As Hastie, Tibshirani, & Friedman (2009, sec. 7.10.2: “The Wrong and Right Way to Do Cross-validation”) explain, if the whole data are used to select or fine-tune a statistical model, subsequent cross-validation of the model is intrinsically misleading, because the model is selected to fit the whole data, including the part of the data that remains when each fold is removed.

The following example is similar in spirit to one employed by Hastie et al. (2009). Suppose that we randomly generate \(n = 1000\) independent observations for a response variable variable \(y \sim N(\mu = 10, \sigma^2 = 0)\), and independently sample \(1000\) observations for \(p = 100\) “predictors,” \(x_1, \ldots, x_{100}\), each from \(x_j \sim N(0, 1)\). The response has nothing to do with the predictors and so the population linear-regression model \(y_i = \alpha + \beta_1 x_{i1} + \cdots + \beta_{100} x_{i,100} + \varepsilon_i\) has \(\alpha = 10\) and all \(\beta_j = 0\).

```
set.seed(24361) # for reproducibility
D <- data.frame(y = rnorm(1000, mean = 10),
X = matrix(rnorm(1000 * 100), 1000, 100))
head(D[, 1:6])
#> y X.1 X.2 X.3 X.4 X.5
#> 1 10.0316 -1.23886 -0.26487 -0.03539 -2.576973 0.811048
#> 2 9.6650 0.12287 -0.17744 0.37290 -0.935138 0.628673
#> 3 10.0232 -0.95052 -0.73487 -1.05978 0.882944 0.023918
#> 4 8.9910 1.13571 0.32411 0.11037 1.376303 -0.422114
#> 5 9.0712 1.49474 1.87538 0.10575 0.292140 -0.184568
#> 6 11.3493 -0.18453 -0.78037 -1.23804 -0.010949 0.691034
```

Least-squares provides accurate estimates of the regression constant \(\alpha = 10\) and the error variance \(\sigma^2 = 1\) for the “null model” including only the regression constant; moreover, the omnibus \(F\)-test of the correct null hypothesis that all of the \(\beta\)s are 0 for the “full model” with all 100 \(x\)s is associated with a large \(p\)-value:

```
m.full <- lm(y ~ ., data = D)
m.null <- lm(y ~ 1, data = D)
anova(m.null, m.full)
#> Analysis of Variance Table
#>
#> Model 1: y ~ 1
#> Model 2: y ~ X.1 + X.2 + X.3 + X.4 + X.5 + X.6 + X.7 + X.8 + X.9 + X.10 +
#> X.11 + X.12 + X.13 + X.14 + X.15 + X.16 + X.17 + X.18 + X.19 +
#> X.20 + X.21 + X.22 + X.23 + X.24 + X.25 + X.26 + X.27 + X.28 +
#> X.29 + X.30 + X.31 + X.32 + X.33 + X.34 + X.35 + X.36 + X.37 +
#> X.38 + X.39 + X.40 + X.41 + X.42 + X.43 + X.44 + X.45 + X.46 +
#> X.47 + X.48 + X.49 + X.50 + X.51 + X.52 + X.53 + X.54 + X.55 +
#> X.56 + X.57 + X.58 + X.59 + X.60 + X.61 + X.62 + X.63 + X.64 +
#> X.65 + X.66 + X.67 + X.68 + X.69 + X.70 + X.71 + X.72 + X.73 +
#> X.74 + X.75 + X.76 + X.77 + X.78 + X.79 + X.80 + X.81 + X.82 +
#> X.83 + X.84 + X.85 + X.86 + X.87 + X.88 + X.89 + X.90 + X.91 +
#> X.92 + X.93 + X.94 + X.95 + X.96 + X.97 + X.98 + X.99 + X.100
#> Res.Df RSS Df Sum of Sq F Pr(>F)
#> 1 999 974
#> 2 899 888 100 85.2 0.86 0.82
summary(m.null)
#>
#> Call:
#> lm(formula = y ~ 1, data = D)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -3.458 -0.681 0.019 0.636 2.935
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 9.9370 0.0312 318 <2e-16 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.987 on 999 degrees of freedom
```

Next, using the `stepAIC()`

function in the
**MASS** package (Venables &
Ripley, 2002), let us perform a forward stepwise regression to
select a “best” model, starting with the null model, and using AIC as
the model-selection criterion (see the help page for
`stepAIC()`

