# Chapter 6 Statistical models

In this chapter, we will not learn about all the models out there that you may or may not need. Instead, I will show you how can use what you have learned until now and how you can apply these concepts to modeling. Also, as you read in the beginning of the book, R has many many packages. So the model you need is most probably already implemented in some package and you will very likely not need to write your own from scratch.

In the first section, I will discuss the terminology used in this book. Then I will discuss linear regression; showing how linear regression works illsutrates very well how other models work too, without loss of generality. Then I will introduce the concepte of hyper-parameters with ridge regression. This chapter will then finish with an introduction to cross-validation as a way to tune the hyper-parameters of models that features them.

## 6.1 Terminology

Before continuing discussing about statistical models and model fitting it is worthwhile to discuss terminology a little bit. Depending on your background, you might call an explanatory variable a feature or the dependent variable the target. These are the same objects. The matrix of features is usually called a design matrix, and what statisticians call the intercept is what machine learning engineers call the bias. Referring to the intercept by bias is unfortunate, as bias also has a very different meaning; bias is also what we call the error in a model that may cause biased estimates. To finish up, the estimated parameters of the model may be called coefficients or weights. Here again, I don’t like the using weight as weight as a very different meaning in statistics. So, in the remainder of this chapter, and book, I will use the terminology from the statistical litterature, using dependent and explanatory variables (y and x), and calling the estimated parameters coefficients and the intercept… well the intercept (the $$\beta$$s of the model). However, I will talk of training a model, instead of estimating a model.

## 6.2 Fitting a model to data

Suppose you have a variable y that you wish to explain using a set of other variables x1, x2, x3, etc. Let’s take a look at the Housing dataset from the Ecdat package:

library(Ecdat)

data(Housing)

You can read a description of the dataset by running:

?Housing
Housing                 package:Ecdat                  R Documentation

Sales Prices of Houses in the City of Windsor

Description:

a cross-section from 1987

_number of observations_ : 546

_observation_ : goods

Usage:

data(Housing)

Format:

A dataframe containing :

price: sale price of a house

lotsize: the lot size of a property in square feet

bedrooms: number of bedrooms

bathrms: number of full bathrooms

stories: number of stories excluding basement

driveway: does the house has a driveway ?

recroom: does the house has a recreational room ?

fullbase: does the house has a full finished basement ?

gashw: does the house uses gas for hot water heating ?

airco: does the house has central air conditioning ?

garagepl: number of garage places

prefarea: is the house located in the preferred neighbourhood of the city ?

Source:

Anglin, P.M.  and R.  Gencay (1996) “Semiparametric estimation of
a hedonic price function”, _Journal of Applied Econometrics_,
*11(6)*, 633-648.

References:

Verbeek, Marno (2004) _A Guide to Modern Econometrics_, John Wiley
and Sons, chapter 3.

Journal of Applied Econometrics data archive : <URL:
http://qed.econ.queensu.ca/jae/>.

‘Index.Source’, ‘Index.Economics’, ‘Index.Econometrics’,
‘Index.Observations’

or by looking for Housing in the help pane of RStudio. Usually, you would take a look a the data before doing any modeling:

glimpse(Housing)
## Rows: 546
## Columns: 12
## $price <dbl> 42000, 38500, 49500, 60500, 61000, 66000, 66000, 69000, 83800… ##$ lotsize  <dbl> 5850, 4000, 3060, 6650, 6360, 4160, 3880, 4160, 4800, 5500, 7…
## $bedrooms <dbl> 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 4, 1, 2, 3… ##$ bathrms  <dbl> 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1…
## $stories <dbl> 2, 1, 1, 2, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2… ##$ driveway <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, no, ye…
## $recroom <fct> no, no, no, yes, no, yes, no, no, yes, yes, no, no, no, no, n… ##$ fullbase <fct> yes, no, no, no, no, yes, yes, no, yes, no, yes, no, no, no, …
## $gashw <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, n… ##$ airco    <fct> no, no, no, no, no, yes, no, no, no, yes, yes, no, no, no, no…
## $garagepl <dbl> 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 0, 0, 1, 0, 0, 1, 1… ##$ prefarea <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no, no, n…

Housing prices depend on a set of variables such as the number of bedrooms, the area it is located and so on. If you believe that housing prices depend linearly on a set of explanatory variables, you will want to estimate a linear model. To estimate a linear model, you will need to use the built-in lm() function:

model1 <- lm(price ~ lotsize + bedrooms, data = Housing)

lm() takes a formula as an argument, which defines the model you want to estimate. In this case, I ran the following regression:

$\text{price} = \beta_0 + \beta_1 * \text{lotsize} + \beta_2 * \text{bedrooms} + \varepsilon$

where $$\beta_0, \beta_1$$ and $$\beta_2$$ are three parameters to estimate. To take a look at the results, you can use the summary() method (not to be confused with dplyr::summarise()):

summary(model1)
##
## Call:
## lm(formula = price ~ lotsize + bedrooms, data = Housing)
##
## Residuals:
##    Min     1Q Median     3Q    Max
## -65665 -12498  -2075   8970  97205
##
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)
## (Intercept) 5.613e+03  4.103e+03   1.368    0.172
## lotsize     6.053e+00  4.243e-01  14.265  < 2e-16 ***
## bedrooms    1.057e+04  1.248e+03   8.470 2.31e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21230 on 543 degrees of freedom
## Multiple R-squared:  0.3703, Adjusted R-squared:  0.3679
## F-statistic: 159.6 on 2 and 543 DF,  p-value: < 2.2e-16

if you wish to remove the intercept ($$\beta_0$$ in the above equation) from your model, you can do so with -1:

model2 <- lm(price ~ -1 + lotsize + bedrooms, data = Housing)

summary(model2)
##
## Call:
## lm(formula = price ~ -1 + lotsize + bedrooms, data = Housing)
##
## Residuals:
##    Min     1Q Median     3Q    Max
## -67229 -12342  -1333   9627  95509
##
## Coefficients:
##           Estimate Std. Error t value Pr(>|t|)
## lotsize      6.283      0.390   16.11   <2e-16 ***
## bedrooms 11968.362    713.194   16.78   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 21250 on 544 degrees of freedom
## Multiple R-squared:  0.916,  Adjusted R-squared:  0.9157
## F-statistic:  2965 on 2 and 544 DF,  p-value: < 2.2e-16

