A tutorial on tidy cross-validation with R
Analyzing NetHack data, part 1: What kills the players
Analyzing NetHack data, part 2: What players kill the most
Building a shiny app to explore historical newspapers: a step-by-step guide
Classification of historical newspapers content: a tutorial combining R, bash and Vowpal Wabbit, part 1
Classification of historical newspapers content: a tutorial combining R, bash and Vowpal Wabbit, part 2
Curly-Curly, the successor of Bang-Bang
Dealing with heteroskedasticity; regression with robust standard errors using R
Easy time-series prediction with R: a tutorial with air traffic data from Lux Airport
Exporting editable plots from R to Powerpoint: making ggplot2 purrr with officer
Fast food, causality and R packages, part 1
Fast food, causality and R packages, part 2
For posterity: install {xml2} on GNU/Linux distros
Forecasting my weight with R
From webscraping data to releasing it as an R package to share with the world: a full tutorial with data from NetHack
Get text from pdfs or images using OCR: a tutorial with {tesseract} and {magick}
Getting data from pdfs using the pdftools package
Getting the data from the Luxembourguish elections out of Excel
Going from a human readable Excel file to a machine-readable csv with {tidyxl}
Historical newspaper scraping with {tesseract} and R
How Luxembourguish residents spend their time: a small {flexdashboard} demo using the Time use survey data
Imputing missing values in parallel using {furrr}
Intermittent demand, Croston and Die Hard
Looking into 19th century ads from a Luxembourguish newspaper with R
Making sense of the METS and ALTO XML standards
Manipulate dates easily with {lubridate}
Manipulating strings with the {stringr} package
Maps with pie charts on top of each administrative division: an example with Luxembourg's elections data
Missing data imputation and instrumental variables regression: the tidy approach
Modern R with the tidyverse is available on Leanpub
Objects types and some useful R functions for beginners
Pivoting data frames just got easier thanks to `pivot_wide()` and `pivot_long()`
R or Python? Why not both? Using Anaconda Python within R with {reticulate}
Searching for the optimal hyper-parameters of an ARIMA model in parallel: the tidy gridsearch approach
Some fun with {gganimate}
Split-apply-combine for Maximum Likelihood Estimation of a linear model
Statistical matching, or when one single data source is not enough
The best way to visit Luxembourguish castles is doing data science + combinatorial optimization
The never-ending editor war (?)
The year of the GNU+Linux desktop is upon us: using user ratings of Steam Play compatibility to play around with regex and the tidyverse
Using Data Science to read 10 years of Luxembourguish newspapers from the 19th century
Using a genetic algorithm for the hyperparameter optimization of a SARIMA model
Using cosine similarity to find matching documents: a tutorial using Seneca's letters to his friend Lucilius
Using linear models with binary dependent variables, a simulation study
Using the tidyverse for more than data manipulation: estimating pi with Monte Carlo methods
What hyper-parameters are, and what to do with them; an illustration with ridge regression
{disk.frame} is epic
{pmice}, an experimental package for missing data imputation in parallel using {mice} and {furrr}
Building formulae
Functional peace of mind
Get basic summary statistics for all the variables in a data frame
Getting {sparklyr}, {h2o}, {rsparkling} to work together and some fun with bash
Importing 30GB of data into R with sparklyr
Introducing brotools
It's lists all the way down
It's lists all the way down, part 2: We need to go deeper
Keep trying that api call with purrr::possibly()
Lesser known dplyr 0.7* tricks
Lesser known dplyr tricks
Lesser known purrr tricks
Make ggplot2 purrr
Mapping a list of functions to a list of datasets with a list of columns as arguments
Predicting job search by training a random forest on an unbalanced dataset
Teaching the tidyverse to beginners
Why I find tidyeval useful
tidyr::spread() and dplyr::rename_at() in action
Easy peasy STATA-like marginal effects with R
Functional programming and unit testing for data munging with R available on Leanpub
How to use jailbreakr
My free book has a cover!
Work on lists of datasets instead of individual datasets by using functional programming
Method of Simulated Moments with R
New website!
Nonlinear Gmm with R - Example with a logistic regression
Simulated Maximum Likelihood with R
Bootstrapping standard errors for difference-in-differences estimation with R
Careful with tryCatch
Data frame columns as arguments to dplyr functions
Export R output to a file
I've started writing a 'book': Functional programming and unit testing for data munging with R
Introduction to programming econometrics with R
Merge a list of datasets together
Object Oriented Programming with R: An example with a Cournot duopoly
R, R with Atlas, R with OpenBLAS and Revolution R Open: which is fastest?
Read a lot of datasets at once with R
Unit testing with R
Update to Introduction to programming econometrics with R
Using R as a Computer Algebra System with Ryacas