for details):^{1}

```
library("MASS") # for stepAIC()
m.select <- stepAIC(
m.null,
direction = "forward",
trace = FALSE,
scope = list(lower = ~ 1, upper = formula(m.full))
)
summary(m.select)
#>
#> Call:
#> lm(formula = y ~ X.99 + X.90 + X.87 + X.40 + X.65 + X.91 + X.53 +
#> X.45 + X.31 + X.56 + X.61 + X.60 + X.46 + X.35 + X.92, data = D)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -3.262 -0.645 0.024 0.641 3.118
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 9.9372 0.0310 320.80 <2e-16 ***
#> X.99 -0.0910 0.0308 -2.95 0.0032 **
#> X.90 -0.0820 0.0314 -2.62 0.0090 **
#> X.87 -0.0694 0.0311 -2.24 0.0256 *
#> X.40 -0.0476 0.0308 -1.55 0.1221
#> X.65 -0.0552 0.0315 -1.76 0.0795 .
#> X.91 0.0524 0.0308 1.70 0.0894 .
#> X.53 -0.0492 0.0305 -1.61 0.1067
#> X.45 0.0554 0.0318 1.74 0.0818 .
#> X.31 0.0452 0.0311 1.46 0.1457
#> X.56 0.0543 0.0327 1.66 0.0972 .
#> X.61 -0.0508 0.0317 -1.60 0.1091
#> X.60 -0.0513 0.0319 -1.61 0.1083
#> X.46 0.0516 0.0327 1.58 0.1153
#> X.35 0.0470 0.0315 1.49 0.1358
#> X.92 0.0443 0.0310 1.43 0.1533
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 0.973 on 984 degrees of freedom
#> Multiple R-squared: 0.0442, Adjusted R-squared: 0.0296
#> F-statistic: 3.03 on 15 and 984 DF, p-value: 8.34e-05
library("cv")
mse(D$y, fitted(m.select))
#> [1] 0.93063
#> attr(,"casewise loss")
#> [1] "(y - yhat)^2"
```

The resulting model has 15 predictors, a very modest \(R^2 = .044\), but a small \(p\)-value for its omnibus \(F\)-test (which, of course, is entirely spurious because the same data were used to select and test the model). The MSE for the selected model is smaller than the true error variance \(\sigma^2 = 1\), as is the estimated error variance for the selected model, \(\widehat{\sigma}^2 = 0.973^2 = 0.947\).

If we cross-validate the selected model, we also obtain an optimistic estimate of its predictive power (although the confidence interval for the bias-adjusted MSE includes 1):

```
library("cv")
cv(m.select, seed = 2529)
#> R RNG seed set to 2529
#> 10-Fold Cross Validation
#> method: Woodbury
#> criterion: mse
#> cross-validation criterion = 0.95937
#> bias-adjusted cross-validation criterion = 0.95785
#> 95% CI for bias-adjusted CV criterion = (0.87661, 1.0391)
#> full-sample criterion = 0.93063
```

The `"function"`

method of `cv()`

allows us to
cross-validate the whole model-selection procedure, where first argument
to `cv()`

is a model-selection function capable of refitting
the model with a fold omitted and returning a CV criterion. The
`selectStepAIC()`

function, in the **cv**
package and based on `stepAIC()`

, is suitable for use with
`cv()`

:

```
cv.select <- cv(
selectStepAIC,
data = D,
seed = 3791,
working.model = m.null,
direction = "forward",
scope = list(lower = ~ 1, upper = formula(m.full))
)
#> R RNG seed set to 3791
cv.select
#> 10-Fold Cross Validation
#> criterion: mse
#> cross-validation criterion = 1.0687
#> bias-adjusted cross-validation criterion = 1.0612
#> 95% CI for bias-adjusted CV criterion = (0.97172, 1.1506)
#> full-sample criterion = 0.93063
```

The other arguments to `cv()`

are:

`data`

, the data set to which the model is fit;`seed`

, an optional seed for R’s pseudo-random-number generator; as for`cv()`

, if the seed isn’t supplied by the user, a seed is randomly selected and saved;- additional arguments required by the model-selection function, here
the starting
`working.model`

argument, the`direction`

of model selection, and the`scope`

of models considered (from the model with only a regression constant to the model with all 100 predictors).

By default, `cv()`

performs 10-fold CV, and produces an
estimate of MSE for the model-selection procedure even *larger*
than the true error variance, \(\sigma^2 =
1\).

