ECON 721: Lecture Notes on Duration Analysis Petra E. Todd Fall, 213
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Contents 1 Two state Model, possible non-stationary 1 1.1 Hazard function.......................... 1 1.2 Examples............................. 3 1.3 Expected duration........................ 4 1.4 The Exponential Distribution.................. 5 1.5 Show that a distribution with any form of hazard function can be tranformed into a constant hazard.............. 6 1.6 Introducing Covariates...................... 6 1.7 Variety of Parametric Hazard Models.............. 7 1.8 How would you estimate?.................... 1 2 Multi-state models 11 2.1 Stationary Models........................ 11 2.1.1 Alternative way of computing the likelihood if you only have information on number of spells (not on length of spell)............................ 15 2.2 Nonstationary Models...................... 16 2.2.1 Applications (with nonstationarity)........... 16 2.2.2 The Problem of Left-Censored Spells (Nickell, 1979).. 18 2.3 Examples............................. 18 3 Cox s Partial MLE 21 4 Nonparametric Identification - Heckman and Singer (1981) result 23 5 Mixed Proportional Hazard Models 25 i
ii CONTENTS 6 Nonparametric Estimation of the Survivor Function: The Kaplan-Mier Estimator 27 7 Competing risks model 31
Chapter 1 Two state Model, possible non-stationary reference: Lancaster s Econometric Analysis of Transition Data Assume people can be in one of two possible states (e.g. married or unmarried, alive or dead, employed or unemployed, pregnant or not) Probability of leaving between t, t + dt having survived to t Pr(t apple T apple t + dt T t) 1.1 Hazard function Define the hazard function as By Bayes rule h(t) = lim dt! P (t apple T apple t + dt T t) dt P (t apple T apple t + dt T t) = = P (t apple T apple t + dt, T t) P (T t) F (t + dt) F (t) 1 F (t) =S(t) 1
2 CHAPTER 1. TWO STATE MODEL, POSSIBLE NON-STATIONARY The denominator, S(t), is called the survivor function. of exit times, and F (t) istheassociatedcdf. f(t) is the density F (t + dt) h(t) = lim dt! dt F (t) 1 S(t) = F (t) S(t) = f(t) S(t) Therefore, and Z t h(t) = h(s)ds = Z t d log(s(t)) dt d log(s(t)) ds + C ds = logs(s) t + C = logs(t) log S() + C = logs(t)+c
1.2. EXAMPLES 3 The initial condition S() = 1 (everyone survives in the beginning), implies that C =. The survivor function can be written as: S(t) = 1 F (t) =exp{ Z t h(s)ds} Using that h(t) = f(t) S(t) we get f(t) = h(t)s(t) =h(t)exp{. Z t h(s)ds} If exit is certain, then lim S(t) =, t!1 which implies Z t lim t!1 h(s)ds = 1, if not, then h(t) iscalledadefective hazard. 1.2 Examples
4 CHAPTER 1. TWO STATE MODEL, POSSIBLE NON-STATIONARY 1.3 Expected duration What is the expected total duration in a state conditional on surviving to time s? g(t t e(s) = s) = f(t) S(s) Z 1 s p.d.f. t f(t) S(s) dt = 1 S(s) Z 1 s tf(t)dt Integrate by parts: lim m!1 Z m s tf(t)dt = lim tf (t) m s m!1 Z m s F (t)dt Z m = lim {mf (m) sf (s) F (t)dt m + s + m!1 Z m = lim { ms(m)+ss(s)+ S(t)dt} m!1 = ss(s)+ Z 1 s S(t)dt s s Z m s dt}
