Numerical Methods-Lecture XIII: Dynamic Discrete Choice Estimation

Transcription

1 Numerical Methods-Lecture XIII: Dynamic Discrete Choice Estimation (See Rust 1989) Trevor Gallen Fall / 1

2 Introduction Rust (1989) wrote a paper about Harold Zurcher s decisions as the superintendent of maintenance at the Madison (Wisconsin) Metropolitan Bus Company. 10 years of monthly data bus mileage and engine replacement 104 buses 1 Harold 2 / 1

3 Introduction Rust (1989) wrote a paper about Harold Zurcher s decisions as the superintendent of maintenance at the Madison (Wisconsin) Metropolitan Bus Company. 10 years of monthly data bus mileage and engine replacement 104 buses 1 Harold Not interested in Harold per se 3 / 1

4 Introduction Rust (1989) wrote a paper about Harold Zurcher s decisions as the superintendent of maintenance at the Madison (Wisconsin) Metropolitan Bus Company. 10 years of monthly data bus mileage and engine replacement 104 buses 1 Harold Not interested in Harold per se Interested in application of Dynamic Discrete Choice framework 4 / 1

5 Description Observe monthly bus mileage Observe maintenance diary with date, milage, and list of components repaired or replaced Three types of maintenance operations Routine adjustments (brake adjustments, tire rotation) Replacement of individual components when failed Major engine overhauls Model Zurcher s decision to replace bus engines based on observables and unobservables 5 / 1

6 Description Observe monthly bus mileage Observe maintenance diary with date, milage, and list of components repaired or replaced Three types of maintenance operations Routine adjustments (brake adjustments, tire rotation) Replacement of individual components when failed Major engine overhauls Model Zurcher s decision to replace bus engines based on observables and unobservables 6 / 1

7 Model Agent is forward looking Maximizes expected intertemporal payoff Estimate parameters of the models Test whether the agent s behavior is consistent with the model 7 / 1

8 Replacement data-i 8 / 1

9 Replacement data-ii 9 / 1

10 Replacement data-iii 10 / 1

11 Sample Focus on bus groups 1-4 Most recent acquisitions Have data on replacement costs only for this group Utilization fairly constant within each group (necessary for model) 11 / 1

12 Model - I Harold chooses {i 1, i 2,..., i t,...} to maximize his expected utility max E t β t 1 u(x t, ε t, i t ; θ) {i 1,i 2,...,i t,...} t=1 x t is the total mileage on an engine since last replacement ε t are unobservable (to the econometrician) shocks i t is an indicator of engine replacement θ is vector of parameters (to be discussed below) 12 / 1

13 Model - II Cost function c(x, θ 1 ) = m(x, θ 11 ) + µ(x, θ 12 )b(x, θ 13 ) m(x, θ 11 ) is the conditional expectation of normal maintenance and operation expenditure µ(x, θ 12 ) is the conditional probability of an unexpected engine failure b(x, θ 13 ) is the conditional expectation of towing costs, repair costs, and loss of customer goodwill costs resulting from engine failure No data on maintenance and operating costs, so only estimating the sum of costs c(x, θ 1 ) 13 / 1

14 Model - III Stochastic process governing {i t, x t } is the solution to V θ (x t ) = sup E β j t u(x Π j, f j, θ 1 ) x t j=t where utility u is { given by c(x u(x t, i t, θ 1 ) = t, θ 1 ) if i t = 0 [ P P + c(0, θ 1 ) ] if i t = 1 Π = {f t, f t+1,...} is a sequence of decision rules where each f indicates the optimal choice (replace or not) at time t given the entire history of investment i t 1,..., i 1 and mileage x t, x t 1,..., x 1 observed to date P is the scrap value of an engine P is the cost of a new engine 14 / 1

15 Model - IV Evolution of x t is given by stochastic process: p(x t+1 x t, i t, θ 2 ) θ 2 exp[θ 2 (x t+1 x t )] if i t = 0 & x t+1 x t = θ 2 exp[θ 2 (x t+1 )] if i t = 1 & 0 o/w x t+1 0 Without investment, next period mileage is drawn from exponential CDF 1 exp[θ 2 (x t+1 x t )] With investment, next period mileage is drawn from exponential CDF 1 exp[θ 2 (x t+1 0)] 15 / 1

16 Model - V Bellman: V θ (x t ) = EV θ (x t, i t ) = i t = f (x t, θ) = max u(x t, i t, θ 1 ) + βev θ (x t, i t ) i t {0,1} o V θ (y)p(dy x t, i t, θ 2 ) { 1 if xt > γ(θ 1, θ 2 ) 0 if x t γ(θ 1, θ 2 ) Where γ(θ 1, θ 2 ) is the investment cut-off ( optimal stopping barrier ) given by the unique solution to ( P P)(1 β) = γ(θ1,θ 2 ) 0 [1 β exp{ θ 2 (1 β)y}] c(y, θ 1) ydy 16 / 1

17 Model - VI Making the model stochastic: i t = f (x t, θ) + ε t ε t known to agent but unknown to the econometrician 17 / 1

