2
$\begingroup$

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 $E(x_t)$?

$\endgroup$
3
$\begingroup$

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 $f$.

$\endgroup$
  • 1
    $\begingroup$ Thanks Ulysses. It makes sense. I also realized that if we consider the process $dx_t=\theta(f(t)+\frac{f'(t)}{\theta} -x_t)dt+\sigma dB_t$ then $m_t=x_0\exp(-\theta t)+f(t)-f(0)\exp(-\theta t)$. Do you know if there a closed form solution to x_t in the general case? $\endgroup$ – ght Jun 12 '15 at 14:16
  • 1
    $\begingroup$ If $f\equiv \mu$, then $F=\frac{\mu}{\theta}(e^{\theta p}-1)$. That is, you do not need the limit for $p\rightarrow 0$. $\endgroup$ – Gordon Jun 12 '15 at 14:19
  • 1
    $\begingroup$ There is a general solution, in integral form, for $x_t$. See Formula (3.35) in the book "Interest rate models - theory and practice - second edition" by Brigo. $\endgroup$ – Gordon Jun 12 '15 at 14:23
  • $\begingroup$ @Gordon: indeed, I never used that $p$ is the proper period, can be any value at which the function repeats itself. $\endgroup$ – Ulysses Jun 12 '15 at 14:24
  • $\begingroup$ @Gordon: Unfortunately, I don't have that book. Is there any other open source reference? If not could you please write the expression? Thanks! $\endgroup$ – ght Jun 12 '15 at 14:28
1
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.