Bayesian Logistic Regression

Bayesian logistic regression is the Bayesian counterpart to a common tool in machine learning, logistic regression. The goal of logistic regression is to predict a one or a zero for a given training item. An example might be predicting whether someone is sick or ill given their symptoms and personal information.

In our example, we’ll be working to predict whether someone is likely to default with a synthetic dataset found in the RDatasets package. This dataset, Defaults, comes from R’s ISLR package and contains information on borrowers.

To start, let’s import all the libraries we’ll need.

# Import Turing and Distributions.
using Turing, Distributions

# Import RDatasets.
using RDatasets

# Import MCMCChains, Plots, and StatsPlots for visualizations and diagnostics.
using MCMCChains, Plots, StatsPlots

# We need a logistic function, which is provided by StatsFuns.
using StatsFuns: logistic

# Functionality for splitting and normalizing the data
using MLDataUtils: shuffleobs, stratifiedobs, rescale!

# Set a seed for reproducibility.
using Random

Data Cleaning & Set Up

Now we’re going to import our dataset. The first six rows of the dataset are shown below so you can get a good feel for what kind of data we have.

# Import the "Default" dataset.
data = RDatasets.dataset("ISLR", "Default");

# Show the first six rows of the dataset.
first(data, 6)
6×4 DataFrame
Row Default Student Balance Income
Cat… Cat… Float64 Float64
1 No No 729.526 44361.6
2 No Yes 817.18 12106.1
3 No No 1073.55 31767.1
4 No No 529.251 35704.5
5 No No 785.656 38463.5
6 No Yes 919.589 7491.56

Most machine learning processes require some effort to tidy up the data, and this is no different. We need to convert the Default and Student columns, which say “Yes” or “No” into 1s and 0s. Afterwards, we’ll get rid of the old words-based columns.

# Convert "Default" and "Student" to numeric values.
data[!, :DefaultNum] = [r.Default == "Yes" ? 1.0 : 0.0 for r in eachrow(data)]
data[!, :StudentNum] = [r.Student == "Yes" ? 1.0 : 0.0 for r in eachrow(data)]

# Delete the old columns which say "Yes" and "No".
select!(data, Not([:Default, :Student]))

# Show the first six rows of our edited dataset.
first(data, 6)
6×4 DataFrame
Row Balance Income DefaultNum StudentNum
Float64 Float64 Float64 Float64
1 729.526 44361.6 0.0 0.0
2 817.18 12106.1 0.0 1.0
3 1073.55 31767.1 0.0 0.0
4 529.251 35704.5 0.0 0.0
5 785.656 38463.5 0.0 0.0
6 919.589 7491.56 0.0 1.0

After we’ve done that tidying, it’s time to split our dataset into training and testing sets, and separate the labels from the data. We separate our data into two halves, train and test. You can use a higher percentage of splitting (or a lower one) by modifying the at = 0.05 argument. We have highlighted the use of only a 5% sample to show the power of Bayesian inference with small sample sizes.

We must rescale our variables so that they are centered around zero by subtracting each column by the mean and dividing it by the standard deviation. Without this step, Turing’s sampler will have a hard time finding a place to start searching for parameter estimates. To do this we will leverage MLDataUtils, which also lets us effortlessly shuffle our observations and perform a stratified split to get a representative test set.

function split_data(df, target; at=0.70)
    shuffled = shuffleobs(df)
    return trainset, testset = stratifiedobs(row -> row[target], shuffled; p=at)

features = [:StudentNum, :Balance, :Income]
numerics = [:Balance, :Income]
target = :DefaultNum

trainset, testset = split_data(data, target; at=0.05)
for feature in numerics
    μ, σ = rescale!(trainset[!, feature]; obsdim=1)
    rescale!(testset[!, feature], μ, σ; obsdim=1)

# Turing requires data in matrix form, not dataframe
train = Matrix(trainset[:, features])
test = Matrix(testset[:, features])
train_label = trainset[:, target]
test_label = testset[:, target];

Model Declaration

Finally, we can define our model.

logistic_regression takes four arguments:

  • x is our set of independent variables;
  • y is the element we want to predict;
  • n is the number of observations we have; and
  • σ is the standard deviation we want to assume for our priors.

Within the model, we create four coefficients (intercept, student, balance, and income) and assign a prior of normally distributed with means of zero and standard deviations of σ. We want to find values of these four coefficients to predict any given y.

The for block creates a variable v which is the logistic function. We then observe the likelihood of calculating v given the actual label, y[i].

# Bayesian logistic regression (LR)
@model function logistic_regression(x, y, n, σ)
    intercept ~ Normal(0, σ)

    student ~ Normal(0, σ)
    balance ~ Normal(0, σ)
    income ~ Normal(0, σ)

    for i in 1:n
        v = logistic(intercept + student * x[i, 1] + balance * x[i, 2] + income * x[i, 3])
        y[i] ~ Bernoulli(v)


Now we can run our sampler. This time we’ll use NUTS to sample from our posterior.

# Retrieve the number of observations.
n, _ = size(train)

# Sample using NUTS.
m = logistic_regression(train, train_label, n, 1)
chain = sample(m, NUTS(), MCMCThreads(), 1_500, 3)
Chains MCMC chain (1500×16×3 Array{Float64, 3}):

Iterations        = 751:1:2250
Number of chains  = 3
Samples per chain = 1500
Wall duration     = 12.31 seconds
Compute duration  = 10.06 seconds
parameters        = intercept, student, balance, income
internals         = lp, n_steps, is_accept, acceptance_rate, log_density, hamiltonian_energy, hamiltonian_energy_error, max_hamiltonian_energy_error, tree_depth, numerical_error, step_size, nom_step_size

Summary Statistics
  parameters      mean       std      mcse    ess_bulk    ess_tail      rhat       Symbol   Float64   Float64   Float64     Float64     Float64   Float64   ⋯

   intercept   -4.1705    0.3876    0.0077   2551.1094   2722.6008    1.0010   ⋯
     student   -0.1265    0.6255    0.0120   2734.0165   2441.8077    1.0008   ⋯
     balance    1.5259    0.2432    0.0045   2898.3615   2694.7740    1.0005   ⋯
      income    0.1884    0.3209    0.0063   2568.5319   2885.0031    1.0003   ⋯
                                                                1 column omitted

  parameters      2.5%     25.0%     50.0%     75.0%     97.5%
      Symbol   Float64   Float64   Float64   Float64   Float64

   intercept   -4.9664   -4.4240   -4.1536   -3.9062   -3.4507
     student   -1.3541   -0.5488   -0.1317    0.3016    1.1136
     balance    1.0717    1.3610    1.5227    1.6844    2.0200
      income   -0.4345   -0.0272    0.1895    0.4069    0.8178

The sample() call above assumes that you have at least nchains threads available in your Julia instance. If you do not, the multiple chains will run sequentially, and you may notice a warning. For more information, see the Turing documentation on sampling multiple chains.

Since we ran multiple chains, we may as well do a spot check to make sure each chain converges around similar points.


Looks good!

We can also use the corner function from MCMCChains to show the distributions of the various parameters of our logistic regression.

# The labels to use.
l = [:student, :balance, :income]

# Use the corner function. Requires StatsPlots and MCMCChains.
corner(chain, l)