==============================================================================
Causality and Invertibility
==============================================================================

Causality
==============================================================================
Suppose :math:`\smash{\{Y_t\}}` is an :math:`\smash{AR(1)}` process:

.. math::

   \begin{align*}
   Y_t & = \phi Y_{t-1} + \varepsilon_t.
   \end{align*}

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- We have shown that when :math:`\smash{|\phi| < 1}`,
  :math:`\smash{\{Y_t\}}` is stationary.

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- What if :math:`\smash{|\phi| > 1}`?

Causality
==============================================================================
Let's run the :math:`\smash{AR}` recursion forward:

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.. math::

   \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*}

Causality
==============================================================================
The previous sum converges, so :math:`\smash{Y_t}` is stationary.

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- However it is not a function of past :math:`\smash{\varepsilon_t}`.


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

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- An :math:`\smash{AR(1)}` is causal only if :math:`\smash{|\phi|
  < 1}`.

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- However it is stationary as long as :math:`\smash{|\phi|
  \neq 1}`.

Causality of :math:`\smash{AR(p)}`
==============================================================================
Suppose :math:`\smash{\{Y_t\}}` is an :math:`\smash{AR(p)}` process
with lag polynomial :math:`\smash{\phi(L)}`.

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- If all roots of :math:`\smash{\phi(L)}` are inside or outside the
  unit circle, :math:`\smash{\{Y_t\}}` is stationary.

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- If any root of :math:`\smash{\phi(L)}` is on the unit circle,
  :math:`\smash{\{Y_t\}}` is not stationary.

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- If all roots of :math:`\smash{\phi(L)}` are outside the unit circle,
  :math:`\smash{\phi(L)^{-1}}` exists and :math:`\smash{\{Y_t\}}` is
  stationary and causal.

Invertibility
==============================================================================
Suppose :math:`\smash{\{Y_t\}}` is an :math:`\smash{MA(q)}` process:

.. math::

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

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

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- We say :math:`\smash{\{Y_t\}}` is *invertible* if
  :math:`\smash{\theta(L)^{-1}}` exists.

Invertibility
==============================================================================
The :math:`\smash{MA(q)}` lag polynomial can be factored as

.. math::

   \smash{\theta(L) = 1+\theta_1 L^1 + \ldots + \theta_q L^q =
   (1-\eta_1 L)\cdots(1-\eta_q L).}

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- :math:`\smash{\left\{\frac{1}{\eta_i}\right\}_{i=1}^q}` are the
  roots of :math:`\smash{\theta(L)}`.

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Suppose :math:`\smash{|\eta_i| < 1 \,\, \forall i}`. Then

.. math::

   \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*}

Stationarity/Invertibility
==============================================================================
We previously showed that an :math:`\smash{MA(q)}` is *always*
stationary, regardless of the roots of :math:`\smash{\theta(L)}`.

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- It is only invertible if all of the roots of
  :math:`\smash{\theta(L)}` lie outside the unit circle.

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- In this case

.. math::

   \smash{\varepsilon_t = \theta(L)^{-1} Y_t}.

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- That is, :math:`\smash{\{\varepsilon_t\}}` is a causal function of
  :math:`\smash{\{Y_t\}}`.

Inverting an :math:`\smash{MA(q)}`
==============================================================================
Given an :math:`\smash{MA(q)}` process,

.. math::

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

suppose, without loss of generality,

- :math:`\smash{|\eta_i| < 1}` for :math:`\smash{i=1,\ldots,m}`

- :math:`\smash{|\eta_i| > 1}` for :math:`\smash{i=m+1,\ldots,q}`.

Inverting an :math:`\smash{MA(q)}`
==============================================================================
Create a new process

.. math::

   \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

.. math::

   \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*}

Inverting an :math:`\smash{MA(q)}`
==============================================================================
It can be shown that :math:`\smash{\widetilde{Y}_t}` has the same first
and second moments as :math:`\smash{Y_t}`.

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- :math:`\smash{\widetilde{Y}_t}` is known as the invertible represenation
  of the :math:`\smash{MA(q)}` process.

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- Note that every :math:`\smash{MA(q)}` process has an invertible
  representation so long as none of the roots of
  :math:`\smash{\theta(L)}` lie on the unit circle.

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- If an invertible representation exists, it is unique.

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

.. math::

   \begin{align}
   \phi(L) Y_t & = c + \theta(L) \varepsilon_t.
   \end{align}

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- If none of the roots of :math:`\smash{\phi(L)}` lie on the unit
  circle, :math:`\smash{\{Y_t\}}` is stationary.

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- If all of the roots of :math:`\smash{\phi(L)}` lie outside the unit
  circle, :math:`\smash{\{Y_t\}}` is causal.

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- If none of the roots of :math:`\smash{\theta(L)}` lie on the unit
  circle, :math:`\smash{\{Y_t\}}` has a unique invertible
  representation.