Causality and Invertibility

Causality

Suppose \(\smash{\{Y_t\}}\) is an \(\smash{AR(1)}\) process:

\[\begin{align*} Y_t & = \phi Y_{t-1} + \varepsilon_t. \end{align*}\]
  • We have shown that when \(\smash{|\phi| < 1}\), \(\smash{\{Y_t\}}\) is stationary.
  • What if \(\smash{|\phi| > 1}\)?

Causality

Let’s run the \(\smash{AR}\) recursion forward:

\[\begin{split}\begin{align*} Y_{t-1} & = \frac{1}{\phi} Y_t - \frac{1}{\phi} \varepsilon_t = \frac{1}{\phi} \left(\frac{1}{\phi} Y_{t+1} - \frac{1}{\phi} \varepsilon_{t+1}\right) - \frac{1}{\phi} \varepsilon_t \\ & = - \frac{1}{\phi} \varepsilon_t - \left(\frac{1}{\phi}\right)^2 \varepsilon_{t+1} + \left(\frac{1}{\phi}\right)^2 Y_{t+1} \\ & \vdots \\ & = - \sum_{j=0}^{\infty} \left(\frac{1}{\phi}\right)^{j+1} \varepsilon_{t+j} \\ & = - \left(\sum_{j=0}^{\infty} \left(\frac{1}{\phi}\right)^{j+1} L^{-j}\right) \varepsilon_t. \end{align*}\end{split}\]

Causality

The previous sum converges, so \(\smash{Y_t}\) is stationary.

  • However it is not a function of past \(\smash{\varepsilon_t}\).

Causality

A process \(\smash{\{X_t\}}\) is a causal function of \(\smash{\{W_t\}}\) if \(\smash{\exists \psi(L) = \psi_0 + \psi_1L^1 + \ldots}\) such that \(\smash{x_t = \psi(L) w_t}\) and \(\smash{\sum_{j=0}^{\infty} |\psi_j| < \infty}\).

  • An \(\smash{AR(1)}\) is causal only if \(\smash{|\phi| < 1}\).
  • However it is stationary as long as \(\smash{|\phi| \neq 1}\).

Causality of \(\smash{AR(p)}\)

Suppose \(\smash{\{Y_t\}}\) is an \(\smash{AR(p)}\) process with lag polynomial \(\smash{\phi(L)}\).

  • If all roots of \(\smash{\phi(L)}\) are inside or outside the unit circle, \(\smash{\{Y_t\}}\) is stationary.
  • If any root of \(\smash{\phi(L)}\) is on the unit circle, \(\smash{\{Y_t\}}\) is not stationary.
  • If all roots of \(\smash{\phi(L)}\) are outside the unit circle, \(\smash{\phi(L)^{-1}}\) exists and \(\smash{\{Y_t\}}\) is stationary and causal.

Invertibility

Suppose \(\smash{\{Y_t\}}\) is an \(\smash{MA(q)}\) process:

\[\smash{Y_t = \mu + \theta(L) \varepsilon_t,}\]

where \(\smash{\theta(L) = 1+\theta_1 L^1 + \ldots + \theta_q L^q}\).

  • We say \(\smash{\{Y_t\}}\) is invertible if \(\smash{\theta(L)^{-1}}\) exists.

Invertibility

The \(\smash{MA(q)}\) lag polynomial can be factored as

\[\smash{\theta(L) = 1+\theta_1 L^1 + \ldots + \theta_q L^q = (1-\eta_1 L)\cdots(1-\eta_q L).}\]
  • \(\smash{\left\{\frac{1}{\eta_i}\right\}_{i=1}^q}\) are the roots of \(\smash{\theta(L)}\).

Suppose \(\smash{|\eta_i| < 1 \,\, \forall i}\). Then

\[\begin{split}\begin{align*} (1-\eta_i L)^{-1} & = \sum_{j=0}^{\infty} \eta^j_i L^j \,\,\,\, \forall i \\ \theta(L)^{-1} & = \left(\sum_{j=0}^{\infty} \eta^j_1 L^j\right) \cdots \left(\sum_{j=0}^{\infty} \eta^j_q L^j\right). \end{align*}\end{split}\]

Stationarity/Invertibility

We previously showed that an \(\smash{MA(q)}\) is always stationary, regardless of the roots of \(\smash{\theta(L)}\).

  • It is only invertible if all of the roots of \(\smash{\theta(L)}\) lie outside the unit circle.
  • In this case
\[\smash{\varepsilon_t = \theta(L)^{-1} Y_t}.\]
  • That is, \(\smash{\{\varepsilon_t\}}\) is a causal function of \(\smash{\{Y_t\}}\).

Inverting an \(\smash{MA(q)}\)

Given an \(\smash{MA(q)}\) process,

\[\smash{Y_t = \mu + \theta(L) \varepsilon_t, \,\,\,\, \varepsilon_t \stackrel{i.i.d.}{\sim} WN(0,\sigma^2)},\]

suppose, without loss of generality,

  • \(\smash{|\eta_i| < 1}\) for \(\smash{i=1,\ldots,m}\)
  • \(\smash{|\eta_i| > 1}\) for \(\smash{i=m+1,\ldots,q}\).

Inverting an \(\smash{MA(q)}\)

Create a new process

\[\smash{\widetilde{Y}_t = \mu + \tilde{\theta}(L) \tilde{\varepsilon}_t, \,\,\,\, \tilde{\varepsilon}_t \stackrel{i.i.d.}{\sim} WN(0,\sigma^2 \eta^2_{m+1} \cdots \eta^2_q),}\]

where

\[\begin{split}\begin{align*} \tilde{\theta}(L) & = 1+\tilde{\theta}_1 L^1 + \ldots + \tilde{\theta}_q L^q \\ & = \left(1-\eta_1 L\right) \cdots \left(1-\eta_mL\right) \cdot \left(1-\frac{1}{\eta_{m+1}}L\right) \cdots \left(1 - \frac{1}{\eta_q}L\right). \end{align*}\end{split}\]

Inverting an \(\smash{MA(q)}\)

It can be shown that \(\smash{\widetilde{Y}_t}\) has the same first and second moments as \(\smash{Y_t}\).

  • \(\smash{\widetilde{Y}_t}\) is known as the invertible represenation of the \(\smash{MA(q)}\) process.
  • Note that every \(\smash{MA(q)}\) process has an invertible representation so long as none of the roots of \(\smash{\theta(L)}\) lie on the unit circle.
  • If an invertible representation exists, it is unique.

Causality and Invertibility of an \(\smash{ARMA(p,q)}\)

The notions of stationarity, causality and invertibility extend to an \(\smash{ARMA(p,q)}\) process:

\[\begin{align} \phi(L) Y_t & = c + \theta(L) \varepsilon_t. \end{align}\]
  • If none of the roots of \(\smash{\phi(L)}\) lie on the unit circle, \(\smash{\{Y_t\}}\) is stationary.
  • If all of the roots of \(\smash{\phi(L)}\) lie outside the unit circle, \(\smash{\{Y_t\}}\) is causal.
  • If none of the roots of \(\smash{\theta(L)}\) lie on the unit circle, \(\smash{\{Y_t\}}\) has a unique invertible representation.