I have a stochastic process (Ornstein-Uhlenbeck) defined as:

$X(t) = e^{-at}(\int_0^t e^{a \tau} dW(\tau) + X_0)$

Where $W(t)$ is the Wiener process, and $X_0$ is the initial value of my process.

I want to derive the covariance function, which obviously can be found easily online, however the derivations I found skipped a lot of steps without much explanation.

Assuming that $s < t$ then after a few steps I get to:

$EX(t)X(s) = e^{-a(t+s)} E[\int_0^s e^{a\tau}dW(\tau) \int_0^s e^{a\sigma}dW(\sigma) + X_0^2]$

At this point I figure I can combine the integrals and then take the expectation of just $dW(\tau)dW(\sigma)$ since the rest is non-random, and an integral is just an infinite sum. Doing this gets me to:

$EX(t)X(s) = e^{-a(t+s)} (\int_0^s \int_0^s e^{a(\tau+\sigma)} E[dW(\tau) dW(\sigma)] + EX_0^2)$

From this point I figure I can use the fact that for a Wiener process we have that $EdW(t)dW(s)$ is $0$ if $t \neq s$ and is equal to $dt$ if $t=s$. This should then simplify to a single integral which can then be resolved easily.

Is this line of thinking correct? Or is there something I'm missing.

Edit: Finishing it how I described above, I ended up with:

$EX(t)X(s) = \frac{e^{-a(t+s)}}{2a}(e^{2as} -1 + 2a EX_0^2)$

  • 1
    $\begingroup$ After correcting a few errors in the equation I was given, I was able to get the same answer as that given on Wikipedia (assuming $X_0$ is constant and therefore has $0$ variance). So I assume my logic is correct. Would be good if someone more knowledgeable could confirm, cheers. $\endgroup$ – Patty Jun 12 '17 at 3:59
  • $\begingroup$ I've written a more rigorous answer bellow. Let me know if it answers your question. $\endgroup$ – byouness Jun 5 '18 at 16:06

You are correct.

The computation boils down to figuring out the experession of: $$f(s,t) = \mathbb{E}\left[\left(\int_0^t e^{au}dW_u\right) \left(\int_0^s e^{av}dW_v\right) \right]$$

We can suppose $s \leq t$ without any loss of generality.

Developing, then using the fact that the brownian motion has independent increments, and finally Ito's isometry, we can write:

$$\begin{aligned} f(s,t) & = \mathbb{E}\left[\int_0^s e^{au}dW_u \int_0^s e^{au}dW_u \right] + \mathbb{E}\left[\int_s^t e^{au}dW_u \int_0^s e^{au}dW_u \right] \\ & = \mathbb{E}\left[\left(\int_0^s e^{au}dW_u \right)^2\right] + 0 \ \ \text{ (independent increments)}\\ & = \int_0^s e^{2au} du \ \ \ \text{ (Ito's isometry)} \\ & = \frac{1}{2a} (e^{2as} - 1) \end{aligned}$$

  • $\begingroup$ Apologies for never responding, I forgot the login for my account. Thanks a lot for your help, and have some very delayed rep :) $\endgroup$ – Patty Sep 23 '18 at 9:00
  • $\begingroup$ You're welcome. Always glad to help. $\endgroup$ – byouness Sep 24 '18 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.