# How to interpret mean of this estimated AR(1) process

I estimated an AR(1) process, my data looks like this:

Making usual unit root test, they suggest that an estimated AR(1) from this data is stationary. Estimating the AR(1) over this data, these are the results:

z test of coefficients:

Estimate Std. Error z value  Pr(>|z|)
ar1        0.728652   0.085601  8.5121 < 2.2e-16 ***
intercept 20.176618   0.809543 24.9235 < 2.2e-16 ***


The purpose of this estimation is to get the stationary mean of the process, i.e.:

$$x_t=c+\phi x_{t-1}+\varepsilon_t$$

$$E\{x_t\}=\frac{c}{1-\phi}$$

With the estimated values this turns out to be, given the significance of estimators:

$$E\{x_t\}=\frac{20.176618}{1-0.728652}\approx74.3569$$

Which clearly is outside the range the values data take, and is much larger than expected. What is wrong? Maybe I misinterpreted the unit root tests or something similar?

• Be careful. Some estimation routines do not give you estimates $\hat c$ and $\hat \phi$ for $y_t = c + \phi y_{t-1} + e_t$ but instead returns the estimates of $\mu = c/(1-\phi)$. This could very well be the case since your series seems to be reverting to $20$ the intercept estimate. May 31, 2021 at 22:40
• @JesperforPresident Thanks for your answer, I'm using R's arima() from stats package, and also checked with Arima() from forecast, do you know if those have the mentioned by you representation? Also if that's the case, then $\hat\mu=20.176618$? May 31, 2021 at 22:45

I made the same mistake several years ago. When in doubt read the documentation or do a simulation. Here is a simulation

T <- 10000
y <- rep(0,T)
c <- 5.6
phi <- 0.72
y[1] <- rnorm(1) + c/(1-phi)
for (t in 2:T)
{
y[t] <- phi*y[t-1] + c + rnorm(1)
}

arima(y,order=c(1,0,0))


for which arima call return 20 as intercept

Coefficients: ar1 intercept 0.7213 20.0361 s.e. 0.0069 0.0358

so the intercept reported is

$$\mu = \frac{c}{1-\phi}$$