1.4. THE EXPONENTIAL DISTRIBUTION 5 Thus, expected durations as of time s and as of time are: e(s) = s + 1 S(s) e() = Z 1 S(t)dt Z 1 s S(t)dt Result: Mean duration is the integral of the survivor function. 1.4 The Exponential Distribution When the hazard is independent of how long the state has been occupied, the integral is Z t h(s)ds = t The survivor function is S(t) = exp{ t}, >,t and the pdf is f(t) = exp{ t} This is the exponential probability density function. distributed exponental with parameter. We can say that T is
6 CHAPTER 1. TWO STATE MODEL, POSSIBLE NON-STATIONARY 1.5 Show that a distribution with any form of hazard function can be tranformed into a constant hazard Pr(T t) =S(t) =exp{ Z t h(s)ds} Let Z(T ) = Z T h(s)ds Because h(s) >, Z(T )isamonotonictransformation. Pr(T t) =Pr(Z(T ) Z(t)) = exp{ Z(t)}, This expression is the survivor function for a random variable at point Z(t), so we can conclude that Z has an exponential distribution. This shows that the integrated hazard is a unit exponential variate. A distribution with any form of the hazard function can be transformed into a constant hazard (i.e. exponential form) by a suitable transformation of the time scale. 1.6 Introducing Covariates time-invariant: sex, race, perhaps education time-varying: age, business cycle e ects, month
1.7. VARIETY OF PARAMETRIC HAZARD MODELS 7 h(t; x) = = P (t apple T apple t + dt T t, x) dt f(t; x) S(t; x) and S(t; x) = exp{ Z t h(s; x)ds} We could do all the estimation within x cells. Alternatively, we could specify a functional form for how the hazard function depends on observables x, suchas S(t; x) = exp{ Z t h(s; x)ds} =exp{ x } 1.7 Variety of Parametric Hazard Models Would like to use economic theory to guide in selection of appropriate functional forms, depending on context Weibull family f(t) = t 1 exp{ ( t) } h(t) = t 1 S(t) = exp{ ( t) } = exp{ x} This hazard depends on time (unless =1)butisrestrictedtobemonotonic. =1givestheexponentialhazard.
8 CHAPTER 1. TWO STATE MODEL, POSSIBLE NON-STATIONARY Proportional hazard h(x, t) = k 1 (x)k 2 (t) This form implies that hazards for two people with x = x 1 and x = x 2 are in the same ratio for all t, but not necessarily monotonic. k 2 is called the baseline hazard. Akeyadvantageofthismodelisthatwecanestimatek 1 (x) without having to specify k 2 (t).(moreonthislater) If k 1 (x) =k 1 (x(t)), then only get proportional hazard model if covariates vary same way for all. Box-Cox Model h(t) = k 1 (x) exp{ t ( ) }
1.7. VARIETY OF PARAMETRIC HAZARD MODELS 9 Box-Cox transformation: t ( ) = t 1 t (1) = t 1 t 1 e ln t 1 lim = lim!! e ln t ln t = lim (by l hopital srule)! 1 = lnt As!, we get h(t) = k 1 (x) exp{ t ( ) } = k 1 (x) exp{ ln t} = k 1 (x) t Weibull The Box-Cox generates Weibull as a special case. Still, monotonic though dt dt = t 1, which is of constant sign Generalized Box-Cox Flinn and Heckman (1982) suggest using instead h(t) =k 1 (x) exp{ 1 t ( 1) + 2 t ( 2) } (for example, 1 =1and 2 =2). This form now allows for a nonmonotonic hazard.
1CHAPTER 1. TWO STATE MODEL, POSSIBLE NON-STATIONARY 1.8 How would you estimate? If you observe all exit times, the likelihood is L(t; 1, 2,, 1, 2) = n i=1[1 S(t i )] = n i=1[1 exp{ Z t h(s, x)ds}] If there is no closed form solution for the survival function, then evaluating the likelihood requires numerical integration.
