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I was reading some notes and something made me confuse about characteristic polynomial in ARIMA model. Suppose that I have the following AR(2) model

$$x_t=0.6x_{t-1}-0.8x_{t-2}+\epsilon_t=x_t(1+0.6B-0.8B^2 )=a_t$$

Now came the doubt, for me the right way to find the roots should be use the form $$1+0.6x-0.8x^2=0$$ but in some notes I found $$x^2 +0.6x-0.8=0$$

It's not the first case the right way?

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None of the two is right, but the first is less wrong than the second.

First, the polynomial is in terms of the backshift operator, which is why you use $1+0.6B-0.8B^2$ or, substituting $x$ for $B$, $1+0.6x-0.8x^2$.

Second, you got the signs wrong. The model equation is $$ x_t=0.6x_{t-1}-0.8x_{t-2}+\epsilon_t $$ which is equivalent to $$ x_t-0.6x_{t-1}+0.8x_{t-2}=\epsilon_t $$ or $$ x_t(1-0.6B+0.8B^2)=\epsilon_t. $$ Thus the polynomial is $1-0.6B+0.8B^2$ or $1-0.6x+0.8x^2$. Set it to zero to find the roots.

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  • $\begingroup$ Yeah, I forgot to change the signs, but the second form is wrong right? $\endgroup$ – user72621 Sep 4 '17 at 14:53
  • $\begingroup$ Yes, the second form is wrong because the degrees of the polynomial are misplaced. $\endgroup$ – Richard Hardy Sep 4 '17 at 15:07

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