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:
- 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.
- 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}
- 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.)
Best Answer
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.