Causality and Invertibility¶
Causality¶
Suppose \(\smash{\{Y_t\}}\) is an \(\smash{AR(1)}\) process:
- 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:
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:
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
- \(\smash{\left\{\frac{1}{\eta_i}\right\}_{i=1}^q}\) are the roots of \(\smash{\theta(L)}\).
Suppose \(\smash{|\eta_i| < 1 \,\, \forall i}\). Then
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
- That is, \(\smash{\{\varepsilon_t\}}\) is a causal function of \(\smash{\{Y_t\}}\).
Inverting an \(\smash{MA(q)}\)¶
Given an \(\smash{MA(q)}\) process,
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
where
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:
- 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.