CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 1 Chapter7FunctionalFormandStructuralChange 7.1Usingbinaryvariables Example1 Earnings equation for married women lnearnings=β1 +β2age+β3age2 +β4education+β5kids+ε kids= braceleftBigg = 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 Heredi takes value either 1 or 0 depending on the category. We may use several dummy variables. Example2 (Seasonal effects) Suppose yt is a quarterly data yt =β1 +β′Xt +δ1D1t +δ2D2t +δ3D3t +εt. Here Xt does not contain 1. D1t braceleftBigg = 1, if spring = 0, otherwise D2t braceleftBigg = 1, if summer = 0, otherwise D3t braceleftBigg = 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 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. Example3 (Threshold effects and categorical variables) income=β1 +β2age+δBB+δMM +δPP +ε B braceleftBigg = 1, Bachelor’s degree only = 0, otherwise M braceleftBigg = 1, Master’s and Bachelor’s degrees = 0, otherwise P braceleftBigg = 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 Example4 (Spine regression) Suppose that E[Income|age] = α0 +β0 if t?2 >age≥t?1 E[Income|age] = α1 +β1 if age≥t?2 (t1 <t?2) CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 3 Introduce d1 braceleftBigg = 1, if age≥t? 1 = 0, otherwise d2 braceleftBigg = 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 +β2t?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. 7.2Nonlinearityinthevariables 1. Log linear model lny =β1 + summationdisplayβk lnxk +ε The coefficients βk are elasticities ?lny ?lnxk = ?y/y ?xk/xk =βk CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 4 2. Semilog model lny =β1 +β2x+ε Ifx=t,β2 is the average growth rate ofy.Ifxis a dummy variable,yis 100×β2% higher (lower ifβ2 <0) whenx= 1. 3. Polynomial equation y =β1 +β2x+β3x2 +···+βpxp?1 +ε. 4. Interaction model Example5 D =β1 +β2S+β3W +β4SW +ε D : breaking distance W : road wetness S : speed ?E[D|S,W] ?S =β2 +β4W Ifβ4 >0,the marginal effect of higher speed on breaking distance is increased when the road is wetter. Example6 Y =β1 +β2D2 +β3D3 +β4D2D3 +β5X +ε Y : log annual salary of a bank clerk in HK D2 braceleftBigg = 1 if fluent in English = 0 otherwise D3 braceleftBigg = 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 5 7.3Structuralchange 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 bracketleftBigg y 1 y2 bracketrightBigg = bracketleftBigg X 1 0 0 X2 bracketrightBiggbracketleftBigg β 1 β2 bracketrightBigg + bracketleftBigg ε 1 ε2 bracketrightBigg b= (X′X)?1X′y = parenleftBigg X′ 1X1 0 0 X′2X2 parenrightBiggparenleftBigg X′ 1y1 X′2y2 parenrightBigg = parenleftBigg (X′ 1X1) ?1 0 0 (X′2X2)?1 parenrightBiggparenleftBigg X′ 1y1 X′2y2 parenrightBigg = parenleftBigg b 1 b2 parenrightBigg Thus e = y?Xb= parenleftBigg y 1 y2 parenrightBigg ? parenleftBigg X 1b1 X2b2 parenrightBigg = parenleftBigg e 1 e2 parenrightBigg : unrestricted residual and e′e=e′1e1 +e′2e2. 2. Restricted model bracketleftBigg y 1 y2 bracketrightBigg = bracketleftBigg X 1 X2 bracketrightBigg β+ bracketleftBigg ε 1 ε2 bracketrightBigg ory =X?β+ε b? = (X?′X?)?1X?′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 6 where R= bracketleftBig I ?I bracketrightBig andr = 0. We may use eitherF (Chow test) or Wald test for this null hypothesis. F = (Rb?r) ′ parenleftBigR(X′X)?1R′parenrightBig?1 (Rb?r)/K s2 = (e ?′e? ?e′e)/K e′e/(n1 +n2 ?2K) ～ F (K,n1 +n2 ?2K) ifεi ～iidN parenleftBig 0,σ2 parenrightBig . K : # of columns inX2 (= # of restrictions) X = parenleftBigg X 1 0 0 X2 parenrightBigg W = KF d→χ2 (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 = parenleftBigg Z 1 Z2 parenrightBigg γ+ parenleftBigg X 1 0 0 X2 parenrightBigg β+ε. Whenβ is unrestricted. Let δ = parenleftBigg γ β parenrightBigg . Then, the null is written as H0 :Rδ =r when R= bracketleftbigg 0K 1 IK 2 ?I K3 bracketrightbigg and rK 2×1 = 0. Thus, we proceed as before. In this case, F ～F (K2,n1 +n2 ?(K1 + 2K2)). CHAPTER 7 FUNCTIONAL FORM AND STRUCTURAL CHANGE 7 2. Test of structural change with unequal variances Suppose that ε1i ～iid(0,σ21) and ε2i ～iid(0,σ22) (σ21 negationslash=σ22). In additions, assume (X1,ε1) and (X2,ε2) are independent. Wald test for the null hypothesis H0 :β1 =β2 is defined by W = parenleftBig? β1 ? ?β2 parenrightBig′ bracketleftBig s21 (X′1X1)?1 +s22 (X′2X2)?1 bracketrightBig?1 parenleftBig? β1 ? ?β2 parenrightBig as n→∞,W d→χ2 (K2). The formula follows because under the null E parenleftBig? β1 ? ?β2 parenrightBig = 0 Asy.Var parenleftBig? β1 ? ?β2 parenrightBig =σ21plim parenleftBiggX′ 1X1 n1 parenrightBigg?1 +σ22plim parenleftBiggX′ 2X2 n2 parenrightBigg?1 . The second equality uses the independence assumption. 3. Insufficient observation Suppose that n2 <K. This implies thate2 cannot be calculated. In this case, use F = (e ?′e? ?e′ 1e1)/n2 e′1e1/(n1 ?K) . (We are usinge=e1) Fisher (1970, Econometrica) show that F ～F (n2,n1 ?K).