1
$\begingroup$

Suppose one has a stochastic differential equation: $$dX_t = f(X_t) dt + g(X_t)d\eta(t)$$ where $\eta$ solves the Ornstein-Uhlenbeck process:

$$d\eta(t) = \lambda \eta(t) dt + \sigma dW(t)$$

Suppose $g \in L^2(\Omega \times [0,T])$. How, if it all, can one calculate $E\left[\int_0^T g(s) d\eta(s) \right]$?

I'm working on a problem with colored noise and I'm wondering if there's feasible way to compute this expectation, or if not, possibly to transform it to an Ito integral so that I can have expectation 0. I haven't come across anything that deals with a stochastic integral of this form

$\endgroup$
4
  • $\begingroup$ there should be a dt in the second equation next to $\eta(t)$? $\endgroup$
    – muaddib
    Commented May 28, 2015 at 0:55
  • $\begingroup$ Yes, thank you. I have edited it $\endgroup$
    – Brenton
    Commented May 28, 2015 at 2:13
  • $\begingroup$ Also, not really sure why there's a downvote? $\endgroup$
    – Brenton
    Commented May 28, 2015 at 2:16
  • $\begingroup$ So Ornstein Uhlenbeck has an explicit solution, you should be able to right that integral in terms of it. Cheers. $\endgroup$
    – muaddib
    Commented May 28, 2015 at 2:31

1 Answer 1

0
$\begingroup$

We let $\theta:=-\lambda>0$, then the OU is solved by

$$\eta_t = \eta_0\,e^{-\theta t} + \mu\,(1-e^{-\theta t}) + \sigma \int_0^t e^{-\theta (t-s)}\, dW_s$$

and we have

$$dX_t = f(X_t) dt + g(X_t)d\eta(t)$$

$$= \left[f(X_{t})+g(X_{t})\left(\eta u_0\,e^{-\theta t} + \mu\,(1-e^{-\theta t})\right)\right] dt + g(X_{t})e^{-\theta (t-s)}\sigma dW(t).$$

Suppose $g \in L^2(\Omega \times [0,T])$. How, if it all, can one calculate $E\left[\int_0^T g(s) d\eta(s) \right]$?

The Ito integral term has zero expectation as explained here Dealing with a term coming from Ito formula.

A sufficient condition for the integral $\int_0^t f(\omega, s)\, dB_s$ to be a martingale on $[0,T]$ is that

  1. $f(\omega,s)$ is adapted, measurable in s, and
  2. $\mathbb{E}\left(\int_0^T f^2(\omega,s)\,ds\right) < \infty$. In this case, indeed, $\mathsf{E} \left(\int_0^T f(\omega,s)\, dB_s\right)=0$.

So we are just left with the drift term

$$E\left[\int_0^T g(s) d\eta(s) \right]=E\left[\int_0^T g(s)\left(\eta_0\,e^{-\theta t} + \mu\,(1-e^{-\theta t}) \right) ds \right].$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .