Logistic Regression and Model Comparison with Bambi and ArviZ

Adults dataset

The adults dataset is comprised of census data from 1994 in United States.
The goal is to use demographic variables to predict whether an individual makes more than $50,000 per year.

The following is a description of the variables in the dataset.

  • age: Individual’s age
  • workclass: Labor class.
  • fnlwgt: It is not specified, but we guess it is a final sampling weight.
  • education: Education level as a categorical variable.
  • educational_num: Education level as numerical variable. It does not reflect years of education.
  • marital_status: Marital status.
  • occupation: Occupation.
  • relationship: Relationship with the head of household.
  • race: Individual’s race.
  • sex: Individual’s sex.
  • capital_gain: Capital gain during unspecified period of time.
  • capital_loss: Capital loss during unspecified period of time.
  • hs_week: Hours of work per week.
  • native_country: Country of birth.
  • income: Income as a binary variable (either below or above 50K per year).

We are only using the following variables in this example: income, sex, race, age, and hs_week. This subset is comprised of both categorical and numerical variables which allows us to visualize how to incorporate both types in a logistic regression model while helping to keep the analysis simpler.

import arviz as az
import bambi as bmb
import matplotlib.pyplot as plt
import matplotlib.lines as mlines
import numpy as np
import pandas as pd
import seaborn as sns
import warnings

from scipy.special import expit as invlogit
# Disable a FutureWarning in ArviZ at the moment of running the notebook
az.style.use("arviz-darkgrid")
warnings.simplefilter(action='ignore', category=FutureWarning)
data = bmb.load_data("adults")
data.info()
data.head()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 32561 entries, 0 to 32560
Data columns (total 5 columns):
 #   Column   Non-Null Count  Dtype 
---  ------   --------------  ----- 
 0   income   32561 non-null  object
 1   sex      32561 non-null  object
 2   race     32561 non-null  object
 3   age      32561 non-null  int64 
 4   hs_week  32561 non-null  int64 
dtypes: int64(2), object(3)
memory usage: 1.2+ MB
income sex race age hs_week
0 <=50K Male White 39 40
1 <=50K Male White 50 13
2 <=50K Male White 38 40
3 <=50K Male Black 53 40
4 <=50K Female Black 28 40

Categorical variables are presented as from type object. In this step we convert them to category.

categorical_cols = data.columns[data.dtypes == object].tolist()
for col in categorical_cols:
    data[col] = data[col].astype("category")
data.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 32561 entries, 0 to 32560
Data columns (total 5 columns):
 #   Column   Non-Null Count  Dtype   
---  ------   --------------  -----   
 0   income   32561 non-null  category
 1   sex      32561 non-null  category
 2   race     32561 non-null  category
 3   age      32561 non-null  int64   
 4   hs_week  32561 non-null  int64   
dtypes: category(3), int64(2)
memory usage: 604.7 KB

Instead of going straight to fitting models, we’re going to do a some exploratory analysis of the variables in the dataset. First we have some plots, and then some conclusions about the information in the plots.

# Just a utilitary function to truncate labels and avoid overlapping in plots
def truncate_labels(ticklabels, width=8):
    def truncate(label, width):
        if len(label) > width - 3:
            return label[0 : (width - 4)] + "..."
        else:
            return label

    labels = [x.get_text() for x in ticklabels]
    labels = [truncate(lbl, width) for lbl in labels]

    return labels
fig, axes = plt.subplots(3, 2, figsize=(12, 15))
sns.countplot(x="income", color="C0", data=data, ax=axes[0, 0], saturation=1)
sns.countplot(x="sex", color="C0", data=data, ax=axes[0, 1], saturation=1);
sns.countplot(x="race", color="C0", data=data, ax=axes[1, 0], saturation=1);
axes[1, 0].set_xticklabels(truncate_labels(axes[1, 0].get_xticklabels()))
axes[1, 1].hist(data["age"], bins=20);
axes[1, 1].set_xlabel("Age")
axes[1, 1].set_ylabel("Count")
axes[2, 0].hist(data["hs_week"], bins=20);
axes[2, 0].set_xlabel("Hours of work / week")
axes[2, 0].set_ylabel("Count")
axes[2, 1].axis('off');

Highlights

  • Approximately 25% of the people make more than 50K a year.
  • Two thirds of the subjects are males.
  • The great majority of the subjects are white, only a minority are black and the other categories are very infrequent.
  • The distribution of age is skewed to the right, as one might expect.
  • The distribution of hours of work per week looks weird at first sight. But what is a typical workload per week? You got it, 40 hours :).

We only keep the races black and white to simplify the analysis. The other categories don’t appear very often in our data.

Now, we see the distribution of income for the different levels of our explanatory variables. Numerical variables are binned to make the analysis possible.

data = data[data["race"].isin(["Black", "White"])]
data["race"] = data["race"].cat.remove_unused_categories()
age_bins = [17, 25, 35, 45, 65, 90]
data["age_binned"] = pd.cut(data["age"], age_bins)
hours_bins = [0, 20, 40, 60, 100]
data["hs_week_binned"] = pd.cut(data["hs_week"], hours_bins)
fig, axes = plt.subplots(3, 2, figsize=(12, 15))
sns.countplot(x="income", color="C0", data=data, ax=axes[0, 0])
sns.countplot(x="sex", hue="income", data=data, ax=axes[0, 1])
sns.countplot(x="race", hue="income", data=data, ax=axes[1, 0])
sns.countplot(x="age_binned", hue="income", data=data, ax=axes[1, 1])
sns.countplot(x="hs_week_binned", hue="income", data=data, ax=axes[2, 0])
axes[2, 1].axis("off");

Some quick and gross info from the plots

  • The probability of making more than \$50k a year is larger if you are a Male.
  • A person also has more probability of making more than \$50k/yr if she/he is White.
  • For age, we see the probability of making more than \$50k a year increases as the variable increases, up to a point where it starts to decrease.
  • Also, the more hours a person works per week, the higher the chance of making more than \$50k/yr. There’s a big jump in that probability when the hours of work per week jump from the (20, 40] bin to the (40, 60] one.

Some data preparation before fitting our model. Here we standardize numerical variables age and hs_week because it may help sampler convergence. Also, we compute their second and third power. These powers will be sequantialy added to the model.

age_mean = np.mean(data["age"])
age_std = np.std(data["age"])
hs_mean = np.mean(data["hs_week"])
hs_std = np.std(data["hs_week"])

data["age"] = (data["age"] - age_mean) / age_std
data["age2"] = data["age"] ** 2
data["age3"] = data["age"] ** 3
data["hs_week"] = (data["hs_week"] - hs_mean) / hs_std
data["hs_week2"] = data["hs_week"] ** 2
data["hs_week3"] = data["hs_week"] ** 3

data = data.drop(columns=["age_binned", "hs_week_binned"])

This is how our data looks like before fitting the models.

data.head()
income sex race age hs_week age2 age3 hs_week2 hs_week3
0 <=50K Male White 0.024207 -0.037250 0.000586 0.000014 0.001388 -0.000052
1 <=50K Male White 0.827984 -2.222326 0.685557 0.567630 4.938734 -10.975479
2 <=50K Male White -0.048863 -0.037250 0.002388 -0.000117 0.001388 -0.000052
3 <=50K Male Black 1.047195 -0.037250 1.096618 1.148374 0.001388 -0.000052
4 <=50K Female Black -0.779569 -0.037250 0.607728 -0.473766 0.001388 -0.000052

The model

We will use a logistic regression model to estimate the probability of making more than \$50K as a function of age, hours of work per week, sex, race and education level.

If we have a binary response variable \(Y\) and a set of predictors or explanatory variables \(X_1, X_2, \cdots, X_p\) the logistic regression model can be defined as follows:

\[\log{\left(\frac{\pi}{1 - \pi}\right)} = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \cdots + \beta_p X_p\]

where \(\pi = P(Y = 1)\) (a.k.a. probability of success) and \(\beta_0, \beta_1, \cdots \beta_p\) are unknown parameters. The term on the left side is the logarithm of the odds ratio or simply known as the log-odds. With little effort, the expression can be re-arranged to express our probability of interest, \(\pi\), as a function of the betas and the predictors.

\[ \pi = \frac{e^{\beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p}}{1 + e^{\beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p}} = \frac{1}{1 + e^{-(\beta_0 + \beta_1 X_1 + \cdots + \beta_p X_p)}} \]

We need to specify a prior and a likelihood in order to draw samples from the posterior distribution. We could use sociological knowledge about the effects of age and education on income, but instead, let’s use the default prior specification in Bambi.

