Sale!

# Problem Set 5 ECON 6343: Econometrics III

\$30.00

Category:

Problem Set 5
ECON 6343: Econometrics III

Directions: Answer all questions. Each student must turn in their own copy, but you may work
in groups. Clearly label all answers. Show all of your code. Turn in jl-file(s), output files and
writeup via GitHub. Your writeup may simply consist of comments in jl-file(s). If applicable, put
the names of all group members at the top of your writeup or jl-file.
You may need to install and load the following package:
DataFramesMeta
You will also need to load the following previously installed packages:
Optim
HTTP
GLM
LinearAlgebra
Random
Statistics
DataFrames
CSV
1
In this problem set, we will explore a simplified version of the Rust (1987, Econometrica) bus
engine replacement model. Let’s start by reading in the data.
using DataFrames
using CSV
using HTTP
url = “https://raw.githubusercontent.com/OU-PhD-Econometrics/fall-2020/
master/ProblemSets/PS5-ddc/busdataBeta0.csv”
Static estimation
1. Reshape the data into “long” panel format, calling your long dataset df long. I have included code on how to do this in the PS5starter.jl file that accompanies this problem
set.
2. The model we would like to estimate is Harold Zurcher’s decision to run buses in his fleet.
Zurcher’s flow utility of running (i.e. not replacing) a bus is
u1 (x1t
,b) = θ0 +θ1x1t +θ2b (1)
where x1t
is the mileage on the bus’s odometer (in 10,000s of miles) and b is a dummy
variable indicating whether the bus is branded (meaning its manufacturer is high-end). The
choice set is {0,1} where 0 denotes replacing the engine.
Estimate the θ parameters assuming Zurcher is completely myopic. This amounts to estimating a simple binary logit model. (Note: you may estimate this any way you wish. I
would recommend using the GLM package, but you may also use Optim with your own log
likelihood function.)
Dynamic estimation
Now I will walk you through how to estimate the dynamic version of this model using backwards
recursion. With discount factor β, the differenced conditional value function for running the bus
(relative to replacing it) is
v1t (xt
,b)−v0t (xt
,b) = θ0 +θ1x1t +θ2b+β
Z
Vt+1 (xt+1,b)dF (xt+1|xt) (2)
where Vt+1 is the value function and the integral is over transitions in the mileage states xt
.
We will approximate the integral with a summation, which means that we will specify a discrete
mass function for f (xt+1|xt). This probability mass function depends on the current odometer
reading (x1t), whether the engine is newly replaced (i.e. dt−1 = 0), and on the value of another
state variable x2 which measures the usage intensity of the bus’s route (i.e. high values of x2 imply
a low usage intensity and vice versa).
We discretize the mileage transitions into 1,250-mile bins (i.e. 0.125 units of x1t). We specify
x2 as a discrete uniform distribution ranging from 0.25 to 1.25 with 0.01 unit increments.
2
Formally, we are discretely (but not discreetly!) approximating an exponential distribution:
fj(x1,t+1|x1,t
, x2) =



e
−x2(x1,t+1−x1t) −e
−x2(x1,t+1+0.125−x1t)
if j = 1 and x1,t+1 ≥ x1,t
e
−x2(x1,t+1) −e
−x2(x1,t+1+0.125)
if j = 0 and x1,t+1 ≥ 0
0 otherwise
(3)
You will not need to program (3); I will provide code for this part.
Under this formulation, (2) can be written as
v1t (xt
,b)−v0t (xt
,b) = θ0 +θ1x1t +θ2b+
β ∑x1,t+1
Vt+1 (xt+1,b)[ f1 (x1,t+1|x1,t
, x2)− f0 (x1,t+1|x1,t
, x2)] (4)
Finally, we can simplify (4) since we know that Vt+1 = log
∑k exp
vk,t+1
when we assume
that unobserved utility is drawn from a T1EV distribution (as we do here):
v1t (xt
,b)−v0t (xt
,b) = θ0 +θ1x1t +θ2b+
β ∑x1,t+1
log
1+exp(v1,t+1 (xt+1,b)−v0,t+1 (xt+1,b))
×
[ f1 (x1,t+1|x1,t
, x2)− f0 (x1,t+1|x1,t
, x2)]
(5)
Estimation of our dynamic model now requires two steps:
Solving the model First, we need to solve the value functions for a given value of our parameters
θ. The way we do this is by backwards recursion. We know that Vt+1 = 0 in our final period (i.e.
when t = T). Then we work backwards to obtain the future value at every possible state in our
model. This will include many states that do not actually show up in our data.
Estimating the model Second, once we’ve solved the value functions, we use maximum likelihood to estimate the parameters θ. The log likelihood function in this case is simply
` =
N

