DYNAMIC ECONOMETRIC MODELS: AUTOREGRESSIVE AND DISTRIBUTED-LAG MODELS Reference : Gujarati, Chapter 17 A distributed-lag model — the regression model includes the lagged values of the explanatory variables, for example, Y t = β 0 + β + β 1X t + β + β 2X t−1 + β + β 3X t−2 + u + ut A dynamic model — the regression model includes one or more lagged values of the dependent variable among its explanatory variables, for example, Y t = β 0 + β + β 1X t + β + β 2Y t−1 + β + β 3Y t−2 + u + ut
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The Role of Lag For the following distributed-lag model Y t = α+β 0X t+β 1X t−1+β 2X t−2+· · ·+β k X t−k +ut (1) The coefficient β 0 is the short-run, or impact,
multiplier. These coefficients and their partial sums are called interim, or intermediate, multiplier. k
i=0
β i = β 0 + β 1 + β 2 + · · · + β k = β
is the long-run, or total, distributed-lag
multiplier. We define
β i = k = β i=0 β i be “standardized’ β i which give the proportion of the β i∗
β i
long-run, or total, impact felt by a time period i.
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Example: The consumption function. Suppose a person’s consumption function is Y t = α + 0.4X t + 0.3X t−1 + 0.2X t−2 + ut where Y t is consumption expenditure and X is annual income. If he receives a salary increase of $2000 in annual pay, he will increase consumption expenditure by $800 in the first year, by another $600 in the next year, and by another $400 in the following year. By the end of the third year, his consumption expenditure will be increased by $1800. The short-run multiplier is 0.4 and the long-run multiplier is 0.9. If we divide each β i by 0.9, we obtain respectively, 0.44, 0.33 and 0.23, which indicate that 44 percent of the total impact of a unit change in X on Y is felt immediately, 77 percent after one year, and 100 percent by the end of the second year. 6-3
Estimation of Distributed-lag Model Equation (1) is called a finite (lag)
distributed-lag model while the following model is called an infinite (lag) model: Y t = α + β 0X t + β 1X t−1 + β 2X t−2 + · · · + ut . Two approaches to estimate the parameters: • ad hoc estimation, and • a priori restriction on the β ’s.
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(2)
Ad Hoc Approach First, use OLS to regress Y t on X t, then regress Y t on X t and X t−1, then regress Y t on X t, X t−1 and X t−2 and so on. The precedure stops when β becomes statistically insignificant. Consider the following example with SAS program: DATA a ; infile ’c:\ec5103\G17_23.dat’ firstobs = 2 ; INPUT Year Y X ; X1 = lag(X) ; X2 = lag(X1) ; X3 = lag(X2) ; X4 = lag(X3) ; label Y = ’Expenditure’ X = ’Sale’ ; proc reg data=a; model Y = X ; model Y = X X1 ; model Y = X X1 X2 ; model Y = X X1 X2 X3 ; run ; 6-5
Model: MODEL1 Dependent Variable: Y Analysis of Variance
Expenditure
Sum of Mean Source DF Squares Square F Value Prob>F Model 1 40493.24354 40493.24354 857.570 0.0001 Error 17 802.71596 47.21859 C Total 18 41295.95949 Root MSE 6.87158 R-square 0.9806 Dep Mean 117.08053 Adj R-sq 0.9794 C.V. 5.86910 Parameter Estimates Parameter Standard T for H0: Variable DF Estimate Error Parameter=0 Prob > |T| INTERCEP 1 -26.001181 5.13397793 -5.065 0.0001 X 1 0.876207 0.02992072 29.284 0.0001
Variable INTERCEP X
DF 1 1
Variable Label Intercept Sale
Model: MODEL2 Dependent Variable: Y Root MSE 6.50264 Dep Mean 117.08053
Variable INTERCEP X X1
DF 1 1 1
Parameter Estimate -20.932059 0.474206 0.392746
Expenditure R-square Adj R-sq Standard Error 5.67585062 0.23444150 0.22736735
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0.9836 0.9816 T for H0: Parameter=0 -3.688 2.023 1.727
