QuantEcon DataScience

Introduction to Economic Modeling and Data Science

Case Study: Recidivism

Co-authors

Prerequisites

Outcomes

  • See an end-to-end data science exercise
  • Application of regression
In [1]:
# Uncomment following line to install on colab
#! pip install qeds fiona geopandas xgboost gensim folium pyLDAvis descartes
In [2]:
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
import seaborn as sns

from sklearn import (
    linear_model, metrics, neural_network, pipeline, preprocessing, model_selection
)

%matplotlib inline
# activate plot theme
import qeds
qeds.themes.mpl_style();

Introduction to Recidivism

Recidivism is the tendency for an individual who has previously committed a crime to commit another crime in the future.

One key input to a judge’s sentencing decision is how likely a given convict is to re-offend, or recidivate.

In an effort to assist the legal system with sentencing guidelines, data scientists have attempted to predict an individual’s risk of recidivism from known observables.

Some are concerned that this process may exhibit prejudice, either through biased inputs or through statistical discrimination.

For example,

  1. Biased inputs: Imagine that a judge often writes harsher sentences to people of a particular race or gender. If an algorithm is trained to reproduce the sentences of this judge, the bias will be propagated by the algorithm.
  2. Statistical discrimination: Imagine that two variables (say race and income) are correlated, and one of them (say income) is correlated with the risk of recidivism. If income is unobserved, then an otherwise unbiased method would discriminate based on race, even if race has nothing to say about recidivism after controlling for income.

This has given rise to serious discussions about the moral obligations data scientists have to those who are affected by their tools.

We will not take a stance today on our moral obligations, but we believe this is an important precursor to any statistical work with public policy applications.

One predictive tool used by various courts in the United States is called COMPAS (Correctional Offender Management Profiling for Alternative Sanctions).

We will be following a Pro Publica article that analyzes the output of COMPAS.

The findings of the article include:

  • Black defendants were often predicted to be at a higher risk of recidivism than they actually were.
  • White defendants were often predicted to be less risky than they were.
  • When controlling for prior crimes, future recidivism, age, and gender, black defendants were 45 percent more likely to be assigned higher risk scores than white defendants.
  • Black defendants were twice as likely as white defendants to be misclassified as being a higher risk of violent recidivism.
  • Even when controlling for prior crimes, future recidivism, age, and gender, black defendants were 77 percent more likely to be assigned higher risk scores than white defendants.

Data Description

The authors of this article filed a public records request with the Broward County Sheriff’s office in Florida.

Luckily for us, they did a significant amount of the legwork which is described in this methodology article.

We download the data below.

In [3]:
data_url = "https://raw.githubusercontent.com/propublica/compas-analysis"
data_url += "/master/compas-scores-two-years.csv"

df = pd.read_csv(data_url)
df.head()
Out[3]:
id name first last compas_screening_date sex dob age age_cat race ... v_decile_score v_score_text v_screening_date in_custody out_custody priors_count.1 start end event two_year_recid
0 1 miguel hernandez miguel hernandez 2013-08-14 Male 1947-04-18 69 Greater than 45 Other ... 1 Low 2013-08-14 2014-07-07 2014-07-14 0 0 327 0 0
1 3 kevon dixon kevon dixon 2013-01-27 Male 1982-01-22 34 25 - 45 African-American ... 1 Low 2013-01-27 2013-01-26 2013-02-05 0 9 159 1 1
2 4 ed philo ed philo 2013-04-14 Male 1991-05-14 24 Less than 25 African-American ... 3 Low 2013-04-14 2013-06-16 2013-06-16 4 0 63 0 1
3 5 marcu brown marcu brown 2013-01-13 Male 1993-01-21 23 Less than 25 African-American ... 6 Medium 2013-01-13 NaN NaN 1 0 1174 0 0
4 6 bouthy pierrelouis bouthy pierrelouis 2013-03-26 Male 1973-01-22 43 25 - 45 Other ... 1 Low 2013-03-26 NaN NaN 2 0 1102 0 0

5 rows × 53 columns

We summarize some of the variables that we will use.

  • first: An individual’s first name
  • last: An individual’s last name
  • sex: An individual’s sex
  • age: An individual’s age
  • race: An individual’s race. It takes values of Caucasian, Hispanic, African-American, Native American, Asian, or Other
  • priors_count: Number of previous arrests
  • decile_score: The COMPAS risk score
  • two_year_recid: Whether the individual had been jailed for a new crime in next two years

Descriptive Statistics

The first thing we do with our data is to drop any classes without “enough” observations.

One of our focuses will be on inter-race differences in scores and recidivism, so we only keep data on races with at least 500 observations in our data set.

Just be aware that this kind of seemingly and even genuinely benign or “technical” decision can still perpetuate inequality by exclusion.

For example, Asians are a small minority, so they’re not really present in the data, and therefore they’re absent from the policy discussion — we have no inferential knowledge on how COMPAS scores work for them.

In [4]:
race_count = df.groupby(["race"])["name"].count()
at_least_500 = list(race_count[race_count > 500].index)
print("The following race have at least 500 observations:", at_least_500)
df = df.loc[df["race"].isin(at_least_500), :]
The following race have at least 500 observations: ['African-American', 'Caucasian', 'Hispanic']

Next, we explore the remaining data using plots and tables.

Age, Sex, and Race

Let’s look at how the dataset is broken down into age, sex, and race.

In [5]:
def create_groupcount_barplot(df, group_col, figsize, **kwargs):
    "call df.groupby(group_col), then count number of records and plot"
    counts = df.groupby(group_col)["name"].count().sort_index()

    fig, ax = plt.subplots(figsize=figsize)
    counts.plot(kind="bar", **kwargs)

    ax.spines["right"].set_visible(False)
    ax.spines["top"].set_visible(False)
    ax.set_xlabel("")
    ax.set_ylabel("")

    return fig, ax
In [6]:
age_cs = ["Less than 25", "25 - 45", "Greater than 45"]
df["age_cat"] = pd.Categorical(df["age_cat"], categories=age_cs, ordered=True)
fig, ax = create_groupcount_barplot(df, "age_cat", (14, 8), color="DarkBlue", rot=0)
In [7]:
sex_cs = ["Female", "Male"]
df["sex"] = pd.Categorical(df["sex"], categories=sex_cs, ordered=True)
create_groupcount_barplot(df, "sex", (6, 8), color="DarkBlue", rot=0)
Out[7]:
(<Figure size 432x576 with 1 Axes>,
 <matplotlib.axes._subplots.AxesSubplot at 0x7fa6a690e3c8>)
In [8]:
race_cs = ["African-American", "Caucasian", "Hispanic"]
df["race"] = pd.Categorical(df["race"], categories=race_cs, ordered=True)
create_groupcount_barplot(df, "race", (12, 8), color="DarkBlue", rot=0)
Out[8]:
(<Figure size 864x576 with 1 Axes>,
 <matplotlib.axes._subplots.AxesSubplot at 0x7fa6a6886080>)

From this, we learn that our population is mostly between 25-45, male, and is mostly African-American or Caucasian.

Recidivism

We now look into how recidivism is split across groups.

In [9]:
recid = df.groupby(["age_cat", "sex", "race"])["two_year_recid"].mean().unstack(level="race")
recid
Out[9]:
race African-American Caucasian Hispanic
age_cat sex
Less than 25 Female 0.449704 0.310345 0.411765
Male 0.645806 0.541254 0.536364
25 - 45 Female 0.382278 0.423948 0.333333
Male 0.533074 0.433699 0.375000
Greater than 45 Female 0.227273 0.239766 0.217391
Male 0.425101 0.289157 0.216667

In the table, we see that the young have higher recidivism rates than the old, except for among Caucasian females.

Also, African-American males are at a particularly high risk of recidivism even as they get older.