The likelihood is the product of \(n\) Bernoulli trials, \(\prod_{i=1}^{n}{p_i^y(1-p_i)^{1-y_i}}\) where \(p_i = P(Y=1)\).

In our case, we have

\[Y = \left\{ \begin{array}{ll} 1 & \textrm{if the person makes more than 50K per year} \\ 0 & \textrm{if the person makes less than 50K per year} \end{array} \right. \]

\[\pi = P(Y=1)\]

But this is a Bambi example, right? Let’s see how Bambi can helps us to build a logistic regression model.

Model 1:

\[ \log{\left(\frac{\pi}{1 - \pi}\right)} = \beta_0 + \beta_1 X_1 + \beta_2 X_2 + \beta_3 X_3 + \beta_4 X_4 \]

Where:

\[ \begin{split} X_1 &= \displaystyle \frac{\text{Age} - \text{Age}_{\text{mean}}}{\text{Age}_{\text{std}}} \\ X_2 &= \displaystyle \frac{\text{Hours\_week} - \text{Hours\_week}_{\text{mean}}}{\text{Hours\_week}_{\text{std}}} \\ X_3 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is male} \\ 0 & \textrm{if the person is female} \end{array} \right. \\ X_4 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is white} \\ 0 & \textrm{if the person is black} \end{array} \right. \end{split} \]

model1 = bmb.Model("income['>50K'] ~ sex + race + age + hs_week", data, family="bernoulli")
fitted1 = model1.fit(draws=1000, idata_kwargs={"log_likelihood": True})
Modeling the probability that income==>50K
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [Intercept, sex, race, age, hs_week]
100.00% [4000/4000 00:20<00:00 Sampling 2 chains, 0 divergences] 
Sampling 2 chains for 1_000 tune and 1_000 draw iterations (2_000 + 2_000 draws total) took 21 seconds.
az.plot_trace(fitted1);
az.summary(fitted1)
mean sd hdi_3% hdi_97% mcse_mean mcse_sd ess_bulk ess_tail r_hat
Intercept -2.635 0.062 -2.757 -2.525 0.001 0.001 2457.0 1739.0 1.0
sex[Male] 1.018 0.037 0.948 1.087 0.001 0.001 2141.0 1572.0 1.0
race[White] 0.630 0.058 0.532 0.751 0.001 0.001 3060.0 1566.0 1.0
age 0.578 0.015 0.554 0.608 0.000 0.000 1837.0 1281.0 1.0
hs_week 0.504 0.015 0.477 0.533 0.000 0.000 2047.0 1568.0 1.0

Model 2

\[ \log{\left(\frac{\pi}{1 - \pi}\right)} = \beta_0 + \beta_1 X_1 + \beta_2 X_1^2 + \beta_3 X_2 + \beta_4 X_2^2 + \beta_5 X_3 + \beta_6 X_4 \]

Where:

$$ \[\begin{aligned} X_1 &= \displaystyle \frac{\text{Age} - \text{Age}_{\text{mean}}}{\text{Age}_{\text{std}}} \\ X_2 &= \displaystyle \frac{\text{Hours\_week} - \text{Hours\_week}_{\text{mean}}}{\text{Hours\_week}_{\text{std}}} \\ X_3 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is male} \\ 0 & \textrm{if the person is female} \end{array} \right. \\ X_4 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is white} \\ 0 & \textrm{if the person is black} \end{array} \right. \end{aligned}\]

$$

model2 = bmb.Model("income['>50K'] ~ sex + race + age + age2 + hs_week + hs_week2", data, family="bernoulli")
fitted2 = model2.fit(idata_kwargs={"log_likelihood": True})
Modeling the probability that income==>50K
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [Intercept, sex, race, age, age2, hs_week, hs_week2]
100.00% [4000/4000 00:29<00:00 Sampling 2 chains, 0 divergences] 
Sampling 2 chains for 1_000 tune and 1_000 draw iterations (2_000 + 2_000 draws total) took 30 seconds.
az.plot_trace(fitted2);
az.summary(fitted2)
mean sd hdi_3% hdi_97% mcse_mean mcse_sd ess_bulk ess_tail r_hat
Intercept -2.282 0.065 -2.406 -2.166 0.001 0.001 2037.0 1330.0 1.0
sex[Male] 1.006 0.038 0.939 1.074 0.001 0.001 2192.0 1628.0 1.0
race[White] 0.702 0.061 0.590 0.818 0.001 0.001 2084.0 1343.0 1.0
age 1.069 0.024 1.028 1.117 0.001 0.000 1720.0 1406.0 1.0
age2 -0.538 0.018 -0.570 -0.503 0.000 0.000 1730.0 1161.0 1.0
hs_week 0.499 0.022 0.455 0.538 0.001 0.000 1665.0 1431.0 1.0
hs_week2 -0.088 0.009 -0.103 -0.072 0.000 0.000 1687.0 1577.0 1.0

