5
$\begingroup$

Consider the Ornstein-Uhlenbeck process, $U(t)$, whose evolution follows: $$ \mathrm{d}U(t) = -\theta U(t) \mathrm{d}t + \sigma \mathrm{d}W(t), $$ where $\theta \in (0,2)$ is the mean-reversion rate, $\sigma >0$ is the dispersion rate, and $\{W(t)|t\geq0\}$ is a standard Brownian motion. Note that this is a zero-mean OU process. Now consider a new variable, $V(t)$, which is a function of $U(t)$ and a geometric Brownian motion, $X(t)$: $$ V(t) = \ln{\left[\frac{X(t) + U(t)}{X(t)}\right]}, $$ where $\{X(t),X(t)+U(t)>0|t\geq0\}$, and $X(t)$ follows: $$ \mathrm{d}X(t) = \mu X(t) \mathrm{d}t + \eta X(t) \mathrm{d}Z(t), $$ where $\mu, \eta >0$ are the drift and dispersion rates, respectively, and $\{Z(t)|t\geq0\}$ is a standard Brownian motion independent of $W(t)$ for all time.

Is it possible to show that $U(t)$ and $V(t)$ are perfectly positively correlated? Or, more importantly, is it possible to determine the conditions under which they are perfectly positively correlated? Simulations have shown me that the two are almost perfectly correlated, but I lack a formal proof.

$\endgroup$
6
$\begingroup$

They are not perfectly positively correlated: Even when random variables are deterministically related (which would require $X$ to be deterministic in this case), perfect correlation requires them to be related via an affine transformation. This would require a relationship of the form:

$$V(t) = \ln \Big[ 1+\frac{U(t)}{X(t)} \Big] = a + b U(t),$$

where $a \in \mathbb{R}$ and $b>0$. Solving for the process $X$ gives:

$$X(t) = \frac{U(t)}{\exp(a + b U(t))-1}.$$

This is inconsistent with your specification that $X$ is a geometric Brownian motion. However, note that if your geometric Brownian motion process has a large mean and small variance (such that it is approximately constant at a mean value $\mu_X$ that is much bigger than $U(t)$) then you would have $X(t) \approx \mu_X \gg U(t)$ which gives the approximation:

$$V(t) = \ln \Big[ 1+\frac{U(t)}{X(t)} \Big] \approx \frac{U(t)}{X(t)} \approx \frac{1}{\mu_X} \cdot U(t),$$

so in this case you could get something that is close to an affine transform, and so you would get something close to perfect correlation.

$\endgroup$
  • $\begingroup$ Great answer, Ben. Thanks very much for your thoughtful reply. Instead of showing that $U(t)$ and $V(t)$ are perfectly positively correlated, which is clearly not possible, is there any way to demonstrate that their correlation is approximately equal to 1? $\endgroup$ – Jeff Jul 2 '18 at 11:20
  • $\begingroup$ @Jeff: I have edited to elaborate on when you would get correlation close to one. $\endgroup$ – Ben Jul 2 '18 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.