Bayesian Linear Regression

Turing is powerful when applied to complex hierarchical models, but it can also be put to task at common statistical procedures, like linear regression. This tutorial covers how to implement a linear regression model in Turing.

Set Up

We begin by importing all the necessary libraries.

# Import Turing.
using Turing

# Package for loading the data set.
using RDatasets

# Package for visualization.
using StatsPlots

# Functionality for splitting the data.
using MLUtils: splitobs

# Functionality for constructing arrays with identical elements efficiently.
using FillArrays

# Functionality for normalizing the data and evaluating the model predictions.
using StatsBase

# Functionality for working with scaled identity matrices.
using LinearAlgebra

# For ensuring reproducibility.
using StableRNGs: StableRNG
setprogress!(false)

We will use the mtcars dataset from the RDatasets package. mtcars contains a variety of statistics on different car models, including their miles per gallon, number of cylinders, and horsepower, among others.

We want to know if we can construct a Bayesian linear regression model to predict the miles per gallon of a car, given the other statistics it has. Let us take a look at the data we have.

# Load the dataset.
data = RDatasets.dataset("datasets", "mtcars")

# Show the first six rows of the dataset.
first(data, 6)
6×12 DataFrame
Row Model MPG Cyl Disp HP DRat WT QSec VS AM Gear Carb
String31 Float64 Int64 Float64 Int64 Float64 Float64 Float64 Int64 Int64 Int64 Int64
1 Mazda RX4 21.0 6 160.0 110 3.9 2.62 16.46 0 1 4 4
2 Mazda RX4 Wag 21.0 6 160.0 110 3.9 2.875 17.02 0 1 4 4
3 Datsun 710 22.8 4 108.0 93 3.85 2.32 18.61 1 1 4 1
4 Hornet 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
5 Hornet Sportabout 18.7 8 360.0 175 3.15 3.44 17.02 0 0 3 2
6 Valiant 18.1 6 225.0 105 2.76 3.46 20.22 1 0 3 1 
size(data)
(32, 12)

The next step is to get our data ready for testing. We’ll split the mtcars dataset into two subsets, one for training our model and one for evaluating our model. Then, we separate the targets we want to learn (MPG, in this case) and standardize the datasets by subtracting each column’s means and dividing by the standard deviation of that column. The resulting data is not very familiar looking, but this standardization process helps the sampler converge far easier.

# Remove the model column.
select!(data, Not(:Model))

# Split our dataset 70%/30% into training/test sets.
trainset, testset = map(DataFrame, splitobs(StableRNG(468), data; at=0.7, shuffle=true))

# Turing requires data in matrix form.
target = :MPG
train = Matrix(select(trainset, Not(target)))
test = Matrix(select(testset, Not(target)))
train_target = trainset[:, target]
test_target = testset[:, target]

# Standardize the features.
dt_features = fit(ZScoreTransform, train; dims=1)
StatsBase.transform!(dt_features, train)
StatsBase.transform!(dt_features, test)

# Standardize the targets.
dt_targets = fit(ZScoreTransform, train_target)
StatsBase.transform!(dt_targets, train_target)
StatsBase.transform!(dt_targets, test_target);

Model Specification

In a traditional frequentist model using OLS, our model might look like:

\[ \mathrm{MPG}_i = \alpha + \boldsymbol{\beta}^\mathsf{T}\boldsymbol{X_i} \]

where \(\boldsymbol{\beta}\) is a vector of coefficients and \(\boldsymbol{X}\) is a vector of inputs for observation \(i\). The Bayesian model we are more concerned with is the following:

\[ \mathrm{MPG}_i \sim \mathcal{N}(\alpha + \boldsymbol{\beta}^\mathsf{T}\boldsymbol{X_i}, \sigma^2) \]

where \(\alpha\) is an intercept term common to all observations, \(\boldsymbol{\beta}\) is a coefficient vector, \(\boldsymbol{X_i}\) is the observed data for car \(i\), and \(\sigma^2\) is a common variance term.

For \(\sigma^2\), we assign a prior of truncated(Normal(0, 100); lower=0). This is consistent with Andrew Gelman’s recommendations on noninformative priors for variance. The intercept term (\(\alpha\)) is assumed to be normally distributed with a mean of zero and a variance of three. This represents our assumptions that miles per gallon can be explained mostly by our assorted variables, but a high variance term indicates our uncertainty about that. Each coefficient is assumed to be normally distributed with a mean of zero and a variance of 10. We do not know that our coefficients are different from zero, and we don’t know which ones are likely to be the most important, so the variance term is quite high. Lastly, each observation \(y_i\) is distributed according to the calculated mu term given by \(\alpha + \boldsymbol{\beta}^\mathsf{T}\boldsymbol{X_i}\).