18 Model - VII C(x t ) Choice set. A finite set of allowable values of the control variable i t when state variable is x t ε t = {e t (i) i C(x t )} A #C(x t )-dimensional vector of state variables observed by agent by not by the econometrician. x t = {x t (1),..., x t (K)} K dimensional vector of state variables observed by both the agent and the econometrician. u(x t, i t, θ 1 ) + ε t (i) Realized single period utility of decision i when state variable is x t, ε t ). θ 1 is a vector of unknown parameters to be estimated p(x t+1, ε t+1 x t, ε t, i t, θ 2, θ 3 ) Markov transitional denisty for state variable (x t, ε t ) when alternative i t is selected. θ 1 and θ 2 are vectors of unknown parameters to be estimated. θ = (β, θ 1, θ 2, θ 3 ) The complete 1 + K 1 + K 2 + K 3 vector of parameters to be estimated 18 / 1

19 Model - VIII Infinite horizon, discounted Harold decision problem: V θ (x t, ε t ) = sup E β j t (u(x Π j, f j, θ 1 ) + ε j (f j )) x t, ε t, θ 2, θ 3 j=t The optimal value function V θ is the unique solution to V θ (x t, ε t ) = with the decision rule i t = f (x t, ε t, θ) arg max [u(x t, i t, θ 1 ) + ε t (i) + βev θ (x t, ε t, i t )] i t C(x t) max [u(x t, i t, θ 1 ) + ε t (i t ) + βev θ (x t, ε t, i t )] i t C(x t) 19 / 1

20 Model - IX Problems: ε t appears nonlinearly. Have to integrate out over ε t to obtain choice probabilities Dimensionality: grid approach to estimating ε would still be too large to be computationally tractable (especially in 1989). 20 / 1

21 Model - X Conditional Independence Assumption: P(x t+1, ε t+1 x t, ε t, i, θ 2, θ 3 ) = q(ε t+1 x t+1, θ 2 )P(x t+1 x t, i, θ 3 ) 21 / 1

22 Estimation-I Three likelihood functions. Two partial likelihoods: T l 1 (x 1,..., x T, i 1,..., i T x 0, i 0, θ) = p(x t x t 1, i t 1, θ 3 ) l 2 (x 1,..., x T, i 1,..., i T θ) = And the full likelihood function: l f (x 1,..., x T, i 1,..., i T x 0, i 0, θ) = Three stages of estimation. t=1 T P(i t x t, θ) t=1 T P(i t x t, θ)p(x t x t 1, i t 1, θ 3 ) t=1 22 / 1

23 Estimation-II Recall, have value function which is given by the functional equation EV θ (x, i) = log exp(u(y, j, θ 1 ) + βev θ (y, j)) p(dy x, i, θ 3) y j C(y) Goal is to estimate θ using a nested fixed point algorithm: For each θ, compute EV θ using a fixed point algorithm Outer hill climbing algorithm searches for the value of θ which maximizes the likelihood function. 23 / 1

24 Estimation-III First estimate the parameters θ 3 of the transition probability { g(xt+1 x p(x t+1 x t, i t, θ 3 ) = t, θ 3 ) if i t = 1 g(x t+1 0, θ 3 ) if i t = 0 Using l 1 (x 1,..., x T, i 1,..., i T x 0, i 0 θ) = T p(x t x t 1, i t 1, θ 3 ) t=1 g is a multinomial distribution on the set {0, 1, 2} corresponding to monthly mileage intervals [0, 5000), [5000, 10000), [10000, ) so θ 3j = Pr{x t+1 = x t + j x t, i t = 0}, j = 0, 1 This first stage doesn t require estimation of EV θ! 24 / 1

25 Estimation-IV 25 / 1

26 Estimation-V 26 / 1

27 Estimation-VI Given θ 3, and using l 2 we can obtain estimates for β, θ 1, and RC (RC = P P): where with EV θ (x, i) = and l 2 (x 1,..., x T, i 1,...i, T ) θ) = T P(i t x t, θ) t=1 exp{u(x, i, θ 1 ) + βev θ (x, i)} P(i x, θ) = j C(x) exp{u(x, j, θ 1) + βev θ (x, j)} y log j C(y) u(x t, i t, θ 1 ) + ε t (i) = exp(u(y, j, θ 1 ) + βev θ (y, j)) p(dy x, i, θ 3) { RC c(0, θ1 ) + ε t (1) if i = 1 c(x t, θ 1 ) + ε t (0) if i = 0 27 / 1

28 Estimation-VII θ 1 are the parameters of the cost function Compare different (parsimonious) specifications Linear and square root forms do well 28 / 1

29 Estimation-VIII 29 / 1

30 Estimation-IX Next look at myopic replacement rule replace only when operating costs c(x t, θ 1 ) exceed current cost of replacement RC + c(0, θ 1 ) This model is rejected β =.999 produces a statistically significantly better fit of the model to the data 30 / 1

31 Estimation-VIII 31 / 1

32 Takeaways So what? Why not just do a logit? 32 / 1

33 Takeaways So what? Why not just do a logit? Rust estimates Harold s problem at a micro level 33 / 1

34 Takeaways So what? Why not just do a logit? Rust estimates Harold s problem at a micro level Changes in interest rates, costs, etc. he can still deal with 34 / 1

35 Takeaways So what? Why not just do a logit? Rust estimates Harold s problem at a micro level Changes in interest rates, costs, etc. he can still deal with Not myopic 35 / 1

36 Takeaways So what? Why not just do a logit? Rust estimates Harold s problem at a micro level Changes in interest rates, costs, etc. he can still deal with Not myopic Good example of indirectly inferring underlying parameters from agent behavior 36 / 1