i=1
1

j=0
T

t=1
di jt logPi jt (6)
where
Pi1t =
exp(v1t −v0t)
1+exp(v1t −v0t)
Pi0t = 1−Pi1t
(7)
3. Now estimate the θ’s assuming that Zurcher discounts the future with discount factor β =
0.9. I will walk you through specific steps for how to do this:
(a) Read in the data for the dynamic model. This can be found at the same URL as listed
at the top of p. 2, but remove the “Beta0” from the CSV filename.
Rather than reshaping the data to “long” format as in question 1, we want to keep the
data in “wide” format. Thus, columns :Y1 through :Y20 should be converted to an
array labeled Y which has dimension 1000 × 20 where N = 1000 and T = 20. And
similarly for columns starting with :Odo and :Xst. Variables :Xst* and :Zst keep
track of which discrete bin of the fj’s the given observation falls into.
(b) Construct the state transition matrices, which are the fj’s in (3). To do so, simply
run the following code:
zval,zbin,xval,xbin,xtran = create_grids()
zval and xval are the grids defined at the bottom of p. 2, which respectively correspond to the route usage and odometer reading. zbin and xbin are the number of
bins in zval and xval, respectively. xtran is a (zbin*xbin)×xbin Markov transition
matrix1
that gives the probability of falling into each x1,t+1 bin given values of x1,t and
x2, according to the formula in (3).
(c) Compute the future value terms for all possible states of the model.
• First, initialize the future value array, which should be a 3-dimensional array of
zeros. The size of the first dimension should be the total number of grid points
(i.e. the number of rows of xtran). The second dimension should be 2, which is
the possible outcomes of :Branded. The third dimension should be T +1.
• Now write four nested for loops over each of the possible states:
– Loop backwards over t from T +1 to 1
– Loop over the two possible brand states {0,1}
– Loop over the possible permanent route usage states (i.e. from 1 to zbin)
– Loop over the possible odometer states (i.e. from 1 to xbin)
• Inside all of the for loops, make the following calculations
– Create an object that marks the row of the transition matrix that we need to
be looking at (based on the loop values of the two gridded state variables).
This will be x + (z-1)*xbin (where x indexes the mileage bin and z indexes the route usage bin), given how the xtran matrix was constructed in the
create grids() function.
– Create the conditional value function for driving the bus (v1t) based on the
values of the state variables in the loop (not the values observed in the data).
For example, for the mileage (x1t), you should plug in xval[x] rather than
:Odo.
The difficult part of the conditional value function is the discrete summation
over the state transitions. For this, you need to grab the appropriate row (and
all columns) of the xtran matrix, and then take the dot product with that and
the all possible x1t rows of the FV matrix.
1A Markov transition matrix is a matrix where each row sums to 1 and moving from e.g. column 1 to column 4
4
You should end up with something like
xtran[row,:]’*FV[(z-1)*xbin+1:z*xbin,b+1,t+1]
where b indexes the branded dummy and t indexes time periods.
– Now create the conditional value function for replacing the engine (v0t). For
this, we repeat the same process as with v1t except the θ’s are normalized to
be 0. The code for the expected future value is the same as for v1t with the
exception that mileage resets to 0 after replacement, so instead of grabbing
xtran[row,:] we want xtran[1+(z-1)*xbin,:].
– Finally, update the future value array in period t by storing β log(exp(v0t) +exp(v1t))
in the tth slice of the 3rd dimension of the array. This will be the new future
value term for period t −1. Remember to set β = 0.9
(d) Construct the log likelihood using the future value terms from the previous step and
only using the observed states in the data. This will entail a for loop over buses and
time periods.
• Initialize the log likelihood value to be 0. (We will iteratively add to it as we loop
over observations in the data)
• Create a variable that indexes the state transition matrix rows for the case where the
bus has been replaced. This will be the same 1+(z-1)*xbin as in the conditional
value function v0t above. However, we need to plug in :Zst from the data rather
than a hypothetical value z.
• Create a variable that indexes the state transition matrix rows for the case where
the bus has not been replaced. This will be the same x + (z-1)*xbin as in v1t
above, except we substitute :Xst and :Zst for x and z.
• Now create the flow utility component of v1t −v0t using the actual observed data
on mileage and branding.
• Next, we need to add the appropriate discounted future value to round out our
calculation of v1t −v0t
. Here, we can difference the fj’s as in (5). You should get
something like
(xtran[row1,:].-xtran[row0,:])’*FV[row0:row0+xbin-1,B[i]+1,t+1]
• Finally, create the choice probabilities for choosing each option as written in (7)
and then create the log likelihood according to the summation in (6).
(e) Wrap all of the code you wrote in (c) and (d) into a function and set up the function so
that it can be passed to Optim. For example, you will need to return the negative of the
log likelihood and you will need to have the first argument be the θ vector that we are
trying to estimate
(f) On the same line as the function, prepend the function declaration with the macros so
that your code says @views @inbounds function myfun() rather than function
myfun(). This will give you more performant code. On my machine, it cut the computation time in half.
5
(g) Wrap all of your code in an empty function as you’ve done with other problem sets
(h) Try executing your script to estimate the likelihood function. This took about 4 minutes
on my machine when I started from the estimates of the static model in Question 2.
(i) Pat yourself on the back and grab a beverage of your choice, because that was a lot of
work!
6

Problem Set 5 ECON 6343: Econometrics III
\$30.00
Hello
Can we help?