CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE Chapter 7 Functional Form and Structural Change 7.1 Using binary variable Example 1 Earnings equation for married women In earnings=B,+ B2age+ B3age+ Education Skids +a =0. no kids =1. kids Variable Coefficient se Age0.200600.08862.39 -0.00231470.00098688-2.345 Education0.0674720.0252482.672 Kid 351190.14753-2380 The earnings of women with children are 35% less than those without Model with one dummy variable Here di takes value either l or 0 depending on the category. We may use several dummy variables Example 2(Seasonal effects) quarterly de yt=B,+BXL+5,DIt+&2D2t +53 D3t+Et Here x does not contain 1 spreng 0. othe if if fa 0, di denote seasonal effects Alternatively, we may use BXt+61D1t+62D2+63D3x+64D
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 1 Chapter 7 Functional Form and Structural Change 7.1 Using binary variables Example 1 Earnings equation for married women ln earnings = β1 + β2age + β3age2 + β4 education + β5kids + ε kids = = 0, no kids = 1, kids Variable Coefficient s.e. t Age 0.20056 0.08386 2.392 Age2 —0.0023147 0.00098688 —2.345 Education 0.067472 0.025248 2.672 Kids —0.35119 0.14753 —2.380 The earnings of women with children are 35% less than those without. Model with one dummy variable yi = X ′ iβ + δdi + εi Here di takes value either 1 or 0 depending on the category. We may use several dummy variables. Example 2 (Seasonal effects) Suppose yt is a quarterly data yt = β1 + β ′Xt + δ1D1t + δ2D2t + δ3D3t + εt . Here Xt does not contain 1. D1t = 1, if spring = 0, otherwise D2t = 1, if summer = 0, otherwise D3t = 1, if fall = 0, otherwise δi denote seasonal effects. Alternatively, we may use yt = β ′Xt + δ1D1t + δ2D2t + δ3D3t + δ4D4t + εt
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE But we should not use =B1+BXt+61D1t+62D2+63D3+64D+t since we run into a multicollinearity problem Example 3(Threshold effects and categorical variables) income=B,+B2age+SBB+SMM+OpP+E n{ 1, Bachelor's degree only 0. otherwise M=,Master's and Bachelor's degrees 0. otherwise 1, Ph. D, Master's and Bachelor's degrees 0. otherwise High school E Incomelage, HS=B,+B2age Bachelor 's E[Incomelage, B=B1+B2age +8B Master's EIncomelage, M=B1+B2age +8B+SM E [Incomelage, P=B,+B2age+8B+5M+8p SB: marginal effect of Bachelor's degree SM: marginal effect of Master's degree Sp: marginal effect of Ph. D's degree Example 4(Spine regression Suppose that E[ncomelagel= a0+Bo if t*>age>t E[ncomelagel=a+B if age >t2(t1<t2)
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 2 But we should NOT use yt = β1 + β ′Xt + δ1D1t + δ2D2t + δ3D3t + δ4D4t + εt since D1t + D2t + D3t + D4t = 1, we run into a multicollinearity problem. Example 3 (Threshold effects and categorical variables) income = β1 + β2age + δBB + δMM + δP P + ε B = 1, Bachelor’s degree only = 0, otherwise M = 1, Master’s and Bachelor’s degrees = 0, otherwise P = 1, Ph.D, Master’s and Bachelor’s degrees = 0, otherwise High school E [Income|age, HS] = β1 + β2age Bachelor’s E [Income|age, B] = β1 + β2age + δB Master’s E [Income|age, M] = β1 + β2age + δB + δM Ph.D E [Income|age, P] = β1 + β2age + δB + δM + δP δB : marginal effect of Bachelor’s degree δM : marginal effect of Master’s degree δP : marginal effect of Ph.D’s degree Example 4 (Spine regression) Suppose that E [Income|age] = α 0 + β 0 if t ∗ 2 > age ≥ t ∗ 1 E [Income|age] = α 1 + β 1 if age ≥ t ∗ 2 (t1 < t∗ 2 )
