornstein-Uhlenbeck, mean reversion, exponential decay, integrating factor.