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I would like to compute the expectation of a modified Ornstein-Uhlenbeck process of the form $$ dx_t = \theta(\mu_t-x_t)dt + \sigma x_t dW_t \ ,$$ where $\kappa, \sigma>0$ and $W_t$ is a Brownian motion. $\mu_t$ is itself a stochastic process and is a solution to the SDE: $$ d\mu_t = \beta\mu_tdB_t \ ,$$ where $\beta>0$ and $B_t$ is a Brownian motion independent from $W_t.$

I read in this blog post cited in this question how to procede when $\mu$ is not a stochastic process. I started with the same approach and I obtained the following ODE: $$ d\big(x_t e^{-\sigma W_t +\frac{1}{2}\sigma^2t}\big)=e^{-\sigma W_t +\frac{1}{2}\sigma^2t}\theta(\mu_t-x_t)dt\ ,$$ which can be rewritten as: $$ \frac{dy_t}{dt}=\theta\mu_te^{-\sigma W_t +\frac{1}{2}\sigma^2t}-\theta y_t\ ,$$ where $y_t$ is defined as $$ y_t = x_t e^{-\sigma W_t +\frac{1}{2}\sigma^2t}\ .$$ I tried to use methods for linear homogeneous ODEs and I came up with the following solution: $$y_t=y_0e^{-\theta t}+e^{-\theta t}\int_0^te^{\theta s}\mu_se^{-\sigma W_s +\frac{1}{2}\sigma^2s}ds\ .$$ The I substitute the dynamics for $\mu_t$, which I know and I get: $$y_t=y_0e^{-\theta t}+e^{-\theta t}\int_0^te^{\theta s}e^{-\sigma W_s +\beta B_s +\frac{1}{2}(\sigma^2-\beta^2)s}ds\ .$$ Now, integration by parts lead to: $$y_t=\frac{2\theta e^{-\sigma W_t +\beta B_t +\frac{1}{2}(\sigma^2-\beta^2)t}}{\sigma^2-\beta^2+2\theta}+e^{-\theta t}\bigg(y_0 -\frac{2\theta e^{-\sigma W_t +\beta B_t}}{\sigma^2-\beta^2+2\theta}\bigg)\ .$$ In turn this implies: $$ x_t = \frac{2\theta e^{\beta B_t-\frac{1}{2}\beta^2t}}{\sigma^2-\beta^2+2\theta}+e^{-(\theta+\frac{1}{2}\sigma^2) t +\sigma W_t}y_0 -e^{-(\theta+\frac{1}{2}\sigma^2) t}\frac{2\theta e^{\beta B_t}}{\sigma^2-\beta^2+2\theta}\ .$$

I'm wondering if what I wrote makes sense and how to explain the condition $\sigma^2-\beta^2+2\theta\neq0$ that emerges. Thanks in advance.

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  • $\begingroup$ If I abbreviate $Z_t=e^{-\sigma W_t+\beta B_t}$ and $\alpha=(\sigma^2-\beta^2)/2$ and for simplicity take $y_0=0$ then you are saying that $$ e^{-\theta t}\int_0^te^{\theta s}Z_s\,ds=\theta Z_t\frac{e^{\alpha t}}{\alpha+\theta}-e^{-\theta t}\theta Z_t\frac{1}{\alpha+\theta} $$ Looks strange because the Ito rule applied to the LHS gives $$ Z_t-\theta e^{-\theta t}\int_0^te^{\theta s}Z_s\,ds. $$ $\endgroup$
    – Kurt G.
    Apr 15, 2022 at 14:28
  • $\begingroup$ @KurtG. I think that you're missing an $e^{\alpha s}$ inside the first integral or in the definition of $Z_t$. But it's definitely more clear written like this $\endgroup$
    – Zwei
    Apr 15, 2022 at 15:09
  • $\begingroup$ Yes correct. I missed that. Still the LHS has bounded variation while the two naked $Z_t$s on the RHS clearly have not. Something went wrong. $\endgroup$
    – Kurt G.
    Apr 15, 2022 at 15:18
  • $\begingroup$ @KurtG. thanks for spotting that! $\endgroup$
    – Zwei
    Apr 15, 2022 at 15:22
  • $\begingroup$ No problem. A similar question about the unmodified OU process was posted recently, $\endgroup$
    – Kurt G.
    Apr 15, 2022 at 15:26

1 Answer 1

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The expectation is $E[y_t]=e^{At}y_0$ where $y_t=[x_t,\mu_t]'$ and $A=\small\begin{bmatrix}-\theta&\theta\\0&0\end{bmatrix}$. This is the solution to the ODE $z'=Az$. So $dz_1=-\theta z_1dt+\theta z_2dt,\,dz_2=0$ and therefore $z_2(t)=\mu_0,\,\forall t\geq 0$ and $$dz_1=-\theta z_1dt+\theta \mu_0dt\implies z_1+\theta z_1dt=\theta \mu_0dt\\\implies z_1(t)=z_1(0)e^{-\theta t}+\theta \mu_0\int_{[0,t]}e^{-\theta(t-s)}ds$$ which simplifies into $E[x_t]=z_1(t)=z_1(0)e^{-\theta t}+\mu_0(1-e^{-\theta t})$ where $z_1(0)=x_0$.

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