*In this blog post, I present a paper that has really interested me for a long time. This is part2,
where I will briefly present the model of the paper, and try to play around with the data.
If you haven’t, I suggest you read
part 1 where I provide more context.*

Welcome to part 2 of this series, which might or might not have a part 3. I have been quite busy with this paper and especially with reinforcement learning these past couple of weeks, but in the meantime, other things have taken some of my time, so who knows if I’ll keep discussing this paper.

Before going into the data, let me describe the model very broadly. The problem is as follows: each month, Harold Zurcher must decide whether to simply perform some basic maintenance on the buses he’s responsible for, or he can decide to completely replace the engine. Let his utility function be as follows:

\[ u(x_t, i_t, \theta_1) = \left\{ \begin{array}{lcl} -c(x_t, \theta_1) & \text{if} & i_t = 0, \\ -[\overline{P} - \underline{P} + c(0, \theta_1)] & \text{if} & i_t = 1,\\ \end{array}\right. \]

where \(x_t\) is the state variable, the reading of the odometer at month \(t\), \(i_t\) is Harold Zurcher’s decision at time \(t\). \(i_t = 0\) is the decision to keep the engine, \(i_t = 1\) is the decision to replace. Each time the engine is replaced, the state variable \(x_t\) regenerates to 0. That is why John Rust, the paper’s author, calls the problem under study a regenerative optimal stopping model. If \(i_t = 0\) (keep the engine) is chosen, then the cost of normal maintenance is \(c(x_t, \theta_1)\), if \(i_t = 1\) (change the engine) then the cost is \(\overline{P}\), which is the price of the new engine. However, it is still possible to sell the old engine for scrap value, \(\underline{P}\). The replacement cost is equal to \(c(0, \theta_1)\). \(\theta_1\) is a vector of parameters of the cost function to estimate. Because Harold Zurcher is forward looking, and does not want to simply maximize the current month’s utility, he seeks to maximize his intertemporal utility function. The optimal policy would be the solution to the following equation:

\[ V_{\theta} = \max E\left\{ \sum_{j = t}^\infty \beta^{j-t}u(x_j, f_j, \theta_1) | x_t\right\} \]

This is a so-called value function, which is the total reward at the solution of the problem.

The state variable evolves according to a stochastic process given by the following transition probability:

\[ p(x_{t+1} | x_t, i_t, \theta_2) = \left\{ \begin{array}{lllll} \theta_2 \exp\{\theta_2(x_{t+1} - x_t)\} & \text{if} & i_t = 0 & \text{and} & x_{t+1} \geq x_t \\ \theta_2 \exp\{\theta_2(x_{t+1})\} & \text{if} & i_t = 0 & \text{and} & x_{t+1} \geq 0 \\ 0 & \text{otherwise}\\ \end{array}\right. \]

\(\theta_2\) is the parameter of the exponential distribution, another parameter to estimate. I’ll stop with one more equation, the Bellman equation:

\[ V_\theta(x_t) = \max_{i_t \in C(x_t)} [u(x_t, i_t, \theta_1) + \beta EV_\theta(x_t, i_t)] \]

where \(C(x_t) = {0, 1}\) is the action set. The value function is the unique solution to this Bellman equation.

As you can see, this is quite complex (and I have not detailed everything!) but the advantage of models is that one can estimate its structural parameters and put a dollar value on the expected replacement cost, \(\overline{P} - \underline{P}\) in addition to validating the very first hypothesis of the paper; does Harold Zurcher behave optimally?

