On Sat, 31 Mar 2012, Daniel Bencik wrote:
I tried to google on the internet what the inverted roots problem is, as you
recently told me about the ARMA-GARCH problem I had.
Following up our discussion, the "best" model for my problem (under given
circumstances) is AR(7)-GARCH(1,1)-t but the common inverse roots show up
again. Can you please tell me what this means?
The Eviews estimation result is http://eubie.sweb.cz/gretl_forum/ar7.PNG
There's no such a thing as an "inverted roots" problem. The problem I
mentioned in my earlier message was the common roots problem, which may
arise when you have both an AR part and an MA part. A classic reference is
Mcleod (1999), "Necessary and Sufficient Condition for Nonsingular Fisher
Information Matrix in ARMA and Fractional ARIMA Models", The American
Statistician, although the problem was known long before that.
In short: suppose you have a difference equation
A(L) y_t = C(L) x_t
(of which an arma model is a special case, with x_t a white noise
process), where the order of A() is p and the order of C() is q. Of course
the above equality still holds if you apply the polynomial (1-bL) on both
sides of the equation, whatever the value of b.
This means that, for each value of b, you have an equivalent
P(L) y_t = Q(L) x_t
where P(L) = A(L) (1-bL) and Q(L) (1-bL). In turn, this means that for
every ARMA(p,q) process there exists an uncountable infinity of equivalent
ARMA(p+1, q+1) representations. These representations are called
If you read the argument backwards, when you apply an ARMA(p,q) model to a
data series, you may have common factors between the A() and C()
polynomials: if so, then your representation is just one of the possible
equivalent redundant representations and you'll be much better of with an
ARMA(p-1, q-1) model.
Of course, all this does not apply, for obvious reasons, to pure AR or MA
models (like in your case).
Hope this helps,
Riccardo (Jack) Lucchetti
Dipartimento di Economia
Università Politecnica delle Marche
(formerly known as Università di Ancona)