An Ornstein Uhlenbeck process $x_t$ satisfies the following stochastic differential equation:
$$dx_t = \theta (\mu – x_t ) dt + \sigma dW_t $$
where $\theta,\mu, \sigma > 0$, and $W_t$ denotes the Wiener process. When $\mu, \sigma$ and $\theta$ are constant, the analytic solution is:
$$x_t = x_0 e^{-\theta t} + \mu (1 – e^{-\theta t}) + \displaystyle\int_{0}^{t} \sigma e^{\theta(s-t)} dW_s$$
and the expectation is given by:
$$E(x_t) = x_o e^{-\theta t} + \mu (1 – e^{-\theta t})$$
However, I want to solve this SDE, and find the expectation, when $\mu = \mu(t)$, i.e. $\mu$ is a function of $t$.
Does anyone know of a solution or a reference for where a solution may be found? Thanks!
Best Answer
The stochastic differential equation solved by $y_t=\mathrm{e}^{\theta t}x_t$ indicates that $$ \mathrm{e}^{\theta t}x_t = x_0 + \int_{0}^{t} \theta\, \mathrm{e}^{\theta s}\mu(s)\,\mathrm{d}s +\sigma \int_{0}^{t} \mathrm{e}^{\theta s}\, \mathrm{d}W_s, $$ hence $$ E(x_t) = x_0 \mathrm{e}^{-\theta t} + \int_{0}^t\theta\, \mathrm{e}^{\theta (s-t)}\mu(s)\,\mathrm{d}s. $$ A more direct way to compute the expectation uses the fact that the function $u$ defined by $u(t)=E(x_t)$ is the unique solution of the ordinary differential equation $$ u'(t)=\theta\cdot(\mu(t)-u(t)),\quad u(0)=x_0. $$