# SDE with respect to an Ornstein-Uhlenbeck process

I have come across the following equation $$$$dX_t=-\lambda X_t dt +dU_t\quad (1)$$$$ where $$U_t$$ is an Ornstein Uhlenbeck process: $$dU_t=-\theta U_tdt + \sigma dW_t$$ The context in which this was introduced is an applied modeling setting, so their only interest was to integrate this equation in a time discrete way. This is no problem: First you create a sample path of $$U_t$$ using the Euler-Maruyama method and then similarly integrate $$X_t$$: $$X_{t+\Delta t}-X_t=-\lambda X_t *\Delta t +\Delta U_t=-\lambda X_t *\Delta t+(U_{t+\Delta t}-U_t)$$ I want to know more about the analytical solution of (1) though.

1. Am I correct, that this is technically not an SDE? The most general equation allowed for that seems to be $$dX_t=a(t,X_t)dt+b(t,X_t)dW_t$$
2. Is equation (1) well-defined? I only know that the Ito integral is defined with respect to semi martingales. Is $$U_t$$ a semi martingale? Can you even meaningfully integrate this equation if it is not?
3. You could rewrite (1) as $$$$dX_t=-(\lambda X_t +\theta U_t)dt +\sigma dW_t$$$$Note that this is not an answer to question 1. since $$-(\lambda X_t +\theta U_t)\neq a(t,X_t)$$. We could then discretize a second way: $$X_{t+\Delta t}-X_t=-(\lambda X_t +\theta U_t)*\Delta t +\sigma\Delta W_t$$ I assume the two different Euler Maruyama discretizations converge to the same process, right? (At let distributionally speaking)
• Your equation in 3. along with the original equation for $U$ makes an SDE system with a single one-dimensional Brownian motion, so it is still in the context of the standard theory. Commented Jan 26, 2022 at 19:11
• Great thanks! Would it be a lot harder to treat something like this:$$dX_t=-\lambda X_tdt+\alpha U_tdt + d\widetilde{W_t}$$ $$dU_t=-\theta U_tdt + \sigma dW_t$$ where $\widetilde{W_t}$ is another Wiener process independent of $W_t$? Commented Jan 26, 2022 at 19:32
• This makes it more difficult to apply methods beyond the Euler-Maruyama method. One gets nontrivial interaction or connection terms for the components of the Brownian motion. Commented Jan 26, 2022 at 20:04
• Would you agree that the solution to (1) would be $$X_t=-\theta\int_0^t\exp(-\lambda(t-s))U_sds+\sigma\int_0^t\exp(-\lambda(t-s))dW_s$$ and is there something to watch out for in the process of deducing it? (For example an implicit presence of $W_t$ in $U_t$ when using Ito). Could you also say that the solution to the equation in my comment above is $$X_t=\alpha\int_0^t\exp(-\lambda(t-s))U_sds+\sigma\int_0^t\exp(-\lambda(t-s))d\widetilde{W_s}$$ Commented Jan 26, 2022 at 20:35
• That is correct, integrating factors that only depend on time behave the same as for ODE, no extra terms from the Ito theorem. Commented Jan 26, 2022 at 20:39

You can interpret the given SDE $$dX_t=-\lambda X_t dt +dU_t$$ simply by integrating: $$X_t - X_0 = -\lambda \int_0^t X_s ds + U_t - U_0$$

You may even solve the SDE for $$U_t$$ by using an integration factor, to arrive at: $$U_t = U_0e^{-\theta t} + \sigma \int_0^t e^{-\theta (t-s)} dW_s$$ So that ultimately, $$X_t$$ is given by: $$X_t - X_0 = -\lambda \int_0^t X_s ds + U_0e^{-\theta t} - U_0 + \sigma \int_0^t e^{-\theta (t-s)} dW_s$$ There is no ambiguity in this expression, as the first integral is a Riemann integral, while the second is a usual Itô integral.

Am I correct, that this is technically not an SDE? The most general equation allowed for that seems to be $$dX_t=a(t,X_t)dt+b(t,X_t)dW_t$$

As a consequence of the Bichteler-Dellacherie theorem, the most general Itô integrators are semi-martingales; thus the most general SDEs that can be interpreted in the Itô sense are those driven by semi-martingales. The above is one example.

Is $$U_t$$ a semi-martingale?

Yes; this can be seen by noting it is an Itô process. Thus, $$U_t$$ is an admissible stochastic integrator.

• Thanks that helps a lot! I see now that $U_t$ is an Ito process and therefore a semimartingale. I don't quite see your explanation (ii) though: The closed form solution is the sum of a decreasing process and $$\sigma\exp(-\lambda t)\int_0^t\exp(\lambda s)dW_s$$ The integral is a martingale, but the factor infront of it destroys that property right? The decomposition might be more complicated. Maybe even impossible to write it in closed form? Commented Jan 26, 2022 at 21:23
• @courageousmartingale Sorry, you're right. I've deleted that bit as it's not quite correct. You can still witness $U_t$ as a semi-martingale from its Itô dynamics, so I'll leave it at that Commented Jan 26, 2022 at 21:38
• @courageousmartingale If you do want to fix that issue, you may notice that the pre-factor makes this process into a supermartingale. Thus, it admits a Doob-Meyer decomposition, after which you may conclude from the closed-form solution that $U_t$ is indeed a semi-martingale Commented Jan 26, 2022 at 21:45
• Ah yes very nice! Do you happen to have a comment on my question number 1.? Would you consider (1) an SDE even though the drift is not strictly a function $a(t,X_t)$ but instead has an $\mathcal{F}_t$ measurable part $U_t$? Do you just treat the rest of the equation as an SDE and then the $U_tdt$ part in an ODE sense? Commented Jan 26, 2022 at 22:01
• @courageousmartingale It's fine, I think, since $a$ is a function of $X$ through $U$. There doesn't seem a need, in my mind, to be so restrictive, though. Any process with, say, continuous paths should be $dt$-integrable. Commented Jan 27, 2022 at 2:59