Variable Drift Ornstein–Uhlenbeck Process

Question

The Ornstein–Uhlenbeck process is defined as the stochastic process that solves the following SDE:

$dx_t = \theta (\mu-x_t)\,dt + \sigma\, dW_t$

where $\theta>0$, $\mu$ and $\sigma>0$ are parameters and $W_t$ is Brownian motion. It is well known the solution to this equation. In particular, it is known that

$E(x_t)=x_0 e^{-\theta t}+\mu(1-e^{-\theta t})$

and

$\operatorname{cov}(x_s,x_t) = \frac{\sigma^2}{2\theta}\left( e^{-\theta(t-s)} - e^{-\theta(t+s)} \right).$

It can be easily seen that $\lim_{t\to+\infty}E(x_t)=\mu$ and that $\lim_{t\to+\infty}Var(x_t)=\frac{\sigma^2}{2\theta}$. Assume that $f(t)$ is a well behaved function. What is it known about the process

$dx_t = \theta (f(t)-x_t)\,dt + \sigma\, dW_t$?

Is there a closed form expression for $x_t$ as in the constant case?

In particular, assume that $f(t)$ is periodic with certain period $\tau$. What is the limit of $E(x_t)$?

Answer 1

You can just take expectations on both sides of your SDE/corresponding integral equation and obtain an ODE on the expectation function $m_t = \Bbb E[x_t]$: $$ \dot m = \theta(f - m) $$ which you can easily solve using ansatz $m_t = c_t \mathrm e^{-\theta t}$ which brings you to $$ m_t = x_0\mathrm e^{-\theta t} + \theta\cdot\int_0^tf(s)\mathrm e^{\theta(s-t)}\mathrm ds $$ so for $x_0 = 0$ you get a truncated version of convolution $m = f*\exp$.

Now, assume for simplicity that $x_0 = 0$, that would not matter for the asymptotic analysis of periodic $f$ anyways. Let's $p>0$ be the period of $f$, then for any integer $n$ we have $$ \begin{align} m(np) &= \theta\mathrm e^{-\theta np}\cdot \sum_{k=0}^{n-1}\int\limits_{kp}^{(k+1)p}f(s)\mathrm e^{\theta s}\mathrm ds = \theta \mathrm e^{-\theta np}\cdot \sum_{k=0}^{n-1}F\mathrm e^{\theta kp} \\ &= \theta F\cdot\frac{1 - \mathrm e^{-\theta np}}{\mathrm e^{\theta p} - 1} \to \frac{\theta F}{\mathrm e^{\theta p} - 1} \end{align} $$ where $$ F = \int_0^pf(s)\mathrm e^{\theta s}\mathrm ds. $$ Notice that if $f \equiv \mu$ then we need to take a limit at $p\to 0$ in ratio, so we get $\mathrm e^{\theta p} -1\sim \theta p$ and $F \sim \mu p$ so that ratio is $\mu$, which confirms the case of constant $f$.

Answer 2

For the general solution in the case where $f$ is not a constant, note that, from the SDE \begin{align*} dx_t = \theta(f(t)-x_t)dt + \sigma dW_t, \end{align*} we obtain that \begin{align*} d\big(e^{\theta t} x_t \big) = \theta e^{\theta t} f(t)dt + \sigma e^{\theta t} dW_t. \end{align*} Then \begin{align*} e^{\theta t} x_t = x_0 + \int_0^t \theta e^{\theta s} f(s)ds + \sigma \int_0^t e^{\theta s} dW_s. \end{align*} That is, \begin{align*} x_t = x_0e^{-\theta t} + \int_0^t \theta e^{-\theta (t-s)} f(s)ds + \sigma \int_0^t e^{-\theta (t-s)} dW_s. \end{align*}