Writing MA and AR representations I have to determine if
$$(1 - 1.1B + 0.8B^2)Y_t = (1 - 1.7B + 0.72B^2)a_t$$
is stationary, invertible or both.
I have shown that $\Phi(B) = 1 - 1.1B + 0.8B^2 = 0$ when $B_{1,2} = 0.6875 \pm 0.8817i$, whose moduli are both larger than 1, hence is stationary. Similarly, I have shown that $\Theta(B) = 1 - 1.7B + 0.72B^2 = 0$, when $B_1 = 1.25 > 1$ and $B_2 = 1.11 > 1$, hence is invertible.
I also need to express the model as a MA and AR representation if it exists; which they do as I have already shown. However, to write as an MA process, I would need to write as:
$$Y_t = \frac{1 - 1.7B + 0.72B^2}{1 - 1.1B + 0.8B^2}a_t$$
and for an AR process as:
$$\frac{1 - 1.1B + 0.8B^2}{1 - 1.7B + 0.72B^2}Y_t = a_t$$
However, I am confused on how to do this given the division of the quadratic expressions. Should I use long division or is there some expansion formula I should be using?
 A: Try the partial fraction decomposition:
\begin{align}
\frac{1}{(1 - \alpha B)(1 - \beta B)} & = \frac{\alpha/(\alpha - \beta)}{1 - \alpha B} +  \frac{\beta/(\beta - \alpha)}{1 - \beta B} \\
& = (\alpha - \beta)^{-1} \left( \alpha(1 - \alpha B)^{-1} - \beta (1 - \beta B)^{-1} \right) \\
& = (\alpha - \beta)^{-1} \left( \alpha \sum_{k = 0}^\infty \alpha^k B^k - \beta \sum_{k = 0}^\infty \beta^k B^k \right) \\
& = (\alpha - \beta)^{-1} \sum_{k = 0}^\infty (\alpha^{k+1} - \beta^{k+1}) B^k
\end{align}
and apply it to both cases like:
\begin{align}
\frac{1 + c B + d B^2}{(1 - \alpha B)(1 - \beta B)} & = (1 + c B + d B^2) (\alpha - \beta)^{-1} \sum_{k = 0}^\infty (\alpha^{k+1} - \beta^{k+1}) B^k \\
& = (\alpha - \beta)^{-1} \sum_{k = 0}^\infty (\alpha^{k+1} - \beta^{k+1}) (1 + c B + d B^2) B^k \\
& = (\alpha - \beta)^{-1} \sum_{k = 0}^\infty (\alpha^{k+1} - \beta^{k+1}) (B^k + c B^{k+1} + d B^{k+2})
\end{align}
and by distributing and reindexing the summations, we have
\begin{align}
& = (\alpha - \beta)^{-1} \left( \sum_{k = 0}^\infty (\alpha^{k+1} - \beta^{k+1}) B^k + \sum_{k = 1}^\infty c (\alpha^k - \beta^k) B^k + \sum_{k = 2}^\infty d (\alpha^{k-1} - \beta^{k-1}) B^k) \right) \\
& = (\alpha - \beta)^{-1} \left( (\alpha - \beta) + (\alpha^2 - \beta^2 + c(\alpha - \beta)) B + \sum_{k = 2}^\infty [(\alpha^{k+1} - \beta^{k+1}) + c (\alpha^k - \beta^k) + d (\alpha^{k-1} - \beta^{k-1})] B^k \right) \\
& =  1 + (\alpha + \beta + c) B + (\alpha - \beta)^{-1} \sum_{k = 2}^\infty [(\alpha^{k+1} - \beta^{k+1}) + c (\alpha^k - \beta^k) + d (\alpha^{k-1} - \beta^{k-1})] B^k
\end{align}
Assuming I haven't made any mistakes, this gives the AR representation when $\alpha = 0.9$, $\beta = 0.8$, $c = -1.1$, and $d = 0.8$ and gives the MA process when $\alpha, \beta = 0.68750 \pm 0.88167 i$, $c = -1.7$, and $d = 0.72$. Perhaps you could even simplify this more using the difference of nth powers formula (certainly you could cancel the $(\alpha - \beta)^{-1}$ term this way, but I don't know if you would call the result "simpler."
