Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

An Ornstein Uhlenbeck process $x_t$ satisfies the following stochastic differential equation:

${\large dx_t = \theta (\mu - x_t ) dt + \sigma dW_t }$

where $\theta,\mu, \sigma > 0$, and $W_t$ denotes the Wiener process. When $\mu, \sigma$ and $\theta$ are constant, the analytic solution is:

${\large x_t = x_0 e^{-\theta t} + \mu (1 - e^{-\theta t}) + \displaystyle\int_{0}^{t} \sigma e^{\theta(s-t)} dW_s}$

and the expectation is given by:

${\large 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!

share|cite|improve this question
up vote 5 down vote accepted

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 $$ u'(t)=\theta\cdot(\mu(t)-u(t)),\quad u(0)=x_0. $$

share|cite|improve this answer

It looks like what you want is the Hull-White model, generally used for stochastic rate processes.

A good analysis of this model can be found for example in the Brigo & Mercurio, although it is very much finance oriented.

I'm guessing the Oksendal would contain the information you need as well.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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