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Copyright 1996 Lawrence C. Marsh Arithmetic Lag Structure y t
α +
β 0 x t +
β 1
x t-1 +
β 2
x t-2
+ . . . + β n x t-n + e t y t =
α
+ (n+1)
γ x t + n γ x t-1
+ (n-1)
γ x t-2 + . . . + γ x t-n + e
t Step 1: impose the restriction: β #
γ Step 2: factor out the unknown coefficient, γ .
t =
α
+ γ [ (n+1) x t + n x t-1 +
(n-1) x t-2 + . . . + x t-n
] + e t 15.6 Copyright 1996 Lawrence C. Marsh Arithmetic Lag Structure Step 3: Define z t . y t = α
+ γ [ (n+1) x t + n x t-1 +
(n-1) x t-2 + . . . + x t-n
] + e t z t =
[ (n+1) x t + n x t-1 + (n-1)
x t-2
+ . . . + x t-n
] Step 5: Run least squares regression on: y t
α
+ γ z t + e
t Step 4: Decide number of lags, n. For n = 4: z t = [ 5 x t + 4 x t-1 + 3 x t-2 +
2 x t-3 + x t-4
] 15.7
64 Copyright 1996 Lawrence C. Marsh Arithmetic Lag Structure i β
β 0 = (n+1) γ β 1 = n γ β 2 = (n-1)
γ β n = γ
. . 0 1 2 . . . . . n n+1
. . . . lag structure 15.8
Polynomial Lag Structure proposed by Shirley Almon (1965) the lag weights fit a polynomial where i = 1, . . . , n p = 2 and n = 4 For example, a quadratic polynomial: β 0 = γ 0 β 1 = γ 0
+ γ 1 + γ 2 β 2 = γ 0
+ 2 γ 1 + 4 γ 2 β 3 = γ 0
+ 3 γ 1 + 9 γ 2 β 4 = γ 0
+ 4 γ 1 + 16 γ 2 n = the length of the lag where i = 1, . . . , n β i = γ 0 + γ 1 i + γ 2 i
+...+ γ p i
2 p β i =
γ 0 + γ 1 i + γ 2 i 2 15.9
Copyright 1996 Lawrence C. Marsh Polynomial Lag Structure y t
α +
β 0 x t +
β 1
x t-1 +
β 2
x t-2
+
β 3
x t-3 + β 4
x t-4 + e
t y t =
α + γ 0
x t +
(γ 0
+ γ 1 + γ 2 ) x t-1 + ( γ 0 + 2
γ 1
+ 4 γ 2 ) x t-2 + ( γ 0 + 3
γ 1
+ 9 γ 2 ) x t-3 + ( γ 0 + 4
γ 1
+ 16 γ 2 ) x t-4 + e t Step 2: factor out the unknown coefficients: γ 0 , γ 1 , γ 2 . y t = α
+ γ 0 [x t + x t-1
+ x t-2
+ x t-3
+ x t-4
] +
γ 1
[x t + x t-1 +
2 x t-2 + 3 x t-3 + 4 x t-4
] +
γ 2
[x t + x t-1 +
4 x t-2 + 9 x t-3 + 16 x t-4
] + e t Step 1: impose the restriction: β i = γ 0 + γ 1 i + γ 2 i
15.10
Polynomial Lag Structure Step 3: Define z t0
, z t1 and z t2 for
γ 0 , γ 1 , and γ 2 . y t = α
+ γ 0 [x t + x t-1
+ x t-2
+ x t-3
+ x t-4
] +
γ 1
[x t + x t-1 +
2 x t-2 + 3 x t-3 + 4 x t-4
] +
γ 2
[x t + x t-1 +
4 x t-2 + 9 x t-3 + 16 x t-4
] + e t z t0 =
[x t + x t-1
+ x t-2
+ x t-3
+ x t-4
] z t1 =
[x t + x
t-1 +
2 x t-2 + 3 x t-3 + 4 x t- 4
]
z t2 =
[x t + x t-1
+ 4 x t-2 +
9 x t-3 + 16 x t- 4 ]
15.11 Copyright 1996 Lawrence C. Marsh Polynomial Lag Structure Step 4: Regress y t on z t0 , z
t1 and z
t2 . y t =
α
+ γ 0 z t0 + γ 1 z t1 + γ 2 z t2 + e t Step 5: Express β i ‘s in terms of γ 0 , γ 1 , and γ 2 . ^ ^ ^
^ β 0 = γ 0 β 1 = γ 0
+ γ 1 + γ 2 β 2 = γ 0
+ 2 γ 1 + 4 γ 2 β 3 = γ 0
+ 3 γ 1 + 9 γ 2 β 4 = γ 0
+ 4 γ 1 + 16 γ 2 ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 15.12 Copyright 1996 Lawrence C. Marsh Polynomial Lag Structure . . . . . 0 1 2 3 4 i β
Figure 15.3 β 0 β 1 β 2 β 3 β 4 15.13 65 Copyright 1996 Lawrence C. Marsh Geometric Lag Structure y t
α +
β 0 x t +
β 1 x t-1 +
β 2 x t-2 + . . . + e t infinite distributed lag model: y t = α +
Σ β i x t-i + e
t i =0
∞ (15.3.1)
geometric lag structure: β i = β φ
i
where |
φ | < 1
and
βφ i > 0 . 15.14 Copyright 1996 Lawrence C. Marsh Geometric Lag Structure y t
α +
β 0 x t +
β 1 x t-1 +
β 2 x t-2 +
β 3 x t-3 + . . . + e t y
