3

$U(t)=e^{-\mu t}W(\frac{\sigma^2e^{2\mu t}}{2\mu})$. 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.

ztyh
  • 211

1 Answers1

5

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*}

  • Thanks, but I just don't see why the identity gives the wrong answer... – ztyh Aug 24 '15 at 21:59
  • 1
    The identity $W(\frac{\sigma^2e^{2\mu t}}{2\mu})=W(\frac{\sigma^2e^{2\mu t}e^{2\mu s}}{2\mu })e^{-\mu s}$ is not correct. These two random variables have the same mean and variance, but are not the same random variable. In particular, their covariance with another random variable is not usually going to be the same. –  Aug 24 '15 at 22:08
  • I wondered as well that may be. Thank you. – ztyh Aug 24 '15 at 22:19