or if you want to use all the columns inside Housing, replacing the column names by .:

model3 <- lm(price ~ ., data = Housing)

summary(model3)
##
## Call:
## lm(formula = price ~ ., data = Housing)
##
## Residuals:
##    Min     1Q Median     3Q    Max
## -41389  -9307   -591   7353  74875
##
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)
## (Intercept) -4038.3504  3409.4713  -1.184 0.236762
## lotsize         3.5463     0.3503  10.124  < 2e-16 ***
## bedrooms     1832.0035  1047.0002   1.750 0.080733 .
## bathrms     14335.5585  1489.9209   9.622  < 2e-16 ***
## stories      6556.9457   925.2899   7.086 4.37e-12 ***
## drivewayyes  6687.7789  2045.2458   3.270 0.001145 **
## recroomyes   4511.2838  1899.9577   2.374 0.017929 *
## fullbaseyes  5452.3855  1588.0239   3.433 0.000642 ***
## gashwyes    12831.4063  3217.5971   3.988 7.60e-05 ***
## aircoyes    12632.8904  1555.0211   8.124 3.15e-15 ***
## garagepl     4244.8290   840.5442   5.050 6.07e-07 ***
## prefareayes  9369.5132  1669.0907   5.614 3.19e-08 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 15420 on 534 degrees of freedom
## Multiple R-squared:  0.6731, Adjusted R-squared:  0.6664
## F-statistic: 99.97 on 11 and 534 DF,  p-value: < 2.2e-16

You can access different elements of model3 with $, because the result of lm() is a list (you can check this claim with typeof(model3): print(model3$coefficients)
##  (Intercept)      lotsize     bedrooms      bathrms      stories  drivewayyes
## -4038.350425     3.546303  1832.003466 14335.558468  6556.945711  6687.778890
##   recroomyes  fullbaseyes     gashwyes     aircoyes     garagepl  prefareayes
##  4511.283826  5452.385539 12831.406266 12632.890405  4244.829004  9369.513239

but I prefer to use the {broom} package, and more specifically the tidy() function, which converts model3 into a neat data.frame:

results3 <- broom::tidy(model3)