= α + β( x t + φ x t-1 +
φ 2 x t-2 +
φ 3 x t-3 + . . .) + e t infinite unstructured lag: infinite geometric lag: Substitute β i
β φ i
β 0 = β β 1 = β φ
β 2 = β φ 2 β 3 =
β φ 3
. .. 15.15
Copyright 1996 Lawrence C. Marsh Geometric Lag Structure interim multiplier (3-period) : impact multiplier : long-run multiplier : y t = α + β( x t + φ x t-1 +
φ 2 x t-2 +
φ 3 x t-3 + . . .) + e t β
+ β
φ +
β
φ 2 β(1
+ φ + φ 2 + φ 3 + . . . ) =
β 1− φ
15.16 Copyright 1996 Lawrence C. Marsh Geometric Lag Structure β i
. . . . . 0 1 2 3 4 i β
= β φ
β 2 = β φ 2 β 3 =
β φ 3 β 4 =
β φ 4 β 0 =
β geometrically weights 15.17
Geometric Lag Structure Problem: How to estimate the infinite number of geometric lag coefficients ??? y t = α + β( x t + φ x t-1 +
φ 2 x t-2 +
φ 3 x t-3 + . . .) + e t Answer:
Use the Koyck transformation. 15.18
Copyright 1996 Lawrence C. Marsh The Koyck Transformation y t
α +
β( x t + φ x t-1 +
φ 2 x t-2 +
φ 3 x t-3 + . . .) + e t y
−
φ y t-1
= α(1− φ)
+ β x t + (e
t
− φ e t-1 ) Lag everything once, multiply by φ and subtract from original: φ y t-1 =
φ
α + β(φ
x t-1
+ φ 2 x t-2
+ φ 3 x t-3
+ . . .) + φ e t-1 15.19
66 Copyright 1996 Lawrence C. Marsh The Koyck Transformation y t
φ y t-1 =
α(1− φ) +
β x t + (e t
− φ
e t-1 ) y t =
α(1− φ) +
φ y t-1 + β x t + (e
t
− φ e t-1 ) Solve for y t
by adding φ y t-1
to both sides: y t
= δ 1 + δ 2 y t-1
+ δ 3 x t + ν t 15.20 Copyright 1996 Lawrence C. Marsh The Koyck Transformation y t
= α(1− φ)
+ φ y t-1 +
β x t + (e t
− φ
e t-1 ) y t =
δ 1 + δ 2 y t-1 +
δ 3 x t +
ν t Defining δ 1 = α(1− φ) ,
δ 2 = φ , and δ 3
β , use ordinary least squares: β =
δ 3
^ ^ φ = δ 2 ^ ^ α = δ 1 / (1− δ 2 ) ^ ^ ^ The original structural parameters can now be estimated in terms of these reduced form parameter estimates. 15.21
Geometric Lag Structure β 0
β β 1 = β φ
β 2 = β φ 2
β 3 = β φ 3
. . . ^ ^ ^ ^ ^
^ ^ ^
^ ^ ^
y t = α +
β( x t + φ x t-1 +
φ 2 x t-2 +
φ 3 x t-3 + . . .) + e t ^
^ ^ ^ ^ ^ y t =
α +
β 0 x t +
β 1 x t-1 +
β 2 x t-2 +
β 3 x t-3 + . . . + e t ^
^
^
^ ^ 15.22
Copyright 1996 Lawrence C. Marsh Durbin’s h-test for autocorrelation T − 1 1 − ( T − 1)[se(b 2 )] 2 h = 1 − d 2 h = Durbin’s h-test statistic d = Durbin-Watson test statistic se(b
2 ) = standard error of the estimate b 2 T = sample size Estimates inconsistent if geometric lag model is autocorrelated, but Durbin-Watson test is biased in favor of no autocorrelation. 15.23
y t = α + β x*
t + e
t
Adaptive Expectations y t = credit card debt x* t = expected (anticipated) income ( x* t is not observable) 15.24
Copyright 1996 Lawrence C. Marsh Adaptive Expectations x* t
t-1 =
λ (x
t-1 - x*
t-1 ) adjust expectations based on past realization: 15.25
67 Copyright 1996 Lawrence C. Marsh Adaptive Expectations x* t
t-1 =
λ (x
t-1 - x*
t-1 )
x* t = λ x t-1 + (1- λ ) x* t-1
rearrange to get: λ x t-1 = [x*
t - (1- λ ) x*
t-1 ] or 15.26 Copyright 1996 Lawrence C. Marsh Adaptive Expectations y t
α +
β x*
t + e
t
Lag this model once and multiply by ( 1 − λ ) : y t = α λ - ( 1 −
λ ) y t-1 + β [x* t - ( 1 −
λ ) x* t-1 ] + e t - ( 1 −
λ ) e t-1 subtract this from the original to get: ( 1
λ ) y t-1
= ( 1 −
λ ) α + ( 1 − λ ) β x*
t-1 +
( 1 − λ ) e t-1
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