Lag Operators¶

Lag Operator¶

Given a sequence of values, \(\smash{y_1, y_2, \ldots, y_t}\), indexed by time, the lag operator, \(\smash{L}\), is defined as

\[\smash{Ly_t = y_{t-1}.}\]

The lag operator shifts a time value \(\smash{y_t}\) back by one period.

\(\smash{y_t}\) can be thought of as the input of the operator and \(\smash{y_{t-1}}\) as the output.

The lag operator can be applied to all values in a series \(\smash{\{y_t\}_{t=1}^T}\) and the result is a new series shifted back by one period: \(\smash{\{y_t\}_{t=0}^{T-1}}\).

Lag Operator¶

Applying the lag operator twice:

\[\smash{L(Ly_t) = Ly_{t-1} = y_{t-2}.}\]

We write \(\smash{L(Ly_t)}\) as \(\smash{L^2 y_t}\).

Applying recursively:

\[\smash{L^k y_t = y_{t-k}.}\]

We will define \(\smash{L^0 = 1}\).

Useful Properties of the Lag Operator¶

The lag operator is commutative:

\[\smash{L(\beta y_t) = \beta L y_t.}\]

The lag operator is distributive:

\[\smash{L(x_t+y_t) = L z_t = z_{t-1} = x_{t-1} + y_{t-1} = L x_t + L y_t,}\]

where \(\smash{z_t = x_t + y_t}\).

The lag of a constant is the same constant:

\[\smash{Lc = c.}\]

First-Order Difference Equation¶

Suppose we have a first-order difference equation:

\[\smash{y_t = \phi y_{t-1} + w_t.}\]

In terms of the lag operator

\[\smash{(1-\phi L)y_t = w_t.}\]

We can write

\[\smash{\phi(L)y_t = w_t,}\]

where \(\smash{\phi(L) = (1-\phi L)}\).

First-Order Difference Equation¶

Suppose the operator \(\smash{\phi(L) = (1-\phi L)}\) has an inverse:

\[\smash{\phi(L)^{-1} = (1-\phi L)^{-1}.}\]

The inverse is the operator such that

\[\smash{(1-\phi L)^{-1}(1-\phi L) = 1.}\]

If an inverse operator exists,

\[\smash{y_t = \phi(L)^{-1} w_t = (1-\phi L)^{-1} w_t.}\]

Recursive Substitution of First-Order Difference Equation¶

Applying recursive substitution to the first-order difference equation:

\[\begin{split}\begin{align} y_t & = \phi y_{t-1} + w_t \\ & = \phi(\phi y_{t-2} + w_{t-1}) + w_t \\ & = w_t + \phi w_{t-1} + \phi^2 y_{t-2} \\ & = w_t + \phi w_{t-1} + \phi^2 (\phi y_{t-3} + w_{t-2}) \\ & = w_t + \phi w_{t-1} + \phi^2 w_{t-2} + \phi^3 y_{t-3} \\ & \vdots \\ & = \sum_{i=0}^{\infty} \phi^i w_{t-i} = \sum_{i=0}^{\infty} \phi^i L^i w_t. \end{align}\end{split}\]

The infinite recursive substitution can only be performed if \(\smash{|\phi| < 1}\).

Inverse of Lag Operator¶

Restating the previous result, for \(\smash{|\phi| < 1}\):

\[\smash{y_t = \left(\sum_{i=0}^{\infty} \phi^i L^i\right) w_t.}\]

Substituting:

\[\smash{w_t = (1-\phi L) y_t = (1-\phi L) \left(\sum_{i=0}^{\infty} \phi^i L^i\right) w_t.}\]

So when \(\smash{|\phi| < 1}\):

\[\smash{(1-\phi L)^{-1} = \sum_{i=0}^{\infty} \phi^i L^i.}\]

That is, \(\smash{\sum_{i=0}^{\infty} \phi^i L^i}\) is the inverse operator of \(\smash{(1-\phi L)}\).

\(\smash{p}\) th-Order Difference Equation¶

Suppose we have a \(\smash{p}\) th-order difference equation:

\[\smash{y_t = \phi_1 y_{t-1} + \phi_2 y_{t-2} + \ldots + \phi_p y_{t-p} + w_t.}\]

In terms of the lag operator

\[\smash{(1-\phi_1 L - \phi_2 L^2 - \ldots - \phi_p L^p)y_t = w_t.}\]

We can write

\[\smash{\phi(L)y_t = w_t,}\]

where \(\smash{\phi(L) = (1-\phi_1 L - \phi_2 L^2 - \ldots - \phi_p L^p)}\).

