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)


3 Answers 3


$\newcommand{\Cov}{\operatorname{Cov}}\newcommand{\Var}{\operatorname{Var}}$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

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

\begin{align} \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. \end{align}

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.

  • $\begingroup$ @Piotr Milos: Yes it was exactly what I had in mind. Regards $\endgroup$
    – The Bridge
    Jan 5, 2012 at 21:16
  • 1
    $\begingroup$ 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 $dW_t$. $\endgroup$
    – Grzenio
    Jan 8, 2012 at 18:16
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    $\begingroup$ 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. $\endgroup$ Jan 9, 2012 at 9:00
  • $\begingroup$ Do you know if there is a formula for the distribution of the hitting time of process $A_t$? $\endgroup$
    – Arthur B
    May 21, 2015 at 14:13
  • $\begingroup$ Is there a stationary solution to this (the unconditional stationary covariance)? $\endgroup$
    – mathtick
    Mar 13 at 20:25

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 ?



I'm pretty sure the actual solution is given in Ornstein and Uhlenbeck 1930.

Since the O-U process is the velocity of a free particle undergoing Brownian motion, then you are asking for the the distribution of its displacement. In the limit, the displacement process is a Brownian motion process having variance $\frac{2\sigma}{\theta}$.

I came to this question looking to confirm my understanding, but you should definitely take a look at the original 1930 paper.


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