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Sampling Plans and Initial Condition Problems For Continuous Time Duration Models James J. Heckman University of Chicago Econ 312, Spring 2019 Heckman Sampling Plans Sampling Plans and Initial Condition Problems For Continuous Time Duration


  1. Sampling Plans and Initial Condition Problems For Continuous Time Duration Models James J. Heckman University of Chicago Econ 312, Spring 2019 Heckman Sampling Plans

  2. Sampling Plans and Initial Condition Problems For Continuous Time Duration Models James J. Heckman University of Chicago Econ 312, Spring 2019 Heckman Sampling Plans

  3. Sampling plans and initial condition problems: Duration Models Consider a random sample of unemployment spells in progress. For sampled spells, one of the following duration times may be observed: • time in state up to sampling date ( T b ) (recall of time spent) • time in state after sampling date ( T a ) (prospective sampling forward) • total time in completed spell observed at origin of sample ( T c = T a + T b ) Heckman Sampling Plans

  4. Duration of spells beginning after the origin date of the sample, denoted T d , are not subject to initial condition problems. The intake rate at time − t b (assuming sample occurs at time 0: the proportion of the population entering a spell at − t b . Assume: • A time homogenous environment, i.e. constant intake rate, k ( − t b ) = k , ∀ b • A model without observed or unobserved explanatory variables. • No right censoring, so T c = T a + T b • Underlying distribution f ( x ) is nondefective � ∞ • m = 0 ( x ) dx < ∞ Heckman Sampling Plans

  5. The proportion of the population experiencing a spell at t = 0, the origin date of the sample, is � ∞ � ∞ P 0 = k ( − t b )(1 − F ( t b )) dt b = k (1 − F ( t b )) dt b 0 0 � ∞ � � t b (1 − F ( t b )) | ∞ = k 0 − t b d (1 − F ( t b )) 0 � ∞ = k t b f ( t b ) dt b = km 0 where 1 − F ( t b ) is the probability the spell lasts from − t b to 0 (or equivalently, from 0 to − t b ). Heckman Sampling Plans

  6. So the density of a spell of length t b interrupted at the beginning of the sample ( t = 0) is proportion surviving til t = 0 from batch t b g ( t b ) = total surviving til t = 0 k ( − t b )(1 − F ( t b )) = 1 − F ( t b ) = � = f ( t b ) P 0 m Notice: g is the distribution of T b in the population constructed by sampling rule of source population. Distinguish from F : cdf of the true population. G : cdf of the sampled spells. Heckman Sampling Plans

  7. The probability that a spell lasts until t c given that it has lasted from − t b to 0, is the conditioned density of t c given 0 < t b < t c . f ( t c ) f ( t c | t > t b > 0) = 1 − F ( t b ); t c ≥ t b ≥ 0 So the density of a spell in the sampled population that lasts, t c is � t c g ( t c ) = f ( t c | t ≥ t b ) f ( t ≥ t b ) dt b 0 � t c f ( t c ) m dt b = f ( t c ) t c = m 0 Heckman Sampling Plans

  8. Likewise, the density of a sampled spell that lasts until t a is � ∞ g ( t a ) = f ( t a + t b | t b ) Pr ( t ≥ t a ≥ 0)) dt b 0 � ∞ f ( t a + t b ) = dt b m 0 � ∞ 1 = f ( t b ) dt b m t a 1 − F ( t a ) = m (Stationarity, mirror images have some densities). So the functional form of f ( t b ) = f ( t a ): Consequences of stationarity. Heckman Sampling Plans

  9. Some useful results that follow from this model: 1 If f ( t ) = θ e − t θ , then g ( t b ) = θ e − t b θ and g ( t a ) = θ e − t a θ . Proof : θ e − t θ → m = 1 f ( t ) = θ, 1 − e − t θ → g ( t a ) = 1 − F ( t ) = θ e − t θ F ( t ) = m Heckman Sampling Plans

  10. 2 (1 + σ 2 1 E ( T a ) = m m 2 ). Proof : � � 1 − G ( t a ) E ( T a ) = t a f ( t a ) dt a = t a dt a m � 1 1 � 1 � 2 t 2 2 t 2 a (1 − F ( t a )) | ∞ = 0 − a d (1 − F ( t a )) m � 1 1 a F ( t a ) dt a = 1 2 t 2 2 m [ var ( t a ) + E 2 ( t a )] = m 1 2 m [ σ 2 + m 2 ] = Heckman Sampling Plans

  11. 2 (1 + σ 2 1 E ( T b ) = m m 2 ). Proof : See proof of Proposition 2. 2 E ( T c ) = m (1 + σ 2 m 2 ). Proof : � t 2 c F ( t c ) dt c = 1 m ( var ( t c ) + E 2 ( t c )) E ( T c ) = m → E ( T c ) = 2 E ( T a ) = 2 E ( T b ) , E ( T c ) > m unless σ 2 = 0 Heckman Sampling Plans

