Parole dataset logistic regression.

parole.csv

parole.R

library(caTools)
library(ROCR)

parole <- read.csv('parole.csv')

table(parole$violator)

unique(parole$state)
unique(parole$crime)

parole$state <- as.factor(parole$state)
parole$crime <- as.factor(parole$crime)

set.seed(144)

split = sample.split(parole$violator, SplitRatio = 0.7)

train <- subset(parole, split == TRUE)
test <- subset(parole, split == FALSE)

mod1 <- glm(violator ~ ., data=train, family='binomial')
summary(mod1)

# Our model predicts that a parolee who committed multiple offenses has 5.01 times higher odds of being a violator than a parolee who did not commit multiple offenses but is otherwise identical.
exp(mod1$coefficients['multiple.offenses'])
# or
exp(1.6119919)

# According to the model, what are the odds this individual is a violator?
# -4.2411574 + 0.3869904*male + 0.8867192*race - 0.0001756*age + 0.4433007*state2 + 0.8349797*state3 - 3.3967878*state4 - 0.1238867*time.served + 0.0802954*max.sentence + 1.6119919*multiple.offenses + 0.6837143*crime2 - 0.2781054*crime3 - 0.0117627*crime4.
# This parolee has male=1, race=1, age=50, state2=0, state3=0, state4=0, time.served=3, max.sentence=12, multiple.offenses=0, crime2=1, crime3=0, crime4=0. We conclude that log(odds) = -1.700629.
B <- mod1$coefficients
x <- c(1, 1, 50, 0, 0, 0, 3, 12, 0, 1, 0, 0)

# What are the odds? 0.1825685
logit <- B[1] + B[2]*x[1] + B[3]*x[2] + B[4]*x[3] + B[5]*x[4] + B[6]*x[5] + B[7]*x[6] + B[8]*x[7] + B[9]*x[8] + B[10]*x[9] + B[11]*x[10] + B[12]*x[11]
odds <- exp(logit)

# What is the probability this individual is a violator?
p <- oods / (1 + odds)

predtest <- predict(mod1, type='response', newdata=test)

# What is the maximum predicted probability of a violation? 0.9072791
predtest[order(predtest, decreasing=T)][1]

table(test$violator, predtest >= 0.5)
#     FALSE TRUE
# 0   167   12
# 1    11   12

# Sensitivity (true positive rate) = TP / (TP+FN): 0.5217391
12 / (12 + 11)

# Specificity (true negative rate) = TN / (TN+FP): 0.9329609
167 / (167 + 12)

# Accuracy: 0.8861386
(167 + 12) / (167 + 11 + 12 + 12)

# Try against a baseline model, which predicts every parolee as a non-violator.
table(test$violator, test$violator == FALSE)

#      FALSE TRUE
# 0     0  179
# 1    23    0

# Accuracy: 0.8861386
179 / (179 + 23)

# The parole board would assign more cost to false negatives (individuals released for parole who then violate).
# If we lower the threshold, more false positives will result, and less false negatives. So, more individuals will be denied parole, even though they're false positives (not likely to violate parole).
# The board assigns more cost to a false negative than a false positive, and should therefore use a logistic regression cutoff less than 0.5.

ROCRpred <- prediction(predtest, test$violator)
as.numeric(performance(ROCRpred, 'auc')@y.values)
# AUC on test set: 0.8945834