I know that the long run variance of the standard OU process is
$\lim_{s\rightarrow \infty}\mbox{Var}(P_{t+s}|P_t) = \frac{\sigma^2}{2\theta}$
I'm using the geometric version of the process. I used the simulation equation
$P_{t}=P_{t-1}+\theta\left(\mu-P_{t-1}\right)+\sigma P_{t-1}\epsilon_{t-1}$.
SRKX posted a derivation of the variance for the geometric process:
$\mbox{Var}[St]=S^2_0(\exp(σ2t)−1)\exp(−2θt)$
This doesn't appear to reach any long run value, the variance just keeps growing:
$\lim_{s\rightarrow \infty}\mbox{Var}(P_{t+s}|P_t) = \infty $.
What's wrong with just modifying the variance term of the OU process to be
$\mbox{Var}\left(P_{t+s}|P_{t}\right)=\frac{\sigma^{2}P_t^2}{2\theta}\left(1-\exp\left(-2\theta s\right)\right)$?
That way, the long run variance is
$\lim_{s\rightarrow \infty}\mbox{Var}(P_{t+s}|P_t) = \frac{\sigma^2P^2_t}{2\theta}$.
Is this correct?
The reason this is important is because it affects the threshold calculation for buying a mean reverting asset. See: https://quant.stackexchange.com/a/7639/3043