Risk Scores

Each individual in the dataset was assigned a decile_score ranging from 1 to 10.

This score represents the perceived risk of recidivism with 1 being the lowest risk and 10 being the highest.

We show a bar plot of all decile scores below.

In [10]:
create_groupcount_barplot(df, "decile_score", (12, 8), color="DarkBlue", rot=0)
Out[10]:
(<Figure size 864x576 with 1 Axes>,
 <matplotlib.axes._subplots.AxesSubplot at 0x7fa6a6292cf8>)

How do these scores differ by race?

In [11]:
dfgb = df.groupby("race")
race_count = df.groupby("race")["name"].count()

fig, ax = plt.subplots(3, figsize=(14, 8))

for (i, race) in enumerate(["African-American", "Caucasian", "Hispanic"]):
    (
        (dfgb
            .get_group(race)
            .groupby("decile_score")["name"].count() / race_count[race]
        )
        .plot(kind="bar", ax=ax[i], color="#353535")
    )
    ax[i].set_ylabel(race)
    ax[i].set_xlabel("")
    # set equal y limit for visual comparison
    ax[i].set_ylim(0, 0.32)

fig.suptitle("Score Frequency by Race")
Out[11]:
Text(0.5, 0.98, 'Score Frequency by Race')

While Caucasians and Hispanics both see the majority of their score distribution on low values, African-Americans are almost equally likely to receive any score.

Risk Scores and Recidivism

Now we can explore the relationship between the risk score and actual two year recidivism.

The first measure we look at is the frequency of recidivism by decile score – these numbers tell us what percentage of people assigned a particular risk score committed a new crime within two years of being released.

In [12]:
df.groupby("decile_score")["two_year_recid"].mean()
Out[12]:
decile_score
1     0.220392
2     0.309112
3     0.375887
4     0.426593
5     0.478723
6     0.564228
7     0.590988
8     0.681363
9     0.698795
10    0.770889
Name: two_year_recid, dtype: float64

Let’s also look at the correlation.

In [13]:
df[["decile_score", "two_year_recid"]].corr()
Out[13]:
decile_score two_year_recid
decile_score 1.000000 0.346797
two_year_recid 0.346797 1.000000

As the risk score increases, the percentage of people committing a new crime does as well, with a positive correlation (~0.35).

This is good news – it means that the score is producing at least some signal about an individual’s recidivism risk.

One of the key critiques from Pro Publica, though, was that the inaccuracies were nonuniform — that is, the tool was systematically wrong about certain populations.

Let’s now separate the correlations by race and see what happens.

In [14]:
recid_rates = df.pivot_table(index="decile_score", columns="race", values="two_year_recid")

recid_rates
Out[14]:
race African-American Caucasian Hispanic
decile_score
1 0.228643 0.208517 0.244898
2 0.302799 0.313019 0.318584
3 0.419075 0.340659 0.313953
4 0.459740 0.396491 0.346154
5 0.482192 0.460581 0.538462
6 0.559896 0.572165 0.567568
7 0.592500 0.615385 0.470588
8 0.682451 0.719298 0.500000
9 0.707895 0.693878 0.550000
10 0.793706 0.703125 0.666667

Or, in plotted form,

In [15]:
fig, ax = plt.subplots(3, sharex="all")

for (i, _race) in enumerate(["African-American", "Caucasian", "Hispanic"]):
    _rr_vals = recid_rates[_race].values

    ax[i].bar(np.arange(1, 11), _rr_vals, color="#c60000")
    ax[i].bar(np.arange(1, 11), 1 - _rr_vals, bottom=_rr_vals, color="#353535")
    ax[i].set_ylabel(_race)
    ax[i].spines["left"].set_visible(False)
    ax[i].spines["right"].set_visible(False)
    ax[i].spines["top"].set_visible(False)
    ax[i].spines["bottom"].set_visible(False)
    ax[i].yaxis.tick_right()
    ax[i].xaxis.set_ticks_position("none")

fig.suptitle("Recidivism Rates by Race")
Out[15]:
Text(0.5, 0.98, 'Recidivism Rates by Race')

Regression

In what follows, we will be doing something slightly different than what was done in the Pro Publica article.

First, we will explore what happens when we try to predict the COMPAS risk scores using the observable data that we have.

Second, we will use binary probability models to predict whether an individual is at risk of recidivism.

We will do this first using the COMPAS risk scores, and then afterwards we will try to write our own model based on raw observables, like age, race and sex.

Preprocessing

We would like to use some features that are inherently non-numerical such as sex, age group, and race in our model.

Before we can do that, we need to encode these string values as numerical values so our machine learning algorithms can understand them – an econometrician would call this, creating dummy variables.

sklearn can automatically do this for us using OneHotEncoder.

Essentially, we make one column for each possible value of a categorical variable and then we set just one of these columns equal to a 1 if the observation has that column’s category, and set all other columns to 0.

Let’s do an example.

Imagine we have the array below.

In [16]:
sex = np.array([["Male"], ["Female"], ["Male"], ["Male"], ["Female"]])

The way to encode this would be to create the array below.

In [17]:
sex_encoded = np.array([
    [0.0, 1.0],
    [1.0, 0.0],
    [0.0, 1.0],
    [0.0, 1.0],
    [1.0, 0.0]
])

Using sklearn it would be:

In [18]:
ohe = preprocessing.OneHotEncoder(sparse=False)
sex_ohe = ohe.fit_transform(sex)

# This should shows 0s!
sex_ohe - sex_encoded
Out[18]:
array([[0., 0.],
       [0., 0.],
       [0., 0.],
       [0., 0.],
       [0., 0.]])

We will use this encoding trick below as we create our data.

Predicting COMPAS Scores

First, we proceed by creating the X and y inputs into a manageable format.

We encode the categorical variables using the OneHotEncoder described above, and then merge that with the non-categorical data.

Finally, we split the data into training and validation (test) subsets.

In [19]:
def prep_data(df, continuous_variables, categories, y_var, test_size=0.15):

    ohe = preprocessing.OneHotEncoder(sparse=False)

    y = df[y_var].values
    X = np.zeros((y.size, 0))

    # Add continuous variables if exist
    if len(continuous_variables) > 0:
        X = np.hstack([X, df[continuous_variables].values])

    if len(categories) > 0:
        X = np.hstack([X, ohe.fit_transform(df[categories])])

    X_train, X_test, y_train, y_test = model_selection.train_test_split(
        X, y, test_size=test_size, random_state=42
    )

    return X_train, X_test, y_train, y_test

As we proceed, our goal will be to see which variables are most important for predicting the COMPAS scores.

As we estimate these models, one of our metrics for success will be mean absolute error (MAE).

In [20]:
def fit_and_report_maes(mod, X_train, X_test, y_train, y_test, y_transform=None, y_inv_transform=None):
    if y_transform is not None:
        mod.fit(X_train, y_transform(y_train))
    else:
        mod.fit(X_train, y_train)

    yhat_train = mod.predict(X_train)
    yhat_test = mod.predict(X_test)

    if y_transform is not None:
        yhat_train = y_inv_transform(yhat_train)
        yhat_test = y_inv_transform(yhat_test)

    return dict(
        mae_train=metrics.mean_absolute_error(y_train, yhat_train),
        mae_test=metrics.mean_absolute_error(y_test, yhat_test)
    )

Let’s begin with a simple linear model which uses just prior arrests.

In [21]:
X_train, X_test, y_train, y_test = prep_data(
    df, ["priors_count"], [], "decile_score"
)

fit_and_report_maes(linear_model.LinearRegression(), X_train, X_test, y_train, y_test)
Out[21]:
{'mae_train': 2.162527833108664, 'mae_test': 2.1917544845291337}

This simple model obtains a MAE of about 2 for both the test data and training data.

This means, on average, that our model can predict the COMPAS score (which ranges from 1-10) within about 2 points.

