Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X_t$ be an Ornstein-Uhlenbeck process solving $dx_t = \theta (\mu-x_t)dt + \sigma dW_t$. The solution is known and given by: $$ x_t = x_0 e^{-\theta t} + \mu(1-e^{-\theta t}) + \int_0^t \sigma e^{\theta (s-t)} dW_s$$

Is there a closed-form formula (both SDE and actual solution) for time integral $\int_0^t X_t dt$?

(I know there is a lot of literature on interest theory that analyzes the expectation of this kind of integral, but this is not something I am after)

share|improve this question
add comment

2 Answers

Let us denote $A_t = \int_0^t X_s ds$. $A_t$ is a Gaussian random variable, so it is enough to calculate its mean and variance. This goes by using Fubini's theorem.

For simplicity let us assume that $x_0 = 0, \mu=0$. Then $\mathbb{E} X_t =0$ and

$\mathbb{E} A_t = \mathbb{E} \int_0^t X_s ds = \int_0^t \mathbb{E} X_s ds = 0$.

$Var(A_t) = \mathbb{E} A_t^2 = \mathbb{E} \int_0^t \int_0^t X_s X_u ds du = \int_0^t \int_0^t Cov(X_s, X_u) ds du = 2 \int_0^t \int_0^u Cov(X_s, X_u) ds du.$

Now it is enough to use $Cov(X_s, X_u) = \frac{\sigma^2}{2\theta}\left( e^{-\theta(u-s)} - e^{-\theta(u+s)} \right)$ valid for $s\leq u$.

This solution is more or less what The Bridge suggest. One can go a step further and calculate $Cov(A_t, A_s)$ and $\mathbb{E}A_t$ which is enough to fully characterise that process.

share|improve this answer
    
@Piotr Milos: Yes it was exactly what I had in mind. Regards –  The Bridge Jan 5 '12 at 21:16
    
So you think there is no easy representation with respect to the Brownian Motion $\{W_t\}$? Because without it I don't know how to use it with other processes, e.g. what is the covariation between $A_t$ and some other process driven by $dW_t$. –  Grzenio Jan 8 '12 at 18:16
    
I would guess so. First notice that, when $Y_t=\int_0^t f(t,s)dW_s$, for some deterministic function $f$, then $Var(Y_u,Y_v)= \int_{0}^{v\wedge u} f(v,s)f(u,s)ds. Now it is enough to guess $f$ such that we get the required covariance. The question goes further. If such a function exists for any gaussian process. My guess is that yes and I suspect that this may follow by the reproducing kernel Hilbert spaces but I do not have time to check this at a moment. –  Piotr Miłoś Jan 9 '12 at 9:00
add comment

Hi Grzenio,

Using Stochastic Fubini's theorem I think you can re-express this integral in an Itô form and more precisely in a Wiener integral form whih are known to be gaussian. So you can derive the law of this random variable, is it what you meant by "closed-form" formula ?

Regards

share|improve this answer
add comment

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.