Hmisc (version 5.3-0)

areg: Additive Regression with Optimal Transformations on Both Sides using Canonical Variates

Description

Expands continuous variables into restricted cubic spline bases and categorical variables into dummy variables and fits a multivariate equation using canonical variates. This finds optimum transformations that maximize \(R^2\). Optionally, the bootstrap is used to estimate the covariance matrix of both left- and right-hand-side transformation parameters, and to estimate the bias in the \(R^2\) due to overfitting and compute the bootstrap optimism-corrected \(R^2\). Cross-validation can also be used to get an unbiased estimate of \(R^2\) but this is not as precise as the bootstrap estimate. The bootstrap and cross-validation may also used to get estimates of mean and median absolute error in predicted values on the original y scale. These two estimates are perhaps the best ones for gauging the accuracy of a flexible model, because it is difficult to compare \(R^2\) under different y-transformations, and because \(R^2\) allows for an out-of-sample recalibration (i.e., it only measures relative errors).

Note that uncertainty about the proper transformation of y causes an enormous amount of model uncertainty. When the transformation for y is estimated from the data a high variance in predicted values on the original y scale may result, especially if the true transformation is linear. Comparing bootstrap or cross-validated mean absolute errors with and without restricted the y transform to be linear (ytype='l') may help the analyst choose the proper model complexity.

Usage

areg(x, y, xtype = NULL, ytype = NULL, nk = 4,
     B = 0, na.rm = TRUE, tolerance = NULL, crossval = NULL)
# S3 method for areg
print(x, digits=4, ...)
# S3 method for areg
plot(x, whichx = 1:ncol(x$x), ...)
# S3 method for areg
predict(object, x, type=c('lp','fitted','x'),
                       what=c('all','sample'), ...)

Value

a list of class "areg" containing many objects

Details

areg is a competitor of ace in the acepack package. Transformations from ace are seldom smooth enough and are often overfitted. With areg the complexity can be controlled with the nk parameter, and predicted values are easy to obtain because parametric functions are fitted.

If one side of the equation has a categorical variable with more than two categories and the other side has a continuous variable not assumed to act linearly, larger sample sizes are needed to reliably estimate transformations, as it is difficult to optimally score categorical variables to maximize \(R^2\) against a simultaneously optimally transformed continuous variable.

References

Breiman and Friedman, Journal of the American Statistical Association (September, 1985).

See Also

cancor,ace, transcan

Examples

set.seed(1)

ns <- c(30,300,3000)
for(n in ns) {
  y <- sample(1:5, n, TRUE)
  x <- abs(y-3) + runif(n)
  par(mfrow=c(3,4))
  for(k in c(0,3:5)) {
    z <- areg(x, y, ytype='c', nk=k)
    plot(x, z$tx)
	title(paste('R2=',format(z$rsquared)))
    tapply(z$ty, y, range)
    a <- tapply(x,y,mean)
    b <- tapply(z$ty,y,mean)
    plot(a,b)
	abline(lsfit(a,b))
    # Should get same result to within linear transformation if reverse x and y
    w <- areg(y, x, xtype='c', nk=k)
    plot(z$ty, w$tx)
    title(paste('R2=',format(w$rsquared)))
    abline(lsfit(z$ty, w$tx))
 }
}

par(mfrow=c(2,2))
# Example where one category in y differs from others but only in variance of x
n <- 50
y <- sample(1:5,n,TRUE)
x <- rnorm(n)
x[y==1] <- rnorm(sum(y==1), 0, 5)
z <- areg(x,y,xtype='l',ytype='c')
z
plot(z)
z <- areg(x,y,ytype='c')
z
plot(z)

if (FALSE) {		
# Examine overfitting when true transformations are linear
par(mfrow=c(4,3))
for(n in c(200,2000)) {
  x <- rnorm(n); y <- rnorm(n) + x
    for(nk in c(0,3,5)) {
    z <- areg(x, y, nk=nk, crossval=10, B=100)
    print(z)
    plot(z)
    title(paste('n=',n))
  }
}
par(mfrow=c(1,1))

# Underfitting when true transformation is quadratic but overfitting
# when y is allowed to be transformed
set.seed(49)
n <- 200
x <- rnorm(n); y <- rnorm(n) + .5*x^2
#areg(x, y, nk=0, crossval=10, B=100)
#areg(x, y, nk=4, ytype='l', crossval=10, B=100)
z <- areg(x, y, nk=4) #, crossval=10, B=100)
z
# Plot x vs. predicted value on original scale.  Since y-transform is
# not monotonic, there are multiple y-inverses
xx <- seq(-3.5,3.5,length=1000)
yhat <- predict(z, xx, type='fitted')
plot(x, y, xlim=c(-3.5,3.5))
for(j in 1:ncol(yhat)) lines(xx, yhat[,j], col=j)
# Plot a random sample of possible y inverses
yhats <- predict(z, xx, type='fitted', what='sample')
points(xx, yhats, pch=2)
}

# True transformation of x1 is quadratic, y is linear
n <- 200
x1 <- rnorm(n); x2 <- rnorm(n); y <- rnorm(n) + x1^2
z <- areg(cbind(x1,x2),y,xtype=c('s','l'),nk=3)
par(mfrow=c(2,2))
plot(z)

# y transformation is inverse quadratic but areg gets the same answer by
# making x1 quadratic
n <- 5000
x1 <- rnorm(n); x2 <- rnorm(n); y <- (x1 + rnorm(n))^2
z <- areg(cbind(x1,x2),y,nk=5)
par(mfrow=c(2,2))
plot(z)

# Overfit 20 predictors when no true relationships exist
n <- 1000
x <- matrix(runif(n*20),n,20)
y <- rnorm(n)
z <- areg(x, y, nk=5)  # add crossval=4 to expose the problem

# Test predict function
n <- 50
x <- rnorm(n)
y <- rnorm(n) + x
g <- sample(1:3, n, TRUE)
z <- areg(cbind(x,g),y,xtype=c('s','c'))
range(predict(z, cbind(x,g)) - z$linear.predictors)