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Let $C(B)$ be a $\infty$-order polynomial: $$ C(B) = \sum_{k=0}^\infty \alpha_k B^k$$

Show that $$C(B) = C(1) + (1-B)C^*(B)$$ where $C^*(B)$ is a another $\infty$-order polynomial.

This comes from the prove of the Engle-Granger Representation Theorem in their original paper

Here, the polynomials $C(B), C^*(B)$ are moving-average polynomials in time series analysis.

I can't seem to understand how its derived... any help?

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To prove this, you need to show every coefficient on the left is equal to the analogous coefficient on the right. To do so, you can use induction, so start with $k=0$, which is trivial, and proceed as you would in the iteration if an induction proof.

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Suppose $1$ is in the convergent domain of $C$. \begin{align} C(B)-C(1) &= \sum_{k=0}^\infty \alpha_k B^k-\sum_{k=0}^\infty \alpha_k \\ &=\sum_{k=0}^\infty \alpha_k (B^k-1) \\ &=\sum_{k=0}^\infty \alpha_k (B-1)\sum_{n=0}^{k-1}B^n \\ &=(B-1)\sum_{k=0}^\infty \sum_{n=0}^{k-1}\alpha_k B^n \\ &=(B-1)\sum_{n=0}^\infty B^n\sum_{k=n+1}^\infty \alpha_k \\ &= (1-B)C^*(B) \end{align} where $$ C^*(B)=-\sum_{n=0}^\infty B^n\sum_{k=n+1}^\infty \alpha_k $$ $\sum_{k=n+1}^\infty \alpha_k$ exists because $\sum_{k=0}^\infty \alpha_k$ converges.

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  • $\begingroup$ I can't seem to figure out how you went from lines two to lines three. Any hints? Thanks a lot. $\endgroup$ Sep 8 '16 at 0:51

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