While the MAE is about 2, knowing what the errors on our prediction model look like is often very useful.

Below, we create a histogram which shows the distribution of these errors. In our case, we take the difference between predicted value and actual value, so a positive value means that we overpredicted the COMPAS score and a negative value means we underpredicted it.

In [22]:
lr_model = linear_model.LinearRegression()
lr_model.fit(X_train, y_train)

yhat_train = lr_model.predict(X_train)
yhat_test = lr_model.predict(X_test)

fig, ax = plt.subplots(1, 2, figsize=(12, 4), sharey="all")

ax[0].hist(yhat_train - y_train, density=True)
ax[0].set_title("Training Data")
ax[1].hist(yhat_test - y_test, density=True)
ax[1].set_title("Test Data")
Out[22]:
Text(0.5, 1.0, 'Test Data')

In both cases, the long left tails of errors suggest the existence of relevant features which would improve our model.

The first thing we might consider investigating is whether there are non-linearities in how the number of priors enters the COMPAS score.

First, we try using polynomial features in our exogenous variables.

In [23]:
X_train, X_test, y_train, y_test = prep_data(
    df, ["priors_count"], [], "decile_score"
)

# Transform data to quadratic
pf = preprocessing.PolynomialFeatures(2, include_bias=False)
X_train = pf.fit_transform(X_train)
X_test = pf.fit_transform(X_test)

fit_and_report_maes(linear_model.LinearRegression(), X_train, X_test, y_train, y_test)
Out[23]:
{'mae_train': 2.1204058017552914, 'mae_test': 2.1179838134597335}

We don’t see a very significant increase in performance, so we also try using log on the endogenous variables.

In [24]:
X_train, X_test, y_train, y_test = prep_data(
    df, ["priors_count"], [], "decile_score"
)

fit_and_report_maes(
    linear_model.LinearRegression(), X_train, X_test, y_train, y_test,
    y_transform=np.log, y_inv_transform=np.exp
)
Out[24]:
{'mae_train': 2.2550821558610115, 'mae_test': 2.3332184125647917}

Still no improvement… The next natural thing is to add more features to our regression.

In [25]:
X_train, X_test, y_train, y_test = prep_data(
    df, ["priors_count"], ["age_cat", "race", "sex"], "decile_score"
)

fit_and_report_maes(linear_model.LinearRegression(), X_train, X_test, y_train, y_test)
Out[25]:
{'mae_train': 1.807656365060343, 'mae_test': 1.8277010173497898}

By allowing for indicator variables on age, race, and sex, we are able to slightly improve the MAE. The errors also seem to have a less extreme tail.

In [26]:
X_train, X_test, y_train, y_test = prep_data(
    df, ["priors_count"], ["age_cat", "race", "sex"], "decile_score"
)

lr_model = linear_model.LinearRegression()
lr_model.fit(X_train, y_train)

yhat_train = lr_model.predict(X_train)
yhat_test = lr_model.predict(X_test)

fig, ax = plt.subplots(1, 2, figsize=(12, 4), sharey="all")

ax[0].hist(yhat_train - y_train, density=True)
ax[0].set_title("Training Data")
ax[1].hist(yhat_test - y_test, density=True)
ax[1].set_title("Test Data")
Out[26]:
Text(0.5, 1.0, 'Test Data')

The coefficients are listed below:

In [27]:
names = [
    "priors_count", "Less than 25", "25-45", "Greater than 45", "African-American",
    "Caucasian", "Hispanic", "Female", "Male"
]
for (_name, _coef) in zip(names, lr_model.coef_):
    print(_name, ": ", _coef)
priors_count :  0.2799923049684575
Less than 25 :  -0.1026275580794656
25-45 :  -1.7189018948181372
Greater than 45 :  1.821529452897601
African-American :  0.6102522379830793
Caucasian :  -0.10156939689442336
Hispanic :  -0.5086828410886551
Female :  0.04293582427466207
Male :  -0.04293582427466212

What stands out to you about these coefficients?

See exercise 1 in the exercise list

Binary Probability Models

Binary probability models are used to model “all or nothing” outcomes, like the occurrence of an event.

Their output is the probability that an event of interest occurs.

With this probability in hand, the researcher chooses an acceptable cutoff (perhaps 0.5) above which the event is predicted to occur.

Note

Binary probability models can be thought of as a special case of classification.

In classification, we are given a set of features and asked to predict one of a finite number of discrete labels.

We will learn more about classification in an upcoming lecture!

In our example, we will be interested in how the COMPAS scores do at predicting recidivism and how their ability to predict depends on race or sex.

To assist us in evaluating the performance of various models we will use a new metric called the confusion matrix.

Scikit-learn knows how to compute this metric and also provides a good description of what is computed.

Let’s see what they have to say.

In [28]:
help(metrics.confusion_matrix)
Help on function confusion_matrix in module sklearn.metrics.classification:

confusion_matrix(y_true, y_pred, labels=None, sample_weight=None)
    Compute confusion matrix to evaluate the accuracy of a classification
    
    By definition a confusion matrix :math:`C` is such that :math:`C_{i, j}`
    is equal to the number of observations known to be in group :math:`i` but
    predicted to be in group :math:`j`.
    
    Thus in binary classification, the count of true negatives is
    :math:`C_{0,0}`, false negatives is :math:`C_{1,0}`, true positives is
    :math:`C_{1,1}` and false positives is :math:`C_{0,1}`.
    
    Read more in the :ref:`User Guide <confusion_matrix>`.
    
    Parameters
    ----------
    y_true : array, shape = [n_samples]
        Ground truth (correct) target values.
    
    y_pred : array, shape = [n_samples]
        Estimated targets as returned by a classifier.
    
    labels : array, shape = [n_classes], optional
        List of labels to index the matrix. This may be used to reorder
        or select a subset of labels.
        If none is given, those that appear at least once
        in ``y_true`` or ``y_pred`` are used in sorted order.
    
    sample_weight : array-like of shape = [n_samples], optional
        Sample weights.
    
    Returns
    -------
    C : array, shape = [n_classes, n_classes]
        Confusion matrix
    
    References
    ----------
    .. [1] `Wikipedia entry for the Confusion matrix
           <https://en.wikipedia.org/wiki/Confusion_matrix>`_
           (Wikipedia and other references may use a different
           convention for axes)
    
    Examples
    --------
    >>> from sklearn.metrics import confusion_matrix
    >>> y_true = [2, 0, 2, 2, 0, 1]
    >>> y_pred = [0, 0, 2, 2, 0, 2]
    >>> confusion_matrix(y_true, y_pred)
    array([[2, 0, 0],
           [0, 0, 1],
           [1, 0, 2]])
    
    >>> y_true = ["cat", "ant", "cat", "cat", "ant", "bird"]
    >>> y_pred = ["ant", "ant", "cat", "cat", "ant", "cat"]
    >>> confusion_matrix(y_true, y_pred, labels=["ant", "bird", "cat"])
    array([[2, 0, 0],
           [0, 0, 1],
           [1, 0, 2]])
    
    In the binary case, we can extract true positives, etc as follows:
    
    >>> tn, fp, fn, tp = confusion_matrix([0, 1, 0, 1], [1, 1, 1, 0]).ravel()
    >>> (tn, fp, fn, tp)
    (0, 2, 1, 1)

In [29]:
def report_cm(mod, X_train, X_test, y_train, y_test):
     return dict(
         cm_train=metrics.confusion_matrix(y_train, mod.predict(X_train)),
         cm_test=metrics.confusion_matrix(y_test, mod.predict(X_test))
     )

We will start by using logistic regression using only decile_score as a feature and then examine how the confusion matrices differ by race and sex.