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE Introduce di 1,age≥t 0. otherwise d2 1,iage≥t 0. otherwise income=B1+ B2age+21d1+81dlage+22d2+52d2age+a I>age≥t, 1+2age+1+61ag (1+71)+(2+61)age+ Iage≥t income= B,+B,age+1+hage +12+orage+E (1+1+72)+(2+61+62)age+ The model is piecewise continuous if B1+B2t=(1+m1)+(2+61)t (1+m1)+(2+61)t=(1+1+72)+(2+61+62)t 71+61t1=0 72+6212=0. Putting this constraint into the model, we obtain income=B,+B2age +Sdi(age-ti)+82d2(age-t2)+E 7.2 Nonlinearity in the variables 1. Log linear model lny=月1+∑Aln The coefficients B, are elasticities aIn a In Bk
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 3 Introduce d1 = 1, if age ≥ t ∗ 1 = 0, otherwise d2 = 1, if age ≥ t ∗ 2 = 0, otherwise and use the model income = β1 + β2age + γ1d1 + δ1d1age + γ2d2 + δ2d2age + ε. If t ∗ 2 > age ≥ t ∗ 1 , income = β1 + β2age + γ1 + δ1age + ε = (β1 + γ1 ) + (β2 + δ1) age + ε If age ≥ t ∗ 2 , income = β1 + β2age + γ1 + δ1age + γ2 + δ2age + ε = (β1 + γ1 + γ2 ) + (β2 + δ1 + δ2) age + ε The model is piecewise continuous if β1 + β2 t ∗ 1 = (β1 + γ1 ) + (β2 + δ1)t ∗ 1 (β1 + γ1 ) + (β2 + δ1)t ∗ 2 = (β1 + γ1 + γ2 ) + (β2 + δ1 + δ2)t ∗ 2 or γ1 + δ1t ∗ 1 = 0 γ2 + δ2t ∗ 2 = 0. Putting this constraint into the model, we obtain income = β1 + β2age + δ1d1 (age − t ∗ 1 ) + δ2d2 (age − t ∗ 2 ) + ε. 7.2 Nonlinearity in the variables 1. Log linear model ln y = β1 + βk ln xk + ε The coefficients βk are elasticities ∂ ln y ∂ ln xk = ∂y/y ∂xk/xk = βk
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 2. Semilog model Iny=61+B2t+E If =t, B, is the average growth rate of y. If r is a dummy variable, y is 100 x B2% higher(lower if B20, the marginal effect of higher speed on breaking distance is increased when the road is wetter Example 6 Y=1+B2D2+3D3+4D2D3+BX+e Y: log annual salary of a bank clerk in HK D23= if Huent in English 1 if fluent in Putonghua 0 otherwise X: work experience B4: effect of fluency in both English and Putonghua on salary
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 4 2. Semilog model ln y = β1 + β2x + ε If x = t, β2 is the average growth rate of y. If x is a dummy variable, y is 100×β2% higher (lower if β2 0, the marginal effect of higher speed on breaking distance is increased when the road is wetter. Example 6 Y = β1 + β2D2 + β3D3 + β4D2D3 + β5X + ε Y : log annual salary of a bank clerk in HK D2 = 1 if fluent in English = 0 otherwise D3 = 1 if fluent in Putonghua = 0 otherwise X : work experience β4 : effect of fluency in both English and Putonghua on salary
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 7.3 Structural change Examples of events that led to structural change Oil shock of 1973 Emergency of Deng XiaoPing ata to Y2, X2: data for period 2 We want to test whether coefficient vector undergoes changes from period 1 to period 2 1. Unrestricted model that allows structural change X1 0B 0X2|A2 E2 b=(X'X)Xy XIXi 0 0X2X2(X2 (X1X1) 0 91 0 (X2X2) Thus -x=(0)( X161 X.bo el unrestricted residual ee=betel 2. Restricted model XI + El 2 e"=y-Xb: restricted residual The null of no structural change can be written as RB=r
