| Library: | times |
| See also: | acf pacf acfplot pacfplot |
| Quantlet: | eacf | |
| Description: | The EACF allows for the identification of ARIMA models (differencing is not necessary). The quantlet generates a table of the extended (sample) autocorrelation function (EACF) for the time series y. You have to specify the maximal number of AR lags (p) and MA lags (q). Every row of the output table gives the ACF up to q lags for the residuals of an AR regression with k<=p lags. Furthermore, the simplified EACF is also tabulated. If an autocorrelation is significant (see the note), the entry is 1. Otherwise the entry is 0. For identification, look for the vertex of a triangle of zeros. You can immediately read the order of y from the table. |
| Usage: | eacf(y,p,q) | |
| Input: | ||
| y | T x 1 vector of observed time series | |
| p | integer; the maximal lag length for which the AR regression is fitted | |
| q | integer; the maximal lag length of the ACF for every fitted AR(k) regression (k <= p) | |
library("times") ; loads the quantlets from times library
tiaoma = read("tiaoma") ; Tiao's example 3.1
eacf(tiaoma,6,12)
The Tables show the EACF and the simplified EACF for the generated MA(1) example
of Tiao (2001), p.68. The vertex of a triangle of zeros is at p=0 and q=1.
We conclude that the process is MA(1). The first Table is identical to the
one in Tiao, p.70. The Table for the simplified EACF is slightly different.
The reason is, that we use Bartlett's formula instead of the simple approximation
given in Tiao (2001), p.68.
==================================================================================
EACF
==================================================================================
q= 0 1 2 3 4 5 6 7 8 9 10 11 12
==================================================================================
p=0 -0.47 -0.05 +0.16 -0.17 +0.11 -0.04 -0.08 +0.08 +0.06 -0.23 +0.17 -0.04 -0.10
p=1 -0.52 -0.20 +0.08 -0.06 +0.08 -0.03 -0.10 +0.07 +0.03 -0.22 +0.13 -0.04 -0.12
p=2 -0.10 -0.34 -0.18 +0.01 -0.01 -0.09 -0.01 -0.02 -0.02 -0.21 -0.10 -0.10 +0.02
p=3 -0.20 -0.38 -0.20 -0.03 +0.00 -0.07 -0.05 -0.02 -0.03 -0.18 +0.08 -0.12 +0.04
p=4 +0.00 -0.18 -0.14 +0.07 -0.08 -0.06 -0.01 +0.02 -0.01 -0.15 -0.09 -0.15 +0.04
p=5 +0.00 -0.01 -0.40 -0.16 +0.09 -0.05 -0.01 +0.02 -0.02 -0.14 +0.06 -0.01 -0.06
p=6 -0.24 +0.08 -0.15 -0.27 +0.04 +0.04 +0.15 +0.00 -0.05 -0.13 +0.06 -0.02 -0.04
==================================================================================
q= 0 1 2 3 4 5 6 7 8 9 10 11 12
==================================================================================
p=0 1 0 0 0 0 0 0 0 0 0 0 0 0
p=1 1 0 0 0 0 0 0 0 0 0 0 0 0
p=2 0 1 0 0 0 0 0 0 0 0 0 0 0
p=3 1 1 0 0 0 0 0 0 0 0 0 0 0
p=4 0 0 0 0 0 0 0 0 0 0 0 0 0
p=5 0 0 1 0 0 0 0 0 0 0 0 0 0
p=6 1 0 0 1 0 0 0 0 0 0 0 0 0
==================================================================================
library("times") ; loads the quantlets from times library
dax = read("dax") ; monthly DAX 1979:1-2000:10
eacf(log(dax),6,12)
The Tables show the EACF and the simplified EACF for the logarithmized DAX. The
vertex of a triangle of zeros is at p=1 and q=0. We conclude that the differenced
process---the return of the DAX---is white noise.
==================================================================================
EACF
==================================================================================
q= 0 1 2 3 4 5 6 7 8 9 10 11 12
==================================================================================
p=0 +0.99 +0.97 +0.96 +0.94 +0.93 +0.91 +0.90 +0.88 +0.86 +0.85 +0.83 +0.82 +0.80
p=1 +0.03 +0.03 -0.03 +0.01 -0.07 -0.05 -0.02 -0.02 -0.04 +0.13 +0.04 +0.02 -0.04
p=2 -0.50 +0.03 -0.03 -0.04 -0.06 -0.09 +0.01 -0.05 -0.03 +0.13 -0.03 +0.02 -0.03
p=3 +0.13 -0.31 -0.02 -0.06 -0.08 -0.02 -0.08 -0.01 -0.02 +0.14 +0.06 +0.04 -0.03
p=4 +0.20 -0.26 +0.01 -0.04 -0.09 -0.11 -0.03 -0.05 -0.01 +0.13 -0.02 -0.05 +0.02
p=5 +0.47 +0.19 +0.22 +0.07 -0.07 -0.06 -0.03 -0.01 -0.03 +0.12 -0.03 -0.04 +0.00
p=6 -0.38 +0.33 -0.01 -0.06 -0.10 -0.10 -0.08 -0.07 -0.03 +0.12 +0.04 -0.05 -0.04
==================================================================================
q= 0 1 2 3 4 5 6 7 8 9 10 11 12
==================================================================================
p=0 1 1 1 1 1 1 1 1 1 1 1 1 1
p=1 0 0 0 0 0 0 0 0 0 1 0 0 0
p=2 1 0 0 0 0 0 0 0 0 0 0 0 0
p=3 1 1 0 0 0 0 0 0 0 0 0 0 0
p=4 1 1 0 0 0 0 0 0 0 0 0 0 0
p=5 1 1 1 0 0 0 0 0 0 0 0 0 0
p=6 1 1 0 0 0 0 0 0 0 0 0 0 0
==================================================================================