Sum of Phi's for AR(p)

[Bühlmann]

The textbook says that it is necessary for the sum of phi's to be less than one for stationarity. However, I think one of the NEAS postings said the absolute value for the sum must be less than one.
Is the absolute value required?

[NewTubaBoy]

Hmmm... interesting. I'm not sure of the answer for this. Good catch. Most of the examples that we've seen had AR coefficeints as positive numbers <1 and usually only one term. My guess is that it would have to be the sum of the absolute values of the coefficients would have to be less than unity. Not that the sum must have absolute value less than one.

Anyone else have an idea?

[Peter Lemonjello]

Hmm.

According to the text, (page 535), a necessary condition for stationarity of an AR process is that the sum of phi's be less than 1. This is so that mu will be well defined.

The way I'm looking at it right now, mu would be well defined as long as the sum is not exactly 1.

So, I guess I don't fully understand this necessary but not sufficient condition.

[Abraham Weishaus]

NEAS is wrong. See textbook p. 546 equations A17.18-A17.20. An AR(2) system with phi_1=-1, phi_2=-0.75 is stationary (it satifies all 3 necessary and sufficient conditions) yet the absolute value of the sum is 1.75.

[Peter Lemonjello]

I'm not clear on this, yet. A couple of questions:

1. Dr. Weishaus - Did you mean "phi" above? The section of text you reference does deal with the invertibility of MA processes and is talking in terms of "thetas", but, they should be convertible to "phi"'s.

2. On page 535, it says "mu = delta / (1 - phi_1 - ... phi_p)
This gives a necessary condition for the stationarity of the process, that is, phi_1 + phi_2 + ... + phi_p < 1."

How does the necessary condition follow from the definition of mu? Suppose the phi's summed to 10, wouldn't mu still be well defined? If the sum were 1, it wouldn't, of course, but otherwise, it seems ok. What am I missing?

3. (tangentially related, at least) On page 545, P&R say, "If y_t is a stationary process, then phi^-1(B) must converge. This requires that the roots of the characteristic equation Phi(B) = 0 all be outside the unit circle."

Why?

[Bühlmann]

Thanks a lot guys! I apprecite the effort to look at this even though we are kind of confused.

[Abraham Weishaus]

I corrected the theta's to phi's. The conditions for invertibility of MA are identical to the conditions for invertibility of AR.

You cannot mechanically divide 1/(1-sum phi) and use it just because the denominator is not zero. An analogy is the sum of a geometric series.
sum x^n = 1/(1-x) for |x|<1, but does not converge for x>1, regardless of the fact that 1/(1-x) is defined.

[Bühlmann]

In Practice Problem Set #4, Question 4.9. They give the times series y(t) = -2.0y(t-1) + 6 + error term. They ask for the mean of the series. The solution says that the time series is not stationary and thus has no mean because the coefficient, -2.0, is greater in absolute value than one.
So I guess the conclusion then is that NEAS is correct based on A17.15 in the textbook for the analogous condition for AR(1)?

[Colymbosathon ecplecticos]

Think about what happens over time with your example. Eventually the process wanders around and becomes "big" (maybe that's 10 or 20 or so.) Once that happens the next term is about twice as big (in absolute value), the term after that twice as big again, and so on ... series diverges, no mean.

You have to wait until the process gets "big" for this argument to work because of the constant 6 that you add each time.

[NewTubaBoy]

Alright, so I understand this if all the AR coefficients are positive. So am I right in saying that in some cases if the AR coefficients are negative and "large" that it can be stationary such as in Dr. Weishaus' first example... but in other cases like the most recent example from NEAS material it isn't? How can we tell for sure which is the case?