Also by default, when the number of folds is 10 or fewer,
`cv()`

saves details data about the folds. In this example,
the `compareFolds()`

function reveals that the variables
retained by the model-selection process in the several folds are quite
different:

```
compareFolds(cv.select)
#> CV criterion by folds:
#> fold.1 fold.2 fold.3 fold.4 fold.5 fold.6 fold.7 fold.8 fold.9 fold.10
#> 1.26782 1.12837 1.04682 1.31007 1.06899 0.87916 0.88380 0.95026 1.21070 0.94130
#>
#> Coefficients by folds:
#> (Intercept) X.87 X.90 X.99 X.91 X.54 X.53 X.56
#> Fold 1 9.9187 -0.0615 -0.0994 -0.0942 0.0512 0.0516
#> Fold 2 9.9451 -0.0745 -0.0899 -0.0614 0.0587 0.0673
#> Fold 3 9.9423 -0.0783 -0.0718 -0.0987 0.0601 0.0512
#> Fold 4 9.9410 -0.0860 -0.0831 -0.0867 0.0570 -0.0508
#> Fold 5 9.9421 -0.0659 -0.0849 -0.1004 0.0701 0.0511 -0.0487 0.0537
#> Fold 6 9.9633 -0.0733 -0.0874 -0.0960 0.0555 0.0629 -0.0478
#> Fold 7 9.9279 -0.0618 -0.0960 -0.0838 0.0533 -0.0464
#> Fold 8 9.9453 -0.0610 -0.0811 -0.0818 0.0497 -0.0612 0.0560
#> Fold 9 9.9173 -0.0663 -0.0894 -0.1100 0.0504 0.0524 0.0747
#> Fold 10 9.9449 -0.0745 -0.0906 -0.0891 0.0535 0.0482 -0.0583 0.0642
#> X.40 X.45 X.65 X.68 X.92 X.15 X.26 X.46 X.60
#> Fold 1 -0.0590 -0.0456 0.0658 0.0608
#> Fold 2 0.0607 0.0487
#> Fold 3 -0.0496 -0.0664 0.0494
#> Fold 4 -0.0597 0.0579 -0.0531 0.0519 -0.0566 -0.0519
#> Fold 5 0.0587 0.0527 -0.0603
#> Fold 6 -0.0596 0.0552 0.0474
#> Fold 7 0.0572 0.0595
#> Fold 8 0.0547 -0.0617 0.0453 0.0493 -0.0613 0.0591 0.0703 -0.0588
#> Fold 9 -0.0552 0.0573 -0.0635 0.0492 -0.0513 0.0484 -0.0507
#> Fold 10 -0.0558 0.0529 0.0710
#> X.61 X.8 X.28 X.29 X.31 X.35 X.70 X.89 X.17
#> Fold 1 -0.0490 0.0616 -0.0537 0.0638
#> Fold 2 0.0671 0.0568 0.0523
#> Fold 3 -0.0631 0.0616
#> Fold 4 0.0659 -0.0549 0.0527 0.0527
#> Fold 5 0.0425 0.0672 0.0613 0.0493
#> Fold 6 0.0559 -0.0629 0.0498 0.0487
#> Fold 7 0.0611 0.0472
#> Fold 8 -0.0719 0.0586
#> Fold 9 0.0525
#> Fold 10 -0.0580 0.0603
#> X.25 X.4 X.64 X.81 X.97 X.11 X.2 X.33 X.47
#> Fold 1 0.0604 0.0575
#> Fold 2 0.0478 0.0532 0.0518
#> Fold 3 0.0574 0.0473
#> Fold 4 0.0628
#> Fold 5 0.0518
#> Fold 6 0.0521
#> Fold 7 0.0550
#> Fold 8
#> Fold 9 0.0556 0.0447
#> Fold 10 0.0516
#> X.6 X.72 X.73 X.77 X.79 X.88
#> Fold 1 0.0476
#> Fold 2 0.0514
#> Fold 3
#> Fold 4 -0.0473
#> Fold 5 0.0586 0.07
#> Fold 6 -0.0489
#> Fold 7
#> Fold 8
#> Fold 9
#> Fold 10
```

`mpg`

to
`horsepower`

for the `Auto`

data, saving the
results in `m.1`

through `m.10`

. We then used
`cv()`

to compare the cross-validated MSE for the 10 models,
discovering that the 7th degree polynomial had the smallest MSE (by a
small margin); repeating the relevant graph:
If we then select the 7th degree polynomial model, intending to use
it for prediction, the CV estimate of the MSE for this model will be
optimistic. One solution is to cross-validate the process of using CV to
select the “best” model—that is, to apply CV to CV recursively. The
function `selectModelList()`

, which is suitable for use with
`cv()`

, implements this idea.

Applying `selectModelList()`

to the `Auto`

polynomial-regression models, and using 10-fold CV, we obtain:

```
recursiveCV.auto <- cv(
selectModelList,
Auto,
working.model = models(m.1, m.2, m.3, m.4, m.5,
m.6, m.7, m.8, m.9, m.10),
save.model = TRUE,
seed = 2120
)
#> R RNG seed set to 2120
recursiveCV.auto
#> 10-Fold Cross Validation
#> criterion: mse
#> cross-validation criterion = 19.105
#> bias-adjusted cross-validation criterion = 19.68
#> full-sample criterion = 18.746
recursiveCV.auto$selected.model
#>
#> Call:
#> lm(formula = mpg ~ poly(horsepower, 7), data = Auto)
#>
#> Coefficients:
#> (Intercept) poly(horsepower, 7)1 poly(horsepower, 7)2
#> 23.45 -120.14 44.09
#> poly(horsepower, 7)3 poly(horsepower, 7)4 poly(horsepower, 7)5
#> -3.95 -5.19 13.27
#> poly(horsepower, 7)6 poly(horsepower, 7)7
#> -8.55 7.98
cv(m.7, seed = 2120) # same seed for same folds
#> R RNG seed set to 2120
#> 10-Fold Cross Validation
#> method: Woodbury
#> criterion: mse
#> cross-validation criterion = 18.898
#> bias-adjusted cross-validation criterion = 18.854
#> full-sample criterion = 18.078
```

As expected, recursive CV produces a larger estimate of MSE for the selected 7th degree polynomial model than CV applied directly to this model.

