One period models Method II For working persons Labor Supply Optimal Wage-Hours Fixed Cost Models. Labor Supply. James Heckman University of Chicago

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1 Labor Supply James Heckman University of Chicago April 23, / 77

2 One period models: (L < 1) U (C, L) = C α 1 α b = taste for leisure increases ( ) L ϕ 1 + b ϕ α, ϕ < 1 2 / 77

3 MRS at zero hours of work (Reservation Wage or Virtual Price): ( ) U at L = 1, C = A R = L ( ) L = 1, C = A U C R = b Lϕ 1 C α 1 R = b A α 1 ln R = ln b + (1 α) ln A 3 / 77

4 Set: ln b = X β + ε b Assume: ε b N ( ) 0, σb 2 Assume: ln W ε b (X, A, W ) ε b 4 / 77

5 Assume wage is observed for everyone. Probability that a person with assets A, X, and Wage W works: Pr (ln R ln W X, A) = Pr(X β + (1 α) ln A + ε b ln W X, A) ( ) εb ln W X β (1 α) ln A = Pr σ b σ b = Φ (C) where C ln W X β (1 α) ln A σ b A > 0 5 / 77

6 Let D = 1 D = 0 } if person works D = 1 [ln W ln R] otherwise Pr(ln R ln W X, A) = Pr(D = 1 X, A) Take Grouped Data: Each cell has common values of W i, X i and A i. Set P ) i = Φ (Ĉi P i = cell proportion working i inverse exists: C i = ln W i X i β (1 α) ln A i σ b Ĉ i = Φ 1 ( Pi ) (table lookup) 6 / 77

7 Run Regression: Ĉ i on ln W i X i β (1 α) ln A i σ b Coefficient on ln W i is 1 σ b Coefficient on X is β σ b Coefficient on ln A is 1 α σ b Do for Logit ( ) ε Pr z = ez σ b 1 + e z 7 / 77

8 Linear Probability Model ( ) ε z Pr z = σ b z U z L z L ε σ b z U 8 / 77

9 Micro Data Analogue: Sample size I, (Assumes we have symmetric ε around zero): L = I Φ (C i (2D i 1)) i=1 ( β, σb, α) = arg max ln L consistent, asymptotically normal. (Likelihood is concave) Assumes we know wage for all persons, including those who work, but we don t. Can be nonparametric about F εb (Cosslett, Manski, Matzkin) 9 / 77

10 Digression D = Zγ V, D = 1(Zγ > V ), assume Var(V ) = 1. Can be nonparametric about V. Normality is not needed. Assume Z i, Z j are continuous: Pr (D = 1 Z) = F V (Zγ) Pr (D = 1 Zγ) Z i Pr (D = 1 Zγ) Z j = f V (Zγ) γ i f V (Zγ) γ j = γ i γ j We can identify the coefficients up to scale. Back to text. 10 / 77

11 Method II Don t know wage, but ln W = Zγ + U ln R = X β + (1 α) ln A + ε ( U ε ) ( 0 N 0, σ UU σ εu σ εu σ εε ) ln R ln W 0 D = 0 Y 1 X β (1 α) ln A + Zγ (ε U) = ln W ln R 11 / 77

12 (X, ln A, Z) (ε U) (ε U) N (0, σ εε + σ UU 2σ εu ) Var (ε U) = (σ ) 2 σ σ εε + σ UU 2σ εu Pr (Y 1 0 X, A, Z) = Pr (D = 1 X, ln A, Z) 12 / 77

13 C Pr (D = 1 X, ln A, Z) ( X β (1 α) ln A + Zγ = Pr ε U ) σ σ ( ) X β (1 α) ln A + Zγ = Φ = Φ(C) σ X β (1 α) ln A + Zγ σ If Z and X distinct from each other and A, estimate γ σ, β σ, 1 α σ, can t estimate σ, get relative values. 13 / 77

14 Suppose X and Z have some elements in common; X c = Z c elements in common X d, Z d are distinct elements in X, Z Y 1 σ = X dβ d X c (β c γ c ) + Z dγ d (1 α) + ln A + ε U σ σ σ σ σ identify β d σ, β c γ c, γ d σ σ, 1 α σ (The leading example of variables in common is education.) Allows U to be correlated with ε. (Method II may be required anyway.) 14 / 77

