Author: VIBASS

# Practical 3: Bayesian polynomial regression

#### VIBASS

# 1 Fitting a Bayesian polynomial model

    ## Load some required libraries
    library(extraDistr)    # functions related with the inverse-Gamma distribution
    library(LaplacesDemon) # functions related with the multivariate Student dist.

    ##
    ## Attaching package: 'LaplacesDemon'

    ## The following objects are masked from 'package:extraDistr':
    ##
    ##     dbern, dcat, ddirichlet, dgpd, dgpois, dinvchisq, dinvgamma,
    ##     dlaplace, dpareto, pbern, plaplace, ppareto, qbern, qcat, qlaplace,
    ##     qpareto, rbern, rcat, rdirichlet, rgpd, rinvchisq, rinvgamma,
    ##     rlaplace, rpareto

## 1.1 Introduction

The `Auto` dataset of the `ISLR` library contains information on several characteristics of 392 vehicles. Specifically, we will want to model the relationship between the efficiency of these cars, measured as the number of miles they are able to travel per gallon of fuel (variable `mpg`), and their power, quantified by the variable `horsepower`.

We begin by visualizing the relationship between the two variables

    data("Auto", package = "ISLR")
    plot(mpg ~ horsepower, data = Auto)

The relationship between both variables is clearly nonlinear, possibly quadratic, so we will fit a (Bayesian) quadratic regression model to this data set to describe this relationship.

## 1.2 Fitting a (frequentist) quadratic regression model

We are going to fit now a regression model which describes a quadratic function of `horsepower` following a frequentist approach and tools. Previously, we will standardize `horsepower` in order to obtain more manageable estimates of the coefficients of the model and their variability. Let’s do it:

    # Standardization of horsepower
    Auto$horse.std <- (Auto$horsepower - mean(Auto$horsepower))/sd(Auto$horsepower)
    # Quadratic fit
    Fit1 <- lm(mpg ~ poly(horse.std, degree = 2, raw = TRUE), data = Auto)
    summary(Fit1)

    ##
    ## Call:
    ## lm(formula = mpg ~ poly(horse.std, degree = 2, raw = TRUE), data = Auto)
    ##
    ## Residuals:
    ##      Min       1Q   Median       3Q      Max
    ## -14.7135  -2.5943  -0.0859   2.2868  15.8961
    ##
    ## Coefficients:
    ##                                          Estimate Std. Error t value Pr(>|t|)
    ## (Intercept)                               21.6274     0.2852   75.83   <2e-16
    ## poly(horse.std, degree = 2, raw = TRUE)1  -8.0478     0.2953  -27.25   <2e-16
    ## poly(horse.std, degree = 2, raw = TRUE)2   1.8231     0.1809   10.08   <2e-16
    ##
    ## (Intercept)                              ***
    ## poly(horse.std, degree = 2, raw = TRUE)1 ***
    ## poly(horse.std, degree = 2, raw = TRUE)2 ***
    ## ---
    ## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
    ##
    ## Residual standard error: 4.374 on 389 degrees of freedom
    ## Multiple R-squared:  0.6876, Adjusted R-squared:  0.686
    ## F-statistic:   428 on 2 and 389 DF,  p-value: < 2.2e-16

    # Plot of the fitted curve
    plot(mpg ~ horse.std, data = Auto)
    where <- seq(min(Auto$horse.std), max(Auto$horse.std), length = 100)
    lines(where, predict(Fit1, data.frame(horse.std = where)), col = 2, lwd = 2)

The fitted curve adequately describes the relationship between both variables. Moreover, every term of the polynomial seem to have a significant effect, that is, all of them seem necessary to appropriately describe the variation of `mpg` as a function of `horse.std`.

## 1.3 Fitting a Bayesian quadratic regression model

Let us now fit the Bayesian version of the above model. We will use, for convenience, Jeffreys’ prior for \(P(\boldsymbol{\beta},\sigma^2)\propto \sigma^{-2}\). Therefore, we have: \[\pi(\sigma^2\mid \mathcal{D})\sim IG(\sigma^2\mid(n-(p+1))/2,s_e^2/2)\]

    plot(seq(13, 25, by = 0.1),
         dinvgamma(seq(13, 25, by = 0.1), (nrow(Auto)-3)/2,sum(residuals(Fit1)^2)/2),
         type = "l", ylab = "Posterior density", xlab = "sigma2")

This distribution yields a posterior mean of \((s^2_e/2)/((n-(p+1))/2-1)\) and a mode of \((s^2_e/2)/((n-(p+1))/2+1)\), that is:

    # posterior mean
    sum(residuals(Fit1)^2)/2/((nrow(Auto)-3)/2-1)

    ## [1] 19.23005

    # posterior mode
    sum(residuals(Fit1)^2)/2/((nrow(Auto)-3)/2+1)

    ## [1] 19.03332

    # Frequentist point estimate of sigma2
    sum(Fit1$residuals^2)/(nrow(Auto)-3)

    ## [1] 19.13118

The Bayesian posterior point estimates are quite close to the frequentist estimate, although the posterior distribution of \(\sigma^2\) allows to explore also the uncertainty of this parameter.