We can equivalently call `cv()`

with the list of models as
the first argument and set `recursive=TRUE`

:

Next, let’s apply model selection to Mroz’s logistic regression for
married women’s labor-force participation, also discussed in the
introductory vignette on cross-validating regression models. First,
recall the logistic regression model that we fit to the
`Mroz`

data:

```
data("Mroz", package = "carData")
m.mroz <- glm(lfp ~ ., data = Mroz, family = binomial)
summary(m.mroz)
#>
#> Call:
#> glm(formula = lfp ~ ., family = binomial, data = Mroz)
#>
#> Coefficients:
#> Estimate Std. Error z value Pr(>|z|)
#> (Intercept) 3.18214 0.64438 4.94 7.9e-07 ***
#> k5 -1.46291 0.19700 -7.43 1.1e-13 ***
#> k618 -0.06457 0.06800 -0.95 0.34234
#> age -0.06287 0.01278 -4.92 8.7e-07 ***
#> wcyes 0.80727 0.22998 3.51 0.00045 ***
#> hcyes 0.11173 0.20604 0.54 0.58762
#> lwg 0.60469 0.15082 4.01 6.1e-05 ***
#> inc -0.03445 0.00821 -4.20 2.7e-05 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> (Dispersion parameter for binomial family taken to be 1)
#>
#> Null deviance: 1029.75 on 752 degrees of freedom
#> Residual deviance: 905.27 on 745 degrees of freedom
#> AIC: 921.3
#>
#> Number of Fisher Scoring iterations: 4
```

Applying stepwise model selection Mroz’s logistic regression, using
BIC as the model-selection criterion (via the argument
`k=log(nrow(Mroz))`

to `stepAIC()`

) selects 5 of
the 7 original predictors:

```
m.mroz.sel <- stepAIC(m.mroz, k = log(nrow(Mroz)),
trace = FALSE)
summary(m.mroz.sel)
#>
#> Call:
#> glm(formula = lfp ~ k5 + age + wc + lwg + inc, family = binomial,
#> data = Mroz)
#>
#> Coefficients:
#> Estimate Std. Error z value Pr(>|z|)
#> (Intercept) 2.9019 0.5429 5.35 9.0e-08 ***
#> k5 -1.4318 0.1932 -7.41 1.3e-13 ***
#> age -0.0585 0.0114 -5.13 2.9e-07 ***
#> wcyes 0.8724 0.2064 4.23 2.4e-05 ***
#> lwg 0.6157 0.1501 4.10 4.1e-05 ***
#> inc -0.0337 0.0078 -4.32 1.6e-05 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> (Dispersion parameter for binomial family taken to be 1)
#>
#> Null deviance: 1029.75 on 752 degrees of freedom
#> Residual deviance: 906.46 on 747 degrees of freedom
#> AIC: 918.5
#>
#> Number of Fisher Scoring iterations: 3
BayesRule(Mroz$lfp == "yes",
predict(m.mroz.sel, type = "response"))
#> [1] 0.31873
#> attr(,"casewise loss")
#> [1] "y != round(yhat)"
```

Bayes rule applied to the selected model misclassifies 32% of the
cases in the `Mroz`

data.

Cross-validating the selected model produces a similar, slightly larger, estimate of misclassification, about 33%:

```
cv(m.mroz.sel, criterion = BayesRule, seed = 345266)
#> R RNG seed set to 345266
#> 10-Fold Cross Validation
#> method: exact
#> criterion: BayesRule
#> cross-validation criterion = 0.33068
#> bias-adjusted cross-validation criterion = 0.33332
#> 95% CI for bias-adjusted CV criterion = (0.2997, 0.36695)
#> full-sample criterion = 0.31873
```

Is this estimate of predictive performance optimistic?

We proceed to apply the model-selection procedure by cross-validation, producing more or less the same result:

```
m.mroz.sel.cv <- cv(
selectStepAIC,
Mroz,
seed = 6681,
criterion = BayesRule,
working.model = m.mroz,
AIC = FALSE
)
#> R RNG seed set to 6681
m.mroz.sel.cv
#> 10-Fold Cross Validation
#> criterion: BayesRule
#> cross-validation criterion = 0.33068
#> bias-adjusted cross-validation criterion = 0.33452
#> 95% CI for bias-adjusted CV criterion = (0.3009, 0.36815)
#> full-sample criterion = 0.31873
```

Setting `AIC=FALSE`

in the call to `cv()`

uses
the BIC rather than the AIC as the model-selection criterion. As it
turns out, exactly the same predictors are selected when each of the 10
folds are omitted, and the several coefficient estimates are very
similar, as we show using `compareFolds()`