glimpse(results3)
## Rows: 12
## Columns: 5
## $term <chr> "(Intercept)", "lotsize", "bedrooms", "bathrms", "stories", … ##$ estimate  <dbl> -4038.350425, 3.546303, 1832.003466, 14335.558468, 6556.9457…
## $std.error <dbl> 3409.4713, 0.3503, 1047.0002, 1489.9209, 925.2899, 2045.2458… ##$ statistic <dbl> -1.184451, 10.123618, 1.749764, 9.621691, 7.086369, 3.269914…
## $p.value <dbl> 2.367616e-01, 3.732442e-22, 8.073341e-02, 2.570369e-20, 4.37… I explicitely write broom::tidy() because tidy() is a popular function name. For instance, it is also a function from the {yardstick} package, which does not do the same thing at all. Since I will also be using {yardstick} I prefer to explicitely write broom::tidy() to avoid conflicts. Using broom::tidy() is useful, because you can then work on the results easily, for example if you wish to only keep results that are significant at the 5% level: results3 %>% filter(p.value < 0.05) ## # A tibble: 10 × 5 ## term estimate std.error statistic p.value ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 lotsize 3.55 0.350 10.1 3.73e-22 ## 2 bathrms 14336. 1490. 9.62 2.57e-20 ## 3 stories 6557. 925. 7.09 4.37e-12 ## 4 drivewayyes 6688. 2045. 3.27 1.15e- 3 ## 5 recroomyes 4511. 1900. 2.37 1.79e- 2 ## 6 fullbaseyes 5452. 1588. 3.43 6.42e- 4 ## 7 gashwyes 12831. 3218. 3.99 7.60e- 5 ## 8 aircoyes 12633. 1555. 8.12 3.15e-15 ## 9 garagepl 4245. 841. 5.05 6.07e- 7 ## 10 prefareayes 9370. 1669. 5.61 3.19e- 8 You can even add new columns, such as the confidence intervals: results3 <- broom::tidy(model3, conf.int = TRUE, conf.level = 0.95) print(results3) ## # A tibble: 12 × 7 ## term estimate std.error statistic p.value conf.low conf.high ## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 (Intercept) -4038. 3409. -1.18 2.37e- 1 -10736. 2659. ## 2 lotsize 3.55 0.350 10.1 3.73e-22 2.86 4.23 ## 3 bedrooms 1832. 1047. 1.75 8.07e- 2 -225. 3889. ## 4 bathrms 14336. 1490. 9.62 2.57e-20 11409. 17262. ## 5 stories 6557. 925. 7.09 4.37e-12 4739. 8375. ## 6 drivewayyes 6688. 2045. 3.27 1.15e- 3 2670. 10705. ## 7 recroomyes 4511. 1900. 2.37 1.79e- 2 779. 8244. ## 8 fullbaseyes 5452. 1588. 3.43 6.42e- 4 2333. 8572. ## 9 gashwyes 12831. 3218. 3.99 7.60e- 5 6511. 19152. ## 10 aircoyes 12633. 1555. 8.12 3.15e-15 9578. 15688. ## 11 garagepl 4245. 841. 5.05 6.07e- 7 2594. 5896. ## 12 prefareayes 9370. 1669. 5.61 3.19e- 8 6091. 12648. Going back to model estimation, you can of course use lm() in a pipe workflow: Housing %>% select(-driveway, -stories) %>% lm(price ~ ., data = .) %>% broom::tidy() ## # A tibble: 10 × 5 ## term estimate std.error statistic p.value ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 (Intercept) 3025. 3263. 0.927 3.54e- 1 ## 2 lotsize 3.67 0.363 10.1 4.52e-22 ## 3 bedrooms 4140. 1036. 3.99 7.38e- 5 ## 4 bathrms 16443. 1546. 10.6 4.29e-24 ## 5 recroomyes 5660. 2010. 2.82 5.05e- 3 ## 6 fullbaseyes 2241. 1618. 1.38 1.67e- 1 ## 7 gashwyes 13568. 3411. 3.98 7.93e- 5 ## 8 aircoyes 15578. 1597. 9.75 8.53e-21 ## 9 garagepl 4232. 883. 4.79 2.12e- 6 ## 10 prefareayes 10729. 1753. 6.12 1.81e- 9 The first . in the lm() function is used to indicate that we wish to use all the data from Housing (minus driveway and stories which I removed using select() and the - sign), and the second . is used to place the result from the two dplyr instructions that preceded is to be placed there. The picture below should help you understand: You have to specify this, because by default, when using %>% the left hand side argument gets passed as the first argument of the function on the right hand side. Since version 4.2, R now also natively includes a placeholder, _: Housing |> select(-driveway, -stories) |> lm(price ~ ., data = _) |> broom::tidy() ## # A tibble: 10 × 5 ## term estimate std.error statistic p.value ## <chr> <dbl> <dbl> <dbl> <dbl> ## 1 (Intercept) 3025. 3263. 0.927 3.54e- 1 ## 2 lotsize 3.67 0.363 10.1 4.52e-22 ## 3 bedrooms 4140. 1036. 3.99 7.38e- 5 ## 4 bathrms 16443. 1546. 10.6 4.29e-24 ## 5 recroomyes 5660. 2010. 2.82 5.05e- 3 ## 6 fullbaseyes 2241. 1618. 1.38 1.67e- 1 ## 7 gashwyes 13568. 3411. 3.98 7.93e- 5 ## 8 aircoyes 15578. 1597. 9.75 8.53e-21 ## 9 garagepl 4232. 883. 4.79 2.12e- 6 ## 10 prefareayes 10729. 1753. 6.12 1.81e- 9 For the example above, I’ve also switched from %>% to |>, or else I can’t use the _ placeholder. The advantage of the _ placeholder is that it disambiguates .. So here, the . is a placeholder for all the variables in the dataset, and _ is a placeholder for the dataset. ## 6.3 Diagnostics Diagnostics are useful metrics to assess model fit. You can read some of these diagnostics, such as the $$R^2$$ at the bottom of the summary (when running summary(my_model)), but if you want to do more than simply reading these diagnostics from RStudio, you can put those in a data.frame too, using broom::glance(): glance(model3) ## # A tibble: 1 × 12 ## r.squared adj.r.…¹ sigma stati…² p.value df logLik AIC BIC devia…³ ## <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> ## 1 0.673 0.666 15423. 100. 6.18e-122 11 -6034. 12094. 12150. 1.27e11 ## # … with 2 more variables: df.residual <int>, nobs <int>, and abbreviated ## # variable names ¹​adj.r.squared, ²​statistic, ³​deviance You can also plot the usual diagnostics plots using ggfortify::autoplot() which uses the {ggplot2} package under the hood: library(ggfortify) autoplot(model3, which = 1:6) + theme_minimal() which=1:6 is an additional option that shows you all the diagnostics plot. If you omit this option, you will only get 4 of them. You can also get the residuals of the regression in two ways; either you grab them directly from the model fit: resi3 <- residuals(model3) or you can augment the original data with a residuals column, using broom::augment(): housing_aug <- augment(model3) Let’s take a look at housing_aug: glimpse(housing_aug) ## Rows: 546 ## Columns: 18 ##$ price      <dbl> 42000, 38500, 49500, 60500, 61000, 66000, 66000, 69000, 838…
## $lotsize <dbl> 5850, 4000, 3060, 6650, 6360, 4160, 3880, 4160, 4800, 5500,… ##$ bedrooms   <dbl> 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 2, 3, 4, 1, 2,…
## $bathrms <dbl> 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2,… ##$ stories    <dbl> 2, 1, 1, 2, 1, 1, 2, 3, 1, 4, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1,…
## $driveway <fct> yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, yes, no, … ##$ recroom    <fct> no, no, no, yes, no, yes, no, no, yes, yes, no, no, no, no,…
## $fullbase <fct> yes, no, no, no, no, yes, yes, no, yes, no, yes, no, no, no… ##$ gashw      <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no, no,…
## $airco <fct> no, no, no, no, no, yes, no, no, no, yes, yes, no, no, no, … ##$ garagepl   <dbl> 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 3, 0, 0, 0, 0, 0, 1, 0, 0, 1,…
## $prefarea <fct> no, no, no, no, no, no, no, no, no, no, no, no, no, no, no,… ##$ .fitted    <dbl> 66037.98, 41391.15, 39889.63, 63689.09, 49760.43, 66387.12,…
## $.resid <dbl> -24037.9757, -2891.1515, 9610.3699, -3189.0873, 11239.5735,… ##$ .hat       <dbl> 0.013477335, 0.008316321, 0.009893730, 0.021510891, 0.01033…
## $.sigma <dbl> 15402.01, 15437.14, 15431.98, 15437.02, 15429.89, 15437.64,… ##$ .cooksd    <dbl> 2.803214e-03, 2.476265e-05, 3.265481e-04, 8.004787e-05, 4.6…
## $.std.resid <dbl> -1.56917096, -0.18823924, 0.62621736, -0.20903274, 0.732539… A few columns have been added to the original data, among them .resid which contains the residuals. Let’s plot them: ggplot(housing_aug) + geom_density(aes(.resid)) Fitted values are also added to the original data, under the variable .fitted. It would also have been possible to get the fitted values with: fit3 <- fitted(model3) but I prefer using augment(), because the columns get merged to the original data, which then makes it easier to find specific individuals, for example, you might want to know for which housing units the model underestimates the price: total_pos <- housing_aug %>% filter(.resid > 0) %>% summarise(total = n()) %>% pull(total) we find 261 individuals where the residuals are positive. It is also easier to extract outliers: housing_aug %>% mutate(prank = cume_dist(.cooksd)) %>% filter(prank > 0.99) %>% glimpse() ## Rows: 6 ## Columns: 19 ##$ price      <dbl> 163000, 125000, 132000, 175000, 190000, 174500
## $lotsize <dbl> 7420, 4320, 3500, 9960, 7420, 7500 ##$ bedrooms   <dbl> 4, 3, 4, 3, 4, 4
## $bathrms <dbl> 1, 1, 2, 2, 2, 2 ##$ stories    <dbl> 2, 2, 2, 2, 3, 2
## $driveway <fct> yes, yes, yes, yes, yes, yes ##$ recroom    <fct> yes, no, no, no, no, no
## $fullbase <fct> yes, yes, no, yes, no, yes ##$ gashw      <fct> no, yes, yes, no, no, no
## $airco <fct> yes, no, no, no, yes, yes ##$ garagepl   <dbl> 2, 2, 2, 2, 2, 3
## $prefarea <fct> no, no, no, yes, yes, yes ##$ .fitted    <dbl> 94826.68, 77688.37, 85495.58, 108563.18, 115125.03, 118549.…
## $.resid <dbl> 68173.32, 47311.63, 46504.42, 66436.82, 74874.97, 55951.00 ##$ .hat       <dbl> 0.02671105, 0.05303793, 0.05282929, 0.02819317, 0.02008141,…
## $.sigma <dbl> 15144.70, 15293.34, 15298.27, 15159.14, 15085.99, 15240.66 ##$ .cooksd    <dbl> 0.04590995, 0.04637969, 0.04461464, 0.04616068, 0.04107317,…
## $.std.resid <dbl> 4.480428, 3.152300, 3.098176, 4.369631, 4.904193, 3.679815 ##$ prank      <dbl> 0.9963370, 1.0000000, 0.9945055, 0.9981685, 0.9926740, 0.99…

