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.
Best Answer
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 $z_1(0)=x_0$.