I need to show that the Ornstein-Uhlenbeck process, $$ dX_t = -\theta X_tdt + dB(t) $$ Where $X_0=0$, $B(t)$ is Brownian motion and $\theta>0$

can be written explicitly as: $$ X_t=B(t) - \theta \exp(-\theta t)\int_0^t\exp(\theta s)B(s)ds $$

I can use Ito's lemma to get it to the form $X_t = \exp(-\theta t)\int_0^t\exp(\theta s)dB(s)$, but this isn't quite what I want, and am not certain how to proceed


1 Answer 1


Hint It follows from Itô's formula that

$$\exp(\theta t) \cdot B_t = \int_0^t \exp(\theta s) \, dB_s + \theta \int_0^t \exp(\theta s) B_s \, ds,$$


$$ \int_0^t \exp(\theta s) \, dB_s = \exp(\theta t) \cdot B_t - \theta \int_0^t \exp(\theta s) B_s \, ds.$$

  • $\begingroup$ I think I just got it. Thanks for the help :) $\endgroup$
    – user140513
    Commented Apr 16, 2014 at 6:18

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