I have come across the following equation \begin{equation}dX_t=-\lambda X_t dt +dU_t\quad (1)\end{equation} 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 \begin{equation}dX_t=-(\lambda X_t +\theta U_t)dt +\sigma dW_t\end{equation}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)
  • $\begingroup$ 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. $\endgroup$ Commented Jan 26, 2022 at 19:11
  • $\begingroup$ 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 $W_t$? $\endgroup$ Commented Jan 26, 2022 at 19:32
  • $\begingroup$ 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. $\endgroup$ Commented Jan 26, 2022 at 20:04
  • $\begingroup$ 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 $$X_t=\alpha\int_0^t\exp(-\lambda(t-s))U_sds+\sigma\int_0^t\exp(-\lambda(t-s))d\widetilde{W_s}$$ $\endgroup$ Commented Jan 26, 2022 at 20:35
  • $\begingroup$ That is correct, integrating factors that only depend on time behave the same as for ODE, no extra terms from the Ito theorem. $\endgroup$ Commented Jan 26, 2022 at 20:39

1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Jan 26, 2022 at 21:23
  • $\begingroup$ @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 $\endgroup$ Commented Jan 26, 2022 at 21:38
  • $\begingroup$ @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 $\endgroup$ Commented Jan 26, 2022 at 21:45
  • $\begingroup$ 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? $\endgroup$ Commented Jan 26, 2022 at 22:01
  • $\begingroup$ @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. $\endgroup$ Commented Jan 27, 2022 at 2:59

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