In [30]:
from patsy import dmatrices
groups = [
    "overall", "African-American", "Caucasian", "Hispanic", "Female", "Male"
]

ind = [
    "Portion_of_NoRecid_and_LowRisk", "Portion_of_Recid_and_LowRisk",
    "Portion_of_NoRecid_and_HighRisk", "Portion_of_Recid_and_HighRisk"
]

fmla = "two_year_recid ~ C(decile_score)"
y,X = dmatrices(fmla, df)
X_train, X_test, y_train, y_test, df_train, df_test = model_selection.train_test_split(
    X,y.reshape(-1),df, test_size=0.25, random_state=42
)


decile_mod = linear_model.LogisticRegression(solver="lbfgs").fit(X_train,y_train)

def cm_tables(pred, y, df):
    output = pd.DataFrame(index=ind, columns=groups)
    for group in groups:
        if group in ["African-American", "Caucasian", "Hispanic"]:
            subset=(df.race==group)
        elif group in ["Female", "Male"]:
            subset=(df.sex==group)
        else:
            subset=np.full(y.shape, True)

        y_sub = y[subset]
        pred_sub = pred[subset]

        cm = metrics.confusion_matrix(y_sub, pred_sub)

        # Compute fraction for which the guess is correct
        total = cm.sum()
        vals = np.array(cm/total)
        output.loc[:, group] = vals.reshape(-1)


    def cond_probs(col, axis):
        d=int(np.sqrt(len(col)))
        pcm = np.array(col).reshape(d,d)
        pcm = pcm/pcm.sum(axis=axis, keepdims=True)
        return(pcm.reshape(-1))

    given_outcome = output.copy()
    given_outcome.index = ["P(LowRisk|NoRecid)","P(HighRisk|NoRecid)","P(LowRisk|Recid)","P(HighRisk|Recid)"]
    given_outcome=given_outcome.apply(lambda c: cond_probs(c,1))

    given_pred = output.copy()
    given_pred.index = ["P(NoRecid|LowRisk)","P(NoRecid|HighRisk)","P(Recid|LowRisk)","P(Recid|HighRisk)"]
    given_pred=given_pred.apply(lambda c: cond_probs(c,0))
    return(output,given_outcome, given_pred)

output, given_outcome, given_pred =cm_tables(decile_mod.predict(X_test),
                                             y_test, df_test)
output
Out[30]:
overall African-American Caucasian Hispanic Female Male
Portion_of_NoRecid_and_LowRisk 0.361815 0.263270 0.475000 0.522581 0.438040 0.342222
Portion_of_Recid_and_LowRisk 0.191514 0.230361 0.136667 0.167742 0.207493 0.187407
Portion_of_NoRecid_and_HighRisk 0.152033 0.140127 0.168333 0.161290 0.141210 0.154815
Portion_of_Recid_and_HighRisk 0.294638 0.366242 0.220000 0.148387 0.213256 0.315556

output contains information on the percent of true negatives, false negatives, false positives, and true positives.

What do you see?

The joint probabilities (of prediction and outcome given race or sex) in the above table are a bit hard to interpret.

Conditional probabilities can be easier to think about.

Let’s look at the probability of outcomes given the prediction as well as race or sex.

In [31]:
given_pred
Out[31]:
overall African-American Caucasian Hispanic Female Male
P(NoRecid|LowRisk) 0.704128 0.652632 0.738342 0.764151 0.756219 0.688525
P(NoRecid|HighRisk) 0.393939 0.386121 0.383178 0.530612 0.493151 0.372607
P(Recid|LowRisk) 0.295872 0.347368 0.261658 0.235849 0.243781 0.311475
P(Recid|HighRisk) 0.606061 0.613879 0.616822 0.469388 0.506849 0.627393

As you can see, the distribution of outcomes conditional on predictions does not vary too much with race.

Moreover, if anything, it discriminates in favor of African-Americans.

The algorithm does appear to overpredict recidivism for women compared to men.

This is an important concern.

We will not discuss it too much though because (1) we will see below that when fairness is looked at in another way, women are favored over men, and (2) the company that produces COMPAS also produces a separate questionnaire and risk score designed only for women.

False Positive and Negative Rates

What if we flip this around and look at the distributions of predictions conditional on outcomes?

Why look at these probabilities?

One reason is that in law, it’s traditionally far worse to punish innocents than let the guilty free. This idea goes at least back to 1760 and Blackstone’s ratio.

It is better that ten guilty persons escape than that one innocent suffer. -William Blackstone

Blackstone’s ratio says that we should be particularly concerned about P(HighRisk | NoRecid).

This probability is also called the false positive rate.

In [32]:
given_outcome
Out[32]:
overall African-American Caucasian Hispanic Female Male
P(LowRisk|NoRecid) 0.653887 0.533333 0.776567 0.757009 0.678571 0.646154
P(HighRisk|NoRecid) 0.346113 0.466667 0.223433 0.242991 0.321429 0.353846
P(LowRisk|Recid) 0.340369 0.276730 0.433476 0.520833 0.398374 0.329134
P(HighRisk|Recid) 0.659631 0.723270 0.566524 0.479167 0.601626 0.670866

Now we see some large disparities by race in the false positive rate (and false negative rate). This is one of the main findings of the Pro Publica article.

In response to Pro Publica, Northpointe, the company that produces COMPAS, argued that COMPAS is not biased because the probabilities of outcomes conditional on predictions (like P(NoRecid|LowRisk)) are approximately equal across races [recidDMB16].

Following [recidKLL+17], we will call a prediction algorithm with this property well-calibrated.

Being well-calibrated is one criteria for fairness of a prediction algorithm.

Pro Publica’s critique focuses on a different criteria – that the the probability of predicted categories conditional on true outcomes should be equal across groups (i.e. P(HighRisk|NoRecid) should be equal across races).

[recidKLL+17] calls a prediction algorithm with this property balanced.

Visualizing Calibration and Balance

We can get a slightly more detailed look at calibration and balance by recognizing that prediction algorithms typically compute a predicted probability, not just a discrete predicted outcome.

The predicted outcome will typically be assigned to the category with the highest predicted probability.

We can examine calibration graphically by plotting the P(recidivism | predicted probability)

In [33]:
import scipy

def calibration_plot(pred, y, df, bins=20):
    fig,ax = plt.subplots(3,2, figsize=(12,6), sharey=True, sharex=True)
    for (g,group) in enumerate(groups):
        if group in ["African-American", "Caucasian", "Hispanic"]:
            subset=(df.race==group)
        elif group in ["Female", "Male"]:
            subset=(df.sex==group)
        else:
            subset=np.full(y.shape,True)
        _ax = ax[np.unravel_index(g, ax.shape)]
        y_sub = y[subset]
        pred_sub = pred[subset]
        mu, edges, n=scipy.stats.binned_statistic(pred_sub,y_sub,'mean',bins=bins)
        se, edges,n=scipy.stats.binned_statistic(pred_sub,y_sub,
                         lambda x: np.std(x)/np.sqrt(len(x)),bins=bins)
        midpts = (edges[0:-1]+edges[1:])/2
        _ax.errorbar(midpts, mu, yerr=1.64*se, fmt='o')
        _ax.set_title(group)
        _ax.set_ylabel("Observed recidivism")
        _ax.set_xlabel("Predicted P(recidivism)")
        x = np.linspace(*_ax.get_xlim())
        _ax.plot(x, x)
        _ax.set_xlim(0.0,1.0)
    fig.tight_layout()
    return(fig,ax)

calibration_plot(decile_mod.predict_proba(X_test)[:,1],
                 df_test["two_year_recid"],
                 df_test);

This figure is one way to visualize how well-calibrated these predictions are.

The dots are binned averages of observed recidivism, conditional on predicted recidivism being in some range.

The error bars represent a 90% confidence interval.

A perfectly calibrated prediction would have these dots all lie along the 45 degree line.

For dots below the 45 degree line, the algorithm is overpredicting recidivism.

