Duration Models Introduction to Single Spell Models James J. - PowerPoint PPT Presentation

Duration Models Introduction to Single Spell Models James J. Heckman University of Chicago Econ 312, Spring 2019 Heckman Duration Models

• The hazard function gives the probability that a spell, denoted by the nonnegative random variable T with distribution g ( t ), will end at t , given that it has lasted until t : g ( t ) h ( t ) = f ( t | T > t ) = 1 − G ( t ) ≥ 0 . • Integrated hazard function (using G (0) = 0 to eliminate c ): � t h ( u ) du = − ln(1 − G ( t )) | t H ( t ) = 0 + c = − ln(1 − G ( t )) . 0 • Working backwards, we can derive g from h : � t 0 h ( u ) du = 1 − e − H ( t ) , 1 − e − G ( t ) = h ( t )[1 − G ( t )] = h ( t ) e − H ( t ) , g ( t ) = � t H ( t ) = h ( u ) du . 0 Heckman Duration Models

• So the survival function, the probability that the spell lasts until t , i.e., T ≥ t , is S ( t ) = 1 − G ( t ) = e − H ( t ) . • The density and hazard function for T may have a number of qualities. If T has a nondefective duration density, then � t lim h ( u ) du → ∞ ⇐ ⇒ S ( ∞ ) = 0 t →∞ 0 • Duration dependence arises when ∂ h ( t ) � = 0. ∂ t • If ∂ h ( t ) > 0 ( < 0), then we have positive (negative) duration ∂ t dependence Heckman Duration Models

• In constructing estimable models, we will often work with the conditional hazard function h ( t | x ( t ) , θ ( t )) , where the regressor vector x ( t ) may include � t • Entire past: x 1 ( t ) = ∞ k 1 ( z 1 ( u )) du � ∞ • or future: x 2 ( t ) = k 2 ( z 2 ( u )) du � ∞ t • or both: x 3 ( t ) = −∞ k 3 ( z 3 ( u ) , t ) du of some variables. Heckman Duration Models

• Associated with the conditional hazard function is the conditional survival function � t S ( t | x ( t ) , θ ( t )) = 1 − G ( t | x ( t ) , θ ( t )) = e − 0 h ( u | x ( u ) ,θ ( u )) du and the conditional density of T g ( t | x ( t ) , θ ( t )) = h ( t | x ( t ) , θ ( t )) · [1 − G ( t | x ( t ) , θ ( t ))] � t h ( t | x ( t ) , θ ( t )) · e − 0 h ( u | x ( u ) ,θ ( u )) du . = • In these models we will assume 1 θ ( t ) independent of x ( t )and θ ∼ µ ( θ ), x ∼ D ( x ) 2 No functional restrictions connecting the conditional distribution of T | θ, x and the marginal distribution of θ , x . Heckman Duration Models

• A common specification of the conditional hazard is the proportional hazard specification: h ( t | x ( t ) , θ ( t )) = ψ ( t ) φ ( x ( t )) η ( θ ( t )) ln h ( t | x ( t ) , θ ( t )) = ln ψ ( t ) + ln φ ( x ( t )) + ln η ( θ ( t )) ψ ( t ) ≥ 0 , φ ( x ( t )) > 0 , η ( θ ( t )) ≥ 0 ∀ t . Heckman Duration Models

Sampling Plans and Initial Condition Problems Heckman Duration Models

• For interrupted spells, one of the following duration times may be observed: • time in state up to sampling date ( T b ) • time in state after sampling date ( T a ) • total time in completed spell observed at origin of sample ( T c = T a + T b ) • Duration of spells beginning after the origin date of the sample, denoted T d , are not subject to initial condition problems. • The intake rate, k ( − t b ), is the proportion of the population entering a spell at − t b . Heckman Duration Models

• 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 is nondefective � ∞ • m = 0 xg ( x ) dx < ∞ Heckman Duration Models

• The proportion of the population experiencing a spell at t = 0, the origin date of the sample, is � ∞ � ∞ P 0 = k ( − t b )(1 − G ( t b )) dt b = k (1 − G ( t b )) dt b 0 0 � � � t b (1 − G ( t b )) | ∞ = k 0 − t b d (1 − G ( t b )) � = k t b g ( t b ) dt b = km , where 1 − G ( t b ) is the probability the spell lasts from − t b to 0 (or equivalently, from 0 to − t b ). Heckman Duration Models