Prob > |T| 0.0020 0.0601 0.1034
Model: MODEL3 Dependent Variable: Y
Expenditure
Analysis of Variance Sum of Mean Source DF Squares Square F Value Prob>F Model 3 40842.46575 13614.15525 450.309 0.0001 Error 15 453.49375 30.23292 C Total 18 41295.95949 Root MSE 5.49845 R-square 0.9890 Dep Mean 117.08053 Adj R-sq 0.9868 Parameter Estimates Parameter Standard T for H0: Variable DF Estimate Error Parameter=0 Prob > |T| INTERCEP 1 -26.300379 5.19035250 -5.067 0.0001 X 1 0.460543 0.19830090 2.322 0.0347 X1 1 0.984994 0.29069565 3.388 0.0041 X2 1 -0.579260 0.21325893 -2.716 0.0159 Model: MODEL4 Dependent Variable: Y Root MSE 5.62999 Dep Mean 117.08053 Parameter Estimates Parameter Variable DF Estimate INTERCEP 1 -27.799462 X 1 0.501062 X1 1 0.944238 X2 1 -0.452785 X3 1 -0.126689
Expenditure R-square Adj R-sq Standard Error 5.96304344 0.21580167 0.30659655 0.31581698 0.22855144
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0.9893 0.9862 T for H0: Parameter=0 -4.662 2.322 3.080 -1.434 -0.554
Prob > |T| 0.0004 0.0358 0.0082 0.1736 0.5881
The Koyck Approach Koyck assumes β follows the relation: β k = β 0 λk .
(3)
with λ, such that 0 < λ < 1, which is known as the rate of decline, or decay, of the distributed lag model
and 1 − λ is known as the speed of adjustment. From (3), we have 1 i=0 1−λ and Equation (2) can be re-written as ∞ β i = β 0
Y t = α + β 0X t + β 0λX t−1 + β 0λ2X t−2 + · · · + ut . (4) Consider time t − 1, we have Y t−1 = α + β 0X t−1 + β 0λX t−2 + β 0λ2X t−3 + · · · + ut−1 . (5) Equation (4) - λ × Equation (5), we have Y t = α(1 − λ) + β 0X t + λY t−1 + ν t where ν t = ut − λut−1. 6-8
(6)
Note 1. In (2), the error term ut is iid N (0, σ 2). However in (6), the error term ν t is correlated with ν t−1. 2. We can use Durbin-Watson test or Durbin h test to test the correlation in ν t.
The Median Lag is the time required for 50% of the total change in Y following a unit sustained change in X . For Koyck model, log 2 Median lag = − log λ λ = 0.2 =⇒
Median lag = 0.4306 and hence it takes
less than half a period to obtain 50% of the total change in Y . λ = 0.8 =⇒
Median lag = 3.1067 and hence it takes
more than three periods to obtain 50% of the total change in Y .
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The Mean Lag is the weighted average of time such that Mean lag =
∞
i=0 kβ k ∞ i=0 β k
if all the β k is positive. For Koyck model, λ Mean lag = 1−λ The Median Lag and the Mean Lag measure the speed with which Y responds to X .
Example : Consider PPCEt = −841+0.7117 PDPIt + 0.2954 PPCEt−1 + et Assume it is the Koyck model, λ = 0.954, the Median lag = 0.5684, and the Mean lag = 0.4192. Hence, PPCE adjust to PDPI within a relatively short time.
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The Adaptive Expectation Model Suppose Y t = β 0 + β 1X t∗ + ut
(7)
where Y
= demand for money
X ∗ = equilibrium, optimum, expected long-run or normal rate of interest As X t∗ is not observable, we propose: ∗ ∗ X t∗ − X t− γ X − X = ( t t−1 ) 1
(8)
where γ such that 0 < γ < 1, is the coefficient of
expectation and X is observable. From (8), we have ∗ X t∗ = γX t + (1 − γ )X t− 1
(9)
Substituting (9) into (7), we have ∗ Y t = β 0 + β 1 [ γX t + (1 − γ )X t− ] + ut 1 ∗ = β 0 + β 1γX t + β 1(1 − γ )X t− + ut (10) 1 6-11
lag (7) one period, multiply it by 1 − γ , and subtract (9), we have Y t = γβ 0 + γβ 1X t + (1 − γ )Y t−1 + ut − (1 − γ )ut−1 = γβ 0 + γβ 1X t + (1 − γ )Y t−1 + ν t
Example (Continued) : PPCEt = −841+0.7117 PDPIt + 0.2954 PPCEt−1 + et The coefficient of expectations, γ = 1 − 0.2954 = 0.7046. About 70% of the discrepancy between actual and expected PDPI is eliminated within a year, a fairly rapid adjustment.