See exercise 2 in the exercise list

In [34]:
# Create calibration plot using training data

The algorithm appears fairly well-calibrated.

It does not seem to be making systematic errors in one direction based on any particular race– but it does appear to be systematic overprediction for females compared to males.

Now, let’s create a figure to examine balance.

Balance is about the distribution of predictions conditional on outcomes, so we will plot histograms of predicted probabilities conditional on realized outcomes.

In [35]:
import seaborn as sns
def balance_hist_plot(pred, y, df, bins=20):
    fig,ax = plt.subplots(3,2, figsize=(12,6), sharey=True, sharex=True)
    for (g,group) in enumerate(groups):
        if group in ["African-American", "Caucasian", "Hispanic"]:
            subset=(df.race==group)
        elif group in ["Female", "Male"]:
            subset=(df.sex==group)
        else:
            subset=np.full(y.shape,True)
        _ax = ax[np.unravel_index(g, ax.shape)]
        y_sub = y[subset]
        pred_sub = pred[subset]
        sns.distplot(pred_sub[y_sub==0], hist=True, bins=bins, kde=False, ax=_ax,
                     label="No recidivate", norm_hist=True, axlabel="Predicted Probability")
        sns.distplot(pred_sub[y_sub==1], hist=True, bins=bins, kde=False, ax=_ax,
                     label="Yes recidivate", norm_hist=True, axlabel="Predicted Probability")
        _ax.set_title(group)

    plt.legend()
    fig.tight_layout()
    return(fig,ax)

balance_hist_plot(decile_mod.predict_proba(X_test)[:,1],
                  df_test["two_year_recid"],
                  df_test);

This figure is somewhat useful, but not for depicting balance especially clearly, so let’s try something else.

To get false positive and false negative rates, we must assign the predicted probabilities to outcomes.

The most common choice would be to predict recidivism if the predicted probability is greater than 0.5.

However, if we want to adjust the false positive and false negative rates, we might want to choose some other threshold and predict recidivism if the predicted probability exceeds this threshold.

Different thresholds will lead to different false negative and false positive rates, so let’s plot these rates as functions of the threshold.

In [36]:
def balance_threshold_plot(pred, y, df, bins=20):
    fig,ax = plt.subplots(2,2, figsize=(12,6), sharey=True,
                          sharex=True)
    x = np.linspace(min(pred), max(pred), bins)
    # get colors defined by theme
    colors=plt.rcParams["axes.prop_cycle"].by_key()["color"]
    for (g, group) in enumerate(groups):
        if group in ["African-American", "Caucasian", "Hispanic"]:
            subset=(df.race==group)
            r = 0
        elif group in ["Female", "Male"]:
            subset=(df.sex==group)
            r = 1
        else:
            continue
        y_sub = y[subset]
        pred_sub = pred[subset]
        _ax = ax[r,0]
        fn = np.array([np.mean(pred_sub[y_sub==1]<xi) for xi in x])
        c1 = sum(y_sub==1)
        sen = np.sqrt(fn*(1-fn)/c1)
        fp = np.array([np.mean(pred_sub[y_sub==0]>xi) for xi in x])
        c0 = sum(y_sub==0)
        sep = np.sqrt(fp*(1-fp)/c0)
        p=_ax.plot(x, fn, color=colors[g])
        _ax.fill_between(x, fn-1.64*sen, fn+1.64*sen, alpha=0.25, color=colors[g])
        _ax.annotate(group, (x[bins//7*g], fn[bins//7*g]), color=colors[g])
        _ax.set_ylabel("False +/- Rate")
        _ax.set_xlabel("Threshold")
        _ax.set_title("False Negative Rate")

        _ax = ax[r,1]
        p=_ax.plot(x, fp, color=colors[g])
        _ax.fill_between(x, fp-1.64*sep, fp+1.64*sep, alpha=0.25, color=colors[g])
        _ax.set_xlabel("Threshold")
        _ax.set_title("False Positive Rate")

    fig.tight_layout()
    return(fig,ax)

balance_threshold_plot(decile_mod.predict_proba(X_test)[:,1],
                       df_test["two_year_recid"],
                       df_test);

From this, we can more easily see the balance problem — regardless of which threshold we choose, African-Americans will have a higher false positive rate than Caucasians.

We have seen that COMPAS scores are well-calibrated conditional on race, but not balanced.

Can we create an alternative prediction that is both well-calibrated and balanced?

Creating an Alternative Prediction

As a starting exercise, let’s predict recidivism using the variables in this dataset other than race and COMPAS score.

Almost all variables in this data are categorical.

Any function of categorical variables can be represented as a linear function of indicator variables and their interactions.

Given that linearity in indicators does not impose any substantiative restriction here, a penalized linear model like lasso seems like a good choice for prediction.

To keep the computation time reasonable, we do not include all interaction and indicator terms here.

To ensure that predicted probabilities are between 0 and 1, we fit a logistic regression with an $ \ell-1 $ penalty.

In [37]:
from sklearn import model_selection, linear_model
from patsy import dmatrices

# charge_desc has many values with one observations, we will
# combine these descriptions into a single "other." This could
# be improved upon by looking at the text of descriptions and
# combining.
df.c_charge_desc = df.c_charge_desc.fillna("")
df["charge_cat"] = df.c_charge_desc
cnt = df.c_charge_desc.value_counts()[df.c_charge_desc]
cnt.index = df.index
df.loc[cnt<10,"charge_cat"] = "other"
df.charge_cat = df.charge_cat.astype('category')
df.sex = df.sex.astype('category')


fmla = "two_year_recid ~ sex*(age + juv_fel_count + juv_misd_count + juv_other_count + C(priors_count) + c_charge_degree + charge_cat)"

y,X = dmatrices(fmla, df)
print("There are {} features".format(X.shape[1]))
X_train, X_test, y_train, y_test, df_train, df_test = model_selection.train_test_split(
    X,pd.Series(y.reshape(-1),index=df.index),df, test_size=0.25, random_state=42
)

lasso_mod=linear_model.LogisticRegressionCV(cv=5,verbose=True,
                                            Cs=10, penalty='l1',
                                            max_iter=100,
                                            scoring="neg_log_loss",
                                            solver="liblinear").fit(X_train, y_train)
There are 260 features
[LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear]
[Parallel(n_jobs=1)]: Using backend SequentialBackend with 1 concurrent workers.
[LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear][LibLinear]
[Parallel(n_jobs=1)]: Done   5 out of   5 | elapsed:    6.4s finished

Let’s look at the regularization parameter chosen and the non-zero coefficients.

In [38]:
# plots illustrating regularization parameter choice
scores=lasso_mod.scores_[1.0].mean(axis=0)
logpenalties=np.log(lasso_mod.Cs_)
nnonzero=(np.abs(lasso_mod.coefs_paths_[1.0])>1e-6).sum(axis=2).mean(axis=0)
colors=plt.rcParams["axes.prop_cycle"].by_key()["color"]
fig, ax1 = plt.subplots()
ax1.plot(logpenalties,scores, color=colors[0])
ax1.set_ylabel("CV log likelihood", color=colors[0])
ax1.set_xlabel("-log(penalty)")
ax1.tick_params('y', colors=colors[0])
ax2 = ax1.twinx()
ax2.plot(logpenalties,nnonzero, color=colors[1])
ax2.set_ylabel("nonzero coefficients", color=colors[1])
ax2.tick_params('y', colors=colors[1])
ax2.grid(b=None);

Let’s also look at the nonzero coefficients. We should be careful about interpreting these, since relatively strong assumptions are needed for lasso to produce consistent coefficient estimates.

Note

Lasso gives accurate predictions under weaker assumptions than needed for consistent coefficient estimates.