:

```
compareFolds(m.mroz.sel.cv)
#> CV criterion by folds:
#> fold.1 fold.2 fold.3 fold.4 fold.5 fold.6 fold.7 fold.8 fold.9 fold.10
#> 0.27632 0.40789 0.34211 0.36000 0.28000 0.37333 0.32000 0.29333 0.33333 0.32000
#>
#> Coefficients by folds:
#> (Intercept) age inc k5 lwg wcyes
#> Fold 1 2.5014 -0.0454 -0.0388 -1.3613 0.5653 0.85
#> Fold 2 3.0789 -0.0659 -0.0306 -1.5335 0.6923 0.79
#> Fold 3 3.0141 -0.0595 -0.0305 -1.3994 0.5428 0.86
#> Fold 4 2.7251 -0.0543 -0.0354 -1.4474 0.6298 1.09
#> Fold 5 2.7617 -0.0566 -0.0320 -1.4752 0.6324 0.74
#> Fold 6 3.0234 -0.0621 -0.0348 -1.4537 0.6618 0.94
#> Fold 7 2.9615 -0.0600 -0.0351 -1.4127 0.5835 0.97
#> Fold 8 2.9598 -0.0603 -0.0329 -1.3865 0.6210 0.69
#> Fold 9 3.2481 -0.0650 -0.0381 -1.4138 0.6093 0.94
#> Fold 10 2.7724 -0.0569 -0.0295 -1.4503 0.6347 0.85
```

In this example, therefore, we appear to obtain a realistic estimate of model performance directly from the selected model, because there is little added uncertainty induced by model selection.

The **cv** package also provides a `cv()`

procedure, `selectTrans()`

, for choosing transformations of
the predictors and the response in regression.

Some background: As Weisberg (2014, sec.
8.2) explains, there are technical advantages to having (numeric)
predictors in linear regression analysis that are themselves linearly
related. If the predictors *aren’t* linearly related, then the
relationships between them can often be straightened by power
transformations. Transformations can be selected after graphical
examination of the data, or by analytic methods. Once the relationships
between the predictors are linearized, it can be advantageous similarly
to transform the response variable towards normality.

Selecting transformations analytically raises the possibility of automating the process, as would be required for cross-validation. One could, in principle, apply graphical methods to select transformations for each fold, but because a data analyst couldn’t forget the choices made for previous folds, the process wouldn’t really be applied independently to the folds.

To illustrate, we adapt an example appearing in several places in
Fox & Weisberg (2019) (for example in
Chapter 3 on transforming data), using data on the prestige and other
characteristics of 102 Canadian occupations circa 1970. The data are in
the `Prestige`

data frame in the **carData**
package:

```
data("Prestige", package = "carData")
head(Prestige)
#> education income women prestige census type
#> gov.administrators 13.11 12351 11.16 68.8 1113 prof
#> general.managers 12.26 25879 4.02 69.1 1130 prof
#> accountants 12.77 9271 15.70 63.4 1171 prof
#> purchasing.officers 11.42 8865 9.11 56.8 1175 prof
#> chemists 14.62 8403 11.68 73.5 2111 prof
#> physicists 15.64 11030 5.13 77.6 2113 prof
summary(Prestige)
#> education income women prestige census
#> Min. : 6.38 Min. : 611 Min. : 0.00 Min. :14.8 Min. :1113
#> 1st Qu.: 8.45 1st Qu.: 4106 1st Qu.: 3.59 1st Qu.:35.2 1st Qu.:3120
#> Median :10.54 Median : 5930 Median :13.60 Median :43.6 Median :5135
#> Mean :10.74 Mean : 6798 Mean :28.98 Mean :46.8 Mean :5402
#> 3rd Qu.:12.65 3rd Qu.: 8187 3rd Qu.:52.20 3rd Qu.:59.3 3rd Qu.:8312
#> Max. :15.97 Max. :25879 Max. :97.51 Max. :87.2 Max. :9517
#> type
#> bc :44
#> prof:31
#> wc :23
#> NA's: 4
#>
#>
```

The variables in the `Prestige`

data set are:

`education`

: average years of education for incumbents in the occupation, from the 1971 Canadian Census.`income`

: average dollars of annual income for the occupation, from the Census.`women`

: percentage of occupational incumbents who were women, also from the Census.`prestige`

: the average prestige rating of the occupation on a 0–100 “thermometer” scale, in a Canadian social survey conducted around the same time.`type`

, type of occupation, and`census`

, the Census occupational code, which are not used in our example.

The object of a regression analysis for the `Prestige`

data (and their original purpose) is to predict occupational prestige
from the other variables in the data set.