Chapter 2 Multi-state models reference: Amemiya, Chapter 11 i jk(t) t =Prob(personiobservedinstatekattimet + t in state j at time t) i If jk (t) = i jk, then the model is stationary. Will start with stationary models, then do nonstationary. 2.1 Stationary Models The probability that a person stays in state j in period (,t)andthenmoves to k in period (t, t + t) (eventa) isgivenby P (A) = (1 j t) t/ t jk t (treating t/ t) asaninteger), where is the probability of exiting j. j = MX k=1 jk jj 11
12 CHAPTER 2. MULTI-STATE MODELS Note that lim (1 1 n!1 n )n = e 1 Now, using stationarity, one can show that lim t! (1 j t) t/ t = exp{ jt} Thus, P (A) = exp{ jt} jk t Because t does not depend on parameters, we can drop it and regard exp{ jt} jk as the likelihood. Example 1: Suppose M=3 and event history is state 1 in period (,t 1 ) state 2 in period (t 1,t 1 + t 2 ) state 3 in period t 1 + t 2 to t 1 + t 2 + t 3 then back to state 1 The likelihood in this case is given by L = exp( 1t 1 ) 12 exp( 2t 2 ) 23 exp( 3t 3 ) 31
2.1. STATIONARY MODELS 13 exp( 1t 1 )istheprobabilityofsurvivingt 1 periods in state 1. 12 is the probability of transiting to state 2 exp( 2t 2 )istheprobabilityofsurvivinginstate2 23 is the probability of transiting to state 3, etc... If instead we observe the person leaving state 3 but do not know where the person went, then replace 31 by 3 L = exp( 1t 1 ) 12 exp( 2t 2 ) 23 exp( 3t 3 ) 3 Now suppose that we terminate observation at time t 1 + t 2 + t 3 without knowing whether a person continues to stay in state 3 or not (right censoring), then would drop term 3 altogether. Example 2: Two State model M =2, 1 = 12, 2 = 21 state #1 is unemployment, state #2 is employment Suppose an individual experiences r completed unemployment spells of length t 1,t 2,...,t r. Then L = r 1e T (contribution of r unemp spells to the likelihood) T = rx j=1 t j For overall likelihood, would also need the corresponding part for employment spells.
14 CHAPTER 2. MULTI-STATE MODELS Here, F (t) = 1 P (T >t)=1 e t (cdf of a r.v. signifying duration of a spell) t f(t) = e (density of observed duration spell) = f(t) 1 F (t) = e t e t hazard rate t = f(t) t 1 F (t) = Prob(leaves unemployment in t, t + t has not left up to t) is the hazard rate. Example #3: Suppose we observe one completed unemployment spell of duration t i for the ith individual: L = N i=1f i (t i )= N i=1 i exp( i t i ) where i is the probability of exiting unemployment and exp( i t i )isthe probability of being unemployed t i periods. Now suppose that individuals 1..n complete their unemployment spells of duration t i but that individuals n +1..N are right censored at time t i. The likelihood is then given by L = n i=1f i (t i ) N i=n+1[1 F i (t i )] Note how the duration model with right censoring is similar to a standard Tobit model. Previously, the likelihood depended on the observed duration t 1,t 2,...,t r only through r (number of spells) and T (total length of time in state). This implies that r and T are su cient statistics, which is a property of a stationary model.
2.1. STATIONARY MODELS 15 2.1.1 Alternative way of computing the likelihood if you only have information on number of spells (not on length of spell) Assume that there are two completed unemployment spells and that the third spell is incomplete. T = total unemployment time Probability of observing two completed spells and one censored spell in total time T is P ( apple t 1 <t, <t 2 apple T t 1,t 3 T t 1 t 2 ) = = Z T Z T f(z 1 ) ( T ) 2 e T 2 z1 applez 1 f(z 2 ) T z 1 f(z 3 )dz 3 z 2 dz 2 dz 1 The probability of observing r completed spells in total time T is Pr(r, T )= The likelihood is given by ( T ) r e T r!, which is poisson L = N i=1 r i i exp{ it i } Now, assume that depends on individuals characteristics x i i =exp{ + x i } Amemiya, Chapter 11, derives MLE estimators for,.