In [39]:
# table of nonzero coefficients
coef = pd.DataFrame(index = X.design_info.column_names, columns=["Value"])
coef.Value = np.transpose(lasso_mod.coef_)
print(sum(np.abs(coef.Value)>1.0e-8))
with pd.option_context('display.max_rows', None):
    display(coef[np.abs(coef.Value)>1.0e-8])
181
Value
Intercept 0.125413
sex[T.Male] 0.540480
C(priors_count)[T.1] 0.133072
C(priors_count)[T.2] 0.596614
C(priors_count)[T.3] 0.818545
C(priors_count)[T.4] 1.237669
C(priors_count)[T.5] 1.014005
C(priors_count)[T.6] 1.076504
C(priors_count)[T.7] 1.387007
C(priors_count)[T.8] 1.809135
C(priors_count)[T.9] 2.082273
C(priors_count)[T.10] 1.417792
C(priors_count)[T.11] 1.754724
C(priors_count)[T.12] 1.774474
C(priors_count)[T.13] 1.634816
C(priors_count)[T.14] 0.676243
C(priors_count)[T.15] 1.328460
C(priors_count)[T.16] 0.279271
C(priors_count)[T.17] 0.496707
C(priors_count)[T.18] 2.677733
C(priors_count)[T.19] 2.145368
C(priors_count)[T.20] 1.235025
C(priors_count)[T.21] 2.827853
C(priors_count)[T.22] 3.418749
C(priors_count)[T.23] 3.201684
C(priors_count)[T.24] 0.632187
C(priors_count)[T.25] 3.546179
C(priors_count)[T.26] 1.360175
C(priors_count)[T.27] 0.092306
C(priors_count)[T.28] 1.547917
C(priors_count)[T.29] 0.513622
C(priors_count)[T.30] 2.572566
C(priors_count)[T.31] 1.434996
C(priors_count)[T.33] 0.758442
C(priors_count)[T.36] 0.005513
C(priors_count)[T.38] 0.685426
c_charge_degree[T.M] -0.256778
charge_cat[T.Agg Battery Grt/Bod/Harm] -0.702957
charge_cat[T.Aggrav Battery w/Deadly Weapon] -0.854300
charge_cat[T.Aggravated Assault W/Dead Weap] -0.408468
charge_cat[T.Aggravated Assault W/dead Weap] -0.970472
charge_cat[T.Aggravated Assault w/Firearm] -0.696535
charge_cat[T.Aggravated Battery] -1.436409
charge_cat[T.Att Burgl Unoccupied Dwel] -0.155216
charge_cat[T.Battery] -0.288576
charge_cat[T.Battery on Law Enforc Officer] 0.532573
charge_cat[T.Battery on a Person Over 65] 1.720185
charge_cat[T.Burglary Conveyance Unoccup] -0.377512
charge_cat[T.Burglary Dwelling Assault/Batt] 0.129765
charge_cat[T.Burglary Unoccupied Dwelling] -0.406856
charge_cat[T.Carrying Concealed Firearm] -0.002864
charge_cat[T.Crim Use of Personal ID Info] -0.528541
charge_cat[T.Criminal Mischief] 1.629331
charge_cat[T.Criminal Mischief Damage <$200] 0.308803
charge_cat[T.Criminal Mischief>$200<$1000] -0.287985
charge_cat[T.Cruelty Toward Child] -0.430886
charge_cat[T.DUI - Enhanced] -2.352112
charge_cat[T.DUI Level 0.15 Or Minor In Veh] -0.438058
charge_cat[T.DUI Property Damage/Injury] -1.430928
charge_cat[T.Dealing in Stolen Property] -0.073381
charge_cat[T.Deliver 3,4 Methylenediox] -0.646395
charge_cat[T.Deliver Cocaine] -0.697759
charge_cat[T.Disorderly Conduct] 0.035615
charge_cat[T.Disorderly Intoxication] -0.861357
charge_cat[T.Driving License Suspended] -0.366977
charge_cat[T.Driving Under The Influence] -0.203970
charge_cat[T.Exposes Culpable Negligence] -0.487167
charge_cat[T.False Ownership Info/Pawn Item] -0.375191
charge_cat[T.Felony Petit Theft] 0.552385
charge_cat[T.Grand Theft (Motor Vehicle)] 0.198255
charge_cat[T.Grand Theft in the 3rd Degree] -0.327947
charge_cat[T.Leave Acc/Attend Veh/More $50] 0.541556
charge_cat[T.Leaving Acc/Unattended Veh] -0.507106
charge_cat[T.Leaving the Scene of Accident] -0.454501
charge_cat[T.Lewd or Lascivious Molestation] -1.591408
charge_cat[T.Lve/Scen/Acc/Veh/Prop/Damage] 0.548930
charge_cat[T.Operating W/O Valid License] -0.587523
charge_cat[T.Petit Theft] 0.205814
charge_cat[T.Petit Theft $100- $300] 0.797360
charge_cat[T.Pos Cannabis W/Intent Sel/Del] -0.059560
charge_cat[T.Poss 3,4 MDMA (Ecstasy)] 0.422941
charge_cat[T.Poss Contr Subst W/o Prescript] 0.030039
charge_cat[T.Poss Pyrrolidinovalerophenone] 4.176535
charge_cat[T.Poss3,4 Methylenedioxymethcath] -2.156065
charge_cat[T.Possess Cannabis/20 Grams Or Less] -0.455053
charge_cat[T.Possession Burglary Tools] 0.007158
charge_cat[T.Possession of Cannabis] -0.670054
charge_cat[T.Possession of Cocaine] 0.436734
charge_cat[T.Possession of Hydrocodone] -0.311079
charge_cat[T.Possession of Oxycodone] 0.163186
charge_cat[T.Prowling/Loitering] 0.600412
charge_cat[T.Resist Officer w/Violence] -0.021553
charge_cat[T.Robbery / No Weapon] -1.437635
charge_cat[T.Susp Drivers Lic 1st Offense] -0.317377
charge_cat[T.Tamper With Witness/Victim/CI] 0.320615
charge_cat[T.Tampering With Physical Evidence] 1.788951
charge_cat[T.Uttering a Forged Instrument] 0.402091
charge_cat[T.Viol Pretrial Release Dom Viol] 1.169413
charge_cat[T.arrest case no charge] -0.544847
charge_cat[T.other] -0.540606
sex[T.Male]:C(priors_count)[T.1] 0.178020
sex[T.Male]:C(priors_count)[T.3] 0.202524
sex[T.Male]:C(priors_count)[T.4] -0.199666
sex[T.Male]:C(priors_count)[T.5] 0.086965
sex[T.Male]:C(priors_count)[T.6] 0.159051
sex[T.Male]:C(priors_count)[T.9] -0.374441
sex[T.Male]:C(priors_count)[T.10] 0.688940
sex[T.Male]:C(priors_count)[T.12] -0.128524
sex[T.Male]:C(priors_count)[T.13] 0.552880
sex[T.Male]:C(priors_count)[T.14] 1.185119
sex[T.Male]:C(priors_count)[T.15] 0.177672
sex[T.Male]:C(priors_count)[T.16] 1.879984
sex[T.Male]:C(priors_count)[T.17] 2.036532
sex[T.Male]:C(priors_count)[T.24] 1.458360
sex[T.Male]:C(priors_count)[T.26] 2.834280
sex[T.Male]:C(priors_count)[T.27] 0.025774
sex[T.Male]:C(priors_count)[T.28] 1.666044
sex[T.Male]:C(priors_count)[T.29] 1.503019
sex[T.Male]:C(priors_count)[T.31] 0.405266
sex[T.Male]:C(priors_count)[T.33] 1.302806
sex[T.Male]:C(priors_count)[T.36] 0.030305
sex[T.Male]:C(priors_count)[T.38] 1.849936
sex[T.Male]:c_charge_degree[T.M] 0.248443
sex[T.Male]:charge_cat[T.Aggrav Battery w/Deadly Weapon] 0.543098
sex[T.Male]:charge_cat[T.Aggravated Assault W/Dead Weap] 0.734866
sex[T.Male]:charge_cat[T.Aggravated Assault W/dead Weap] -0.147720
sex[T.Male]:charge_cat[T.Aggravated Battery / Pregnant] 0.160743
sex[T.Male]:charge_cat[T.Assault] -0.000680
sex[T.Male]:charge_cat[T.Att Burgl Unoccupied Dwel] -0.293595
sex[T.Male]:charge_cat[T.Battery] 0.188605
sex[T.Male]:charge_cat[T.Battery on Law Enforc Officer] -0.261569
sex[T.Male]:charge_cat[T.Battery on a Person Over 65] -2.353921
sex[T.Male]:charge_cat[T.Burglary Conveyance Unoccup] 1.074197
sex[T.Male]:charge_cat[T.Burglary Unoccupied Dwelling] -0.109827
sex[T.Male]:charge_cat[T.Carrying Concealed Firearm] -0.006554
sex[T.Male]:charge_cat[T.Child Abuse] -0.194898
sex[T.Male]:charge_cat[T.Corrupt Public Servant] 2.104213
sex[T.Male]:charge_cat[T.Crimin Mischief Damage $1000+] 1.127054
sex[T.Male]:charge_cat[T.Criminal Mischief Damage <$200] -0.936025
sex[T.Male]:charge_cat[T.DUI Property Damage/Injury] 0.027619
sex[T.Male]:charge_cat[T.Deliver 3,4 Methylenediox] -0.207810
sex[T.Male]:charge_cat[T.Driving License Suspended] -0.148052
sex[T.Male]:charge_cat[T.Driving Under The Influence] -0.192439
sex[T.Male]:charge_cat[T.Driving While License Revoked] -0.477104
sex[T.Male]:charge_cat[T.False Ownership Info/Pawn Item] 1.669741
sex[T.Male]:charge_cat[T.Felony Battery (Dom Strang)] -0.024260
sex[T.Male]:charge_cat[T.Felony Battery w/Prior Convict] -0.760619
sex[T.Male]:charge_cat[T.Felony Driving While Lic Suspd] 0.297832
sex[T.Male]:charge_cat[T.Grand Theft (Motor Vehicle)] -0.043314
sex[T.Male]:charge_cat[T.Grand Theft in the 3rd Degree] 0.572814
sex[T.Male]:charge_cat[T.Leave Acc/Attend Veh/More $50] -1.698415
sex[T.Male]:charge_cat[T.Lewd or Lascivious Molestation] -0.809900
sex[T.Male]:charge_cat[T.Lve/Scen/Acc/Veh/Prop/Damage] -0.385189
sex[T.Male]:charge_cat[T.Neglect Child / No Bodily Harm] -1.014313
sex[T.Male]:charge_cat[T.Operating W/O Valid License] -0.160582
sex[T.Male]:charge_cat[T.Petit Theft] 1.082302
sex[T.Male]:charge_cat[T.Poss Cocaine/Intent To Del/Sel] -1.316571
sex[T.Male]:charge_cat[T.Poss3,4 Methylenedioxymethcath] 1.845378
sex[T.Male]:charge_cat[T.Possess Cannabis/20 Grams Or Less] 0.406446
sex[T.Male]:charge_cat[T.Possession Of Alprazolam] 0.112760
sex[T.Male]:charge_cat[T.Possession Of Heroin] 0.511526
sex[T.Male]:charge_cat[T.Possession Of Methamphetamine] 0.234489
sex[T.Male]:charge_cat[T.Possession of Cannabis] -0.095246
sex[T.Male]:charge_cat[T.Possession of Hydromorphone] -0.704449
sex[T.Male]:charge_cat[T.Possession of Oxycodone] -0.818559
sex[T.Male]:charge_cat[T.Prowling/Loitering] 0.000598
sex[T.Male]:charge_cat[T.Resist/Obstruct W/O Violence] 0.402313
sex[T.Male]:charge_cat[T.Robbery / No Weapon] 0.885658
sex[T.Male]:charge_cat[T.Tampering With Physical Evidence] -1.774577
sex[T.Male]:charge_cat[T.Uttering a Forged Instrument] -0.373681
sex[T.Male]:charge_cat[T.Viol Injunct Domestic Violence] 0.752625
sex[T.Male]:charge_cat[T.arrest case no charge] 0.121360
sex[T.Male]:charge_cat[T.other] 0.221352
age -0.030208
sex[T.Male]:age -0.015452
juv_fel_count 0.399418
sex[T.Male]:juv_fel_count -0.158158
juv_misd_count -0.259268
sex[T.Male]:juv_misd_count 0.321946
juv_other_count 0.317857
sex[T.Male]:juv_other_count -0.229113