A scatterplot matrix (using the `scatterplotMatrix()`

function in the **car** package) of the numeric variables
in the data reveals that the distributions of `income`

and
`women`

are positively skewed, and that some of the
relationships among the three predictors, and between the predictors and
the response (i.e., `prestige`

), are nonlinear:

```
library("car")
#> Loading required package: carData
#>
#> Attaching package: 'car'
#> The following object is masked from 'package:dplyr':
#>
#> recode
scatterplotMatrix(
~ prestige + income + education + women,
data = Prestige,
smooth = list(spread = FALSE)
)
```

The `powerTransform()`

function in the
**car** package transforms variables towards multivariate
normality by a generalization of Box and Cox’s maximum-likelihood-like
approach (Box & Cox, 1964). Several
“families” of power transformations can be used, including the original
Box-Cox family, simple powers (and roots), and two adaptations of the
Box-Cox family to data that may include negative values and zeros: the
Box-Cox-with-negatives family and the Yeo-Johnson family; see Weisberg (2014, Chapter 8), and Fox & Weisberg (2019, Chapter 3) for
details. Because `women`

has some zero values, we use the
Yeo-Johnson family:

```
trans <- powerTransform(cbind(income, education, women) ~ 1,
data = Prestige,
family = "yjPower")
summary(trans)
#> yjPower Transformations to Multinormality
#> Est Power Rounded Pwr Wald Lwr Bnd Wald Upr Bnd
#> income 0.2678 0.33 0.1051 0.4304
#> education 0.5162 1.00 -0.2822 1.3145
#> women 0.1630 0.16 0.0112 0.3149
#>
#> Likelihood ratio test that all transformation parameters are equal to 0
#> LRT df pval
#> LR test, lambda = (0 0 0) 15.739 3 0.00128
```

We thus have evidence of the desirability of transforming
`income`

(by the \(1/3\)
power) and `women`

(by the \(0.16\) power—which is close to the “0”
power, i.e., the log transformation), but not `education`

.
Applying the “rounded” power transformations makes the predictors
better-behaved:

```
P <- Prestige[, c("prestige", "income", "education", "women")]
(lambdas <- trans$roundlam)
#> income education women
#> 0.33000 1.00000 0.16302
names(lambdas) <- c("income", "education", "women")
for (var in c("income", "education", "women")) {
P[, var] <- yjPower(P[, var], lambda = lambdas[var])
}
summary(P)
#> prestige income education women
#> Min. :14.8 Min. :22.2 Min. : 6.38 Min. :0.00
#> 1st Qu.:35.2 1st Qu.:44.2 1st Qu.: 8.45 1st Qu.:1.73
#> Median :43.6 Median :50.3 Median :10.54 Median :3.36
#> Mean :46.8 Mean :50.8 Mean :10.74 Mean :3.50
#> 3rd Qu.:59.3 3rd Qu.:56.2 3rd Qu.:12.65 3rd Qu.:5.59
#> Max. :87.2 Max. :83.6 Max. :15.97 Max. :6.83
scatterplotMatrix(
~ prestige + income + education + women,
data = P,
smooth = list(spread = FALSE)
)
```

Comparing the MSE for the regressions with the original and transformed predictors shows a advantage to the latter:

```
m.pres <- lm(prestige ~ income + education + women, data = Prestige)
m.pres.trans <- lm(prestige ~ income + education + women, data = P)
mse(Prestige$prestige, fitted(m.pres))
#> [1] 59.153
#> attr(,"casewise loss")
#> [1] "(y - yhat)^2"
mse(P$prestige, fitted(m.pres.trans))
#> [1] 50.6
#> attr(,"casewise loss")
#> [1] "(y - yhat)^2"
```

Similarly, component+residual plots for the two regressions, produced
by the `crPlots()`

function in the **car**
package, suggest that the partial relationship of `prestige`

to `income`

is more nearly linear in the transformed data,
but the transformation of `women`

fails to capture what
appears to be a slight quadratic partial relationship; the partial
relationship of `prestige`

to `education`

is close
to linear in both regressions:

Having transformed the predictors towards multinormality, we now
consider whether there’s evidence for transforming the response (using
`powerTransform()`

for Box and Cox’s original method), and we
discover that there’s not:

```
summary(powerTransform(m.pres.trans))
#> bcPower Transformation to Normality
#> Est Power Rounded Pwr Wald Lwr Bnd Wald Upr Bnd
#> Y1 1.0194 1 0.6773 1.3615
#>
#> Likelihood ratio test that transformation parameter is equal to 0
#> (log transformation)
#> LRT df pval
#> LR test, lambda = (0) 32.217 1 1.38e-08
#>
#> Likelihood ratio test that no transformation is needed
#> LRT df pval
#> LR test, lambda = (1) 0.012384 1 0.911
```

The `selectTrans()`

function in the **cv**
package automates the process of selecting predictor and response
transformations. The function takes a `data`

set and
“working” `model`

as arguments, along with the candidate
`predictors`

and `response`

for transformation,
and the transformation `family`

to employ. If the
`predictors`

argument is missing then only the response is
transformed, and if the `response`

argument is missing, only
the supplied predictors are transformed. The default `family`

for transforming the predictors is `"bcPower"`

—the original
Box-Cox family—as is the default `family.y`

for transforming
the response; here we specify `family="yjPower`

because of
the zeros in `women`

. `selectTrans()`

returns the
result of applying a lack-of-fit criterion to the model after the
selected transformation is applied, with the default
`criterion=mse`

