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#1
01-11-2006, 07:22 PM
 Bühlmann Member SOA AAA Join Date: Nov 2005 Studying for nothing ever again Favorite beer: Corona Light Posts: 1,844
Sum of Phi's for AR(p)

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?
#2
01-11-2006, 07:50 PM
 NewTubaBoy Member CAS Join Date: Jul 2004 Location: Boston Studying for CAS 8 Favorite beer: Blue Moon Posts: 1,391

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?
#3
01-11-2006, 08:22 PM
 Peter Lemonjello Member Join Date: May 2005 Posts: 323

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.
#4
01-11-2006, 08:26 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,195

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.

Last edited by Abraham Weishaus; 01-12-2006 at 10:11 AM..
#5
01-11-2006, 09:27 PM
 Peter Lemonjello Member Join Date: May 2005 Posts: 323

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?
#6
01-11-2006, 09:29 PM
 Bühlmann Member SOA AAA Join Date: Nov 2005 Studying for nothing ever again Favorite beer: Corona Light Posts: 1,844

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

Last edited by Bühlmann; 01-11-2006 at 09:32 PM..
#7
01-12-2006, 10:13 AM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 6,195

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.
#8
01-12-2006, 10:45 AM
 Bühlmann Member SOA AAA Join Date: Nov 2005 Studying for nothing ever again Favorite beer: Corona Light Posts: 1,844

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)?
#9
01-12-2006, 10:52 AM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 4,984

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.
#10
01-12-2006, 11:33 AM
 NewTubaBoy Member CAS Join Date: Jul 2004 Location: Boston Studying for CAS 8 Favorite beer: Blue Moon Posts: 1,391

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?

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