Expected value of a mean-reverting process

An Ornstein–Uhlenbeck process obeys dXt=θ(μ−Xt) dt+σ dWt\mathrm{d}X_t = \theta(\mu - X_t)\,\mathrm{d}t + \sigma\,\mathrm{d}W_t with reversion speed θ=0.5\theta = 0.5, long-run level μ=10\mu = 10, and volatility σ=3\sigma = 3. It starts at X0=4X_0 = 4. What is E[X2]E[X_2] (the expected value at time t=2t=2, to four decimals)?

Show hints (2)+
  1. Take expectations: dm/dt=θ(μ−m)\mathrm{d}m/\mathrm{d}t=\theta(\mu-m) (noise drops out).
  2. m(t)=μ+(X0−μ)e−θtm(t)=\mu+(X_0-\mu)e^{-\theta t}; here θt=1\theta t=1.

Answer

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7.7927 (± 0.001)

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