prank is a variable I created with cume_dist() which is a dplyr function that returns the proportion of all values less than or equal to the current rank. For example:

example <- c(5, 4.6, 2, 1, 0.8, 0, -1)
cume_dist(example)
##  1.0000000 0.8571429 0.7142857 0.5714286 0.4285714 0.2857143 0.1428571

by filtering prank > 0.99 we get the top 1% of outliers according to Cook’s distance.

## 6.4 Interpreting models

Model interpretation is essential in the social sciences, but it is also getting very important in machine learning. As usual, the terminology is different; in machine learning, we speak about explainability. There is a very important aspect that one has to understand when it comes to interpretability/explainability: classical, parametric models, and black-box models. This is very well explained in Breiman (2001), an absolute must read (link to paper, in PDF format: click here). The gist of the paper is that there are two cultures of statistical modeling; one culture relies on modeling the data generating process, for instance, by considering that a variable y (independent variable, or target) is a linear combination of input variables x (dependent variables, or features) plus some noise. The other culture uses complex algorithms (random forests, neural networks) to model the relationship between y and x. The author argues that most statisticians have relied for too long on modeling data generating processes and do not use all the potential offered by these complex algorithms. I think that a lot of things have changed since then, and that nowadays any practitioner that uses data is open to use any type of model or algorithm, as long as it does the job. However, the paper is very interesting, and the discussion on trade-off between simplicity of the model and interpretability/explainability is still relevant today.

In this section, I will explain how one can go about interpreting or explaining models from these two cultures.

Also, it is important to note here that the discussion that will follow will be heavily influenced by my econometrics background. I will focus on marginal effects as way to interpret parametric models (models from the first culture described above), but depending on the field, practitioners might use something else (for instance by computing odds ratios in a logistic regression).

I will start by interpretability of classical statistical models.

### 6.4.1 Marginal effects

If one wants to know the effect of variable x on the dependent variable y, so-called marginal effects have to be computed. This is easily done in R with the {marginaleffects} package. Formally, marginal effects are the partial derivative of the regression equation with respect to the variable we want to look at.

library(marginaleffects)

effects_model3 <- marginaleffects(model3)

summary(effects_model3)
##        Term Contrast    Effect Std. Error z value   Pr(>|z|)    2.5 %    97.5 %
## 1   lotsize    dY/dX     3.546     0.3503  10.124 < 2.22e-16     2.86     4.233
## 2  bedrooms    dY/dX  1832.003  1047.0056   1.750 0.08016056  -220.09  3884.097
## 3   bathrms    dY/dX 14335.558  1489.9557   9.621 < 2.22e-16 11415.30 17255.818
## 4   stories    dY/dX  6556.946   925.2943   7.086 1.3771e-12  4743.40  8370.489
## 5  driveway yes - no  6687.779  2045.2459   3.270 0.00107580  2679.17 10696.387
## 6   recroom yes - no  4511.284  1899.9577   2.374 0.01757689   787.44  8235.132
## 7  fullbase yes - no  5452.386  1588.0239   3.433 0.00059597  2339.92  8564.855
## 8     gashw yes - no 12831.406  3217.5970   3.988 6.6665e-05  6525.03 19137.781
## 9     airco yes - no 12632.890  1555.0211   8.124 4.5131e-16  9585.11 15680.676
## 10 garagepl    dY/dX  4244.829   840.5965   5.050 4.4231e-07  2597.29  5892.368
## 11 prefarea yes - no  9369.513  1669.0906   5.614 1.9822e-08  6098.16 12640.871
##
## Model type:  lm
## Prediction type:  response

Let’s go through this: summary(effects_model3) shows the average marginal effects for each of the dependent variables that were used in model3. The way to interpret them is as follows: everything else held constant (often you’ll read the Latin ceteris paribus for this), a unit increase in lotize increases the price by 3.546 units, on average. The everything held constant part is crucial; with marginal effects, you’re looking at just the effect of one variable at a time. For discrete variables, like driveway, this is simpler: imagine you have two equal houses, exactly the same house, one has a driveway and the other doesn’t. The one with the driveway is 6687 units more expensive, on average.

Now it turns out that in the case of a linear model, the average marginal effects are exactly equal to the coefficients. Just compare summary(model3) to effects_model3 to see (and remember, I told you that marginal effects were the partial derivative of the regression equation with respect to the variable of interest. So the derivative of $$\alpha*X_1 + ....$$ with respect to $$X_1$$ will be $$\alpha$$). But in the case of a more complex, non-linear model, this is not so obvious. This is where {marginaleffects} will make your life much easier.

