8.4 Partial Autocorrelation Function
101
Fig. 8.6 The partial
autocorrelations φ j j versus j
of the CO 2 concentration
with trend removed and
differenced with lag 12 to
remove seasonality. The red
lines denote the 95%
confidence levels
component that escaped our earlier attempt to remove seasonality. Summarizing, our
best guess for the dynamics is
y i = φ 1 y i−1 + φ 12 y i−12 + ε i
(8.23)
and we will determine the parameters φ 1 and φ 12 in the next section.
8.5 Determining the Model Coefficients
Once we know that the underlying process is auto-regressive and what coefficients
are relevant we can set up a system of equations from which we then determine the
coefficients φ j . We simply write down copies of (8.23) for i = 13, 14, . . . up to the
last data point n and assemble them in matrix form
⎛
⎜
⎜
⎜
⎝
y 13
y 14
. . .
y n
⎞
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎝
y 12 y 1
y 13 y 2
. . .
. . .
y n−1 y n−12
⎞
⎟
⎟
⎟
⎠
φ 1
φ 12
,
(8.24)
which is of the same form as (7.2) in Chap. 7. We solve it in the least square sense
using the pseudo-inverse from (7.4)
φ 1
φ 12
=
A
t A
−1 A
t
⎛
⎜
⎜
⎜
⎝
y 13
y 14
. . .
y n
⎞
⎟
⎟
⎟
⎠
,
(8.25)
where A is the matrix with the two columns of shifted data points from (8.24).
101
Fig. 8.6 The partial
autocorrelations φ j j versus j
of the CO 2 concentration
with trend removed and
differenced with lag 12 to
remove seasonality. The red
lines denote the 95%
confidence levels
component that escaped our earlier attempt to remove seasonality. Summarizing, our
best guess for the dynamics is
y i = φ 1 y i−1 + φ 12 y i−12 + ε i
(8.23)
and we will determine the parameters φ 1 and φ 12 in the next section.
8.5 Determining the Model Coefficients
Once we know that the underlying process is auto-regressive and what coefficients
are relevant we can set up a system of equations from which we then determine the
coefficients φ j . We simply write down copies of (8.23) for i = 13, 14, . . . up to the
last data point n and assemble them in matrix form
⎛
⎜
⎜
⎜
⎝
y 13
y 14
. . .
y n
⎞
⎟
⎟
⎟
⎠
=
⎛
⎜
⎜
⎜
⎝
y 12 y 1
y 13 y 2
. . .
. . .
y n−1 y n−12
⎞
⎟
⎟
⎟
⎠
φ 1
φ 12
,
(8.24)
which is of the same form as (7.2) in Chap. 7. We solve it in the least square sense
using the pseudo-inverse from (7.4)
φ 1
φ 12
=
A
t A
−1 A
t
⎛
⎜
⎜
⎜
⎝
y 13
y 14
. . .
y n
⎞
⎟
⎟
⎟
⎠
,
(8.25)
where A is the matrix with the two columns of shifted data points from (8.24).