  12. Some Additional Results: f ( t ) h ( t ) = hazard : h ( t ) = 1 − F ( t ) . 1 h ′ ( t ) > 0 → E ( T a ) = E ( T b ) < m . Proof: See Barlow and Proschan. 2 h ′ ( t ) < 0 → E ( T a ) = E ( T b ) > m . Proof : See Barlow and Proschan. Heckman Sampling Plans

  13. Examples Heckman Sampling Plans

  14. Specification of the Distribution Weibull Distribution • Parameters: λ > 0 , k > 0 • Probability Density Function (PDF): � t � t � k − 1 � � k � λ exp − k λ k • Cumulative Density Function: � t � � k � 1 − exp − k • Set of Parameters:   λ 1 , k 1 = 0 . 5 λ 2 , k 1 = 1 . 0   respectively  ,   λ 3 , k 1 = 2 . 0  λ 3 , k 1 = 3 . 0 Heckman Sampling Plans

  15. Basic Distribution Graphs ��� ��� &������ ������������ ��� �� &������ ������������ 3 1 Weibull Distribution λ = 0.1, k = 0.5 Weibull Distribution λ = 0.5, k = 1.0 0.9 Weibull Distribution λ = 0.5, k = 2.0 2.5 s 0.8 Weibull Distribution λ = 1.0, k = 3.0 n o i l t l u u b b 0.7 i e i r 2 t s W i D : n 0.6 l l o u i b t u i e b W 1.5 i 0.5 r t s : i s D l l e e 0.4 p h S t f e 1 o h F 0.3 t f D o C F Weibull Distribution λ = 0.1, k = 0.5 D 0.2 P 0.5 Weibull Distribution λ = 0.5, k = 1.0 0.1 Weibull Distribution λ = 0.5, k = 2.0 Weibull Distribution λ = 1.0, k = 3.0 0 0 0 0.5 1 1.5 2 0 0.5 1 1.5 2 t t Heckman Sampling Plans

  16. Basic Duration Graphs �� ��� �������� ��� &������ ������������ !���"����� �� ��� �������� ��� &������ 10 10 Weibull Distribution λ = 0.1, k = 0.5 Weibull Distribution λ = 0.1, k = 0.5 Integrated Hazard Function of the Distribution: Weibull 9 Weibull Distribution λ = 0.5, k = 1.0 9 Weibull Distribution λ = 0.5, k = 1.0 Hazard Function of the Distribution: Weibull Weibull Distribution λ = 0.5, k = 2.0 Weibull Distribution λ = 0.5, k = 2.0 8 8 Weibull Distribution λ = 1.0, k = 3.0 Weibull Distribution λ = 1.0, k = 3.0 7 7 6 6 5 5 4 4 3 3 2 2 1 1 0 0 0 0.5 1 1.5 2 0 0.5 1 1.5 2 t t Heckman Sampling Plans

  17. Observed and Original Distribution for T b (Example 1) 3 The Observed PDF of Spells (T b ) The Original PDF (Weibull Distribution λ = 0.1, k = 0.5) Observed (T b ) and Original PDFs of the Spells 2.5 2 1.5 1 0.5 0 0 0.5 1 1.5 2 t Heckman Sampling Plans

  18. Observed and Original Distribution for T b (Example 3) 2.5 The Observed PDF of Spells (T b ) The Original PDF (Weibull Distribution λ = 0.5, k = 2.0) Observed (T b ) and Original PDFs of the Spells 2 1.5 1 0.5 0 0 0.5 1 1.5 t Heckman Sampling Plans

  19. Observed and Original Distribution for T b (Example 4) 1.6 The Observed PDF of Spells (T b ) The Original PDF (Weibull Distribution λ = 1.0, k = 3.0) 1.4 Observed (T b ) and Original PDFs of the Spells 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 t Heckman Sampling Plans

  20. Observed and Original Distribution for T c (Example 1) 3 The Observed PDF of Spells (T c ) The Original PDF (Weibull Distribution λ = 0.1, k = 0.5) s l l 2.5 e p S e h t f o 2 s F D P l a n 1.5 i g i r O d n a ) 1 T c ( d e v r e 0.5 s b O 0 0 0.5 1 1.5 2 t Heckman Sampling Plans

  21. Observed and Original Distribution for T c (Example 2) 2 The Observed PDF of Spells (T c ) 1.8 The Original PDF (Weibull Distribution λ = 0.5, k = 1.0) s l l e p 1.6 S e h t 1.4 f o s F D 1.2 P l a n i 1 g i r O d 0.8 n a ) T c 0.6 ( d e v 0.4 r e s b O 0.2 0 0 0.5 1 1.5 2 t � Heckman Sampling Plans

  22. Observed and Original Distribution for T c (Example 3) 2.5 The Observed PDF of Spells (T c ) The Original PDF (Weibull Distribution λ = 0.5, k = 2.0) Observed (T c ) and Original PDFs of the Spells 2 1.5 1 0.5 0 0 0.5 1 1.5 t Heckman Sampling Plans �

  23. Observed and Original Distribution for T c (Example 4) 1.6 The Observed PDF of Spells (T c ) The Original PDF (Weibull Distribution λ = 1.0, k = 3.0) 1.4 Observed (T c ) and Original PDFs of the Spells 1.2 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 t Heckman Sampling Plans

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