Subtracting mean from time series in AR(1) process

Question

In Cryer and Chan the book Timer Series with Application in R. We find the following piece.

Consider the first-order autoregressive process, abbreviated AR(1), in detail.

Assume the series is stationary and satisfies.

$Y_{t} = \phi Y_{t - 1} + e_{t}$

Where $e_{t}$ is a white noise process.

We assume that the process mean has been subtracted out so that the series mean is zero.

I don't understand what this means.

Does this implicitly imply the following logic?

$Y_{t} = \phi Y_{t - 1} + e_{t} - \mu + \mu$

Suppose that:

$W_{t} = \phi W_{t - 1} + e_{t} + \mu$

and that:

$X_{t} = W_{t} - \mu = \phi W_{t - 1} + e_{t} - \mu + \mu$

In this case, the time series $Y_{t}$ is stochastically equal to the following time series $X_{t}$.

Am I correct?

Answer 1

Your calculations are correct but the meaning is simpler. The authors assume that $\mathbb{E}\left[Y_t\right] = 0$. If this were not the case, say $\mathbb{E}\left[Y_t\right] = \mu$, then substitute $Z_t = Y_t - \mu$ for $Y_t$ and follow along with the book.

Answer 2

If you have a stationary AR(1) process given by:

$$W_t = \mu + \phi W_{t-1} + \varepsilon_t$$

Then the mean is:

$$\mathbb{E}(W_t) = \frac{\mu}{1-\phi}$$

Note that the mean is not $\mu$ in general. If you subtract the mean and define a new process:

$$X_t := W_t - \mathbb{E}(W_t) = W_t - \frac{\mu}{1-\phi}$$

Therefore $W_t = X_t + \frac{\mu}{1-\phi}$ and you can substitute this back into the first equation to get:

$$ X_t + \frac{\mu}{1-\phi} = \mu + \phi \left(X_{t-1} + \frac{\mu}{1-\phi}\right) + \varepsilon_t $$

Or:

$$ X_t = \mu \left[1 - \frac{1}{1-\phi} + \frac{\phi}{1-\phi}\right] + \phi X_{t-1} + \varepsilon_t $$

Which is just:

$$X_t = \phi X_{t-1} + \varepsilon_t$$

This is a stationary AR(1) process with mean zero.

The issue with your derivation is that you use the wrong mean and then end up with an equation that doesn't have the same process on both sides ($X_t$ on the left and $W_{t-1}$ on the right).