Derive the variance of X_t, the stationary distribution and the half-life of a shock for the process dX = kappa(theta - X) dt + sigma dW. With kappa = 4 per year and sigma = 20%, give the stationary standard deviation and the half-life in months.

Derive the variance of X_t, the stationary distribution and the half-life of a shock for the process dX = kappa(theta - X) dt + sigma dW. With kappa = 4 per year and sigma = 20%, give the stationary standard deviation and the half-life in months.

Approach: Use the integrating factor solution, apply the Ito isometry to the stochastic integral for the variance, then let the horizon grow and solve the decay equation for the time to halve a deviation.

Var(X_t) = sigma^2 (1 - e^{-2 kappa t})/(2 kappa), the stationary distribution is N(theta, sigma^2/(2 kappa)) and the half-life is ln 2 / kappa, which here give a stationary standard deviation of 0.0707 and a half-life of 2.08 months. The integrating factor solution is X_t = theta + (X_0 - theta)e^{-kappa t} + sigma ∫_0^t e^{-kappa(t-s)} dW_s, and the Ito isometry turns the variance into sigma^2 ∫_0^t e^{-2 kappa (t-s)} ds = sigma^2 (1 - e^{-2 kappa t})/(2 kappa). Letting t grow, the mean converges to theta and the variance to sigma^2/(2 kappa), and since every X_t is Gaussian the stationary distribution is Gaussian with those parameters. With kappa = 4 and sigma = 0.2 the stationary standard deviation is 0.2/sqrt(8) = 0.0707, and solving e^{-kappa h} = 1/2 gives h = ln 2 / 4 = 0.173 years, or 2.08 months.

Follow-up: How does the half-life estimate behave when kappa is fitted by ordinary least squares on daily data over a two year sample?

Key concepts: integrating factor, stationary distribution, half-life, Ito isometry.