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I have this process: $Y_t = \frac{2}{5}Y_{t-1}+ \frac{9}{20}Y_{t-2} + e_t+ e_{t-1}+\frac{1}{4}e_{t-2}$ which is an ARMA(2,2) model, and I'd like to write it as an ARMA(1,1).

I found the AR polynomial and its roots: $1-\frac{2}{5}x + \frac{9}{20}x^2 = 0 \implies x= \frac{4}{9} \pm \frac{2\sqrt{41}i}{9}$ which lies outside the unit circle so it's stationary.

But after this I'm not sure how to proceed. I tried using the backshift operator notation to write it out like $(1-(2/5)B - (9/20)B^2)Y_t = (1+B+(1/4)B^2)e_t$, but this doesn't change anything because $d=0$. What am I missing here?

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I realized, you can factor: $(-9B+10)(B+2)Y_{t} = (B+2)^2 e_t$ which yields $(-9B+10)Y_t = (B+2)e_t$ which is an ARMA(1,1)

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