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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:

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

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    $\begingroup$ 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
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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:

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

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  • $\begingroup$ 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

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