15 Observe the wage only for working persons ln W = Zγ + U ln R = X β + (1 α) ln A + ε Assume (X, Z, A) (ε, U) Y 1 = ln W ln R = Zγ X β (1 α) ln A + U ε 15 / 77

16 Letting λ(q) = φ(q), we have Φ(q) E (ln W ln W ln R 0, X, Z, A) Zγ X β (1 α) ln A = E ln W σ ε U, X, Z, A σ = Zγ + σ ( ) UU σ Uε Zγ X β (1 α) ln A λ σ σ C (X, A, Z) = Zγ X β (1 α) ln A σ 16 / 77

17 Remembering the Truncated Normal Random variable: Let: Z N(0, 1) E (Z Z q) = λ(q); λ(q) φ(q) 1 Φ(q) = φ(q) Φ( q) E (Z q Z) = E ( Z Z q) = φ( q) 1 Φ( q) = φ(q) Φ(q) λ(q) φ(q) = E (Z Z q) Φ(q) and : E (Z Z q) = φ(q) = λ(q) = λ( q) Φ( q) 17 / 77

18 Two Stage Procedures (1) Probit on Work participation Pr (D = 1 Z, X, A) = Pr (ln W ln R 0 Z, X, A) ( Zγ X β (1 α) ln A = Pr ε U ) σ σ Z, X, A ( ) Zγ X β (1 α) ln A = Φ σ we can estimate C (X, A, Z) (2) Form λ (C) σ = [Var (U ε)] / 77

19 Run Linear Regression Get Consistent Estimates of γ, σ UU σ Uε σ With one exclusion restriction (one variable in Z not in X or ln A, say Z 1 ). 19 / 77

20 Note that using Probit if X d, Z d are distinct elements in X, Z and X c = Z c are elements in common we can identify β d, βc γc, γ σ σ d, σ 1 α σ. Say we recover γ 1 σ (by Probit) Note that we have γ (by Wage Regression on Z and λ) know σ The estimated coefficient on λ is σ UU σ Uε σ know σ UU σ Uε 20 / 77

21 Look at the residuals from equations [ V ln W Zγ + σ ] UU σ Uε λ (C (X, A, Z)) σ ( ) σ UU σ Uε = U (σ σ (σ UU ) 1 UU ) 1 2 λ (C (X, A, Z)) 2 Let : ρ σ UU σ Uε (σ UU ) 1 2 σ V = U ρ (σ UU ) 1 2 λ (C (X, A, Z)) = U E (U ln W ln R 0) E (V ) = 0 E ( V 2) = Var(V ) = Var (U ln W ln R 0) 21 / 77

22 E ( V 2) [ (1 = σ ) ( UU ρ 2 + ρ λc λ )] 2 ) = σ UU + σ UU ρ ( λc 2 λ2 Regress V 2 on know ρ 2 ( λc C 2) Get σ UU and σ UU ρ 2 22 / 77

23 Look at model: 1 Wrong variables appear in wage equation 2 Errors heteroskedastic 3 Omitted variables 23 / 77

24 Recovered Coefficients: γ 1 } (Probit) σ σ γ (Wage Regression ) σ UU σ Uε } (Wage Regression ) σ σ σ UU σ Uε } σ UU (Error 2 Regression ) σ σ UU σ Uε Uε 24 / 77

25 The term σ UU σ Uε σ : σ UU σ Uε σ is a Wage Regression coefficient } ρ σ UU σ Uε (σ UU ) 2 1 σ (Error2 Regression ) σ UU (Error 2 Regression ) 2 estimates of σ UU σ Uε σ σ UU σ Uε σ 25 / 77

26 The term σ : γ 1 } (Probit) σ σ γ (Wage Regression ) σ UU σ Uε (Wage Regression & σ above) ρ σ UU σ Uε (σ UU ) 1 2 σ (Error2 Regression ) σ UU (Error 2 Regression ) 2 estimates of σ To obtain σ εε, we can solve σ (σ ) 2 = σ UU + σ εε 2σ Uε (σ ) σ Uε σ UU = σ εε 26 / 77

