# Quick Start Guide: Optimal Prescriptive Trees

This is an R version of the corresponding OptimalTrees quick start guide.

In this example we will give a demonstration of how to use Optimal Prescriptive Trees (OPT). We will examine the impact of job training on annual earnings using the Lalonde sample from the National Supported Work Demonstration dataset.

First we load in the data:

colnames <- c("treatment", "age", "education", "black", "hispanic", "married",
"nodegree", "earnings_1975", "earnings_1978")
df <- rbind(df_control, df_treated)

  treatment age education black hispanic married nodegree earnings_1975
1         0  23        10     1        0       0        1         0.000
2         0  26        12     0        0       0        0         0.000
3         0  22         9     1        0       0        1         0.000
4         0  34         9     1        0       0        1      4368.413
5         0  18         9     1        0       0        1         0.000
6         0  45        11     1        0       0        1         0.000
earnings_1978
1          0.00
2      12383.68
3          0.00
4      14051.16
5      10740.08
6      11796.47
[ reached 'max' / getOption("max.print") -- omitted 716 rows ]

## Data for prescriptive problems

Prescriptive trees are trained on observational data, and require three distinct types of data:

• X: the features for each observation that can be used as the splits in the tree - this can be a matrix or a dataframe as for classification or regression problems
• treatments: the treatment applied to each observation - this is a vector of the treatment labels similar to the target in a classification problem
• outcomes: the outcome for each observation under the applied treatment - this is a vector of numeric values similar to the target in a regression problem

Refer to the documentation on data preparation for more information on the data format.

In this case, the treatment is whether or not the subject received job training, and the outcome is their 1978 earnings (which we are trying to maximize):

X <- df[, 2:(ncol(df) - 1)]
treatments <- ifelse(df$treatment == 1, "training", "no training") outcomes <- df$earnings_1978


We can now split into training and test datasets:

split <- iai::split_data("prescription_maximize", X, treatments, outcomes,
seed = 2)
train_X <- split$train$X
train_treatments <- split$train$treatments
train_outcomes <- split$train$outcomes
test_X <- split$test$X
test_treatments <- split$test$treatments
test_outcomes <- split$test$outcomes


Note that we have used the default 70%/30% split, but in many prescriptive problems it is desirable to save more data for testing to ensure high-quality reward estimation on the test set.

## Fitting Optimal Prescriptive Trees

We will use a grid_search to fit an optimal_tree_prescription_maximizer (note that if we were trying to minimize the outcomes, we would use optimal_tree_prescription_minimizer):

grid <- iai::grid_search(
iai::optimal_tree_prescription_maximizer(
prescription_factor = 1,
treatment_minbucket = 20,
random_seed = 234,
),
max_depth = 1:5,
)
iai::fit(grid, train_X, train_treatments, train_outcomes)
iai::get_learner(grid)

Optimal Trees Visualization

Here, we have set prescription_factor = 1 to focus the trees on maximizing the outcome, and treatment_minbucket = 10 so that the tree can only prescribe a treatment in a leaf if there are at least 10 subjects in that leaf that received this treatment. This is to ensure that we have sufficient data on how the treatment affects subjects in this leaf before we can prescribe it.

In the resulting tree, the color in each leaf indicates which treatment is deemed to be stronger in this leaf, and the color intensity indicates the size of the difference. The tree contains some interesting insights about the effect of training, for example:

• Node 17 is where the training had the weakest effect, which is for older subjects with high earnings in 1975. This seems to make sense, as these people are likely the least in need of training.
• Node 10 shows that those with low 1975 earnings, at least 9 years of education, and at least 28 years old benefitted greatly from the training.
• Nodes 6 through 8 show that for those with at least 9 year of education and 1975 earnings below $1103, the effectiveness of the training was highly linked to the age of the subject, with older subjects benefitting much more. We can make predictions on new data using predict: pred <- iai::predict(grid, test_X)  This returns the treatment prescribed for each subject as well as the outcome predicted for each subject under the prescribed treatment: pred_treatments <- pred$treatments

 [1] "training"    "no training" "no training" "training"    "training"
[6] "training"    "no training" "training"    "no training" "training"
[11] "training"    "no training" "training"    "training"    "training"
[16] "training"    "no training" "no training" "training"    "no training"
[21] "training"    "no training" "training"    "no training" "training"
[26] "no training" "training"    "training"    "training"    "training"
[31] "training"    "no training" "no training" "no training" "no training"
[36] "training"    "no training" "no training" "training"    "training"
[41] "training"    "no training" "no training" "no training" "no training"
[46] "no training" "training"    "training"    "training"    "training"
[51] "training"    "no training" "training"    "no training" "training"
[56] "no training" "no training" "no training" "training"    "no training"
[ reached getOption("max.print") -- omitted 156 entries ]
pred_outcomes <- pred\$outcomes

