ECON30402 Time Series Econometrics
Outline Solutions for 2013 Exam 1. yt = µ + εt + θ1 εt
(a)
1
−
+ θ2 εt
(1)
2
−
i. Mean E [yt ] = µ
Variance = E [yt − µ]2
var[yt ]
= E [εt + θ1 εt 1 + θ2 εt 2]2 = (1 + θ12 + θ22 )E [εt2 s ] as εt is white noise = (1 + θ12 + θ22 )σ2 −
−
−
Autocovariance cov[yt , yt
j]
−
= E [yt − µ][yt j − µ] = E [εt + θ1 εt 1 + θ2 εt −
=
−
(θ1 + θ1 θ2 )σ 2 θ2 σ
2
−
] [εt
j +
−
θ1 εt
j −1 +
−
θ2 εt
j −2 ]
−
j = 1 j = 2 j > 2
2
0
again using the white noise property of εt . ii. Autocorrela Autocorrelations: tions: cor[yt , yt
j]
−
cov[yt , yt j ] var[yt ] −
=
=
iii. See notes on stationarit stationarity y. iv. M A(∞) given by
θ1 +θ1 θ2 2 2 1+θ1 +θ2
j = 1
θ2 2 2 1+θ1 +θ2
j = 2
0
j > 2
∞
yt = µ +
θs εt
(2)
s
−
s=0
A. Absolute Absolute summability summability condition states
∞
|θs | < ∞
s=0
B. For θ1 = φ 1 + η1 , θs = φ 1 θs
1
−
yt yt
1
−
yt − φ 1 yt
, s > 1, we have
= µ + εt + θ1 εt 1 + φ1 θ1 εt 2 + φ21 θ1 εt 2 + ... = µ + εt 1 + θ1 εt 2 + φ1 θ1 εt 3 + ... = µ − φ1 µ + εt + ( θ1 − φ1 )εt 1 −
−
−
−
−
−
−
=
α + η1 εt
1
−
as required. C. Substituting Substituting θs = φ 1 θs
1
⇒ θs = φ s1
1
−
1
−
θ1 ,
s > 1
D. Therefore
∞
∞
|θs |
= 1+
s=0
φs1
1
−
θ1
s=1
∞
= 1 + |θ1 |
φs1
1
−
s=1
has a finite sum only if |φ1| < 1. Consequently, the stationarity condition for the ARMA(1, 1) is |φ1 | < 1 . (b) Seasonal model yt = α 0 + α1 D2t + α2 D3t + α3 D4t + φ4 yt
4
−
+ εt
(3)
i. The quarterly dummy variables imply a different intercept applies to each quarter of the year, specifically ( α0 + α 1 ), (α0 + α 2 ), (α0 + α 3 ), and α0 for quarters 1, 2, 3 and 4 respectively. The model allows for different E [yt |q ] over q = 1, 2, 3, 4. ii. Prediction. A. Applying the model for t in fourth quarter ⇒
yt+4
= = = =
yt+5
= α0 + α1 + φ4 yt+1
yt+1
yt+2 yt+3
α0 + α1 + φ4 yt
−
α0 + α2 + φ4 yt
−
α0 + α3 + φ4 yt
−
3 2 1
α0 + φ4 yt
B. By comparing these expressions with (3), it is clear that M SE (i) = E [yt+i − yt+i ]2 = E [ε2t+i ] = σ 2 ,
For a horizon of i = 5, M SE (5)
i = 1, 2, 3, 4
= E [yt+5 − yt+5]2 = E [εt+5 + φ4εt+1 ]2 = σ 2 (1 + φ24 )
C. As the horizon i → ∞, prediction for each quarter will approach the respective unconditional mean E [yt |q ] for q = 1, 2, 3, 4. iii. This is bookwork. Students should state that the model (3) is estimated and residuals e t retrieved. Then the auxiliary regression yt
= α0 + α1 D2t + α2D3t + α3 D4t + φ4 yt 4 +ψ1 et 1 + ψ2 et 2 + ψ3 et 3 + ψ4 et 4 + ut −
−
−
−
−
is estimated. [Students may equivalently use e t as the dependent variable here.] The null and alternative hypotheses are H 0 H A
: :
ψ1 = ψ 3 = ψ 3 = ψ 4 = 0
at least one ψj = 0, j = 1, ..., 4
and the test is implemented as a standard F -test (asymptotically valid). The model is designed to capture seasonality in quarterly data; 4 lags provides information about whether all the pattern, including seasonality, has been captured. iv. Application.
2
A. The graph indicates that the series DTCF is highly seasonal, with a persistent pattern where the fourth quarter shows the highest growth of sales, with a large drop in the first quarter and typically small positive values in the second and third quarters. This provides evidence that the means vary over the quarters; otherwise the series appears to be stationary, with the overall level approximately constant and no clearcut change in variance over time. The correlogram after removal of seasonal means continues to show evidence of seasonality, apparently of a seasonal AR form, with roughly geometric decline at lags 4, 8, 12, 16; these are generally significant according to the white noise confidence bands. Some serial correlation is also suggested at nonseasonal lags, especially lags 1 and 3. B. The estimated model reproduces the characteristics of the graph, with significant and negative seasonal dummy variable coefficients for quarters 1, 2 and 3, compared to the base of quarter 4. In line with the correlogram, there and significant stochastic seasonality with φ = 0.68, positive as anticipated. The residual serial correlation test shows no significant autocorrelation at conventional levels of signficance ( p-value 0.11). Despite this, the model cannot capture nonseasonal dynamics, which the correlogram of part i indicated may be present. Therefore, although the serial correlation test is satisfactory, it may be worth investigating whether the addition of nonseasonal AR or M A term(s) improve the model.