27 Suppose we have no exclusion restriction, just regressors. Then we can still estimate γ, σ UU, σ εε provided we substitute other information for exclusion restrictions. b = σ UU σ Uε σ = σ UU σ Uε (σ UU + σ εε 2σ Uε ) 1 2 (coefficient on λ) E ( V 2) ) = σ UU + σ UU ρ ( λc 2 λ2 ( ) 2 σ UU σ ) Uε = σ UU + σ UU ( λc λ2 (σ UU ) 1 2 σ ) = σ UU + b ( λc 2 λ2 σ UU = E ( V 2) ) b ( λc 2 λ2 27 / 77

28 Normalize variables: σ εε = 1 or σ Uε = 0 Example: σ Uε = 0 Then know σ UU can solve for σ εε Alternatively, if σ εε = 1 (σ UU + σ εε ) 1 2 σ UU σ Uε (1 + σ UU 2σ Uε ) 1 2 = known solve for σ Uε, quadratic equation sometimes get unique root. Note crucial role of regressor in getting full identification. 28 / 77

29 Labor Supply Hours of Work Single Period Model More Information: Direct Utility Function for non workers: V 1 (A 1, ϕ) A 1 = unearned income if person works best attainable utility for a person who doesn t work Indirect Utility Function: V 2 (A 2, W, ϕ) (W = wage) best available utility given that he works, (which may be V 1 ). A 2 is unearned income net of money costs of work 29 / 77

30 For person who works: Index Function: If V 2 > V 1 person works Y 1 = V 2 V 1 Y 1 0 person works ( )/ ( ) V2 V2 Y 2 = H = = H (A 2, W, ϕ) W A Roy s Identity: 3 types of labor supply functions: (a) participation (b) E(H H > 0, W, A) (c) E(H W, A) aggregate labor supply 30 / 77

31 None estimates a labor supply function (Hicks-Slutsky). Workers free to choose hours of work. Wage W is independent of hours of work. No fixed costs. Local comparison is global comparison. Consider a simple example based on Heckman (1974), 31 / 77

32 MRS Function: ln R = α 0 + α 1 A + α 2 X + ηh + ε (1) ln W = Zγ + U ln R defines an equilibrium value of time locus. Labor supply H is the value that equates ln W = ln R: ln W = α 0 + α 1 A + α 2 X + ηh + ε H = 1 η (ln W α 0 α 1 A α 2 X ε) The causal effect of ln (wage) on labor supply is 1 η and ε constant). This is a Hicks-Slutsky effect. (holding A, X 32 / 77

33 E.g. H ln W = 1 η = H ln W S }{{} substitution effect + H H, }{{ Y} income effect = WS + (WH) H Y. If η is constant, then as H, for a fixed W, S (more substitution). As W, S + H H, so the Hicks-Slutsky effect declines (net Y labor supply becomes more inelastic in this sense). 33 / 77

34 Figure: Value of Time 34 / 77

35 Define Y 1 = ln W ln R = Zγ + U α 0 α 1 A α 2 X ε = (Zγ α 0 α 1 A α 2 X ) + (U ε). Hours of work then are: Y 2 = H = 1 η Y 1 if Y 1 0 Y 3 = ln W = Zγ + U Var(U ε) = (σ ) 2 35 / 77

36 Population Labor Supply is generated from Y 2 E (H Y 1 > 0, Z, A, X ) = 1 E (ln W ln R ln W ln R > 0, Z, A, X ) η = Zγ α 0 α 1 A α 2 X η + 1 η E (U ε U ε > (Zγ α 0 α 1 A α 2 X )) = Zγ α 0 α 1 A α 2 X η + 1 η E ( U ε σ U ε > (Zγ α ) 0 α 1 A α 2 X ) σ σ 36 / 77

37 Let C Zγ α 0 α 1 A α 2 X σ E (H Y 1 > 0, Z, A, X ) = 1 η E (Y 1 Y 1 > 0, Z, A, X ) = (Zγ α 0 α 1 A α 2 X ) η + 1 η E (U ε U ε > (Zγ α 0 α 1 A α 2 X ), Z, A, X ) 37 / 77

38 = C ( η σ + σ U ε η E U ε σ σ = C η σ + 1 η = C η σ + 1 η σ λ(c) = σ η Cov(U ε, U ε) λ(c) σ ( C + λ(c) ) ) > C, Z, A, X E(ln W Y 1 > 0, Z) = E(ln W ln W ln R > 0, Z) = Zγ + σ E( U σ U ε > C, Z) σ Cov(U, U ε) = Zγ + λ(c) σ 38 / 77