:

```
selectTrans(
data = Prestige,
model = m.pres,
predictors = c("income", "education", "women"),
response = "prestige",
family = "yjPower"
)
#> $criterion
#> [1] 50.6
#> attr(,"casewise loss")
#> [1] "(y - yhat)^2"
#>
#> $model
#> NULL
```

`selectTrans()`

also takes an optional
`indices`

argument, making it suitable for doing computations
on a subset of the data (i.e., a CV fold), and hence for use with
`cv()`

(see `?selectTrans`

for details):

```
cvs <- cv(
selectTrans,
data = Prestige,
working.model = m.pres,
seed = 1463,
predictors = c("income", "education", "women"),
response = "prestige",
family = "yjPower"
)
#> R RNG seed set to 1463
cvs
#> 10-Fold Cross Validation
#> criterion: mse
#> cross-validation criterion = 54.487
#> bias-adjusted cross-validation criterion = 54.308
#> full-sample criterion = 50.6
cv(m.pres, seed = 1463) # untransformed model with same folds
#> R RNG seed set to 1463
#> 10-Fold Cross Validation
#> method: Woodbury
#> criterion: mse
#> cross-validation criterion = 63.293
#> bias-adjusted cross-validation criterion = 63.073
#> full-sample criterion = 59.153
compareFolds(cvs)
#> CV criterion by folds:
#> fold.1 fold.2 fold.3 fold.4 fold.5 fold.6 fold.7 fold.8 fold.9 fold.10
#> 63.453 79.257 20.634 94.569 19.902 55.821 26.555 75.389 55.702 50.215
#>
#> Coefficients by folds:
#> lam.education lam.income lam.women lambda
#> Fold 1 1.000 0.330 0.330 1
#> Fold 2 1.000 0.330 0.169 1
#> Fold 3 1.000 0.330 0.330 1
#> Fold 4 1.000 0.330 0.330 1
#> Fold 5 1.000 0.330 0.000 1
#> Fold 6 1.000 0.330 0.330 1
#> Fold 7 1.000 0.330 0.330 1
#> Fold 8 1.000 0.330 0.000 1
#> Fold 9 1.000 0.330 0.000 1
#> Fold 10 1.000 0.330 0.000 1
```

The results suggest that the predictive power of the transformed
regression is reliably greater than that of the untransformed regression
(though in both case, the cross-validated MSE is considerably higher
than the MSE computed for the whole data). Examining the selected
transformations for each fold reveals that the predictor
`education`

and the response `prestige`

are never
transformed; that the \(1/3\) power is
selected for `income`

in all of the folds; and that the
transformation selected for `women`

varies narrowly across
the folds between the \(0\)th power
(i.e., log) and the \(1/3\) power.

As we mentioned, Hastie et al. (2009, sec. 7.10.2: “The Wrong and Right Way to Do Cross-validation”) explain that honest cross-validation has to take account of model specification and selection. Statistical modeling is at least partly a craft, and one could imagine applying that craft to successive partial data sets, each with a fold removed. The resulting procedure would be tedious, though possibly worth the effort, but it would also be difficult to realize in practice: After all, we can hardly erase our memory of statistical modeling choices between analyzing partial data sets.

Alternatively, if we’re able to automate the process of model
selection, then we can more realistically apply CV mechanically. That’s
what we did in the preceding two sections, first for predictor selection
and then for selection of transformations in regression. In this
section, we consider the case where we both select variable
transformations and then proceed to select predictors. It’s insufficient
to apply these steps sequentially, first, for example, using
`cv()`

with `selectTrans()`

and then with
`selectStepAIC()`

; rather we should apply the whole
model-selection procedure with each fold omitted. The
`selectTransAndStepAIC()`

function, also supplied by the
**cv** package, does exactly that.

To illustrate this process, we return to the `Auto`

data
set:

```
summary(Auto)
#> mpg cylinders displacement horsepower weight
#> Min. : 9.0 Min. :3.00 Min. : 68 Min. : 46.0 Min. :1613
#> 1st Qu.:17.0 1st Qu.:4.00 1st Qu.:105 1st Qu.: 75.0 1st Qu.:2225
#> Median :22.8 Median :4.00 Median :151 Median : 93.5 Median :2804
#> Mean :23.4 Mean :5.47 Mean :194 Mean :104.5 Mean :2978
#> 3rd Qu.:29.0 3rd Qu.:8.00 3rd Qu.:276 3rd Qu.:126.0 3rd Qu.:3615
#> Max. :46.6 Max. :8.00 Max. :455 Max. :230.0 Max. :5140
#>
#> acceleration year origin name
#> Min. : 8.0 Min. :70 Min. :1.00 amc matador : 5
#> 1st Qu.:13.8 1st Qu.:73 1st Qu.:1.00 ford pinto : 5
#> Median :15.5 Median :76 Median :1.00 toyota corolla : 5
#> Mean :15.5 Mean :76 Mean :1.58 amc gremlin : 4
#> 3rd Qu.:17.0 3rd Qu.:79 3rd Qu.:2.00 amc hornet : 4
#> Max. :24.8 Max. :82 Max. :3.00 chevrolet chevette: 4
#> (Other) :365
xtabs( ~ year, data = Auto)
#> year
#> 70 71 72 73 74 75 76 77 78 79 80 81 82
#> 29 27 28 40 26 30 34 28 36 29 27 28 30
xtabs( ~ origin, data = Auto)
#> origin
#> 1 2 3
#> 245 68 79
xtabs( ~ cylinders, data = Auto)
#> cylinders
#> 3 4 5 6 8
#> 4 199 3 83 103
```