# Bayesian linear regression.
@model function linear_regression(x, y)
    # Set variance prior.
    σ² ~ truncated(Normal(0, 100); lower=0)

    # Set intercept prior.
    intercept ~ Normal(0, sqrt(3))

    # Set the priors on our coefficients.
    nfeatures = size(x, 2)
    coefficients ~ MvNormal(Zeros(nfeatures), 10.0 * I)

    # Calculate all the mu terms.
    mu = intercept .+ x * coefficients
    return y ~ MvNormal(mu, σ² * I)
end
linear_regression (generic function with 2 methods)

With our model specified, we can call the sampler. We will use the No U-Turn Sampler (NUTS) here.

model = linear_regression(train, train_target)
chain = sample(StableRNG(468), model, NUTS(), 20_000)
┌ Info: Found initial step size
└   ϵ = 0.2

Chains MCMC chain (20000×26×1 Array{Float64, 3}):

Iterations        = 1001:1:21000
Number of chains  = 1
Samples per chain = 20000
Wall duration     = 14.76 seconds
Compute duration  = 14.76 seconds
parameters        = σ², intercept, coefficients[1], coefficients[2], coefficients[3], coefficients[4], coefficients[5], coefficients[6], coefficients[7], coefficients[8], coefficients[9], coefficients[10]
internals         = 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, lp, logprior, loglikelihood

Use `describe(chains)` for summary statistics and quantiles.

We can also check the densities and traces of the parameters visually using the plot functionality.

plot(chain)

It looks like all parameters have converged.

Comparing to OLS

A satisfactory test of our model is to evaluate how well it predicts. Importantly, we want to compare our model to existing tools like OLS. The code below uses the GLM.jl package to generate a traditional OLS multiple regression model on the same data as our probabilistic model.

# Import the GLM package.
using GLM

# Perform multiple regression OLS.
train_with_intercept = hcat(ones(size(train, 1)), train)
ols = lm(train_with_intercept, train_target)

# Compute predictions on the training data set and unstandardize them.
train_prediction_ols = GLM.predict(ols)
StatsBase.reconstruct!(dt_targets, train_prediction_ols)

# Compute predictions on the test data set and unstandardize them.
test_with_intercept = hcat(ones(size(test, 1)), test)
test_prediction_ols = GLM.predict(ols, test_with_intercept)
StatsBase.reconstruct!(dt_targets, test_prediction_ols);

The function below accepts a chain and an input matrix and calculates predictions. We use the samples of the model parameters in the chain starting with sample 200.

# Make a prediction given an input vector.
function prediction(chain, x)
    p = get_params(chain[200:end, :, :])
    targets = p.intercept' .+ x * reduce(hcat, p.coefficients)'
    return vec(mean(targets; dims=2))
end
prediction (generic function with 1 method)

When we make predictions, we unstandardize them so they are more understandable.

# Calculate the predictions for the training and testing sets and unstandardize them.
train_prediction_bayes = prediction(chain, train)
StatsBase.reconstruct!(dt_targets, train_prediction_bayes)
test_prediction_bayes = prediction(chain, test)
StatsBase.reconstruct!(dt_targets, test_prediction_bayes)

# Show the predictions on the test data set.
DataFrame(; MPG=testset[!, target], Bayes=test_prediction_bayes, OLS=test_prediction_ols)
10×3 DataFrame
Row MPG Bayes OLS
Float64 Float64 Float64
1 30.4 26.981 26.9093
2 19.2 16.3005 16.0834
3 14.3 12.1107 11.9393
4 22.8 26.6502 26.5984
5 22.8 31.536 32.153
6 13.3 9.20312 8.83012
7 18.7 17.1338 17.1669
8 10.4 13.6857 13.821
9 19.7 20.0546 20.0243
10 15.2 17.0071 17.0774

Now let’s evaluate the loss for each method, and each prediction set. We will use the mean squared error to evaluate loss, given by \[ \mathrm{MSE} = \frac{1}{n} \sum_{i=1}^n {(y_i - \hat{y_i})^2} \] where \(y_i\) is the actual value (true MPG) and \(\hat{y_i}\) is the predicted value using either OLS or Bayesian linear regression. A lower SSE indicates a closer fit to the data.

println(
    "Training set:",
    "\n\tBayes loss: ",
    msd(train_prediction_bayes, trainset[!, target]),
    "\n\tOLS loss: ",
    msd(train_prediction_ols, trainset[!, target]),
)

println(
    "Test set:",
    "\n\tBayes loss: ",
    msd(test_prediction_bayes, testset[!, target]),
    "\n\tOLS loss: ",
    msd(test_prediction_ols, testset[!, target]),
)
Training set:
    Bayes loss: 4.914805825351319
    OLS loss: 4.909437479783827
Test set:
    Bayes loss: 14.945528980245133
    OLS loss: 16.70400759383217

We can see from this that both linear regression techniques perform fairly similarly. The Bayesian linear regression approach performs worse on the training set, but better on the test set. This indicates that the Bayesian approach is more able to generalise to unseen data, i.e., it is not overfitting the training data as much.

Back to top