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?


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.

  • $\begingroup$ 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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