(c) Bivariate V AR(1) yt = α + Φ1 yt
1
−
+ εt
(4)
i. V ARMA representation A. Write the system as [I2 − Φ1 L]yt = α + εt where I2 − Φ1 L =
1 − φ11 L −φ12 L −φ21 L 1 − φ22 L
Assuming stationarity, Φ(L) = I − Φ1 L is nonsingular, and [Φ(L)]
1
−
1 = |Φ(L)|
1 − φ22 L φ21 L
φ12 L 1 − φ11 L
.
Multiplying through by the determinant | Φ(L)|, this becomes |Φ(L)| yt
=
1 − φ22 φ21
φ12 1 − φ11
α1 α2
+
1 − φ22 L φ21 L
|Φ(L)| = (1 − φ11 L)(1 − φ22 L) − φ12 φ21 L2
φ12 L 1 − φ11 L
ε1t ε2t
which is the V ARMA representation. B. Univariate ARMA process φ(L)yt = α + θ (L)εt is stationary if all roots of the characteristic equation corresponding to φ(L) are less than one in absolute value. In the V ARMA representation all elements of y t have the same AR polynomial |Φ(L)| , and the V AR is stationary if all characeristic roots correponding to |Φ(L)| are less than one in absolute value. ii. Specific 1.0 −0.2 . Φ1 = 0.3 0.5 A. Then |Φ(L)| = = = 3
1−L 0.2L −0.3L 1 − 0.5L
(1 − L)(1 − 0.5L) + 0 .06L2 1 − 1.5L + 0.56L2
with characteristic equation λ2 − 1.5λ + 0.56λ2 = (λ − 0.8)(λ − 0.7)
with roots λ1 = 0.8, λ2 = 0.7. The V AR is stationary as both | λi | < 1, i = 1, 2. B. The required impulse response functions are given by the V M A coefficient matrices, Θs , for s = 0, 1, 2. We need to compute = Φ21 1.0 −0.2 = 0 .3 0. 5
Θ2
=
1.0 −0.2 0 .3 0. 5
0.94 −0.30 0.45 0.19
The impulse responses for the effects of a unit shock to y 1t are, for s = 0, 1, 2: Effect on y1,t+s Effect on y2,t+s
: :
1, 1, 0.94 0, 0.3, 0.45
The impulse responses for the effects of a unit shock to y 2t are, for s = 0, 1, 2: Effect on y1,t+s Effect on y2,t+s
: :
0, − 0.2, − 0.3 1, 0.5, 0.19
iii. See notes on Cholesky Method. iv. A. Repeated substitution yields yt
= (yt 2 + α + εt 1 ) + α + εt = yt 2 + 2 α + εt + εt 1 −
−
−
−
= (yt 3 + α + εt 2 ) + 2 α + εt + εt = yt 3 + 3 α + εt + εt 1 + εt 2 −
−
−
−
1
−
−
t
= y0 + αt +
εi
i=1
B. From part (a) we have E [yt ]
= y0 + αt = E [yt − E (yt )]2
var [yt ]
2
t
= E
εi
i=1 t
= E
ε2i
as E [εi εj ] = 0, i = j
i=1
= tσ cov[yt , yt
j]
−
2
= E [yt − E (yt )][yt
j −
−
j )]
−
t−j
t
= E
E (yt
εi
εs
s=1
i=1
= E [εt + εt 1 + εt 2 + ... + ε1 ] × [εt j + εt j 1 + ... + ε1 ] since ε t is white noise = (t − j )σ 2 −
−
4
−
−
−
C. From (b), cor[yt , yt
j]
−
=
cov[yt , yt j ] var[yt ] × var[yt
=
−
j]
−
(t − j )σ 2
tσ 2 × (t − j )σ2 t − j
=
=
t − j t
t(t − j ) 1 as t → ∞ for all finite j > 0
→
D. E. Cointegration exists when there a linear combination z1t = y 1t − β 2 y2t − β 3 y3t ∼ I (0)
F. For three series, there could be 2 distinct linear combinations which are I (0), such as = y1t − β 2 y2t = y1t − β 3 y3t
z1t z2t
G. For three variables that are each I (1), the Engle-Granger test examines whether there exists a linear combination of these which is stationary. The intuition is that if such a linear combination exists, then the residuals from a regression such as y3t = α 0 + α1 t + β 1 y1t + β 2 y2t + u3t
should be stationary. A trend is included in the cointegrating regression, as in the EViews output, to take account of linear trends in the individual series. Denoting the residuals as e t , an ADF test is applied to the residuals from this regression as p
∆et = γ et
1
−
+
ψj ∆et
j +
−
vt
j =1
and the hypotheses tested are H 0 H A
: :
γ = 0 γ < 0
The augmentation order p should be sufficient to account for serial correlation in the residuals; no deterministic terms are required, since these are taken account of through the initial regression. For the particular case, the ADF test statistic is -3.529. Using the McKinnon critical values for k = 3 regressors and with a trend in the original equation, this statistic is not significant at even the 10% level. Therefore, the null hypothesis is not rejected and it is concluded that that the evidence does not support cointegration between these series.
5