 [1]  6837.618  6779.621  4697.901 11123.963  5669.544  6837.618  7402.351
[8] 11123.963  3288.427 11123.963  6837.618  4697.901  5669.544  6837.618
[15]  5669.544  6837.618  6779.621  5314.847  6837.618  9646.778 11123.963
[22]  9646.778  6837.618  9646.778  6837.618  4697.901  6837.618  6837.618
[29]  6837.618  6837.618  6837.618  9646.778  3288.427  7402.351  6779.621
[36]  6837.618  9646.778  4697.901  6837.618  5669.544  6837.618  4697.901
[43]  6779.621  4697.901  6779.621  3288.427 11123.963  6837.618  6837.618
[50] 11123.963  6837.618  3288.427  5669.544  6779.621  6837.618  4697.901
[57]  9646.778  9646.778  5669.544  9646.778
[ reached getOption("max.print") -- omitted 156 entries ]

You can also use predict_outcomes to get the predicted outcomes for all treatments:

iai::predict_outcomes(grid, test_X)

   no training  training
1     3106.896  6837.618
2     6779.621  2578.706
3     4697.901  2769.288
4     4754.785 11123.963
5     3351.409  5669.544
6     3106.896  6837.618
7     7402.351  3963.328
8     4754.785 11123.963
9     3288.427  9181.285
10    4754.785 11123.963
11    3106.896  6837.618
12    4697.901  2769.288
13    3351.409  5669.544
14    3106.896  6837.618
15    3351.409  5669.544
16    3106.896  6837.618
17    6779.621  2578.706
18    5314.847  3427.748
19    3106.896  6837.618
20    9646.778  6320.100
21    4754.785 11123.963
22    9646.778  6320.100
23    3106.896  6837.618
24    9646.778  6320.100
25    3106.896  6837.618
26    4697.901  2769.288
27    3106.896  6837.618
28    3106.896  6837.618
29    3106.896  6837.618
30    3106.896  6837.618
[ reached 'max' / getOption("max.print") -- omitted 186 rows ]

## Evaluating Optimal Prescriptive Trees

In prescription problems, it is complicated to evaluate the quality of a prescription policy because our data only contains the outcome for the treatment that was received. Because we don't know the outcomes for the treatments that were not received (known as the counterfactuals), we cannot simply evaluate our prescriptions against the test set as we normally do.

A common approach to resolve this problem is reward estimation, where so-called rewards are estimated for each treatment for each observation. These rewards indicate the relative credit a model should be given for prescribing each treatment to each observation, and thus can be used to evaluate the quality of the prescription policy. For more details on how the reward estimation procedure is conducted, refer to the reward estimation documentation.

We will use a reward_estimator to estimate the rewards (note that we are passing in the test data rather the training data to ensure we get a fair out-of-sample evaluation):

reward_lnr <- iai::reward_estimator(
propensity_estimation_method = "random_forest",
outcome_estimation_method = "random_forest",
reward_estimation_method = "doubly_robust",
random_seed = 1,
)
rewards <- iai::fit_predict(reward_lnr, test_X, test_treatments, test_outcomes)

   no training  training
1  13466.14618  5840.714
2   2950.44706  5562.140
3  10780.62503  4377.647
4  17391.48562  5864.285
5   -599.99004  5225.245
6   6346.62663 15981.110
7   -847.69912  3445.859
8   -469.11793  5503.624
9  -1836.01236  7905.532
10  1586.34820  5100.385
11 17416.11235  4879.462
12 10378.60362  6200.316
13  -337.29477  7499.461
14 17117.64083  5864.029
15  5589.84647  4173.271
16    31.59942  2100.046
17  6388.21546  4067.652
18  1685.01892  5225.815
19  6488.22346  5763.884
20 -2086.81416  2603.910
21  8173.40516 14579.700
22 -2236.90502  6278.348
23   616.79681  4985.877
24 10378.53776 10015.613
25  7603.21607  2371.283
26  -257.77528  2887.120
27 -2080.50426  3777.181
28  8433.54180  7665.401
29  -263.00484  4314.367
30  -941.05120  2423.764
[ reached 'max' / getOption("max.print") -- omitted 186 rows ]

We can now use these reward values to evaluate the prescription in many ways. For example, we might like to see the mean reward achieved across all prescriptions on the test set:

evaluate_reward_mean <- function(treatments, rewards) {
total <- 0.0
for (i in seq(treatments)) {
total <- total + rewards[i, treatments[i]]
}
total / length(treatments)
}

evaluate_reward_mean(pred_treatments, rewards)

[1] 5963.84

For comparison's sake, we can compare this to the mean reward achieved under the actual treatment assignments that were observed in the data:

evaluate_reward_mean(test_treatments, rewards)

[1] 5494.877

We can see that the prescriptive tree policy indeed achieves better results than the actual assignments.