# How to construct appropriately reverting geometric AR(1) process?

Suppose I have a mean-reverting AR(1) type process, $X_{t+1} = X_t + \theta(\mu - X_t) + \epsilon_t$ where $\theta > 0$ and $\mathrm{Var}(\epsilon_t) = \sigma^2$. This process is clearly stationary.

One can consider a geometric version of this, as $Y_{t+1} = Y_{t} + \theta(\mu - Y) + \epsilon_t$, and then $Z_t = \exp(Y_t)$ to get the real space process.

The forecast for the log process is clearly mean reverting around the same mean, but the real process is not. Because of the exponentiation, $E(Z_{t+1} | Y_t) = \exp(Y_{t} + \theta(\mu - Y) + \sigma^2/2)$. The variance of the two-step-ahead forecast for the log process is $\mathrm{Var}(Y_{t+2}|Y_t) = \sigma^2(2-\theta)$, and so the convexity correction for $E(Z_{t+1} | Y_t)$ will involve a term $+\sigma^2(2-\theta)/2$.

This is all very nasty. How can I construct a better geometric AR(1) process so that the forecast converges to a desired constant mean?

Perhaps I need to consider a discretised version of an exponential Ornstein-Uhlenbeck model, such as is done here: http://fxpaul.wordpress.com/2011/06/08/exponential-ornstein-uhlenbeck-process/

Thanks!

I believe that the white noise should be indexed by ${t+1}$. Also this process is not "clearly stationary", since re-arranging gives $$X_{t+1} = \theta\mu + (1-\theta)X_t + \epsilon_{t+1}$$

Assume $\epsilon \sim N(0,2), \mu=0.5, \theta =1.5$. Then a realization of the process is Now change the value to $\theta = 2.03$. A realization using the same series for the shocks is As should be expected, since here $|1-\theta|>1$ and $1-\theta <0$, the process explodes with oscillations. For stationarity, it must be the case that $|1-\theta|<1$.

• Sorry, I should have said that the condition you have given is required. Even if you impose that condition, however, the geometric process is unusually behaved. This doesn't solve my problem. – alpha137 Oct 21 '14 at 2:30

When you are dealing with a stationary NAR(1) model you have a stationary distribution that is the normal distribution. In this case the process is "mean-reverting" in the sense that values above the mean are likely to go down and values below the mean are likely to go up. Applying exponentiation to that process you get a stationary model with a log-normal stationary distribution. The process is then "mode-reverting" in the sense that values above the mode are likely to go down and values below the mode are likely to go up.

This "mode-reversion" property occurs because the underlying process is a symmetric unimodal mean-reverting process, and because it is symmetric unimodal, its mean is its mode. The mode of the stationary distribution of the underlying process is preserved functionally by the exponential transform (i.e., the mode of the exponentiated process is the same as the exponential of the mode of the underlying process).

This aspect of the exponentiated process is not an "inappropriate" aspect of the process - indeed, it is an entirely expected aspect of a process with a skewed stationary distribution. When you have a stationary distribution with positive skew, you probably don't really want mean-reversion, since that is reversion to a value that is relatively high in value.

Exponentiated NAR(1) process: Suppose you have a stationary NAR(1) process with recursion $X_{t} = \mu + \theta (X_{t-1} - \mu) + \varepsilon_{t}$ and $\varepsilon_t \sim \text{IID N}(0, \sigma^2)$. For stationarity we impose the requirement that $|\theta| < 1$ (i.e., the auto-regression coefficient is inside the units circle). This process is mean-reverting with mean $\mu$ and it has the stationary distribution:

$$X_t \sim \text{N} \Big( \mu, \frac{\sigma^2}{1-\theta^2} \Big).$$

Converting this by exponentiation $Z_t \equiv \exp(X_t)$ gives a corresponding stationary process with stationary distribution:

$$Z_t \sim \text{LogN} \Big( \mu, \frac{\sigma^2}{1-\theta^2} \Big).$$

The latter process has mode $\exp(\mu)$ and mean $\exp (\mu + \sigma^2/2(1-\theta^2))$. It is mode-reverting, in the sense that values above the mode will tend to revert downward in the next iteration and values above the mode will tend to revert upward in the next iteration. Values in $\exp(\mu) < Z_t < \exp (\mu + \sigma^2/2(1-\theta^2))$ will tend to go downward in the next iteration and so these revert away from the mean.

As stated, this property is not a drawback or "inappropriate" aspect of the exponentiated process - it is just a natural aspect of the fact that the underlying mean reversion is symmetric, and the exponentiation gives a positively skewed distribution.

What to do if you really want a mean-reverting exponentiated process: I think it's a bad idea, but if you really have your heart set on one, the only way to get it would be to construct a process where the mean reversion is weak when pushing upward but strong when pushing downward. This would be a messy exercise.