Model 3

\[ \log{\left(\frac{\pi}{1 - \pi}\right)} = \beta_0 + \beta_1 X_1 + \beta_2 X_1^2 + \beta_3 X_1^3 + \beta_4 X_2 + \beta_5 X_2^2 + \beta_6 X_2^3 + \beta_7 X_3 + \beta_8 X_4 \]

Where:

\[ \begin{aligned} X_1 &= \displaystyle \frac{\text{Age} - \text{Age}_{\text{mean}}}{\text{Age}_{\text{std}}} \\ X_2 &= \displaystyle \frac{\text{Hours\_week} - \text{Hours\_week}_{\text{mean}}}{\text{Hours\_week}_{\text{std}}} \\ X_3 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is male} \\ 0 & \textrm{if the person is female} \end{array} \right. \\ X_4 &= \left\{ \begin{array}{ll} 1 & \textrm{if the person is white} \\ 0 & \textrm{if the person is black} \end{array} \right. \end{aligned} \]

model3 = bmb.Model(
    "income['>50K'] ~ age + age2 + age3 + hs_week + hs_week2 + hs_week3 + sex + race",
    data,
    family="bernoulli"
)
fitted3 = model3.fit(
    draws=1000, random_seed=1234, target_accept=0.9, idata_kwargs={"log_likelihood": True}
)
Modeling the probability that income==>50K
Auto-assigning NUTS sampler...
Initializing NUTS using jitter+adapt_diag...
Multiprocess sampling (2 chains in 2 jobs)
NUTS: [Intercept, age, age2, age3, hs_week, hs_week2, hs_week3, sex, race]
100.00% [4000/4000 01:15<00:00 Sampling 2 chains, 0 divergences] 
Sampling 2 chains for 1_000 tune and 1_000 draw iterations (2_000 + 2_000 draws total) took 75 seconds.
az.plot_trace(fitted3);
az.summary(fitted3)
mean sd hdi_3% hdi_97% mcse_mean mcse_sd ess_bulk ess_tail r_hat
Intercept -2.145 0.064 -2.270 -2.028 0.001 0.001 3201.0 1540.0 1.0
age 0.963 0.026 0.913 1.009 0.001 0.000 2243.0 1290.0 1.0
age2 -0.894 0.030 -0.946 -0.836 0.001 0.001 1541.0 1229.0 1.0
age3 0.175 0.011 0.153 0.194 0.000 0.000 1653.0 1506.0 1.0
hs_week 0.612 0.025 0.567 0.661 0.001 0.000 2381.0 1300.0 1.0
hs_week2 -0.010 0.010 -0.030 0.010 0.000 0.000 2299.0 1590.0 1.0
hs_week3 -0.035 0.004 -0.042 -0.028 0.000 0.000 1815.0 1572.0 1.0
sex[Male] 0.985 0.038 0.918 1.059 0.001 0.001 2737.0 1549.0 1.0
race[White] 0.681 0.060 0.573 0.798 0.001 0.001 3044.0 1514.0 1.0

Model comparison

We can perform a Bayesian model comparison very easily with az.compare(). Here we pass a dictionary with the InferenceData objects that Model.fit() returned and az.compare() returns a data frame that is ordered from best to worst according to the criteria used. By default, ArviZ uses loo, which is an estimation of leave one out cross validation. Another option is the widely applicable information criterion (WAIC). For more information about the information criteria available and other options within the function see the docs.

models_dict = {
    "model1": fitted1,
    "model2": fitted2,
    "model3": fitted3
}
df_compare = az.compare(models_dict)
df_compare
rank elpd_loo p_loo elpd_diff weight se dse warning scale
model3 0 -13987.197673 9.716205 0.000000 1.000000e+00 89.279906 0.000000 False log
model2 1 -14155.112761 8.147063 167.915088 3.048565e-12 91.305227 19.879825 False log
model1 2 -14915.862090 4.871886 928.664417 0.000000e+00 91.010624 38.923423 False log 
az.plot_compare(df_compare, insample_dev=False);

There is a difference in the point estimations (empty circles) between the model with cubic terms (model 3) and the model with quadratic terms (model 2) but there is some overlap between their interval estimations. This time, we are going to select model 2 and do some extra little work with it because from previous experience with this dataset we know there is no substantial difference between them, and model 2 is simpler. However, as we mention in the final remarks, this is not the best you can achieve with this dataset. If you want, you could also try to add other predictors, such as education level and see how it impacts in the model comparison :).