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 5 7.3 Structural change Examples of events that led to structural change Oil shock of 1973 Emergency of Deng XiaoPing Y1, X1 : data for period 1 Y2, X2 : data for period 2 We want to test whether coefficient vector undergoes changes from period 1 to period 2. 1. Unrestricted model that allows structural change y1 y2 = X1 0 0 X2 β1 β2 + ε1 ε2 b = (X ′X) −1 X ′ y = X′ 1X1 0 0 X′ 2X2 X′ 1 y1 X′ 2 y2 = (X′ 1X1) −1 0 0 (X′ 2X2) −1 X′ 1 y1 X′ 2 y2 = b1 b2 Thus e = y − Xb = y1 y2 − X1b1 X2b2 = e1 e2 : unrestricted residual and e ′ e = e ′ 1 e1 + e ′ 2 e2. 2. Restricted model y1 y2 = X1 X2 β + ε1 ε2 or y = X ∗β + ε b ∗ = (X ∗′X ∗ ) −1 X ∗′y, e ∗ = y − X ∗ b ∗ : restricted residual The null of no structural change can be written as H0 : Rβ = r
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE h R=I-I We may use either F( Chow test)or Wald test for this null hypothesis (Rb-r)(R(X'X)-R)(Rb-r)/K e'e/K ee/(m1+n2-2K) F(K,m1+m2-2)i~iN(0,) of columns in X2(# of restrictions) X10 0 X W= KFX(K) Extensions 1. Change in a subset of coefficients K1×1 22+X22+ We want to test Ho: B1=B2. The model is written as Z1 X10 B+ When B is unrestricted. Let 6 Then, the null is written as Ho: RO=T whe Thus, we proceed as before. In this case FN F(K (K1+2K2)
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 6 where R = I −I and r = 0. We may use either F (Chow test) or Wald test for this null hypothesis. F = (Rb − r) ′ R (X′X) −1 R′ −1 (Rb − r) /K s 2 = (e ∗′e ∗ − e ′ e) /K e ′e/ (n1 + n2 − 2K) ∼ F (K, n1 + n2 − 2K) if εi ∼ iidN 0, σ2 . K : # of columns in X2 (= # of restrictions) X = X1 0 0 X2 W = KF d→ χ 2 (K). Extensions 1. Change in a subset of coefficients Let y1 = Z1 γ K1×1 +X1 β1 K2×1 +ε1 y2 = Z2γ + X2β2 + ε2 We want to test H0 : β1 = β2 . The model is written as y = Z1 Z2 γ + X1 0 0 X2 β + ε. When β is unrestricted. Let δ = γ β . Then, the null is written as H0 : Rδ = r when R = 0 K1 I K2 −I K3 and r K2×1 = 0. Thus, we proceed as before. In this case, F ∼ F (K2, n1 + n2 − (K1 + 2K2))
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 2. Test of structural change with unequal variances Suppose that Eli N iid (0, 02) and E2i N iid (0, 02)(02+02).In additions, assume (X1, E1) and(X2, E2) are independent. Wald test for the null hypothesis B1=B2 is defined by a)(xx)-2+(XX2)](-A2) asn→∞,W→x2(K2). The formula follows because under the null E B. Am()=()+m() The second equality uses the independence assumption 3. Insufficient observation Suppose that n2 K. This implies that e2 cannot be calculated. In this case, use e1e1)/n2 F=飞e1/(mn1-K) We are using e =el Fisher(1970, Econometrica)show that
CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 7 2. Test of structural change with unequal variances Suppose that ε1i ∼ iid (0, σ2 1 ) and ε2i ∼ iid (0, σ2 2 ) (σ 2 1 = σ 2 2 ). In additions, assume (X1, ε1) and (X2, ε2) are independent. Wald test for the null hypothesis H0 : β1 = β2 is defined by W = βˆ 1 − βˆ 2 ′ s 2 1 (X ′ 1X1) −1 + s 2 2 (X ′ 2X2) −1 −1 βˆ 1 − βˆ 2 as n → ∞, W d→ χ 2 (K2). The formula follows because under the null E βˆ 1 − βˆ 2 = 0 Asy.V ar βˆ 1 − βˆ 2 = σ 2 1plim X′ 1X1 n1 −1 + σ 2 2plim X′ 2X2 n2 −1 . The second equality uses the independence assumption. 3. Insufficient observation Suppose that n2 < K. This implies that e2 cannot be calculated. In this case, use F = (e ∗′e ∗ − e ′ 1 e1) /n2 e ′ 1e1/ (n1 − K) . (We are using e = e1) Fisher (1970, Econometrica) show that F ∼ F (n2, n1 − K)