In what follows, I’ll use the `{ReinforcementLearning}`

package to try to find the optimal policy rule.
The optimal policy rule tells us what is the best action at each period. Reinforcement learning is
an approach that is widely used in machine learning to solve problems very similar to the one that
I described above. However, as we shall see, it will fail here, and there’s a very good reason
for that. First, let’s load the data that was prepared last time:

`all_bus_data <- read_csv("https://raw.githubusercontent.com/b-rodrigues/rust/ee15fb87fc4ba5db28d055c97a898b328725f53c/datasets/processed_data/all_buses.csv")`

```
## Parsed with column specification:
## cols(
## bus_id = col_double(),
## date = col_date(format = ""),
## odometer_reading = col_double(),
## replacement = col_double(),
## bus_family = col_character()
## )
```

`head(all_bus_data)`

```
## # A tibble: 6 x 5
## bus_id date odometer_reading replacement bus_family
## <dbl> <date> <dbl> <dbl> <chr>
## 1 4239 1974-12-01 140953 0 a452372
## 2 4239 1975-01-01 142960 0 a452372
## 3 4239 1975-02-01 145380 0 a452372
## 4 4239 1975-03-01 148140 0 a452372
## 5 4239 1975-04-01 150921 0 a452372
## 6 4239 1975-05-01 153839 0 a452372
```

In the paper, the author groups the 4 following bus families, so I’ll be doing the same:

```
family_group <- c("g870", "rt50", "t8h203", "a530875")
group1_4 <- all_bus_data %>%
filter(bus_family %in% family_group)
ggplot(group1_4) +
geom_line(aes(y = odometer_reading, x = date, group = bus_id, col = bus_family)) +
geom_point(aes(y = ifelse(odometer_reading*replacement == 0, NA, odometer_reading*replacement),
x = date), col = "red") +
labs(title = paste0("Odometer readings for bus families ", paste0(family_group, collapse = ", ")),
caption = "The red dots are replacement events.") +
theme(plot.caption = element_text(colour = "white")) +
brotools::theme_blog()
```

`## Warning: Removed 8200 rows containing missing values (geom_point).`

There are 104 buses in this subset of data. Let’s discretize
the odometer reading using the `ntile()`

function. Discretizing the state variable will make
computation faster:

```
group1_4 <- group1_4 %>%
mutate(state_at_replacement = ifelse(replacement == 1, odometer_reading, NA)) %>%
group_by(bus_id) %>%
fill(state_at_replacement, .direction = "down") %>%
ungroup() %>%
mutate(state_at_replacement = odometer_reading - state_at_replacement) %>%
mutate(state_at_replacement = ifelse(is.na(state_at_replacement), odometer_reading, state_at_replacement)) %>%
mutate(state = ntile(state_at_replacement, 50))
```

Let me also save the bus ids in a vector, I’ll need it later:

`buses <- unique(group1_4$bus_id)`

To use the dataset with the `{ReinforcementLearning}`

package, it must first be prepared:

```
group1_4 <- group1_4 %>%
group_by(bus_id) %>%
mutate(next_state = lead(state, 1)) %>%
mutate(replacement = lead(replacement, 1)) %>%
mutate(action = replacement) %>%
select(state, action, reward = replacement, next_state) %>%
mutate(reward = (-1)*reward) %>%
mutate(action = ifelse(is.na(action), 0, action),
reward = ifelse(is.na(reward), 0, reward)) %>%
mutate(next_state = ifelse(is.na(next_state), state + 1, next_state)) %>%
mutate(state = as.character(state),
next_state = as.character(next_state),
action = as.character(action))
```

`## Adding missing grouping variables: `bus_id``

Let’s see how the data looks:

`head(group1_4)`

```
## # A tibble: 6 x 5
## # Groups: bus_id [1]
## bus_id state action reward next_state
## <dbl> <chr> <chr> <dbl> <chr>
## 1 5297 2 0 0 3
## 2 5297 3 0 0 4
## 3 5297 4 0 0 5
## 4 5297 5 0 0 6
## 5 5297 6 0 0 8
## 6 5297 8 0 0 9
```

So when action 0 (do nothing) is chosen, the value of the state is increased by one. If action 1 (replace) is chosen:

```
group1_4 %>%
filter(action == "1") %>%
head
```

```
## # A tibble: 6 x 5
## # Groups: bus_id [6]
## bus_id state action reward next_state
## <dbl> <chr> <chr> <dbl> <chr>
## 1 5297 34 1 -1 1
## 2 5299 42 1 -1 1
## 3 5300 43 1 -1 1
## 4 5301 36 1 -1 1
## 5 5302 30 1 -1 1
## 6 5303 49 1 -1 1
```

The state goes back to 1, and the reward is -1.

