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I was working through my textbook and found this problem that I got stuck at: Consider the AR(2) Model $$X_t = \phi_1X_{t-1}+\phi_2X_{t-2}+\epsilon_t$$ We can assume $\phi_2 > 0$, so the roots of the polynomial are real numbers. Show that the inequalities for the values of $\phi_1$ and $\phi_2$ which ensure it is causal stationary are given by: $$\phi_1 + \phi_2 < 1$$ $$\phi_2 - \phi_1 < 1$$ $$|\phi_1|<1, \phi_2 < 1$$

So here's what I've done so far:

Firstly I rearranged the equation to include the backshift operator $$(1-\phi_1B-\phi_2B^2)X_t = \epsilon_t$$ Then I used the quadratic equation for the roots $$z=\frac{\phi_1±\sqrt{\phi_1^2+4\phi_2}}{-2\phi_2}$$

Except I'm not sure where to go from here. How do I calculate the inequalities that ensure it's causal stationary from here? How do I figure out the conditions?

Thanks heaps in advance!

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In this framework, for causal stationarity, you need $|z|>1$. This means, $$|\phi_1\pm\sqrt{\phi_1^2+4\phi_2}|>2|\phi_2|=2\phi_2$$ Consider $\phi_1>0$ case first. Then, for the first root, we have $\phi_1+\sqrt{\phi_1^2+4\phi_2}>2\phi_2$, which means $\phi_1^2+4\phi_2>4\phi_2^2+\phi_1^2-4\phi_1\phi_2 \rightarrow \phi_2-\phi_1<1$. For the other root, we have $-\phi_1+\sqrt{\phi_1^2+4\phi_2}>2\phi_2$, which yields $\phi_2+\phi_1<1$.

For $\phi_1<0$, you obtain the same inequalities; and these two can be summarized as $\phi_2+|\phi_1|<1$, which also means $|\phi_1|<1$ and $\phi_2<1$. Don't forget that $\phi_2$ is positive.

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