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The Stock Adjustment Model Another reationalization is the Stock Adjustment or Partial Adjustment Model Y t∗ = β 0 + β 1X t + ut
Y t − Y t−1 = I t = δ ( Y t∗ − Y t−1 )
(11) (12)
where Y
= actual capital stock (observable)
Y ∗
= desired level of capital (unobservable)
I t
= investment in time period t
Y t − Y t−1
= actual change in capital stock
Y t∗ − Y t−1 = desired change in capital stock δ
= the coefficient of adjustment (0 < δ < 1)
(12) can be written as Y t = δY t∗ + (1 − δ )Y t−1
From (11) and (13), we have Y t = δ ( β 0 + β 1X t + ut ) + (1 − δ )Y t−1 6-13
(13)
= δβ 0 + δβ 1X t + (1 − δ )Y t−1 + δut (14) (11) is the long-run, or equilibrium, demand for capital stock, and (14) is the short-run demand for capital stock. Once we estimate (14), we can obtain (11).
Example (Continued) : PPCEt = −841+0.7117 PDPIt + 0.2954 PPCEt−1 + et The coefficient of adjustment, δ = 1 − 0.2954 = 0.7046.
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Combination of Adoptive Expectations and Partial Adjustment Model Consider Y t∗ = β 0 + β 1X t∗ + ut
(15)
where Y ∗
= desired level of capital (unobservable)
X ∗
= expected level of output (unobservable)
From (15), (9) and (13), we have Y t = δγβ 0 + δγβ 1X t + [(1 − γ ) + (1 − δ )]Y t−1 −(1 − γ )(1 − δ )Y t−2 + [δut − δ (1 − γ )ut−1] = α0 + α1X t + α2Y t−1 + a3Y t−2 + ν t where ν t = δ [ut − (1 − γ )ut−1].
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(16)
Durbin-h Test for Autocorrelation If the random error terms may follow a first-order autoregressive process such that εt = ρεt−1 + ut where ρ is a parameter such that |ρ| < 1 ut are independent N (0, σ 2). To test: H 0 : ρ = 0 against H 1 : ρ =0 We can use the Durbin-h statistics n h = ρˆ 1 − n[V ar(α ˆ 2)] n 1 ≈ 1 − d 2 1 − n[V ar(α ˆ 2)] where α ˆ 2 is the coefficient of Y t−1 and d is the Durbin-Watson statistic. 6-16
h ∼ AN (0, 1) (h is asymptotically normally distributed with zer mean and unit variance). If |h| < z α/2, conclude H 0 If h > z α/2, conclude H 1 and there is positive first-order autocorrelation, and If h < −z α/2, conclude H 1 and there is negative first-order autocorrelation. For the hypothesis: H 1 : ρ > 0, if h > z α, conclude H 1 and there is positive first-order autocorrelation; Otherwise, conclude H 0 and there is no first-order autocorrelation. For the hypothesis: H 1 : ρ < 0, if h < −z α, conclude H 1 and there is negative first-order autocorrelation; Otherwise, conclude H 0 and there is no first-order autocorrelation.
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Example : Suppose n = 100, d = 1.9 and V ar(α ˆ 2) = 0.005, then 1 h ≈ 1 − × 1.9
100 1 − 100 × 0.005
2 = 0.7071
For the hypothesis: H 1 : ρ = 0, as |h| < 1.96 = z .05/2, we conclude H 0. For the hypothesis: H 1 : ρ > 0, as h < 1.6456 = z .05, we conclude H 0.
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