# I understand why stationarity is a requirement for AR(p) models, but why is it necessary for MA(q) models?

I have (or at least I think I have) a good intuition for why stationarity is a requirement for modeling $AR(p)$ models:

an $AR(p)$ model with coefficients $a_1,....,a_p$ :

$Y_t = a_1Y_{t-1}+a_2Y_{t-2}+....a_qY_{t-q}$

such that one or more of the $|a_i| >1$ , would diverge quickly.

But why is it a requirement for $MA(q)$ models? I don't have an intuitive grasp of why it is a requirement?

• your question is unclear. you seem to start from two places: stationarity and the condition on the coefficient magnitude. what is your question about? also clarify what's "$a$" in your question – Aksakal Jan 18 '18 at 19:44
• @Aksakal the stationarity condition is related to whether $|a| < 1$ or not, so I'm not really starting from different places. I'll edit for clarity. – Skander H. Jan 18 '18 at 23:12

The autoregressive part indeed deals with stationarity. If it is statistically true that $|a| = 1$ there is evidence of a unit root and we get a whole bunch of unit-root associated problems. If $|a| > 1$, you are right, the series will quickly diverge.

For MA it indeed is a bit more tricky. We have to look at a property called invertibility: An invertible MA model is one that can be written as an infinite order AR model that converges so that the AR coefficients converge to 0 as we move infinitely back in time. We’ll demonstrate invertibility for the MA(1) model.

The MA(1) model can be written as $x_t - μ = w_t + θ_1w_{t-1}$.

If we let $z_t = x_t - μ$, then the MA(1) model is

\begin{equation} z_t=w_t+θ_1w_{t−1}. \end{equation} At time $t-1$, the model is $z_{t-1} = w_{t-1} + θ_1w_{t-2}$ which can be reshuffled to

\begin{equation} (2) w_{t−1}=z_{t−1}−θ_1w_{t−2}. \end{equation}

We then substitute relationship (2) for $w_{t-1}$ in equation (1)

\begin{equation} (3) z_t=w_t+θ_1(z_{t−1}−θ_1w_{t−2})=w_t+θ_1z_{t−1}−θ_2w_{t−2} \end{equation} At time t-2, equation (2) becomes

\begin{equation} (4) w_{t−2}=z_{t−2}−θ_1w_{t−3}. \end{equation} We then substitute relationship (4) for $w_{t-2}$ in equation (3)

\begin{equation} z_t=w_t+θ_1z_{t−1}−θ_{21}w_{t−2}=w_t+θ_1z_{t−1}−θ_{21}(z_{t−2}−θ_1w_{t−3})=w_t+θ_1z_{t−1}−θ_{21}z_{t−2}+θ_{31}w_{t−3} \end{equation} If we were to continue (infinitely), we would get the infinite order AR model

\begin{equation} z_t=w_t+θ_1z_{t−1}−θ_{21}z_{t−2}+θ_{31}z_{t−3}−θ_{41}z_{t−4}+… \end{equation} Note however, that if $|θ_1| ≥1$, the coefficients multiplying the lags of z will increase (infinitely) in size as we move back in time. To prevent this, we need |θ1| <1. This is the condition for an invertible MA(1) model.

We see that an AR(1) model can be converted to an infinite order MA model:

\begin{equation} x_t−μ=w_t+ϕ_1w_{t−1}+ϕ_{21}w_{t−2}+⋯+ϕ_{k1}w_{t−k}+⋯=\sum_{j = 0}^\infty ϕ_{j1}w_{t−j} \end{equation} This summation of past white noise terms is known as the causal representation of an AR(1). In other words, xt is a special type of MA with an infinite number of terms going back in time. This is called an infinite order MA or MA(∞). A finite order MA is an infinite order AR and any finite order AR is an infinite order MA.

And thus finally you are back at the stationary requirement for a AR model

• please, use math markup – Aksakal Jan 18 '18 at 19:41
• Yeahyeah, I quickly changed it to a more human-friendly format. Hope the answer helps! – Tom Jan 18 '18 at 19:52
• What is $\theta_{21}$ etc.? – Jarle Tufto Jan 18 '18 at 20:16