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Let $f$ be a variable which evolves according to the above. What does it mean to say that $\Phi$ is a mean-reversion factor?

I mean, I guess it means $f_{t+1} = (1-\Phi)f_t + \epsilon_{t+1}$ and so by substitution, I get something like $f_{t+1} = (1-\Phi)^2f_{t-1} + (1-\Phi)\epsilon_t + \epsilon_{t+1}$ and continuing in this fashion, I guess $f_{t+1} = (1-\Phi)^{t+1}f_0 + \sum \epsilon_k (1-\Phi)^k$

but I still don't get why $\Phi$ is a "mean-reversion factor"? What is the mean? How quickly do we revert to the mean? How far "away" from the mean do we get? How does all of this depend on $\Phi$?

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  • $\begingroup$ In this simple example the mean is zero, or if you prefer $f_t$ has been expressed as the difference between the quantity of interest and it's long term mean, i.e has been "de-meaned". $\Phi$ is the speed of mean-reversion, a small $\Phi$ means a large number of steps $k$ are needed to get back to zero i.e. $(1-\Phi)^k\approx 0$. $\endgroup$ – Alex C Jun 10 '18 at 15:56
  • $\begingroup$ For any given $\Phi>0$ the number $k(\Phi)$ such that $(1-\Phi)^k=0.5$ is usually referred to as the "half-life", i.e. the time it takes for half of the initial deviation from the mean to disappear in the absence of a stochastic input $\epsilon$ $\endgroup$ – Alex C Jun 10 '18 at 16:10
  • $\begingroup$ Please don't cross post math.stackexchange.com/questions/2814775. $\endgroup$ – LocalVolatility Jun 10 '18 at 20:27

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