Probability estimation

In this section we plot age vs the probability of making more than 50K a year given different profiles.

We set hours of work per week at 40 hours and assign a grid from 18 to 75 age. They’re standardized because they were standardized when we fitted the model.

Here we use az.plot_hdi() to get Highest Density Interval plots. We get two bands for each profile. One corresponds to an hdi probability of 0.94 (the default) and the other to an hdi probability of 0.5.

HS_WEEK = (40 - hs_mean) / hs_std
AGE = (np.linspace(18, 75) - age_mean) / age_std

fig, ax = plt.subplots()
handles = []
i = 0

for race in ["Black", "White"]:
    for sex in ["Female", "Male"]:  
        color = f"C{i}"
        label = f"{race} - {sex}"
        handles.append(mlines.Line2D([], [], color=color, label=label, lw=3))
        
        new_data = pd.DataFrame({
            "sex": [sex] * len(AGE),
            "race": [race] * len(AGE), 
            "age": AGE,
            "age2": AGE ** 2,
            "hs_week": [HS_WEEK] * len(AGE),
            "hs_week2": [HS_WEEK ** 2] * len(AGE),
        })
        new_idata = model2.predict(fitted2, data=new_data, inplace=False)
        mean = new_idata.posterior["income_mean"].values

        az.plot_hdi(AGE * age_std + age_mean, mean, ax=ax, color=color)
        az.plot_hdi(AGE * age_std + age_mean, mean, ax=ax, color=color, hdi_prob=0.5)
        i += 1

ax.set_xlabel("Age")
ax.set_ylabel("P(Income > $50K)")
ax.legend(handles=handles, loc="upper left");

The highest posterior density bands show how the probability of earning more than 50K changes with age for a given profile. In all the cases, we see the probability of making more than $50K increases with age until approximately age 52, when the probability begins to drop off. We can interpret narrow portions of a curve as places where we have low uncertainty and spread out portions of the bands as places where we have somewhat higher uncertainty about our coefficient values.

Final remarks

In this notebook we’ve seen how easy it is to incorporate ArviZ into a Bambi workflow to perform model comparison based on information criteria such as LOO and WAIC. However, an attentive reader might have seen that the highest density interval plot never shows a predicted probability greater than 0.5 (which is not good if we expect to predict that at least some people working 40hrs/wk make more than \$50k/yr). You can increase the hours of work per week for the profiles we’ve used and the HDIs will show larger values. But we won’t be seeing the whole picture.

Although we’re using some demographic variables such as sex and race, the cells resulting from the combinations of their levels are still very heterogeneous. For example, we are mixing individuals of all educational levels. A possible next step is to incorporate education into the different models we compared. If any of the readers (yes, you!) is interested in doing so, here there are some notes that may help

  • Education is an ordinal categorical variable with a lot of levels.
    • Explore the conditional distribution of income given education levels.
    • See what are the counts/proportions of people within each education level.
    • Collapse categories (but respect the ordinality!). Try to end up with 5 or less categories if possible.
  • Start with a model with only age, sex, race, hs_week and education. Then incorporate higher order terms (second and third powers for example). Don’t go beyond fourth powers.
  • Look for a nice activity to do while the sampler does its job.
    We know it’s going to take a couple of hours to fit all those models :)

And finally, please feel free to open a new issue if you think there’s something that we can improve.

%load_ext watermark
%watermark -n -u -v -iv -w
Last updated: Thu Jan 05 2023

Python implementation: CPython
Python version       : 3.10.4
IPython version      : 8.5.0

bambi     : 0.9.3
numpy     : 1.23.5
matplotlib: 3.6.2
arviz     : 0.14.0
seaborn   : 0.12.2
pandas    : 1.5.2
sys       : 3.10.4 (main, Mar 31 2022, 08:41:55) [GCC 7.5.0]

Watermark: 2.3.1