MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am studying GARCH processes in Time Series Analysis by Hamilton.

Something that has regularly been used in the book is the assumption that an infinite-order polynomial can be written as the ratio of two finite-order polynomials.

$\pi(L) = \sum_{j=1}^{\infty} \pi_j L^j$

And then

$\pi(L) = \frac{\alpha(L)}{1-\delta(L)} = \frac{\alpha_1 L^1 + \alpha_1 L^2 + ... + \alpha_m L^m}{1 - \delta_1 L^1 - \delta_1 L^2 - ... - \delta_r L^r}$

Followed by "assuming the roots of $1-\delta(L) = 0$ are outside the unit circle".

What is the reasoning behind this transformation? And is the assumption about the roots outside the unit circle required for the infinite order polynomial to have this ratio representation?

share|cite|improve this question
@Lee Mosher: the question could certainly be more clear and I do not claim it is right for this site, but your objection strikes me as strange. There is a clear context mentioned and AFAIK the coefficients in that context will be in linear recurrecne, so it will be a rational. – user9072 Apr 18 '13 at 13:59
@Nathan Wilson: Likely the best thing to do for you is ask this on another site (for example the mathematics or the statistics site on the stackexchange network, see FAQs for details). However, here are two remarks: pi(L) is the fraction of two polynomials since (or let us rather say if) its coefficients fulfil a linear recurrence relation that is you can compute a coefficient by a fixed linear function of a fixed number of preceeding ones. And this should be the case in your context. For the question regarding the roots: you do not need this to express it as a fraction... – user9072 Apr 18 '13 at 14:01
Thank you @quid. – Nathan Wilson Apr 18 '13 at 14:06
...this would be possible in a completely "formal" way (if it is possible at all). However, if one wishes to treat these as expressions as "functions of L" it is (or might be) important that the denominator is never 0 for the L one wants to plug into the expression, relatedly that the series converges. And there might be a condition there that confines the L to the unit circle, so that if it only vanishes outside the unit circle "everything is fine." Yet again, you might get a better response on other sites on the stackexchange network. – user9072 Apr 18 '13 at 14:09
You are welcome! Yet treat the information I gave with some critical distance; this is a bit of guess work on my part. – user9072 Apr 18 '13 at 14:10

I am not sure how much background you have in analysis but it may be helpful to keep in mind that "the assumption about the roots outside the unit circle" is required not "for the infinite order polynomial to have this ratio representation", which seems to be a separate assumption, but rather for the series to converge for all points in the unit disk.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.