Now, let’s look at calibration and balance using similar tables and figures as we did above.

In [40]:
output, given_outcome, given_pred =cm_tables(
    lasso_mod.predict(X_test),
    y_test,
    df_test
)
display(output)
display(given_pred)
display(given_outcome)

calibration_plot(lasso_mod.predict_proba(X_test)[:,1],y_test, df_test)
balance_threshold_plot(lasso_mod.predict_proba(X_test)[:,1],y_test, df_test);
overall African-American Caucasian Hispanic Female Male
Portion_of_NoRecid_and_LowRisk 0.405421 0.328025 0.483333 0.574194 0.561960 0.365185
Portion_of_Recid_and_LowRisk 0.147908 0.165605 0.128333 0.116129 0.083573 0.164444
Portion_of_NoRecid_and_HighRisk 0.190925 0.182590 0.203333 0.193548 0.242075 0.177778
Portion_of_Recid_and_HighRisk 0.255745 0.323779 0.185000 0.116129 0.112392 0.292593
overall African-American Caucasian Hispanic Female Male
P(NoRecid|LowRisk) 0.679842 0.642412 0.703883 0.747899 0.698925 0.672578
P(NoRecid|HighRisk) 0.366423 0.338395 0.409574 0.500000 0.426471 0.359806
P(Recid|LowRisk) 0.320158 0.357588 0.296117 0.252101 0.301075 0.327422
P(Recid|HighRisk) 0.633577 0.661605 0.590426 0.500000 0.573529 0.640194
overall African-American Caucasian Hispanic Female Male
P(LowRisk|NoRecid) 0.732694 0.664516 0.790191 0.831776 0.870536 0.689510
P(HighRisk|NoRecid) 0.267306 0.335484 0.209809 0.168224 0.129464 0.310490
P(LowRisk|Recid) 0.427441 0.360587 0.523605 0.625000 0.682927 0.377953
P(HighRisk|Recid) 0.572559 0.639413 0.476395 0.375000 0.317073 0.622047

As with COMPAS score, our predictions are well-calibrated, but the false negative and false positive rates are not well balanced across racial groups.

See exercise 3 in the exercise list

In [41]:
# Fit your prediction model and plot calibration and balance

Regularizing to Maximize Balance

Trying to improve balance by ad-hoc modifications will be difficult.

Let’s try to do it more systematically.

We usually select models and choose regularization to minimize prediction errors.

We can just as well select models and regularization parameters to optimize some other criteria.

Let’s choose the regularization parameter for lasso to maximize balance.