It is also possible to plot the results:

plot(effects_model3) effects_model3 is a data frame containing the effects for each house in the data set. For example, let’s take a look at the first house:

effects_model3 %>%
filter(rowid == 1)
##    rowid     type     term contrast         dydx    std.error statistic
## 1      1 response  lotsize    dY/dX     3.546303    0.3502195 10.125944
## 2      1 response bedrooms    dY/dX  1832.003466 1046.1608842  1.751168
## 3      1 response  bathrms    dY/dX 14335.558468 1490.4827945  9.618064
## 4      1 response  stories    dY/dX  6556.945711  925.4764870  7.084940
## 5      1 response driveway yes - no  6687.778890 2045.2460319  3.269914
## 6      1 response  recroom yes - no  4511.283826 1899.9577182  2.374413
## 7      1 response fullbase yes - no  5452.385539 1588.0237538  3.433441
## 8      1 response    gashw yes - no 12831.406266 3217.5971931  3.987885
## 9      1 response    airco yes - no 12632.890405 1555.0207045  8.123937
## 10     1 response garagepl    dY/dX  4244.829004  840.8930857  5.048001
## 11     1 response prefarea yes - no  9369.513239 1669.0904968  5.613544
##         p.value     conf.low    conf.high predicted predicted_hi predicted_lo
## 1  4.238689e-24     2.859885     4.232721  66037.98     66043.14     66037.98
## 2  7.991698e-02  -218.434189  3882.441121  66037.98     66038.89     66037.98
## 3  6.708200e-22 11414.265872 17256.851065  66037.98     66042.28     66037.98
## 4  1.391042e-12  4743.045128  8370.846295  66037.98     66039.94     66037.98
## 5  1.075801e-03  2679.170328 10696.387452  66037.98     66037.98     59350.20
## 6  1.757689e-02   787.435126  8235.132526  66037.98     70549.26     66037.98
## 7  5.959723e-04  2339.916175  8564.854903  66037.98     66037.98     60585.59
## 8  6.666508e-05  6525.031651 19137.780882  66037.98     78869.38     66037.98
## 9  4.512997e-16  9585.105829 15680.674981  66037.98     78670.87     66037.98
## 10 4.464572e-07  2596.708842  5892.949167  66037.98     66039.25     66037.98
## 11 1.982240e-08  6098.155978 12640.870499  66037.98     75407.49     66037.98
##    price lotsize bedrooms bathrms stories driveway recroom fullbase gashw airco
## 1  42000    5850        3       1       2      yes      no      yes    no    no
## 2  42000    5850        3       1       2      yes      no      yes    no    no
## 3  42000    5850        3       1       2      yes      no      yes    no    no
## 4  42000    5850        3       1       2      yes      no      yes    no    no
## 5  42000    5850        3       1       2      yes      no      yes    no    no
## 6  42000    5850        3       1       2      yes      no      yes    no    no
## 7  42000    5850        3       1       2      yes      no      yes    no    no
## 8  42000    5850        3       1       2      yes      no      yes    no    no
## 9  42000    5850        3       1       2      yes      no      yes    no    no
## 10 42000    5850        3       1       2      yes      no      yes    no    no
## 11 42000    5850        3       1       2      yes      no      yes    no    no
##    garagepl prefarea    eps
## 1         1       no 1.4550
## 2         1       no 0.0005
## 3         1       no 0.0003
## 4         1       no 0.0003
## 5         1       no     NA
## 6         1       no     NA
## 7         1       no     NA
## 8         1       no     NA
## 9         1       no     NA
## 10        1       no 0.0003
## 11        1       no     NA

rowid is column that identifies the houses in the original data set, so rowid == 1 filters out the first house. This shows you the marginal effects (column dydx computed for this house; but remember, since we’re dealing with a linear model, the values of the marginal effects are constant. If you don’t see the point of this discussion, don’t fret, the next example should make things clearer.

Let’s estimate a logit model and compute the marginal effects. You might know logit models as logistic regression. Logit models can be estimated using the glm() function, which stands for generalized linear models.

As an example, we are going to use the Participation data, also from the {Ecdat} package:

data(Participation)
?Particpation
Participation              package:Ecdat               R Documentation

Labor Force Participation

Description:

a cross-section

_number of observations_ : 872

_observation_ : individuals

_country_ : Switzerland

Usage:

data(Participation)

Format:

A dataframe containing :

lfp labour force participation ?

lnnlinc the log of nonlabour income

age age in years divided by 10

educ years of formal education

nyc the number of young children (younger than 7)

noc number of older children

foreign foreigner ?

Source:

Gerfin, Michael (1996) “Parametric and semiparametric estimation
of the binary response”, _Journal of Applied Econometrics_,
*11(3)*, 321-340.

References:

Davidson, R.  and James G.  MacKinnon (2004) _Econometric Theory
and Methods_, New York, Oxford University Press, <URL:
http://www.econ.queensu.ca/ETM/>, chapter 11.

Journal of Applied Econometrics data archive : <URL:
http://qed.econ.queensu.ca/jae/>.

‘Index.Source’, ‘Index.Economics’, ‘Index.Econometrics’,
‘Index.Observations’

The variable of interest is lfp: whether the individual participates in the labour force or not. To know which variables are relevant in the decision to participate in the labour force, one could train a logit model, using glm():

logit_participation <- glm(lfp ~ ., data = Participation, family = "binomial")

broom::tidy(logit_participation)
## # A tibble: 7 × 5
##   term        estimate std.error statistic  p.value
##   <chr>          <dbl>     <dbl>     <dbl>    <dbl>
## 1 (Intercept)  10.4       2.17       4.79  1.69e- 6
## 2 lnnlinc      -0.815     0.206     -3.97  7.31e- 5
## 3 age          -0.510     0.0905    -5.64  1.72e- 8
## 4 educ          0.0317    0.0290     1.09  2.75e- 1
## 5 nyc          -1.33      0.180     -7.39  1.51e-13
## 6 noc          -0.0220    0.0738    -0.298 7.66e- 1
## 7 foreignyes    1.31      0.200      6.56  5.38e-11

From the results above, one can only interpret the sign of the coefficients. To know how much a variable influences the labour force participation, one has to use marginaleffects():

effects_logit_participation <- marginaleffects(logit_participation)

summary(effects_logit_participation)
##      Term Contrast    Effect Std. Error z value   Pr(>|z|)     2.5 %   97.5 %
## 1 lnnlinc    dY/dX -0.169940    0.04151 -4.0939 4.2416e-05 -0.251300 -0.08858
## 2     age    dY/dX -0.106407    0.01759 -6.0492 1.4560e-09 -0.140884 -0.07193
## 3    educ    dY/dX  0.006616    0.00604  1.0954    0.27335 -0.005222  0.01845
## 4     nyc    dY/dX -0.277463    0.03325 -8.3436 < 2.22e-16 -0.342642 -0.21229
## 5     noc    dY/dX -0.004584    0.01538 -0.2981    0.76563 -0.034725  0.02556
## 6 foreign yes - no  0.283377    0.03984  7.1129 1.1361e-12  0.205292  0.36146
##
## Model type:  glm
## Prediction type:  response