• 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 f ( t b ) = total surviving til t = 0 k ( − t b )(1 − G ( t b )) = 1 − G ( t b ) = � = g ( t b ) P 0 m Heckman Duration Models

• The probability that a spell lasts until t c given that it has lasted from − t b to 0, is g ( t c ) g ( t c | t b ) = 1 − G ( t b ) • So the density of a spell that lasts for t c is � t c f ( t c ) = g ( t c | t b ) f ( t b ) dt b 0 � t c g ( t c ) m dt b = g ( t c ) t c = m 0 Heckman Duration Models

• Likewise, the density of a spell that lasts until t a is � ∞ f ( t a ) = g ( t a + t b | t b ) f ( t b ) dt b 0 � ∞ g ( t a + t b ) = dt b m 0 � ∞ 1 = g ( t b ) dt b m t a 1 − G ( t a ) = m • So the functional form of f ( t b ) ≈ f ( t a ). Heckman Duration Models

• Some useful results that follow from this model: 1 If g ( t ) = θ e − t θ , then f ( t b ) = θ e − t b θ and f ( t a ) = θ e − t a θ . Proof : g ( t ) = θ e − t θ → m = 1 θ, G ( t ) = 1 − e − t θ → f ( t a ) = 1 − G ( t ) = θ e − t θ m 2 (1 + σ 2 2 E ( T a ) = m m 2 ). Heckman Duration Models

• 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 a (1 − G ( t a )) | ∞ 2 t 2 = 0 − a d (1 − G ( t a )) m � 1 1 a g ( 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 Duration Models

2 (1 + σ 2 • E ( T b ) = m m 2 ). • Proof : See proof of Proposition 2. • E ( T c ) = m (1 + σ 2 m 2 ). • Proof : � t 2 c g ( 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 Duration Models

• h ′ ( t ) > 0 → E ( T a ) = E ( T b ) > m . • Proof: See Barlow and Proschan. • h ′ ( t ) < 0 → E ( T a ) = E ( T b ) < m . • Proof : See Barlow and Proschan. Heckman Duration Models

Pitfalls in Using Regression Methods to Analyze Duration Data Heckman Duration Models

1 Density of duration in a spell ( T ) for an individual with fixed characteristics Z is f ( t | Z ). 2 Assume 1 No time elapses between end of one spell and beginning of another, 2 No unobserved heterogeneity components, 1 3 f ( t | Z ) = θ ( Z ) e − θ ( Z ) t , θ ( Z ) = β Z > 0, 4 At origin, t = 0, of sample of length K , everyone begins a spell. 3 The expected length of spell in the population given Z is � ∞ 1 E ( T | Z ) = tf ( t | Z ) dt = θ ( Z ) = β Z . 0 Heckman Duration Models

1 The expected length of a spell in a sample frame of length k , however, is � K � ∞ E ( T | Z , K ) = tf ( t | Z ) dt + K f ( t | Z ) dt 0 K � ∞ � K t θ e − θ t dt + K θ e − θ t dt = 0 K � K �� ∞ � � � − te − θ t | K e − θ t dt θ e − t θ dt = 0 + + K 0 K � � K �� � − 1 − Ke − θ K + θ e − θ t − e − θ t | ∞ � � � = + K � K � 0 − Ke − θ K − 1 θ e − θ K + 1 θ + Ke − θ K = β Z (1 − e − K β Z ) � = β Z . = Heckman Duration Models

• So OLS of T on Z will not estimate β . But as K → ∞ , the selection bias term ( β Ze − K β Z ) disappears. • A widely used method to avoid this bias is to use only completed first spells. • This results in another sort of selection bias, � K 0 tf ( t | Z ) dt E ( T | Z , K , T < K ) = � K 0 f ( t | Z ) dt − Ke − θ K − 1 θ e − θ K + 1 θ = , 1 − e − θ K where recall that β Z = 1 θ. • As K → ∞ , E ( T | Z , K , T < K ) = β Z . Heckman Duration Models