Mean and Variance of Geometric Mean Reverting Process

Question

For the geometric mean-reverting process $dX_{t} = k(\theta - logX_{t})X_{t}dt + \sigma X_{t}dW_{t} $ it is possible to obtain the solution: $log(X_{t}) = e^{-kt}log(X_{0}) + (\theta - \sigma^2/(2k))(1 - e^{-kt}) + \sigma e^{-kt}\int_{o}^t e^{ks}dW_{s} $ through application of Ito's lemma to $log(X_{t})$ and recognising this as an Ornstein-Uhlenbeck process. My question is how to compute the mean and variance of this solution i.e. of $X_{t}$?

Answer 1

Given the solution that you have, you may exponentiate and write: \begin{equation} X_t = X_{0}^{e^{-kt}} f(t)\exp\left( \sigma e^{-kt} \int_{0}^{t}e^{ks} dW_s \right), \end{equation} where $f(t) = \exp\left( \left( \theta -\frac{\sigma^2}{2k}\right) \left(1-e^{-kt}\right) \right)$.

Note that since the integral of a deterministic function against Brownian motion is normally distributed, $Z_t = \sigma e^{-kt} \int_{0}^{t}e^{ks} dW_s$ is Gaussian. It has zero expectation, its variance is given by the Itô isometry: \begin{equation} \mathbb{E} \left[ \left( \int_{0}^{t} \sigma e^{k(s-t)} dW_s\right)^2 \right] = \int_{0}^{t} \sigma^2e^{2k(s-t)} ds = \dfrac{\sigma^2}{2k}\left( 1-e^{-2kt} \right). \end{equation} Since we now know $Z_t\sim N\left(0, \frac{\sigma^2}{2k}\left( 1-e^{-2kt} \right)\right)$, by the moment generating function of a normal distribution we can calculate $\mathbb{E}[e^{uZ_t}] = \exp\left(\frac{\sigma^2 u^2}{4k}\left( 1-e^{-2kt} \right)\right)$.

Then, $u =1$ and $u=2$ will lead you to the first and second moments of $X_t$, which after some gruesome algebra will give you the mean and variance.