I am looking at the following stochastic differential equation:

$$dY_t = - \theta Y_t dt + \sigma dX_t,$$ where $X_t$ is an Ornstein-Uhlenbeck- (OU-)-process and hence satisfies

$$dX_t = - \beta X_t dt + \gamma dW_t.$$

For simplicity, let $Y_0 = 0$ and $X_0 = 0$. (As usual: $W_t$ is Brownian motion. The parameters $\theta, \sigma, \beta, \gamma$ should all be $> 0$.)

Now I am wondering: Is there a simple expression for $Y_t$? I was hoping, it is again an OU-process but in my calculations, I instead get the sum of two OU-processes.

My thoughts so far:

  1. Through variation of parameters, like in the Wikipedia article on the OU-process, I calculate:

\begin{align} d(e^{\theta t} Y_t) &= \theta e^{\theta t} Y_t dt + e^{\theta t}dY_t \\ &= \sigma e^{\theta t} dX_t \\ &= -\sigma \beta e^{\theta t} X_t dt + \sigma \gamma e^{\theta t} dW_t \end{align}

This means:

$$e^{\theta t} Y_t = - \sigma \beta \int_0^t e^{\theta s} X_s ds + \sigma \gamma \int_0^t e^{\theta s} dW_s,$$

and after multiplication with $e^{- \theta t}$, I obtain: $$Y_t = - \sigma \beta \int_0^t e^{-\theta (t-s)} X_s ds + \sigma \gamma \int_0^t e^{- \theta (t-s)} dW_s$$

The second term is again an OU-process. It remains to look at the first term.

  1. Assuming I can swap the integrals (a formal proof is still lacking here):

\begin{align} \int_0^t e^{-\theta (t-s)} (\gamma \int_0^s e^{- \beta (s-r)} dW_r) ds &= \gamma \int_0^t \int_r^t e^{-\theta (t-s) - \beta (s-r)} ds dW_r \\ &= \frac{\gamma}{\theta - \beta} ( \int_0^t e^{-\beta(t-r) }dW_r - \int_0^t e^{-\theta(t-r) }dW_r)\end{align}

  1. Putting together

\begin{align} Y_t &= - \sigma \beta \frac{\gamma}{\theta - \beta} ( \int_0^t e^{-\beta(t-r)} dW_r - \int_0^t e^{-\theta(t-r)} dW_r ) + \sigma \gamma \int_0^t e^{-\theta (t-s)} dW_s \\ &= - \sigma \beta \frac{\gamma}{\theta - \beta} \int_0^t e^{-\beta(t-s)} dW_s + (\sigma \beta \frac{\gamma}{\theta - \beta} + \sigma \gamma) \int_0^t e^{-\theta (t-s)} dW_s \end{align} So it is a sum of two OU-processes with a bit weird factors in front. Is this correct? Can this be simplified?

(Edit: Some typos corrected.)

  • $\begingroup$ Isn't this a "2-dimensional" OU process? In the same sense proposed here: math.stackexchange.com/q/4431021/532409 n i.e. you can consider d(X,Y) = B(X,Y) dt + A d W where A and B are 2x2 matrices $\endgroup$
    – Quillo
    Apr 30, 2022 at 18:28

1 Answer 1


Not sure if your last formula is totally correct. Clearly, since $X_0=0$ we have $$ X_t=\gamma\int_0^te^{-\beta(t-s)}\,dW_s\,. $$ Setting $\theta=0$ and $\sigma=1$ gives $Y_t=X_t$ and $$ Y_t=\int_0^te^{-\beta(t-s)}\,dW_s+(-\sigma+\sigma\gamma)W_t\,. $$ The last term must vanish but it doesn't.

  • 1
    $\begingroup$ Thank you - yes, you are right. I missed a factor gamma when I plugged in $X_s$, fixed it now. $\endgroup$ Aug 10, 2021 at 15:15
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    $\begingroup$ This $\gamma$ I don't understand. When it is zero your $Y_t\equiv 0$ which I don't see happening when you set $\gamma$ to zero in the SDEs. $\endgroup$
    – Kurt G.
    Aug 12, 2021 at 9:26
  • 1
    $\begingroup$ Hi @Kurt G: I disagree. If $\gamma$ is 0, then $X_t$ should be $0$ everywhere, see also the explicit formula for an OU process on Wikipedia, but then $Y_t = 0$ for all $t$, too $\endgroup$ Aug 16, 2021 at 16:09
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    $\begingroup$ @AdaLovelace : seems right. When $\gamma=0$ then $dX_t=-X_t\,dt\,.$ The initial condition $X_0=0$ only allows the solution $X_t\equiv 0\,.$ That $Y_t\equiv 0$ follows by the same reasoning. $\endgroup$
    – Kurt G.
    Aug 16, 2021 at 17:37

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