39 Assume Regressors are available: We can estimate γ from linear regression ln W on Zγ and λ(c) using the known steps: (1) From participation equation, we can use probit to estimate Pr (D = 1 Z, A, X ) = Pr (Y 1 > 0 Z, A, X ) ( ) Zγ α0 α 1 A α 2 X = Φ = Φ(C) We know γ, α σ 0, α σ 1, α σ 2 if X A Z or set of common and σ distinct coefficients depending on X, A, Z elements. We know C. (2) Form λ(c). (3) From the Wage Regression of ln W on Z and λ(c). σ 39 / 77

40 we know γ, Cov(U,U ε) σ. Thus we know Cov(U, U ε) σ = σ UU σ Uε (σ UU + σ εε 2σ Uε ) 1/2. (4) From Error Regression : V 2 on constant and ( λc C 2), we estimate: E (V 2 ) = σ UU + σ UU ρ 2 ( λc λ 2 ) know σ UU, ρ 2 Same position as before. Further identification of parameters is possible due to hours of work: (5) From hours of work data we have a proportionality restriction 40 / 77

41 ( ) σ E(H Y 1 > 0, Z, A, X ) = C + σ η η λ(c) but from employment (participation) equation we know C can estimate σ η 41 / 77

42 (6) Using Cov(U,U ε) from Wage Regression and covariance σ assumptions, we obtain: if σ Uε = Cov(U, U ε) σ UU 0 = σ (σ UU + σ εε ) 1/2 but σ UU was obtained by Error Regression know σ and σ εε By The hours of Work Regression (5) we obtain σ η know η Similarly if σ εε = 1 σ Uε known σ Uε known (sometimes; multiple roots) we have that all parameters are identified. 42 / 77

43 (7) If there is one variable in Z not in (X, A), say Z 1, from coefficient on Z 1 in E(H Y 1 > 0, Z, A, X ), we obtain: = = = E (H Y 1 > 0, Z, A, X ) ( ) σ C + σ η η λ(c) ( ) σ ( λ(c)) C + η ( ) ( σ Zγ α0 α 1 A α 2 X η σ = Z γ η + α 0 α 1 A α 2 X η ( σ + η ) + λ(c) ) λ(c) 43 / 77

44 but from Wage Regression (3), we obtain γ can estimate η ( ) σ but the coefficient of λ(c) is η can estimate σ. Alternatively, we can determine η if Cov(U, ε) = 0 or Var(ε) = / 77

45 Selection Bias in Labor Supply Assume γ j > 0 [( E(H Y 1 > 0, Z, A, X ) = Z j [ ( Z γ + α 0 α 1 A α 2 X σ + η η η = Z j ) ( σ C + λ(c) )] η Z j ) λ( Z γ η + α 0 α 1 A α 2 X η ) σ ] 45 / 77

46 But λ(q) q = φ(q) Φ(q) q φ(q) = q Φ (q) φ2 (q) ( Φ 2 (q) = λ(q) q + λ(q) ) = γ j η 1 η λ( λ + C)γ j ( ) γj = (1 η λ( λ + C)) 0 < 1 λ( λ + C) < 1 < γ j η downward biased H Z j = E(H Y 1 > 0, Z, A, X ) W ln W Z j downward bias. 1 η 46 / 77

47 = = = E (H Y 1 > 0, Z, A, X ) ( ) σ C + σ η η λ(c) ( ) σ ( λ(c)) C + η ( ) ( σ Zγ α0 α 1 A α 2 X η σ = Z γ η + α 0 α 1 A α 2 X η ( σ + η ) + λ(c) ) λ(c) 47 / 77

48 but from Wage Regression (3), we obtain γ can estimate η ( ) but the coefficient of λ(c) σ is η can estimate σ. 48 / 77

49 Aggregate Labor Supply ALS Φ(C) E (H Y 1 > 0) +Φ( C) E (H Y 1 < 0) }{{} 0 [( ) σ ( = Φ(C) C + η λ(c) ) ] + Φ( C) [0] ( ) [ σ = Φ(C)C + 1 ] e C 2 /2 η 2π 49 / 77

