# 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
# Set a seed for reproducibility.
using Random
Random.seed!(0);
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.
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.
= RDatasets.dataset("datasets", "mtcars")
data
# Show the first six rows of the dataset.
first(data, 6)
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.
= map(DataFrame, splitobs(data; at=0.7, shuffle=true))
trainset, testset
# Turing requires data in matrix form.
= :MPG
target = Matrix(select(trainset, Not(target)))
train = Matrix(select(testset, Not(target)))
test = trainset[:, target]
train_target = testset[:, target]
test_target
# Standardize the features.
= fit(ZScoreTransform, train; dims=1)
dt_features transform!(dt_features, train)
StatsBase.transform!(dt_features, test)
StatsBase.
# Standardize the targets.
= fit(ZScoreTransform, train_target)
dt_targets transform!(dt_targets, train_target)
StatsBase.transform!(dt_targets, test_target); StatsBase.
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.
~ truncated(Normal(0, 100); lower=0)
σ²
# Set intercept prior.
~ Normal(0, sqrt(3))
intercept
# Set the priors on our coefficients.
= size(x, 2)
nfeatures ~ MvNormal(Zeros(nfeatures), 10.0 * I)
coefficients
# Calculate all the mu terms.
= intercept .+ x * coefficients
mu 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.
= linear_regression(train, train_target)
model = sample(model, NUTS(), 5_000) chain
┌ Info: Found initial step size
└ ϵ = 0.4
Chains MCMC chain (5000×24×1 Array{Float64, 3}):
Iterations = 1001:1:6000
Number of chains = 1
Samples per chain = 5000
Wall duration = 11.06 seconds
Compute duration = 11.06 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 = 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 ⋯
Symbol Float64 Float64 Float64 Float64 Float64 Flo ⋯
σ² 0.3363 0.1966 0.0049 1707.4258 2146.9929 1. ⋯
intercept 0.0015 0.1269 0.0018 5005.1913 3021.2680 1. ⋯
coefficients[1] -0.1940 0.4816 0.0081 3478.3140 3189.8436 0. ⋯
coefficients[2] 0.0908 0.7582 0.0163 2163.8602 2666.9402 1. ⋯
coefficients[3] -0.0398 0.4820 0.0097 2493.3218 2318.7054 1. ⋯
coefficients[4] 0.0421 0.2428 0.0044 3041.4206 2875.1297 1. ⋯
coefficients[5] -0.3099 0.6777 0.0146 2162.2695 2301.1674 1. ⋯
coefficients[6] 0.2124 0.6094 0.0122 2480.1030 2584.4146 1. ⋯
coefficients[7] -0.0090 0.3701 0.0070 2785.9006 3078.2843 1. ⋯
coefficients[8] 0.3438 0.2926 0.0051 3398.7041 2599.1770 1. ⋯
coefficients[9] 0.0468 0.3272 0.0055 3443.8369 2840.4979 1. ⋯
coefficients[10] -0.2440 0.3766 0.0081 2157.2477 2532.2613 1. ⋯
2 columns omitted
Quantiles
parameters 2.5% 25.0% 50.0% 75.0% 97.5%
Symbol Float64 Float64 Float64 Float64 Float64
σ² 0.1267 0.2108 0.2856 0.4025 0.8489
intercept -0.2442 -0.0761 0.0003 0.0783 0.2517
coefficients[1] -1.1474 -0.5049 -0.2029 0.1119 0.7812
coefficients[2] -1.3950 -0.3863 0.0903 0.5876 1.5853
coefficients[3] -1.0103 -0.3507 -0.0424 0.2677 0.9512
coefficients[4] -0.4410 -0.1115 0.0439 0.1990 0.5236
coefficients[5] -1.6828 -0.7435 -0.3014 0.1110 1.0654
coefficients[6] -0.9840 -0.1726 0.2161 0.5840 1.4416
coefficients[7] -0.7649 -0.2387 -0.0105 0.2257 0.7235
coefficients[8] -0.2453 0.1621 0.3440 0.5277 0.9207
coefficients[9] -0.5998 -0.1605 0.0534 0.2547 0.6937
coefficients[10] -0.9763 -0.4874 -0.2475 -0.0020 0.5119
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.
= hcat(ones(size(train, 1)), train)
train_with_intercept = lm(train_with_intercept, train_target)
ols
# Compute predictions on the training data set and unstandardize them.
= GLM.predict(ols)
train_prediction_ols reconstruct!(dt_targets, train_prediction_ols)
StatsBase.
# Compute predictions on the test data set and unstandardize them.
= hcat(ones(size(test, 1)), test)
test_with_intercept = GLM.predict(ols, test_with_intercept)
test_prediction_ols reconstruct!(dt_targets, test_prediction_ols); StatsBase.
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)
= get_params(chain[200:end, :, :])
p = p.intercept' .+ x * reduce(hcat, p.coefficients)'
targets 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.
= prediction(chain, train)
train_prediction_bayes reconstruct!(dt_targets, train_prediction_bayes)
StatsBase.= prediction(chain, test)
test_prediction_bayes reconstruct!(dt_targets, test_prediction_bayes)
StatsBase.
# Show the predictions on the test data set.
DataFrame(; MPG=testset[!, target], Bayes=test_prediction_bayes, OLS=test_prediction_ols)
Row | MPG | Bayes | OLS |
---|---|---|---|
Float64 | Float64 | Float64 | |
1 | 15.2 | 15.8766 | 15.8851 |
2 | 22.8 | 28.6799 | 28.5684 |
3 | 22.8 | 25.1332 | 25.3859 |
4 | 19.2 | 17.6049 | 17.5875 |
5 | 15.0 | 13.167 | 13.156 |
6 | 15.2 | 17.3455 | 17.361 |
7 | 16.4 | 14.8657 | 14.7729 |
8 | 10.4 | 12.0932 | 12.0596 |
9 | 21.4 | 26.8521 | 26.7321 |
10 | 30.4 | 29.4983 | 29.5952 |
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: 4.276606319093772
OLS loss: 4.275088978337632
Test set:
Bayes loss: 8.674235433939263
OLS loss: 8.558068292722457
As we can see above, OLS and our Bayesian model fit our training and test data set about the same.