As mentioned in the theoretical session, \(\boldsymbol{\beta}\) has the following marginal posterior distribution: \[\pi(\boldsymbol{\beta}\mid \boldsymbol{y})\sim t_{n-(p+1)}(\hat{\boldsymbol{\beta}},s_e^2(\boldsymbol{X}'\boldsymbol{X})^{-1}/(n-(p+1)))\] The mean of this distribution coincides with the coefficients of the linear model `Fit1`, and has as variance-covariance matrix:

    X <- cbind(rep(1, dim(Auto)[1]), Auto$horse.std, Auto$horse.std^2)
    VarCov <- solve(t(X)%*%X)*sum(residuals(Fit1)^2)/(nrow(Auto)-3)
    VarCov

    ##             [,1]        [,2]        [,3]
    ## [1,]  0.08134909  0.03529659 -0.03262829
    ## [2,]  0.03529659  0.08720960 -0.03538686
    ## [3,] -0.03262829 -0.03538686  0.03271174

    # Correlation between coefficients
    cov2cor(VarCov)

    ##            [,1]       [,2]       [,3]
    ## [1,]  1.0000000  0.4190581 -0.6325082
    ## [2,]  0.4190581  1.0000000 -0.6625338
    ## [3,] -0.6325082 -0.6625338  1.0000000

Moreover, we can also plot the bivariate distribution for each pair of coefficients, for example, the joint posterior distribution for the intercept and the linear term:

    # Joint posterior distribution of the coefficients for the intercept and the
    # linear term
    xlims <- c(Fit1$coefficients[1]-3*sqrt(VarCov[1,1]),
                Fit1$coefficients[1]+3*sqrt(VarCov[1,1]))
    ylims <- c(Fit1$coefficients[2]-3*sqrt(VarCov[2,2]),
                Fit1$coefficients[2]+3*sqrt(VarCov[2,2]))
    gridPoints <- expand.grid(seq(xlims[1],xlims[2],by=0.01),
                            seq(ylims[1],ylims[2],by=0.01))
    resul <- matrix(dmvt(as.matrix(gridPoints), mu = Fit1$coefficients[1:2],
                         S = round(VarCov[1:2,1:2],5), df = nrow(Auto)-3),
                    nrow = length(seq(xlims[1], xlims[2], by = 0.01)))
    image(x = seq(xlims[1], xlims[2], by = 0.01), y = seq(ylims[1], ylims[2], by = 0.01),
          z = resul, xlab = "Intercept", ylab = "beta, linear term",
          main = "Bivariate posterior density")

or the joint posterior distribution for the coefficients of the linear and the quadratic terms:

    # Plot the linear and quadratic component
    xlims <- c(Fit1$coefficients[2]-3*sqrt(VarCov[2,2]),
                Fit1$coefficients[2]+3*sqrt(VarCov[2,2]))
    ylims <- c(Fit1$coefficients[3]-3*sqrt(VarCov[3,3]),
                Fit1$coefficients[3]+3*sqrt(VarCov[3,3]))
    gridPoints <- expand.grid(seq(xlims[1],xlims[2],by=0.01),
                            seq(ylims[1],ylims[2],by=0.01))
    resul <- matrix(dmvt(as.matrix(gridPoints), mu = Fit1$coefficients[c(2,3)],
                         S = round(VarCov[c(2,3),c(2,3)], 5), df = nrow(Auto)-3),
                    nrow = length(seq(xlims[1], xlims[2], by = 0.01)))
    image(x = seq(xlims[1], xlims[2], by = 0.01), y = seq(ylims[1], ylims[2], by = 0.01),
          z = resul, xlab = "beta, linear term",
          ylab = "beta, quadratic term",
          main = "Bivariate posterior density")

That is, all three parameters are mutually dependent. Furthermore, we can also quantify the probability that each of these parameters is negative:

    Probs <- vector()
    for(i in 1:3){
      Probs[i] <- pst(0, mu = Fit1$coefficients[i], sigma = sqrt(VarCov[i,i]),
                      nu = nrow(Auto)-3)
    }
    Probs

    ## [1] 1.913574e-235  1.000000e+00  1.098170e-21

These probabilities may be used as Bayesian measures of evidence of the contribution of each variable to the fit. Values of these probabilities close to 0 or 1 indicate the appropriateness of considering that component in the model.

Finally, the posterior mean of every component of \(\boldsymbol{\beta}\) coincides with the frequentist estimate of that coefficient. Therefore, the point estimate (posterior mean) of the quadratic function resulting from the Bayesian analysis exactly coincides with the corresponding curve from the frequentist analysis.

# 2 Time to individual work

We propose below an individual exercise that pursues to consolidate the basic concepts that we have learned in the previous theoretical session and that we have been practicing in this session.

Exercise

The data set `Weights`, included in the `VIBASS` package, is the data set used for the example of the theoretical session. That data set has a categorical variable, `race`, which contains the race of each child in the study. Fit a Bayesian linear regression model (ANCOVA) that explains each child’s weight as a function of age, race and the interaction of these factors. Explore the posterior distribution of the coefficients in the model and assess therefore its convenience in the regression model.