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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:


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: http://quant.stackexchange.com/a/7639/3043

share|improve this question
Where is SRKX's derivation? – chrisaycock Apr 1 '13 at 0:06
quant.stackexchange.com/a/4932/3043, It's a different simulation equation than I'm using, but it's the same stochastic process. – wcampbell Apr 1 '13 at 2:16
Are you sure your modified process has an equilibrium distribution? If $\sigma$ is much larger than $\theta$, intuitively you might think that it doesn't (Imagine when $S_t$ is large, the effective volatility is significantly larger than the mean-reversion). This intuition is actually reinforced by looking at the variance formula for $S_t$. – quasi Apr 1 '13 at 6:37
I agree, but everyone seems to call the process stationary, even using the geometric process. My main question is whether the variance term that I "made up" is an appropriate calculation. It fits the variance of a simulated series fairly closely. – wcampbell Apr 1 '13 at 13:47
If it's stationary, wouldn't it be impossible for its long run variance to depend on an initial condition? Also, maybe I'm missing something, but when SKRX gives an explicit representation of the process, doesn't that directly mean your derivation is off? – quasi Apr 2 '13 at 14:22

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