Let us take this $\operatorname{AR}(p)$ equation
$$\left(1 - \sum_{i=1}^{p} \phi_i L^i \right)X_t = \mu + \epsilon_t$$
as an example.
When $p=0$ I read this to mean
\begin{align*} \mu + \epsilon_t &= \left(1 - \sum_{i=1}^{0} \phi_i L^i \right)X_t \\ & = \left(1 - \phi_1 L^1 - \phi_0L^0 \right)X_t \\ & = X_t - \phi_1 L^1X_t - \phi_0L^0X_t \\ &= X_t - \phi_1 X_{t-1} - \phi_0 X_t \\ &= (1-\phi_0)X_t - \phi_1 X_{t-1} \\ \end{align*}
which could then be solved for $X_t$ to give
$$X_t = \frac{\mu + \epsilon_t + \phi_1 X_{t-1}}{1-\phi_0}.$$
If we take
\begin{align*} \delta &:= \frac{\mu}{1 - \phi_0} \\ \gamma_t &:= \frac{\epsilon_t}{1 - \phi_0} \\ \theta_{t} &:= \frac{\phi_1}{1 - \phi_0} \end{align*}
then we get the form of an $\operatorname{AR}(1)$ model as follows
$$X_t = \delta + \theta_t X_{t-1} + \gamma_t.$$
Prima facie this tells us $\operatorname{AR}(0) = \operatorname{AR}(1)$, but I thought that $\operatorname{AR}(0)$ should reduce to $$X_t = \mu + \epsilon_t.$$ Although I think many people take $\operatorname{AR}(0)$ to mean $$X_t = \epsilon_t$$ but I included $\mu$ in my definition of $\operatorname{AR}(p)$ at the start in a way where it won't fall out due any choice of $p$.
Some possibilities I have considered:
- There exists a convention that $p=0 \implies \sum_{i=1}^{p} \phi_i L^i = 0$.
- I've made a set of mistakes doing the basic algebra (it happens).
- The sources I have found are mistaken.
What is going on here?
