Covariance of Ornstein-Uhlenbeck process

Question

$U(t)=e^{-\mu t}W(\frac{\sigma^2e^{2\mu t}}{2\mu})$. The problem is to find $Cov[U(t),U(t+s)]$.

I used the identity, $W(\frac{\sigma^2e^{2\mu t}}{2\mu})=W(\frac{\sigma^2e^{2\mu t}e^{2\mu s}}{2\mu })e^{-\mu s}$, which resulted in the wrong answer of $\frac{\sigma^2}{2\mu}$. I should be getting $\frac{\sigma^2 e^{-\mu s}}{2\mu}$.

Why is this identity wrong? It seems like it is only a scale transformation, with $\frac{\sigma^2e^{2\mu t}}{2\mu}$ the variable.

Thank you.

Answer 1

If $U(t)=e^{-\mu t}W(\frac{\sigma^2e^{2\mu t}}{2\mu})$, then \begin{eqnarray*}\mbox{Cov}[U(t),U(t+s)]&=&\mbox{Cov}\left[e^{-\mu t}W\left(\frac{\sigma^2e^{2\mu t}}{2\mu}\right),e^{-\mu (t+s)}W\left(\frac{\sigma^2e^{2\mu (t+s)}}{2\mu}\right)\right]\\[10pt] &=& e^{-\mu t} e^{-\mu (t+s)} \mbox{Cov}\left[W\left(\frac{\sigma^2e^{2\mu t}}{2\mu}\right),W\left(\frac{\sigma^2e^{2\mu (t+s)}}{2\mu}\right)\right]\\[10pt] &=& e^{-\mu t} e^{-\mu (t+s)}\,\frac{\sigma^2e^{2\mu t}}{2\mu}\\[10pt] &=& e^{-\mu s}\,\frac{\sigma^2 }{2\mu}. \end{eqnarray*}