16 CHAPTER 2. MULTI-STATE MODELS 2.2 Nonstationary Models Relax assumption that i jk (t) = i jk (constant hazard rate) Distribution function of duration under a nonstationary model F (t) = 1 exp[ f(t) = (t)exp[ Z t Z t (z)dz] (z)dz] We can write the likelihood function in the integral representation as before. Suppose that i (t) = g(x it ; ) We need to specify x it as a continuous function of t. 2.2.1 Applications (with nonstationarity) Tuma, Hannan and Groeneveld (1979) Study marriage duration. Divide sample period into 4 subperiods and assume that the hazard rate is constant between subperiods. i (t) = P x i, t 2 T p p =1, 2, 3, 4 T p is the pth subperiod
2.2. NONSTATIONARY MODELS 17 Lancaster (1979) Studies unemployment duration F (t) = 1 exp( t ) (Weibull distribution, nonstationary because it depends on t) (t) = t 1 (hazard function) > 1 =) @ @t = 1 =) @ @t < 1 =) @ @t > (increasing hazard) = (constanthazard,=exponential) < (decreasing hazard) Lancaster introduced covariates by specifying i (t) = t 1 exp( x i ), x i is constant over time Lancaster found a pattern of decreasing bazard, but he would that the hazard decreased less when more covariates were included. Thus, he was concerned that the finding of negative duration dependence might be due to omitted unobservables. This led him to consider an alternative specification for the hazard rate that explicitly incorporated unobservables. µ i (t) = v i i (t), v i unobservable, assumed to be iid gamma(1, 2 ) v i is a proxy for unobservable, exogenous variables Heckman and Borjas (198) Study of unemployment duration l =lth unemployment spell experienced by an individual il (t) = t 1 exp( lx il + v i ), where v i is an unobservable that needs to be integrated out to obtain the marginal distribution function of duration. Flinn and Heckman (1982)
18 CHAPTER 2. MULTI-STATE MODELS Use a modified version of the Box-Cox hazard: il (t) =exp[ t lx il (t)+c l v i + 1 1 1 If you set 1 =and 2 =, get a Weibull model Here, x il (t) isassumedtobeexogenous. 1 t 2 1 + 2 ] 2 2.2.2 The Problem of Left-Censored Spells (Nickell, 1979) The problem occurs if individuals are not observed at the start of their unemployment spells. 2.3 Examples Three Cases considered in the literature: (i) s observed, t not observed (ii) both s and t observed (iii) t is observed but s is not observed Case (i): Analyzed by Nickell (1979) in studying unemployment Assume that the beginning of the spell (s) is observed, end of the spell (t) not observed. We observe that individual is unemployed at time t. Also, assume that the P [U started in ( s s, s)] does not depend on s (constant entry rate). For su ciently small s, g(s) s = P [U started in ( s s, s) U at ] = P [U at U started in ( s s, s)] sp [U started in ( s s, s)] (Numerator)ds = Pr(U at ) R 1
2.3. EXAMPLES 19 -s Start of spell Time of interview t End of spell The denominator integrates the numerator over all possible dates when the spell could have started. Assuming that P [U started in ( s s, s)] does not depend on s: = = g(s) = P [U at U started in ( s s, s)] s R 1 (Numerator)ds [1 F (s)] s [1 F (s)] s = [1 F (s)]ds ES = R 1 (show by integration by parts) sf(s)ds R 1 so, [1 F (s)]. ES Case (ii): Lancaster (1979) Assume that both s and t are observed Need the joint density g(s, t) = g(t s)g(s). The density g(s) wasderivedabove.
2 CHAPTER 2. MULTI-STATE MODELS Let X denote total unemployment duration. First, evaluate P (X >s+ t X >s) = = = P (X >s+ t,x >s) P (X >s) P (X >s+ t) P (X >s) 1 F (s + t). 1 F (s) (*) Therefore, Pr(X <s+ t x >s)=1 1 F (s + t) 1 F (s) = F (s + t) F (s) 1 F (s) Di erentiating with respect to t gives g(t s) = f(s + t) 1 F (s), which is the Pr(spell ends at time s+t given that it started at time s). Combining with the earlier results, we get g(s, t) = f(s + t) ES. Case (iii): Flinn and Heckman (1982): t is observed but s is not observed obtain g(t) byintegratingg(t, s) withrespecttos : g(t) = 1 ES Z 1 = 1 F (t). ES f(s + t)ds
Chapter 3 Cox s Partial MLE Consider proportional hazard model of the form i (t) = (t)exp{ x i } and assume that data are right censored (observe start of spell but do not observe length of spell for some individuals). Let t i,i =1, 2,...,n be completed durations and let t i,i= n +1,n+2,n+ 3,...,N be censored durations. The likelihood is given by Z t L = n i=1 exp{ x i } (t i )exp[ exp( x i ) N i=n+1 exp[ exp( x i ) Z t (z)dz]. (z)dz] Through some algebraic manipulation, can obtain n i=1 exp( x i ) (t i )exp{ Z 1 [ X h2r(t) exp( x h )] (t)dt}, where R(t) ={i t i t}. 21
22 CHAPTER 3. COX S PARTIAL MLE Cox (1975) suggested and Tsiatis (1981) proved that the likelihood could be decomposed into two components: L = L 1 L " 2 # = n exp( x i ) i=1 P h2r(t) exp( x h ) 2 3 X exp( x h ) (t i ) 5 exp 4 n i=1 h2r(t) @ Z 1 [ X h2r(t) 1 exp( x h )] (t)dta can be ob- and that a consistent and asymptotically normal estimator for tained by maximizing L 1 (the partial MLE). This means that can be estimated without specifying (t). This is a remarkable result, given that L 1 is not a proper likelihood. See Amemiya, Ch. 11, for an interpretation of the di erent components.