In [42]:
# define a custom CV criteria to maximize
def balance_scorer(y_true, prob, df, weights):
    ind = df.isin(y_true.index)
    df_cv = df.loc[y_true.index.values,:]
    b = df_cv.race=="African-American"
    w = df_cv.race=="Caucasian"
    y_pred = 1*(prob>0.5)
    fprb = np.mean(y_pred[(y_true==0) & b])
    fprw = np.mean(y_pred[(y_true==0) & w])
    fnrb = np.mean(y_pred[(y_true==1) & b]==0)
    fnrw = np.mean(y_pred[(y_true==1) & w]==0)
    return(-weights[0]*(fprb-fprw)**2 +
           -weights[1]*(fnrb-fnrw)**2 +
           -weights[2]*(metrics.log_loss(y_true, prob, normalize=True)))

score_params = {"df": df_train, "weights": [10.0, 1.0, 0.0]}
scorer = metrics.make_scorer(balance_scorer, **score_params, needs_proba=True)
grid_cv = model_selection.GridSearchCV(
    estimator=linear_model.LogisticRegression(penalty="l1",
                                              max_iter=100,
                                              solver="liblinear"),
    scoring=scorer,
    cv=5,
    param_grid={'C':
    np.exp(np.linspace(-10,10,10))},
    return_train_score=True,
    verbose=True,
    refit=True,
    iid=True)

balance_mod=grid_cv.fit(X_train,y_train)
Fitting 5 folds for each of 10 candidates, totalling 50 fits
[Parallel(n_jobs=1)]: Using backend SequentialBackend with 1 concurrent workers.
[Parallel(n_jobs=1)]: Done  50 out of  50 | elapsed:   11.8s finished
In [43]:
# plots illustrating regularization parameter choice
def grid_cv_plot(mod, ylabel=""):
    scores=mod.cv_results_["mean_test_score"]
    Cdict=mod.cv_results_["params"]
    logpenalties=np.log([d['C'] for d in Cdict])
    colors=plt.rcParams["axes.prop_cycle"].by_key()["color"]
    fig, ax1 = plt.subplots()
    ax1.plot(logpenalties,scores, color=colors[0])
    ax1.set_ylabel(ylabel, color=colors[0])
    ax1.set_xlabel("-log(penalty)")
    ax1.tick_params('y', colors=colors[0]);
grid_cv_plot(balance_mod,"CV balance score")

We can be perfectly balanced by making the regularization parameter very large.

Unfortunately, this makes all the predictions identical, so these predictions are not so useful.

In [44]:
output, given_outcome, given_pred =cm_tables(
    balance_mod.best_estimator_.predict(X_test),
    y_test,
    df_test
)
display(output)
display(given_pred)
display(given_outcome)
/home/ubuntu/anaconda3/lib/python3.7/site-packages/ipykernel_launcher.py:44: RuntimeWarning: invalid value encountered in true_divide
overall African-American Caucasian Hispanic Female Male
Portion_of_NoRecid_and_LowRisk 0.553329 0.493631 0.611667 0.690323 0.645533 0.52963
Portion_of_Recid_and_LowRisk 0.000000 0.000000 0.000000 0.000000 0.000000 0.00000
Portion_of_NoRecid_and_HighRisk 0.446671 0.506369 0.388333 0.309677 0.354467 0.47037
Portion_of_Recid_and_HighRisk 0.000000 0.000000 0.000000 0.000000 0.000000 0.00000
overall African-American Caucasian Hispanic Female Male
P(NoRecid|LowRisk) 0.553329 0.493631 0.611667 0.690323 0.645533 0.52963
P(NoRecid|HighRisk) NaN NaN NaN NaN NaN NaN
P(Recid|LowRisk) 0.446671 0.506369 0.388333 0.309677 0.354467 0.47037
P(Recid|HighRisk) NaN NaN NaN NaN NaN NaN
overall African-American Caucasian Hispanic Female Male
P(LowRisk|NoRecid) 1.0 1.0 1.0 1.0 1.0 1.0
P(HighRisk|NoRecid) 0.0 0.0 0.0 0.0 0.0 0.0
P(LowRisk|Recid) 1.0 1.0 1.0 1.0 1.0 1.0
P(HighRisk|Recid) 0.0 0.0 0.0 0.0 0.0 0.0

What if we change our CV scoring function to care about both prediction and balance?

In [45]:
score_params = {"df": df_train, "weights": [10.0, 1.0, 5.0]}
grid_cv.set_params(scoring=metrics.make_scorer(balance_scorer, **score_params, needs_proba=True))
bf_mod=grid_cv.fit(X_train,y_train)
grid_cv_plot(bf_mod,"CV balance & fit")

output, given_outcome, given_pred =cm_tables(
    bf_mod.best_estimator_.predict(X_test),
    y_test,
    df_test
)
display(output)
display(given_pred)
display(given_outcome)
calibration_plot(bf_mod.best_estimator_.predict_proba(X_test)[:,1],y_test, df_test)
balance_threshold_plot(bf_mod.best_estimator_.predict_proba(X_test)[:,1],y_test, df_test);
[Parallel(n_jobs=1)]: Using backend SequentialBackend with 1 concurrent workers.
Fitting 5 folds for each of 10 candidates, totalling 50 fits
[Parallel(n_jobs=1)]: Done  50 out of  50 | elapsed:   11.3s finished
overall African-American Caucasian Hispanic Female Male
Portion_of_NoRecid_and_LowRisk 0.405421 0.328025 0.483333 0.574194 0.561960 0.365185
Portion_of_Recid_and_LowRisk 0.147908 0.165605 0.128333 0.116129 0.083573 0.164444
Portion_of_NoRecid_and_HighRisk 0.189747 0.181529 0.200000 0.200000 0.239193 0.177037
Portion_of_Recid_and_HighRisk 0.256924 0.324841 0.188333 0.109677 0.115274 0.293333
overall African-American Caucasian Hispanic Female Male
P(NoRecid|LowRisk) 0.681188 0.643750 0.707317 0.741667 0.701439 0.673497
P(NoRecid|HighRisk) 0.365357 0.337662 0.405263 0.514286 0.420290 0.359223
P(Recid|LowRisk) 0.318812 0.356250 0.292683 0.258333 0.298561 0.326503
P(Recid|HighRisk) 0.634643 0.662338 0.594737 0.485714 0.579710 0.640777
overall African-American Caucasian Hispanic Female Male
P(LowRisk|NoRecid) 0.732694 0.664516 0.790191 0.831776 0.870536 0.689510
P(HighRisk|NoRecid) 0.267306 0.335484 0.209809 0.168224 0.129464 0.310490
P(LowRisk|Recid) 0.424802 0.358491 0.515021 0.645833 0.674797 0.376378
P(HighRisk|Recid) 0.575198 0.641509 0.484979 0.354167 0.325203 0.623622

See exercise 4 in the exercise list

Tradeoffs are Inevitable

We could try to tweak our predictions further to improve balance.

However, motivated in part by this COMPAS example, [recidKLL+17] proved that it is impossible for any prediction algorithm to be both perfectly balanced and well-calibrated.

Improvements in balance necessarily make calibration worse.

References

[recidDMB16] William Dieterich, Christina Mendoza, and Tim Brennan. Compas risk scales: demonstrating accuracy equity and predictive parity. Northpoint Inc, 2016.

[recidKLL+17] Jon Kleinberg, Himabindu Lakkaraju, Jure Leskovec, Jens Ludwig, and Sendhil Mullainathan. Human Decisions and Machine Predictions. The Quarterly Journal of Economics*, 133(1):237–293, 08 2017. URL: https://dx.doi.org/10.1093/qje/qjx032, arXiv:http://oup.prod.sis.lan/qje/article-pdf/133/1/237/24246094/qjx032.pdf, doi:10.1093/qje/qjx032.

Exercises

Exercise 1

Can you develop a model that performs better at mimicking their risk scores?

(back to text)

Exercise 2

We made our calibration plot using a held-out test sample. What do you think would happen if made the calibration plot using the training sample? Check and see.

(back to text)

Exercise 3

Try to improve balance and/or calibration by creating an alternative prediction.

(back to text)

Exercise 4

Modify the cross-validation scoring function to see how it affects calibration and balance.

(back to text)

Download

Launch Notebook