In the last step of this answer, the author writes that to obtain Wold Representation of an AR(2) process, you need to expand the geometric sequences in the fractions on the right hand side:


After expanding the sequences I am stuck with the following:


How can I get rid of the lag operators and get to the MA representation from here?


1 Answer 1


You have a product of two (infinite-order) polynomials and you have to turn it into a single polynomial in "expanded form" by collecting like-terms. Doing this in the present case gives the following moving-average characteristic polynomial:

$$\begin{align} \lambda(L) &= (1+\lambda_{1}L)^{-1} (1+\lambda_{2}L)^{-1} \\[12pt] &= (1+\lambda_{1}L+\lambda_{1}^{2}L^{2}+...+\lambda_{1}^{n}L^{n})(1+\lambda_{2}L+\lambda_{2}^{2}L^{2}+...+\lambda_{2}^{n}L^{n}) \\[12pt] &= 1 + (\lambda_1+\lambda_2) L + (\lambda_1^2 + \lambda_1 \lambda_2 + \lambda_2^2) L^2 + (\lambda_1^3 + \lambda_1^2 \lambda_2 + \lambda_1 \lambda_2^2 + \lambda_2^3) L^3 + \cdots \\[6pt] &= \sum_{i=0}^\infty \Bigg( \sum_{j=0}^i \lambda_1^{i-j} \lambda_2^j \Bigg) L^i. \\[6pt] \end{align}$$

Applying this to the time-series equation then gives the (infinite) MA representation:

$$\begin{align} X_t = \lambda(L) \varepsilon_t &= \sum_{i=0}^\infty \Bigg( \sum_{j=0}^i \lambda_1^{i-j} \lambda_2^j \Bigg) L^i \varepsilon_t \\[6pt] &= \sum_{i=0}^\infty \Bigg( \sum_{j=0}^i \lambda_1^{i-j} \lambda_2^j \Bigg) \varepsilon_{t-i}. \\[6pt] \end{align}$$


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