Chapter 4 Nonparametric Identification - Heckman and Singer (1981) result Ask what features of hazard functions can be identified from the raw data (i.e. G(t x)). Denote unobservables by. Would like to infer properties of G(t x, ) without having to impose strong parametric assumptions, either on µ( ) orh(t x, ). Assume that x(t) is constant. Heckman and Singer (1981) show that if G(t x) exhibits positive duration dependence, then it must be that h(t x, ) also exhibits positive duration dependence over some interval of values in those intervals of t. Also, show that omitting omitted variables leads to tendency towards negative duration dependence. Consider hazard of the form h(t x, ) = (t) (x), 23
24CHAPTER 4. NONPARAMETRIC IDENTIFICATION - HECKMAN AND SINGER (1981 (time- which gives the proportional hazard model with (t) =, x(t) =x invariant unobserved heterogeneity and time invariant regressors). Let h(t x) betheconditionalhazard,notcontrollingforunobservables. Let F (t x, ) andf (t x) betheconditionaldistributions. h(t x) = = R f(t x, )dµ( ) R [1 F (t x, )]dµ( ) R h(t x, )(1 F (t x, ))dµ( ) R [1 F (t x, )]dµ( ) @h(t x) @t = R @h(t x, ) @t (1 F (t x, ))dµ( ) R [1 F (t x, )]dµ( ) +nonpositiveterm. This shows that ignoring unobservables will bias the hazard downwards.
Chapter 5 Mixed Proportional Hazard Models Consider the Cox Proportional Hazard model: h(t X;, )= (t)exp( x) Cox (1972) observed that it is possible to estimate parameters without specifying (t) using the partial likelihood approach discussed earlier. It is a semi parametric estimation approach, because (t) isleftunspecified. Now, suppose we want to introduce unobservables and want to be flexible about the way in which regressors enter. Let the unobservables be distributed, which is unknown to the econometrician. The model that includes unobservables is called the Mixed Proportional Hazard Model. h(t X, U;, ) = (t)exp(z(x, ))exp(u) Elbers and Ridder (1982) and Heckman and Singer (1984) Showed that this model, with regressors, is non parametrically identified under some restrictions on. Elbers and Ridder (1982) assumed that E(exp(U)) < 1 (Heckman and Singer (1984) consider alternative assumptions). 25
26 CHAPTER 5. MIXED PROPORTIONAL HAZARD MODELS Honore (1993) Studied a multi-spell generalization where T 1 and T 2 are the length of multiple spells. He showed the model can be identified under weaker conditions on. Hahn (1994) derives the semi parametric e ciency bound, whichprovidesaboundonthe attainable statistical accuracy (it was first introduced by Stein (1956) and has been derived for a number of di erent models). Because semi parametric estimation must be at least as di cult as any parametric sub model (model allowed under the semi parametric model), it follows that the asymptotic variance of any p N-consistent estimator is no smaller than the supremum of the Cramer-Rao lower bounds for all parametric sub models. The infimum of the information matrix for,theinverseofthe Cramer-Rao lower bound, gives the semi parametric version of the information matrix, called the semi parametric information bound. Hahn (1994) shows that for the single-spell Weibull mixed proportional hazard model, the information matrix is singular, which implies that there cannot exist a p N-consistent estimator. He also shows this to be the case for the multi-spell version of the model. Thus, even though the model is identified, there is no p N-consistent estimator. Ridder and Woutersen (23) Present new conditions for the mixed proportional hazard model under which parameters are identified and under which the information matrix is nonsingular. The paper also presents an estimator that converges at a p N rate. The key additional assumption is that the baseline hazard needs to be bounded away from and 1 near t =.
