The inverse of 1 + aL

Summary

• Given: a lag polynomial \(f\left( L \right)=1+aL\) where \(a\) is a constant and \(L\) is the lag operator.
• \(f\left( L \right)\) has an inverse if and only if \(\left| a \right|<1\) . If this condition is satisfied, then

\[{\left( 1+aL \right)}^{-1}=1-aL+a^2L^2- \ldots =\sum_{j=0}^{∞}{ {\left( -a \right)}^jL^j }\]

• Result: A stationary AR(1) process \(\left( 1-θL \right)y_t=ε_t\) is equivalent to an MA( \(∞\) ) process

\[y_t={\left( 1-θL \right)}^{-1}ε_t=ε_t+θε_t+θ^2ε_t+ \ldots \]

• Result: An MA(1) process \(y_t=\left( 1+αL \right)ε_t\) with \(\left| α \right|<1\) is equivalent to an AR( \(∞\) ) process

\[{\left( 1+αL \right)}^{-1}y_t=ε_t\]

• or

\[y_t-αy_t+α^2y_{t-2}- \ldots =ε_t\]