50 E(H ALS Z, A, X ) C E(H ALS Z, A, X ) Z j = = σ η [ = σ η [Φ(C)] ] Φ(C) + Ce 1 2 C 2 /2 Ce C 2 /2 2π 2π E(H Z, A, X ) C C = γ j Z j η Φ(C) 50 / 77

51 Obviously aggregate labor supply more elastic because of entry or exit: 1 E(H Y 1 > 0, W )P(Y 1 > 0 W ) {E(H Y 1 > 0, W )P(Y 1 > 0 W )} ( ln W ) E(H Y1 > 0, W ) 1 ln W E(H Y 1 > 0, W ), and ln E(H Y 1 > 0, W ) + ln P(Y 1 > 0 W ) ln W ln W > ln E(H Y 1 > 0, W ) ln W Many ways to estimate model. 51 / 77

52 Labor Supply with Optimal Wage-Hours Contracts (Lewis, 1969; Rosen, 1974; Tinbergen, 1951, 1956) Figure: Optimal Wage 52 / 77

53 If Y (h) is earnings, and Y (h) is marginal wage, Virtual Income = Y (h) Y (h) h + A, where h = h (Y (h), Y (h) Y (h) h + A, ν) Any equilibrium calculation use slope at zero hours of work. ln M(0, A) ln W (0) doesn t work Equilibrium: ln M(h, A) = ln W (h) person works Can use estimated ln M(0, A) to price out goods that previously were not purchased. 53 / 77

54 Fixed Cost Models: Fixed Money Cost (Cogan; 1981 Econometrica) Figure: Fixed Cost Models 54 / 77

55 Introduce fixed money cost: given the wage, the worker selects a minimum number of hours. (1) Solve for W d and H d that causes the worker to be indifferent between work and no work V (A F, W d, ϕ) = U(A, 1, ϕ) Solve for W d If no solution, person doesn t work (2) Minimum number of hours H d = V W /V A H d = H d (A F, W d, ϕ) W d = W d (A F, ϕ) H = H(A F, W, ϕ) 55 / 77

56 Index Function Model. Y 1 = H H d Y 2 = H Observe Y 2 only when Y 1 > 0 Example: (assume wage is known). H = X β + W η + ε 1 functional form assumptions H d = X τ + ε 2 Pr (consumer works) = Pr (H H d > 0 X, W ) = Pr(X β + W η + ε 1 X τ ε 2 > 0) = Pr(X (β τ) + W η > ε 2 ε 1 ) σ Var(ε 2 ε 1 ) 56 / 77

57 Assume normality and we can identify β τ σ and η σ From hours of work equation we know η know σ E (H H H d > 0, X, W ) = X β + W η + E (ε 1 X (β τ) + W η > ε 2 ε 1, X, W ) = X β + W η + Cov(ε 1,ε 2 ) λ(c) Know η, β know τ σ ( ) X (β, τ) + W η C = σ 57 / 77

58 Know Var(ε 1 ), know σ know Cov(ε 1, ε 2 ) know Var(ε 2 ) = (σ ) 2 Var(ε 1 ) + 2 Cov(ε 1, ε 2 ) 58 / 77

59 Cogan doesn t measure fixed costs 59 / 77

60 Figure: Fixed vs. Not Fixed 60 / 77

61 Null: Departure from simple proportionality model Cogan s test conditional on function form. 61 / 77

62 Broken line Budget Constraint (2 part prices; negative income tax data) Two cases: A. Know which interval person is in (no measurement error for hours) B. Don t know which branch (income tax data) 62 / 77

63 Figure: Which Branch? 63 / 77

64 Take case 1: (1) M(A, 1, ε) W 1 does a person work? (2) Person is interior in interval (0, h) if M(A + W 1 h, 1 h, ε) W 1 M(A, 1, ε) < W 1 (3) In equilibrium at h if W 2 M(A + W 1 h, 1 h, ε) W 1 (4) Works beyond h if M(A + W 1 h, 1 h, ε) W 2 64 / 77