As you can see, the average marginal effects here are not equal to the estimated coefficients of the model. Let’s take a look at the first row of the data:

Participation[1, ]
##   lfp lnnlinc age educ nyc noc foreign
## 1  no 10.7875   3    8   1   1      no

and let’s now look at rowid == 1 in the marginal effects data frame:

effects_logit_participation %>%
filter(rowid == 1)
##   rowid     type    term contrast         dydx   std.error  statistic
## 1     1 response lnnlinc    dY/dX -0.156661756 0.038522800 -4.0667282
## 2     1 response     age    dY/dX -0.098097148 0.020123709 -4.8747052
## 3     1 response    educ    dY/dX  0.006099266 0.005367036  1.1364310
## 4     1 response     nyc    dY/dX -0.255784406 0.029367783 -8.7096942
## 5     1 response     noc    dY/dX -0.004226368 0.014167283 -0.2983189
## 6     1 response foreign yes - no  0.305630005 0.045174828  6.7654935
##        p.value     conf.low   conf.high predicted predicted_hi predicted_lo lfp
## 1 4.767780e-05 -0.232165056 -0.08115846 0.2596523    0.2595710    0.2596523  no
## 2 1.089711e-06 -0.137538892 -0.05865540 0.2596523    0.2596111    0.2596523  no
## 3 2.557762e-01 -0.004419931  0.01661846 0.2596523    0.2596645    0.2596523  no
## 4 3.046958e-18 -0.313344203 -0.19822461 0.2596523    0.2595755    0.2596523  no
## 5 7.654598e-01 -0.031993732  0.02354100 0.2596523    0.2596497    0.2596523  no
## 6 1.328556e-11  0.217088969  0.39417104 0.2596523    0.5652823    0.2596523  no
##   lnnlinc age educ nyc noc foreign          eps
## 1 10.7875   3    8   1   1      no 0.0005188749
## 2 10.7875   3    8   1   1      no 0.0004200000
## 3 10.7875   3    8   1   1      no 0.0020000000
## 4 10.7875   3    8   1   1      no 0.0003000000
## 5 10.7875   3    8   1   1      no 0.0006000000
## 6 10.7875   3    8   1   1      no           NA

Let’s focus on the first row, where term is lnnlinc. What we see here is the effect of an infinitesimal increase in the variable lnnlinc on the participation, for an individual who has the following characteristics: lnnlinc = 10.7875, age = 3, educ = 8, nyc = 1, noc = 1 and foreign = no, which are the characteristics of this first individual in our data.

So let’s look at the value of dydx for every individual:

dydx_lnnlinc <- effects_logit_participation %>%
filter(term == "lnnlinc")

head(dydx_lnnlinc)
##   rowid     type    term contrast        dydx  std.error statistic      p.value
## 1     1 response lnnlinc    dY/dX -0.15666176 0.03852280 -4.066728 4.767780e-05
## 2     2 response lnnlinc    dY/dX -0.20013939 0.05124543 -3.905507 9.402813e-05
## 3     3 response lnnlinc    dY/dX -0.18493932 0.04319729 -4.281271 1.858287e-05
## 4     4 response lnnlinc    dY/dX -0.05376281 0.01586468 -3.388837 7.018964e-04
## 5     5 response lnnlinc    dY/dX -0.18709356 0.04502973 -4.154890 3.254439e-05
## 6     6 response lnnlinc    dY/dX -0.19586185 0.04782143 -4.095692 4.209096e-05
##      conf.low   conf.high  predicted predicted_hi predicted_lo lfp  lnnlinc age
## 1 -0.23216506 -0.08115846 0.25965227   0.25957098   0.25965227  no 10.78750 3.0
## 2 -0.30057859 -0.09970018 0.43340025   0.43329640   0.43340025 yes 10.52425 4.5
## 3 -0.26960445 -0.10027418 0.34808777   0.34799181   0.34808777  no 10.96858 4.6
## 4 -0.08485701 -0.02266862 0.07101902   0.07099113   0.07101902  no 11.10500 3.1
## 5 -0.27535020 -0.09883692 0.35704926   0.35695218   0.35704926  no 11.10847 4.4
## 6 -0.28959014 -0.10213356 0.40160949   0.40150786   0.40160949 yes 11.02825 4.2
##   educ nyc noc foreign          eps
## 1    8   1   1      no 0.0005188749
## 2    8   0   1      no 0.0005188749
## 3    9   0   0      no 0.0005188749
## 4   11   2   0      no 0.0005188749
## 5   12   0   2      no 0.0005188749
## 6   12   0   1      no 0.0005188749

dydx_lnnlinc is a data frame with all individual marginal effect for the variable lnnlinc. What if we compute the mean of this column?

dydx_lnnlinc %>%
summarise(mean(dydx))
##   mean(dydx)
## 1 -0.1699405

Let’s compare this to the average marginal effects:

summary(effects_logit_participation)
##      Term Contrast    Effect Std. Error z value   Pr(>|z|)     2.5 %   97.5 %
## 1 lnnlinc    dY/dX -0.169940    0.04151 -4.0939 4.2416e-05 -0.251300 -0.08858
## 2     age    dY/dX -0.106407    0.01759 -6.0492 1.4560e-09 -0.140884 -0.07193
## 3    educ    dY/dX  0.006616    0.00604  1.0954    0.27335 -0.005222  0.01845
## 4     nyc    dY/dX -0.277463    0.03325 -8.3436 < 2.22e-16 -0.342642 -0.21229
## 5     noc    dY/dX -0.004584    0.01538 -0.2981    0.76563 -0.034725  0.02556
## 6 foreign yes - no  0.283377    0.03984  7.1129 1.1361e-12  0.205292  0.36146
##
## Model type:  glm
## Prediction type:  response

Yep, it’s the same! This is why we speak of average marginal effects. Now that we know why these are called average marginal effects, let’s go back to interpreting them. This time, let’s plot them, because why not:

plot(effects_logit_participation) So an infinitesimal increase, in say, non-labour income (lnnlinc) of 0.001 is associated with a decrease of the probability of labour force participation by 0.001*17 percentage points.