We previously used the `Auto`

here in a preliminary
example where we employed CV to inform the selection of the order of a
polynomial regression of `mpg`

on `horsepower`

.
Here, we consider more generally the problem of predicting
`mpg`

from the other variables in the `Auto`

data.
We begin with a bit of data management, and then examine the pairwise
relationships among the numeric variables in the data set:

```
Auto$cylinders <- factor(Auto$cylinders,
labels = c("3.4", "3.4", "5.6", "5.6", "8"))
Auto$year <- as.factor(Auto$year)
Auto$origin <- factor(Auto$origin,
labels = c("America", "Europe", "Japan"))
rownames(Auto) <- make.names(Auto$name, unique = TRUE)
Auto$name <- NULL
scatterplotMatrix(
~ mpg + displacement + horsepower + weight + acceleration,
smooth = list(spread = FALSE),
data = Auto
)
```

A comment before we proceed: `origin`

is clearly
categorical and so converting it to a factor is natural, but we could
imagine treating `cylinders`

and `year`

as numeric
predictors. There are, however, only 5 distinct values of
`cylinders`

(ranging from 3 to 8), but cars with 3 or 5
cylinders are rare. and none of the cars has 7 cylinders. There are
similarly only 13 distinct years between 1970 and 1982 in the data, and
the relationship between `mpg`

and `year`

is
difficult to characterize.^{3} It’s apparent that most these variables are
positively skewed and that many of the pairwise relationships among them
are nonlinear.

We begin with a “working model” that specifies linear partial relationships of the response to the numeric predictors:

```
m.auto <- lm(mpg ~ ., data = Auto)
summary(m.auto)
#>
#> Call:
#> lm(formula = mpg ~ ., data = Auto)
#>
#> Residuals:
#> Min 1Q Median 3Q Max
#> -9.006 -1.745 -0.092 1.525 10.950
#>
#> Coefficients:
#> Estimate Std. Error t value Pr(>|t|)
#> (Intercept) 37.034132 1.969393 18.80 < 2e-16 ***
#> cylinders5.6 -2.602941 0.655200 -3.97 8.5e-05 ***
#> cylinders8 -0.582458 1.171452 -0.50 0.61934
#> displacement 0.017425 0.006734 2.59 0.01004 *
#> horsepower -0.041353 0.013379 -3.09 0.00215 **
#> weight -0.005548 0.000632 -8.77 < 2e-16 ***
#> acceleration 0.061527 0.088313 0.70 0.48643
#> year71 0.968058 0.837390 1.16 0.24841
#> year72 -0.601435 0.825115 -0.73 0.46652
#> year73 -0.687689 0.740272 -0.93 0.35351
#> year74 1.375576 0.876500 1.57 0.11741
#> year75 0.929929 0.859072 1.08 0.27974
#> year76 1.559893 0.822505 1.90 0.05867 .
#> year77 2.909416 0.841729 3.46 0.00061 ***
#> year78 3.175198 0.798940 3.97 8.5e-05 ***
#> year79 5.019299 0.845759 5.93 6.8e-09 ***
#> year80 9.099763 0.897293 10.14 < 2e-16 ***
#> year81 6.688660 0.885218 7.56 3.3e-13 ***
#> year82 8.071125 0.870668 9.27 < 2e-16 ***
#> originEurope 2.046664 0.517124 3.96 9.1e-05 ***
#> originJapan 2.144887 0.507717 4.22 3.0e-05 ***
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
#>
#> Residual standard error: 2.92 on 371 degrees of freedom
#> Multiple R-squared: 0.867, Adjusted R-squared: 0.86
#> F-statistic: 121 on 20 and 371 DF, p-value: <2e-16
Anova(m.auto)
#> Anova Table (Type II tests)
#>
#> Response: mpg
#> Sum Sq Df F value Pr(>F)
#> cylinders 292 2 17.09 7.9e-08 ***
#> displacement 57 1 6.70 0.0100 *
#> horsepower 82 1 9.55 0.0021 **
#> weight 658 1 76.98 < 2e-16 ***
#> acceleration 4 1 0.49 0.4864
#> year 3017 12 29.40 < 2e-16 ***
#> origin 190 2 11.13 2.0e-05 ***
#> Residuals 3173 371
#> ---
#> Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
crPlots(m.auto)
```