65 Example: U = C α 1 α ( ) L ϕ 1 + b ϕ α < 1, ϕ < 1 b Lϕ 1 = MRS C α 1 at zero hours work (L = 1) 65 / 77

66 ln b = X β + ε ln b + (ϕ 1) ln(1) (α 1) ln A ln W 1 (doesn t work) E(ε 2 ) = σ 2 ε X β + ε (α 1) ln A ln W 1 ε ln W 1 + (α 1) ln A X β ε ln W 1 + (α 1) ln A X β σ ε σ ε condition for not working estimate: σ ε, α, β 66 / 77

67 (2) Interior in the first branch (0, h) X β + (ϕ 1) ln(1 h) (α 1) ln(a + W 1 h) ln W 1 σ ε ε σ ε X β + (ϕ 1) ln(1) (α 1) ln(a) ln W 1 σ ε ε σ ε Use principle of index function, with variation in h, we identify ϕ and β. 67 / 77

68 (3) Kink Equilibrium: ln W 2 X β + ε + (ϕ 1) ln(1 h) (α 1) ln(a + W 1 h) ln W 1 ln W 2 X β (ϕ 1) ln(1 h) + (α 1) ln(a + W 1 h) σ ε ε σ ε ln W 1 X β (ϕ 1) ln(1 h) + (α 1) ln(a + W 1 h) σ ε 68 / 77

69 Pr(h = h W 1, W 2, X, A) = ( ) ln W1 X β (ϕ 1) ln(1 h) (α 1) ln(a + W 1 h) Φ σ ε ( ) ln W2 X β (ϕ 1) ln(1 h) + (α 1) ln(a + W 1 h) Φ σ ε 69 / 77

70 (4) In second branch interior ( X β + (η 1) ln(1 h) (α 1) ln(a + W1 h) ln W 2 Pr ε ) σ ε σ ε ( ) ln W2 X β (η 1) ln(1 h) + (α 1) ln(a + W 1 h) = Φ solve out hours of work. σ ε 70 / 77

71 Hours of work (standard case) A = nonmarket income W = wage C = W (1 L) + A U = C α 1 α ( ) L ϕ 1 + b ϕ 71 / 77

72 Set ϕ = α, Set h = estimating equation: ( ) Wh + A ln 1 h ( b W ) 1 α 1 A ( b W ) 1 α 1 + W = ln W 1 α X β 1 α ε 1 α 72 / 77

73 Interior in Branch 1: [ ( ) ] W1 h + A E ln W 1, A = ln W 1 1 h 1 α X β 1 α 1 1 α E ε ( X β + (η 1) ln(1 h) ε σ ε (α 1) ln(a + W 1 h) ln W 1 σ ε ( X β + (α 1) ln(1) (α 1) ln A ln W 1 σ ε ) ) 73 / 77

74 Last term is: ) 1 1 2π (e C 1 2/σ2 ε e C2 2/σ2 ε ( ) ( ) 1 α C Φ 1 C σ ε Φ 2 σ ε ( X β + (α 1) ln(1 h) ) C 1 = (α 1) ln(a + (W 1 W 2 )h) ln W 1 σ ε C 2 = X β + (α 1) ln(1) (α 1) ln A ln W 1 σ ε 74 / 77

75 At h = h (with probability P = P 3 )Q P 3 = Pr etc. E(h h = h) = hp 3 ( 1 ln W2 X β (η 1) ln(1 h) σ ε +(α 1) ln(a + W 1 h) 1 σ ε ( ln W1 X β (η 1) ln(1 h) +(α 1) ln(a + W 1 h) ) ε σ ε ) 75 / 77

76 Branch 2 labor supply ( ) W2 h + (W 1 W 2 )h + A ln 1 h ( ) (W1 W 2 )h + A E 1 h h > h = ln W 2 1 α X β 1 α ε 1 α = ln W 2 1 α X β 1 α 1 1 α E ε ( ln W2 + (α 1) ln(a + W 1 h) X β (α 1) ln(1 h) σ ε ) ε σ ε 76 / 77

77 Let ( ) Z (1) W1 h + A = ln 1 h ( ) Z (2) W2 h + (W 1 W 2 )h + A = ln 1 h E(Z (1) W 1, A, 0 < h < h) = ln W 1 1 α X β 1 α 1 E(ε 0 < h < h) 1 α E(Z (2) W 2, A, h > h) = ln W 2 1 α X β 1 α 1 E(ε h > h) 1 α Can estimate by 2 stage methods. 77 / 77