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?

[Bühlmann]

Good point on the intuitive feel for my example. Thanks Coly.

Tuba, I think I'm gonna assume that NEAS is going to give us small, straightforward ARMA's and so I'm gonna go with what I have in my head now. It is too close to the exam for me to spend too much time on it.

[NewTubaBoy]

Agreed... I'm going w/ you. Even for that one problem where the problem had a -2 for the AR coefficient... the only possible choice was E. Even if you used the normal formula delta/1-phi, the solution wasn't included I don't believe. They've said multiple places that the exam is not meant to trick you.

[Peter Lemonjello]

For the AR(1) example Buhlmann's k gave, we know that for an AR(1) process, phi = row sub 1. (See page 529, 17.29). We also know that row sub k, for k > 0, is strictly between -1 and 1 for a stationary process. (Last sentence of 519 text).

So, even though the cited example meets the necessary condition that the sum of phi's are less than 1, it doesn't meet another condition.

I am, however, still stumped on points 2 and 3 of my earlier post.

[NewTubaBoy]

Peter thanks. I now understand why Buhlman's earlier post is not stationary. I rembered the book saying that the sum had to be <1 was a necessary but not sufficient condition. I didn't realize that the other condition was that rho had to be between +/-1. Thanks for the post.

[Bühlmann]

what makes this especially confusing is that the neas postings seem to interchange stationarity and "absolute values of coefficients" (I can site several examples) being less than one.
It would be helpful if there could be a posting addressing this issue.

[NewTubaBoy]

The way I'm going about studying for this exam:

I'm allowed to get 15 questions wrong wrong wrong.

[Peter Lemonjello]

The way I'm going about studying for this exam:

I'm allowed to get 15 questions wrong wrong wrong.

I can't affort to think that way. There are only 15 questions on the exam I'll be taking (CAS Transitional).

[appleapple]

The way I'm going about studying for this exam:

I'm allowed to get 15 questions wrong wrong wrong.

How many questions are there? I was thinking 20.

[NewTubaBoy]

50 multiple choice questions in 2 hours for NEAS. So they've kinda gotta be easy becuase that's not much time per questino.

[Bühlmann]

Tuba, good luck tomorrow buddy.

[NewTubaBoy]

Hey, thanks. You too. We'll kill it.