This is just scratching the surface of interpreting these kinds of models. There are many more types of effects that you can compute and look at. I highly recommend you read the documentation of {marginaleffects} which you can find here. The author of the package, Vincent Arel-Bundock writes a lot of very helpful documentation for his packages, so if model interpretation is important for your job, definitely take a look.

### 6.4.2 Explainability of black-box models

Just read Christoph Molnar’s Interpretable Machine Learning.

Seriously, I cannot add anything meaningful to it. His book is brilliant.

## 6.5 Comparing models

Consider this section more as an illustration of what is possible with the knowledge you acquired at this point. Imagine that the task at hand is to compare two models. We would like to select the one which has the best fit to the data. Let’s first estimate another model on the same data; prices are only positive, so a linear regression might not be the best model, because the model could predict negative prices. Let’s look at the distribution of prices:

ggplot(Housing) +
geom_density(aes(price)) it looks like modeling the log of price might provide a better fit:

model_log <- lm(log(price) ~ ., data = Housing)

result_log <- broom::tidy(model_log)

print(result_log)
## # A tibble: 12 × 5
##    term          estimate  std.error statistic  p.value
##    <chr>            <dbl>      <dbl>     <dbl>    <dbl>
##  1 (Intercept) 10.0       0.0472        212.   0
##  2 lotsize      0.0000506 0.00000485     10.4  2.91e-23
##  3 bedrooms     0.0340    0.0145          2.34 1.94e- 2
##  4 bathrms      0.168     0.0206          8.13 3.10e-15
##  5 stories      0.0923    0.0128          7.20 2.10e-12
##  6 drivewayyes  0.131     0.0283          4.61 5.04e- 6
##  7 recroomyes   0.0735    0.0263          2.79 5.42e- 3
##  8 fullbaseyes  0.0994    0.0220          4.52 7.72e- 6
##  9 gashwyes     0.178     0.0446          4.00 7.22e- 5
## 10 aircoyes     0.178     0.0215          8.26 1.14e-15
## 11 garagepl     0.0508    0.0116          4.36 1.58e- 5
## 12 prefareayes  0.127     0.0231          5.50 6.02e- 8

Let’s take a look at the diagnostics:

glance(model_log)
## # A tibble: 1 × 12
##   r.squared adj.r.squ…¹ sigma stati…²   p.value    df logLik   AIC   BIC devia…³
##       <dbl>       <dbl> <dbl>   <dbl>     <dbl> <dbl>  <dbl> <dbl> <dbl>   <dbl>
## 1     0.677       0.670 0.214    102. 3.67e-123    11   73.9 -122. -65.8    24.4
## # … with 2 more variables: df.residual <int>, nobs <int>, and abbreviated
## #   variable names ¹​adj.r.squared, ²​statistic, ³​deviance

Let’s compare these to the ones from the previous model:

diag_lm <- glance(model3)

diag_lm <- diag_lm %>%
mutate(model = "lin-lin model")

diag_log <- glance(model_log)

diag_log  <- diag_log %>%
mutate(model = "log-lin model")

diagnostics_models <- full_join(diag_lm, diag_log) %>%
select(model, everything()) # put the model column first
## Joining, by = c("r.squared", "adj.r.squared", "sigma", "statistic", "p.value", "df", "logLik", "AIC", "BIC",
## "deviance", "df.residual", "nobs", "model")
print(diagnostics_models)
## # A tibble: 2 × 13
##   model   r.squ…¹ adj.r…²   sigma stati…³   p.value    df  logLik    AIC     BIC
##   <chr>     <dbl>   <dbl>   <dbl>   <dbl>     <dbl> <dbl>   <dbl>  <dbl>   <dbl>
## 1 lin-li…   0.673   0.666 1.54e+4    100. 6.18e-122    11 -6034.  12094. 12150.
## 2 log-li…   0.677   0.670 2.14e-1    102. 3.67e-123    11    73.9  -122.   -65.8
## # … with 3 more variables: deviance <dbl>, df.residual <int>, nobs <int>, and
## #   abbreviated variable names ¹​r.squared, ²​adj.r.squared, ³​statistic

I saved the diagnostics in two different data.frame objects using the glance() function and added a model column to indicate which model the diagnostics come from. Then I merged both datasets using full_join(), a {dplyr} function. Using this approach, we can easily build a data frame with the diagnostics of several models and compare them. The model using the logarithm of prices has lower AIC and BIC (and this higher likelihood), so if you’re worried about selecting the model with the better fit to the data, you’d go for this model.

## 6.6 Using a model for prediction

Once you estimated a model, you might want to use it for prediction. This is easily done using the predict() function that works with most models. Prediction is also useful as a way to test the accuracy of your model: split your data into a training set (used for training) and a testing set (used for the pseudo-prediction) and see if your model overfits the data. We are going to see how to do that in a later section; for now, let’s just get acquainted with predict() and other functions. I insist, keep in mind that this section is only to get acquainted with these functions. We are going to explore prediction, overfitting and tuning of models in a later section.

Let’s go back to the models we trained in the previous section, model3 and model_log. Let’s also take a subsample of data, which we will be using for prediction:

set.seed(1234)

pred_set <- Housing %>%
sample_n(20)

In order to always get the same pred_set, I set the random seed first. Let’s take a look at the data:

