# Expectation of Modified Stochastic Integral

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

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

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

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