Factoring Polynomials¶

In general, a \(\smash{p}\) th-order, real-valued polynomial can be factored as

\[\smash{1-\phi_1 z - \phi_2 z^2 - \ldots - \phi_p z^p = (1-\lambda_1z)(1-\lambda_2z) \cdots (1-\lambda_pz).}\]

\(\smash{\left\{\frac{1}{\lambda_i}\right\}_{i=1}^p}\) are the \(\smash{p}\) roots of the polynomial.

Some of the roots may be complex and some may be identical.

Factoring \(\smash{p}\) th-Order Difference Equation¶

If we factor the \(\smash{p}\) th-order lag polynomial in the same way as a real-valued polynomial:

\[\begin{split}\begin{align*} & (1-\phi_1 L - \phi_2 L^2 - \ldots - \phi_p L^p)y_t \\ & \hspace{0.5in} = (1-\lambda_1 L)(1-\lambda_2 L) \cdots (1-\lambda_p L) y_t = w_t. \end{align*}\end{split}\]

If \(\smash{|\lambda_i| < 1}\),

\[\smash{(1 - \lambda_i L)^{-1} = \sum_{j=0}^{\infty} \lambda_i^j L^j, \,\,\,\,\, \forall i.}\]

Factoring \(\smash{p}\) th-Order Difference Equation¶

In this case,

\[\begin{split}\begin{align*} y_t & = (1-\lambda_1 L)^{-1} (1-\lambda_2 L)^{-1} \cdots (1-\lambda_p L)^{-1} w_t \\ & = \left(\sum_{j=0}^{\infty} \lambda_1^j L^j\right) \left(\sum_{j=0}^{\infty} \lambda_2^j L^j\right) \cdots \left(\sum_{j=0}^{\infty} \lambda_p^j L^j\right) w_t. \end{align*}\end{split}\]

Factoring \(\smash{p}\) th-Order Difference Equation¶

If we define

\[\begin{equation*} \theta(L) = \left(\sum_{j=0}^{\infty} \lambda_1^j L^j\right) \left(\sum_{j=0}^{\infty} \lambda_2^j L^j\right) \cdots \left(\sum_{j=0}^{\infty} \lambda_p^j L^j\right) \end{equation*}\]

then

\[\begin{equation*} y_t = \theta(L) w_t. \end{equation*}\]

Clearly, \(\smash{\phi(L)^{-1} = \theta(L)}\).

Note that the inverse only exists when \(\smash{|\lambda_i| < 1}\), \(\smash{\forall i}\).

This can also be stated as: the inverse only exists when the roots of \(\smash{\phi(L)}\) are greater than unity: \(\smash{\frac{1}{|\lambda_i|} > 1}\), \(\smash{\forall i}\).

Vector Difference Equation¶

We can rewrite the \(\smash{p}\) th-order difference equation as

\[\begin{equation*} {\bf y}_t = \Phi {\bf y}_{t-1} + {\bf w}_t, \end{equation*}\]

where

\[\begin{split}\begin{equation*} {\bf y}_t = \left[\begin{array}{c} y_t \\ y_{t-1} \\ y_{t-2} \\ \vdots \\ y_{t-p+1} \end{array} \right] \,\,\,\, \Phi = \left[\begin{array}{ccccc} \phi_1 & \phi_2 & \ldots & \phi_{p-1} & \phi_p \\ 1 & 0 & \ldots & 0 & 0 \\ 0 & 1 & \ldots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \ldots & 1 & 0 \end{array} \right] \,\,\,\,\, {\bf w}_t = \left[\begin{array}{c} w_t \\ 0 \\ 0 \\ \vdots \\ 0 \end{array} \right]. \end{equation*}\end{split}\]

Vector Difference Equation¶

It turns out that the values \(\smash{\{\lambda_i\}_{i=1}^p}\) are the \(\smash{p}\) eigenvalues of \(\smash{\Phi}\).

So the eigenvalues of \(\smash{\Phi}\) are the inverses of the roots of the lag polynomial \(\smash{\phi(L)}\).

Hence, \(\smash{\phi(L)^{-1}}\) exists if all \(\smash{p}\) roots of \(\smash{\phi(L)}\) lie outside the unit circle or all \(\smash{p}\) eigenvalues of \(\smash{\Phi}\) lie inside the unit circle.