Making an AR(3) model weakly stationary I have a model as:$$r_t=0.05+\frac{7}{6}r_{t-1}+\frac{1}{6}r_{t-2}-\frac{1}{3}r_{t-3}+a_t$$
When checking for stationarity:
$$1-\frac{7}{6}x-\frac{1}{6}x^2+\frac{1}{3}x^3=0$$ I get $x\in \{-2,1,1.5\}$. Since the absolute value of 1 is not greater than 1, this is not stationary.
My question is how can I make this stationary? I can't deduct the mean from $r_t$'s because of the non-stationarity, the mean is not time-invariant.
Any helps would b appreciated!
 A: You can't make your process stationary. It is simply not stationary, because it has a unit root. However, there is a related process which has no unit root, and is stationary.
Your process can be written like this, with the backshift operator $B$:
$$\left(1-\frac{7}{6}B-\frac{1}{6}B^2+\frac{1}{3}B^3\right)r_t = 0.05 + a_t$$
As you computed, the characteristic polynomial has roots 1, -2 and 1.5. You can use synthetic division to compute the result of factoring out the unit root, which you can use here to write:
$$\left(1-\frac{1}{6}B-\frac{1}{3}B^2\right)\left(1-B\right)r_t = 0.05 + a_t$$
You can then define the differenced process $y_t := (1-B)r_t$, so you have:
$$\left(1-\frac{1}{6}B-\frac{1}{3}B^2\right)y_t = 0.05 + a_t$$
The roots of the characteristic polynomial for $y_t$ are just the ones we didn't factor out, -2 and 1.5, so $y_t$ is stationary.
A: If your process contains a unit root, consider differencing it. Instead of $r_t$, you can study
$$y_t:=\Delta r_t=r_t-r_{t-1}.$$
Once you have learned about $y_t$, you can extend inference to $r_t$ by noting that
$$r_t=r_0+y\sum_{\tau=1}^t y_\tau$$
or
$$r_t=r_{t-1}+y_t$$
depending on what is easier to work with / more relevant in your situation.