Now, let’s split the dataset into two: a training dataset and a testing dataset:

```
set.seed(1234)
train_buses <- sample(buses, size = round(length(buses)*.8))
test_buses <- setdiff(buses, train_buses)
```

There will be 83 in the training data and 21 in the testing data:

```
train_data <- group1_4 %>%
filter(bus_id %in% train_buses)
test_data <- group1_4 %>%
filter(bus_id %in% test_buses)
```

We’re finally ready to use the `{ReinforcementLearning}`

package.

```
library(ReinforcementLearning)
model <- ReinforcementLearning(train_data,
s = "state",
a = "action",
r = "reward",
s_new = "next_state")
```

Now what’s the result?

`model`

```
## State-Action function Q
## 0 1
## X30 0 -0.19000
## X31 0 0.00000
## X1 0 0.00000
## X32 0 0.00000
## X2 0 0.00000
## X33 0 -0.10000
## X3 0 0.00000
## X34 0 -0.19000
## X4 0 0.00000
## X35 0 0.00000
## X5 0 0.00000
## X36 0 -0.19000
## X6 0 0.00000
## X37 0 -0.10000
## X7 0 0.00000
## X38 0 0.00000
## X8 0 0.00000
## X39 0 -0.34390
## X9 0 0.00000
## X10 0 0.00000
## X40 0 -0.10000
## X11 0 0.00000
## X41 0 -0.10000
## X12 0 0.00000
## X42 0 -0.34390
## X13 0 0.00000
## X43 0 -0.40951
## X14 0 0.00000
## X44 0 -0.19000
## X45 0 -0.34390
## X15 0 0.00000
## X46 0 -0.27100
## X16 0 0.00000
## X47 0 -0.19000
## X17 0 0.00000
## X48 0 -0.40951
## X18 0 0.00000
## X49 0 -0.34390
## X19 0 0.00000
## X50 0 -0.34390
## X20 0 0.00000
## X21 0 0.00000
## X22 0 0.00000
## X23 0 0.00000
## X24 0 0.00000
## X25 0 0.00000
## X26 0 0.00000
## X27 0 0.00000
## X28 0 0.00000
## X29 0 -0.10000
##
## Policy
## X30 X31 X1 X32 X2 X33 X3 X34 X4 X35 X5 X36 X6 X37 X7 X38 X8 X39 X9 X10
## "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
## X40 X11 X41 X12 X42 X13 X43 X14 X44 X45 X15 X46 X16 X47 X17 X48 X18 X49 X19 X50
## "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
## X20 X21 X22 X23 X24 X25 X26 X27 X28 X29
## "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
##
## Reward (last iteration)
## [1] -48
```

We see that the optimal policy is always to do nothing! This is actually “normal” here, as we are using historical data; and in this data, there is no information on the likelihood of severe engine failure if one does not replace it completely at some point! So the agent does not see the point in replacing the engine; it costs money and does not seem to bring in any benefit!

Another way of using the `{ReinforcementLearning}`

package
is to write a function that simulates the environment. One could write such a function, and add in it
a probability of severe failure with a very big cost. This probability would increase as the state
(number of miles driven) increases as well. With such a function, there would be simulations where
the cost of doing nothing would be very high, and as such, hopefully, the agent would learn that
replacing the engine once might be a better course of action than doing nothing.

This might be the subject of part 3 of this series!

Hope you enjoyed! If you found this blog post useful, you might want to follow me on twitter for blog post updates and watch my youtube channel. If you want to support my blog and channel, you could buy me an espresso or paypal.me, or buy my ebook on Leanpub.