Chapter 6 Nonparametric Estimation of the Survivor Function: The Kaplan-Mier Estimator Allow for right censored exit times No regressors, but could accommodate regressors by doing everything within x cells. Does not allow for unobservables N possibly right-censored exit times M apple N distinct exit times t (1), t (2),t (3),...,t (M), where multiple people can exit at the same time n j =number leaving at time t j. Ŝ(t) = 1 = 1 number leaving before t N n 1 + n 2 +... + n k, k =max such that t j <t N j 27
28CHAPTER 6. NONPARAMETRIC ESTIMATION OF THE SURVIVOR FUNCTION: TH We can write Ŝ(t) = N n 1 n 2.. n k N N n1 N n1 n 2 N n1 n 2 n 3 = N N n 1 N n 1 n 2 N n1 n 2... n n N n 1... n k 1 n 2 n 3 = 1 1... 1 n 1 1 N N n 1 n n N n 1... n k 1, N n 1 n 2... where the ratios represent the number leaving of those who survive (the risk set). Thus, the hazard function is: ˆ j = n j N n 1... n j 1 The survivor function a time t (which would be used to handle right censoring) can be written as: Ŝ(t) = tj <t(1 ˆ j ) (called a product limit estimator) The term ˆ j is the hazard rate. up to that date.) (prob of leaving at date t j given survived The Kaplan Mier estimator of the survivor function looks like a step function:
29 Kaplan-Mier Survivor Function 1. t
3CHAPTER 6. NONPARAMETRIC ESTIMATION OF THE SURVIVOR FUNCTION: TH
Chapter 7 Competing risks model J causes of failure 1..J T j latent failure time from cause j Observe duration to first failure and associated cause. That is, observe time of death and cause of death. (e.g. observe death from cancer or heart disease) (T,I) = {min j (T j ), argmin j (T j )} In application, there may be considerable content in models with regressors. For example, the goal may be to study how smoking, blood pressure and weight a ect the marginal distribution of time to death attributable to heart attack or cancer. In models without regressors, need to make functional form assumptions about the joint distribution of failure times. Heckman and Honore (1989) study identification in models with regressors. 31
32 CHAPTER 7. COMPETING RISKS MODEL For example, in Cox proportional hazard model: S(t x) = exp{ z(t)'(x)} ('(x) usuallye x ), we can assume that each potential failure time has a proportional hazard specification. Can specify joint survivor function of T 1,T 2 conditional on x. S(t 1,t 2 x) =K[exp{ Z 1 (t 1 )' 1 (x)}, exp{ Z 2 (t 2 )' 2 (x)}] Theorem: Assume that (T 1,T 2 )hasthejointsurvivorfunctionasgivenabove. Then ' 1,' 2,Z 1 and Z 2 are identified from the observed minimum of (T 1,T 2 )under the following assumptions: (i) K is c 1 with partial derivatives K 1 and K 2 and for i =1, 2thelimitas n!1of K( 1n, 2n ) is finite for all sequences 1n! 1, 2n! 1forn!1. K is strictly increasing in each of its arguments in all of [ 1] [ 1] (ii) Z 1 (1) = 1,Z 2 (1) = 1,' 1 (x )=1,' 2 (x ) = 1 for some fixed point in the support of x. (iii) The support of {' 1 (x),' 2 (x)} is ( 1) ( 1) (iv) Z 1 and Z 2 are nonnegative, di erentiable, strictly increasing functions, except that we allow them to be 1 for finite t. Sketch of Proof: Define: Q 1 (t) = pr(t 1 >t,t 2 >T 1 ) (die from cause 1 by time T 1 ) Q 2 (t) = pr(t 2 >t,t 1 >T 2 ) (die from cause 2 by time T 2 ) Q 1(t) = K 1 [exp{ Z 1 (t)' 1 (x)} exp{ Z 2 (t 2 )' 2 (x)}]exp{ Z 1 (t)' 1 (x)}z 1(t)' 1 (x)
The ratio of Q 1 at two points x and x (in the support of X) andusingthe assumptions on x (assumption (iii)) gives: 33 = K 1 [exp{ Z 1 (t)' 1 (x)} exp{ Z 2 (t 2 )' 2 (x)}]exp{ Z 1 (t)' 1 (x)}z 1(t)' 1 (x) K 1 [exp{ Z 1 (t)' 1 (x )} exp{ Z 2 (t 2 )' 2 (x )}]exp{ Z 1 (t)' 1 (x )}Z 1(t)' 1 (x ) Take lim above expression=' 1 (x) t! By a symmetric argument, we can identify ' 2 (x). Heckman and Honore also provide approaches to identify K, Z 1 (t),z 2 (t).