Errata (and commentary): An Introduction to Applied Econometrics: A Time Series Approach by Kerry Patterson.

1. Page 225 Section 6.2.8 Difference stationary and trend stationary series

Line 5 of this section should replace “stationary” by “nonstationary”.

2. Page 254 Section 7.2 ARIMA models

There is a typo that has been carried through equations (7.3) and (7.4). The essence of the error is that wherever the coefficient , j = 1, ..., , appears in the expression for

(1 + 1L)-1 it should be replaced by +; and a superscript has been omitted in the equation following equation (7.4). The error is apparent since the expression in the text clearly does not work!

The correct text starting with the sentence before equation (7.2) should read as follows.

“For example, a simple MA(1) model is

Yt =  + (1 + 1L)t(7.2)

This has an alternative representation as an AR() model provided 1 1, in which case the MA model is said to be invertible. Multiplying through by the inverse of (1 + 1L)-1 = 1 + (–1)L + + + ... + +, ..., gives

[1 + (–1)L + + - ... + ]Yt= * + t(7.3)

so that

Yt = * – [+ + ...+ + ...]Yt + t(7.4)

= * – + t

where * = (1 + 1L)-1. The coefficients in the infinite autoregression, , decline monotonically in absolute value given 1 1 and, therefore, an AR(p) model, with p finite so the infinite lag in the autoregression is truncated, may be a reasonable approximation to the original MA model. More generally, the inversion of an ARMA(p, q) model to an ARMA(, 0) requires that the roots of (L) must have modulus greater than 1. In the simple case considered here there is just one root, equal to 1/1.”

Commentary

The expression for the inverse of the MA polynomial, that is (1 + 1L)-1, is motivated by analogy with the more regularly occurring case infinite AR case. That is we have:

for ||  1E(1)

If we want to work backwards and find the infinite expression corresponding to

1/(1 + 1L), we must make the translation in coefficients, that is  = –1. Hence, we then have:

E(2)

= 1/(1 + 1L)E(3)

Two numerical examples will illustrate E(3); recall that we set L = 1 to obtain the sum of the coefficients. If 1 = –0.9, then 1/(1 + 1) = 1/(1 – 0.9) = 10; the first few terms in the infinite expression are 1, 0.9, 0.81, 0.729, 0.656. If 1 = 0.9 then 1/(1 + 1) = 1/(1 + 0.9) = 0.526; and the first few terms in the infinite expression are 1, –0.9, 0.81, –0.729, 0.656. In both cases, although converging to very different values, the sum is slow to converge and the AR approximation would need to take quite a few more terms to be a good approximation to the MA(1) model. Even with 30 terms in the approximation, the sum of the coefficients in the AR representation is only 95% of the total.

Applying E(3) to equation (7.2) in the text, we then have the infinite AR model given by:

E(4)

Taking the lags over to the right-hand side we obtain the model that can be truncated to an AR(r) model, with r finite, for estimation. That is:

E(5)

Pages 765 and 771 of References

The reference to Godfrey, L. G. (1999) should read:

“Godfrey, L. G. (1999), Instrument Relevance in Multivariate Linear Models, Review of Economics and Statistics, Vol. 81, pp. 550-552.”

The reference to Shea, J (1997) should read:

“Godfrey, L. G. (1999), Instrument Relevance in Multivariate Linear Models: A Simple Meausre, Review of Economics and Statistics, Vol. 79, pp. 348-352.”