Chapter 11 Exercises
Stagnant water: splines
Solutions
In the stagnant water example of 11.7.1, extend the linear change point model to a quadratic spline, and compare the parameter estimates, model fit and DIC.
model
{
for (i in 1:N) {
y[i] ~ dnorm(mu[i], tau)
mu[i] <- alpha + beta[1]*x[i] +
beta[2]*(x[i] - theta) * step(x[i] - theta) +
beta[3]*pow(x[i] - theta, 2) * step(x[i] - theta)
}
tau ~ dgamma(0.001, 0.001)
alpha ~ dnorm(0.0, 1.0E-6)
for (j in 1:3) {
beta[j] ~ dnorm(0.0, 1.0E-6)
}
sigma <- 1/sqrt(tau)
theta ~ dunif(-1.3, 1.1)
}
Data:
list(y = c(1.12, 1.12, 0.99, 1.03, 0.92, 0.90, 0.81, 0.83, 0.65, 0.67, 0.60, 0.59, 0.51, 0.44, 0.43, 0.43, 0.33, 0.30, 0.25, 0.24, 0.13, -0.01, -0.13, -0.14, -0.30, -0.33, -0.46, -0.43, -0.65),
x = c(-1.39, -1.39, -1.08, -1.08, -0.94, -0.80, -0.63, -0.63, -0.25, -0.25, -0.12, -0.12, 0.01, 0.11, 0.11, 0.11, 0.25, 0.25, 0.34, 0.34, 0.44, 0.59, 0.70, 0.70, 0.85, 0.85, 0.99, 0.99, 1.19),
N = 29)
Initial values:
list(alpha = 0.2, beta = c(-0.45, 0, 0), tau = 5, theta = 0)
node mean sd MC error 2.5% median 97.5% start sample
alpha 0.5549 0.01352 7.688E-4 0.5303 0.5543 0.5845 501 9500
beta[1] -0.4121 0.01505 7.414E-4 -0.4405 -0.4126 -0.3804 501 9500
beta[2] -0.421 0.08015 0.005119 -0.5661 -0.4259 -0.244 501 9500
beta[3] -0.1332 0.05109 0.002825 -0.2331 -0.1338 -0.03128 501 9500
sigma 0.02003 0.003019 5.693E-5 0.01517 0.01972 0.02691 501 9500
theta -0.06109 0.06504 0.005198 -0.2209 -0.0533 0.04413 501 9500
Dbar = post.mean of -2logL; Dhat = -2LogL at post.mean of stochastic nodes
Dbar Dhat pD DIC
y -150.079 -156.250 6.171 -143.908
total -150.079 -156.250 6.171 -143.908
There appears to be a statistically significant curvature in the regression line (judging by the posterior distribution of the quadratic coefficient beta[3] and the improvement of about 5 in DIC) but this is small in absolute terms, and qualitatively the fit is hardly changed. The uncertainty around the change point location theta is slightly increased.