============================================================================== Lag Operators ============================================================================== Lag Operator ============================================================================== Given a sequence of values, :math:`\smash{y_1, y_2, \ldots, y_t}`, indexed by time, the lag operator, :math:`\smash{L}`, is defined as .. math:: \smash{Ly_t = y_{t-1}.} .. raw:: - The lag operator shifts a time value :math:`\smash{y_t}` back by one period. .. raw:: - :math:`\smash{y_t}` can be thought of as the input of the operator and :math:`\smash{y_{t-1}}` as the output. .. raw:: - The lag operator can be applied to all values in a series :math:`\smash{\{y_t\}_{t=1}^T}` and the result is a new series shifted back by one period: :math:`\smash{\{y_t\}_{t=0}^{T-1}}`. Lag Operator ============================================================================== Applying the lag operator twice: .. math:: \smash{L(Ly_t) = Ly_{t-1} = y_{t-2}.} .. raw:: - We write :math:`\smash{L(Ly_t)}` as :math:`\smash{L^2 y_t}`. .. raw:: - Applying recursively: .. math:: \smash{L^k y_t = y_{t-k}.} .. raw:: - We will define :math:`\smash{L^0 = 1}`. Useful Properties of the Lag Operator ============================================================================== - The lag operator is commutative: .. math:: \smash{L(\beta y_t) = \beta L y_t.} .. raw:: - The lag operator is distributive: .. math:: \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 :math:`\smash{z_t = x_t + y_t}`. .. raw:: - The lag of a constant is the same constant: .. math:: \smash{Lc = c.} First-Order Difference Equation ============================================================================== Suppose we have a first-order difference equation: .. math:: \smash{y_t = \phi y_{t-1} + w_t.} .. raw:: In terms of the lag operator .. math:: \smash{(1-\phi L)y_t = w_t.} .. raw:: We can write .. math:: \smash{\phi(L)y_t = w_t,} where :math:`\smash{\phi(L) = (1-\phi L)}`. First-Order Difference Equation ============================================================================== Suppose the operator :math:`\smash{\phi(L) = (1-\phi L)}` has an inverse: .. math:: \smash{\phi(L)^{-1} = (1-\phi L)^{-1}.} .. raw:: - The inverse is the operator such that .. math:: \smash{(1-\phi L)^{-1}(1-\phi L) = 1.} .. raw:: - If an inverse operator exists, .. math:: \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: .. math:: \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} .. raw:: - The infinite recursive substitution can only be performed if :math:`\smash{|\phi| < 1}`. Inverse of Lag Operator ============================================================================== Restating the previous result, for :math:`\smash{|\phi| < 1}`: .. math:: \smash{y_t = \left(\sum_{i=0}^{\infty} \phi^i L^i\right) w_t.} .. raw:: Substituting: .. math:: \smash{w_t = (1-\phi L) y_t = (1-\phi L) \left(\sum_{i=0}^{\infty} \phi^i L^i\right) w_t.} .. raw:: So when :math:`\smash{|\phi| < 1}`: .. math:: \smash{(1-\phi L)^{-1} = \sum_{i=0}^{\infty} \phi^i L^i.} .. raw:: That is, :math:`\smash{\sum_{i=0}^{\infty} \phi^i L^i}` is the inverse operator of :math:`\smash{(1-\phi L)}`. :math:`\smash{p}` th-Order Difference Equation ============================================================================== Suppose we have a :math:`\smash{p}` th-order difference equation: .. math:: \smash{y_t = \phi_1 y_{t-1} + \phi_2 y_{t-2} + \ldots + \phi_p y_{t-p} + w_t.} .. raw:: In terms of the lag operator .. math:: \smash{(1-\phi_1 L - \phi_2 L^2 - \ldots - \phi_p L^p)y_t = w_t.} .. raw:: We can write .. math:: \smash{\phi(L)y_t = w_t,} where :math:`\smash{\phi(L) = (1-\phi_1 L - \phi_2 L^2 - \ldots - \phi_p L^p)}`. Factoring Polynomials ============================================================================== In general, a :math:`\smash{p}` th-order, real-valued polynomial can be factored as .. math:: \smash{1-\phi_1 z - \phi_2 z^2 - \ldots - \phi_p z^p = (1-\lambda_1z)(1-\lambda_2z) \cdots (1-\lambda_pz).} .. raw:: - :math:`\smash{\left\{\frac{1}{\lambda_i}\right\}_{i=1}^p}` are the :math:`\smash{p}` roots of the polynomial. .. raw:: - Some of the roots may be complex and some may be identical. Factoring :math:`\smash{p}` th-Order Difference Equation ============================================================================== If we factor the :math:`\smash{p}` th-order lag polynomial in the same way as a real-valued polynomial: .. math:: \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*} .. raw:: If :math:`\smash{|\lambda_i| < 1}`, .. math:: \smash{(1 - \lambda_i L)^{-1} = \sum_{j=0}^{\infty} \lambda_i^j L^j, \,\,\,\,\, \forall i.} Factoring :math:`\smash{p}` th-Order Difference Equation ============================================================================== In this case, .. math:: \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*} Factoring :math:`\smash{p}` th-Order Difference Equation ============================================================================== If we define .. math:: \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*} .. raw:: then .. math:: \begin{equation*} y_t = \theta(L) w_t. \end{equation*} .. raw:: - Clearly, :math:`\smash{\phi(L)^{-1} = \theta(L)}`. .. raw:: - Note that the inverse only exists when :math:`\smash{|\lambda_i| < 1}`, :math:`\smash{\forall i}`. .. raw:: - This can also be stated as: the inverse only exists when the roots of :math:`\smash{\phi(L)}` are greater than unity: :math:`\smash{\frac{1}{|\lambda_i|} > 1}`, :math:`\smash{\forall i}`. Vector Difference Equation ============================================================================== We can rewrite the :math:`\smash{p}` th-order difference equation as .. math:: \begin{equation*} {\bf y}_t = \Phi {\bf y}_{t-1} + {\bf w}_t, \end{equation*} .. raw:: where .. math:: \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*} Vector Difference Equation ============================================================================== It turns out that the values :math:`\smash{\{\lambda_i\}_{i=1}^p}` are the :math:`\smash{p}` eigenvalues of :math:`\smash{\Phi}`. .. raw:: - So the eigenvalues of :math:`\smash{\Phi}` are the inverses of the roots of the lag polynomial :math:`\smash{\phi(L)}`. .. raw:: - Hence, :math:`\smash{\phi(L)^{-1}}` exists if all :math:`\smash{p}` roots of :math:`\smash{\phi(L)}` lie *outside* the unit circle or all :math:`\smash{p}` eigenvalues of :math:`\smash{\Phi}` lie *inside* the unit circle.