print(pred_set)
##      price lotsize bedrooms bathrms stories driveway recroom fullbase gashw
## 284  45000    6750        2       1       1      yes      no       no    no
## 101  57000    4500        3       2       2       no      no      yes    no
## 400  85000    7231        3       1       2      yes     yes      yes    no
## 98   59900    8250        3       1       1      yes      no      yes    no
## 103 125000    4320        3       1       2      yes      no      yes   yes
## 326  99000    8880        3       2       2      yes      no      yes    no
## 79   55000    3180        2       2       1      yes      no      yes    no
## 270  59000    4632        4       1       2      yes      no       no    no
## 382 112500    6550        3       1       2      yes      no      yes    no
## 184  63900    3510        3       1       2      yes      no       no    no
## 4    60500    6650        3       1       2      yes     yes       no    no
## 212  42000    2700        2       1       1       no      no       no    no
## 195  33000    3180        2       1       1      yes      no       no    no
## 511  70000    4646        3       1       2      yes     yes      yes    no
## 479  88000    5450        4       2       1      yes      no      yes    no
## 510  64000    4040        3       1       2      yes      no       no    no
## 424  62900    2880        3       1       2      yes      no       no    no
## 379  84000    7160        3       1       1      yes      no      yes    no
## 108  58500    3680        3       2       2      yes      no       no    no
## 131  35000    4840        2       1       2      yes      no       no    no
##     airco garagepl prefarea
## 284    no        0       no
## 101   yes        0       no
## 400   yes        0      yes
## 98     no        3       no
## 103    no        2       no
## 326   yes        1       no
## 79     no        2       no
## 270   yes        0       no
## 382   yes        0      yes
## 184    no        0       no
## 4      no        0       no
## 212    no        0       no
## 195    no        0       no
## 511    no        2       no
## 479   yes        0      yes
## 510    no        1       no
## 424    no        0      yes
## 379    no        2      yes
## 108    no        0       no
## 131    no        0       no

If we wish to use it for prediction, this is easily done with predict():

predict(model3, pred_set)
##       284       101       400        98       103       326        79       270
##  51143.48  77286.31  93204.28  76481.82  77688.37 103751.72  66760.79  66486.26
##       382       184         4       212       195       511       479       510
##  86277.96  48042.41  63689.09  30093.18  38483.18  70524.34  91987.65  54166.78
##       424       379       108       131
##  55177.75  77741.03  62980.84  50926.99

This returns a vector of predicted prices. This can then be used to compute the Root Mean Squared Error for instance. Let’s do it within a tidyverse pipeline:

rmse <- pred_set %>%
mutate(predictions = predict(model3, .)) %>%
summarise(sqrt(sum(predictions - price)**2/n()))

The root mean square error of model3 is 3646.0817347.

I also used the n() function which returns the number of observations in a group (or all the observations, if the data is not grouped). Let’s compare model3 ’s RMSE with the one from model_log:

rmse2 <- pred_set %>%
mutate(predictions = exp(predict(model_log, .))) %>%
summarise(sqrt(sum(predictions - price)**2/n()))

Don’t forget to exponentiate the predictions, remember you’re dealing with a log-linear model! model_log’s RMSE is 1.2125133^{4} which is lower than model3’s. However, keep in mind that the model was trained on the whole data, and then the prediction quality was assessed using a subsample of the data the model was trained on… so actually we can’t really say if model_log’s predictions are very useful. Of course, this is the same for model3. In a later section we are going to learn how to do cross validation to avoid this issue.

Just as a side note, notice that I had to copy and paste basically the same lines twice to compute the predictions for both models. That’s not much, but if I wanted to compare 10 models, copy and paste mistakes could have sneaked in. Instead, it would have been nice to have a function that computes the RMSE and then use it on my models. We are going to learn how to write our own functions and use it just like if it was another built-in R function.

## 6.7 Beyond linear regression

R has a lot of other built-in functions for regression, such as glm() (for Generalized Linear Models) and nls() for (for Nonlinear Least Squares). There are also functions and additional packages for time series, panel data, machine learning, bayesian and nonparametric methods. Presenting everything here would take too much space, and would be pretty useless as you can find whatever you need using an internet search engine. What you have learned until now is quite general and should work on many type of models. To help you out, here is a list of methods and the recommended packages that you can use:

Model Package Quick example
Robust Linear Regression MASS rlm(y ~ x, data = mydata)
Nonlinear Least Squares stats2 nls(y ~ x1 / (1 + x2), data = mydata)3
Logit stats glm(y ~ x, data = mydata, family = "binomial")
Probit stats glm(y ~ x, data = mydata, family = binomial(link = "probit"))
K-Means stats kmeans(data, n)4
PCA stats prcomp(data, scale = TRUE, center = TRUE)5
Multinomial Logit mlogit Requires several steps of data pre-processing and formula definition, refer to the Vignette for more details.
Cox PH survival coxph(Surv(y_time, y_status) ~ x, data = mydata)6
Time series Several, depending on your needs. Time series in R is a vast subject that would require a very thick book to cover. You can get started with the following series of blog articles, Tidy time-series, part 1, Tidy time-series, part 2, Tidy time-series, part 3 and Tidy time-series, part 4
Panel data plm plm(y ~ x, data = mydata, model = "within|random")
Machine learning Several, depending on your needs. R is a very popular programming language for machine learning. This book is a must read if you need to do machine learning with R.
Nonparametric regression np Several functions and options available, refer to the Vignette for more details.

This table is far from being complete. Should you be a Bayesian, you’d want to look at packages such as {rstan}, which uses STAN, an external piece of software that must be installed on your system. It is also possible to train models using Bayesian inference without the need of external tools, with the {bayesm} package which estimates the usual micro-econometric models. There really are a lot of packages available for Bayesian inference, and you can find them all in the related CRAN Task View.

## 6.8 Hyper-parameters

Hyper-parameters are parameters of the model that cannot be directly learned from the data. A linear regression does not have any hyper-parameters, but a random forest for instance has several. You might have heard of ridge regression, lasso and elasticnet. These are extensions of linear models that avoid over-fitting by penalizing large models. These extensions of the linear regression have hyper-parameters that the practitioner has to tune. There are several ways one can tune these parameters, for example, by doing a grid-search, or a random search over the grid or using more elaborate methods. To introduce hyper-parameters, let’s get to know ridge regression, also called Tikhonov regularization.

### 6.8.1 Ridge regression

Ridge regression is used when the data you are working with has a lot of explanatory variables, or when there is a risk that a simple linear regression might overfit to the training data, because, for example, your explanatory variables are collinear. If you are training a linear model and then you notice that it generalizes very badly to new, unseen data, it is very likely that the linear model you trained overfit the data. In this case, ridge regression might prove useful. The way ridge regression works might seem counter-intuititive; it boils down to fitting a worse model to the training data, but in return, this worse model will generalize better to new data.

The closed form solution of the ordinary least squares estimator is defined as:

$\widehat{\beta} = (X'X)^{-1}X'Y$

where $$X$$ is the design matrix (the matrix made up of the explanatory variables) and $$Y$$ is the dependent variable. For ridge regression, this closed form solution changes a little bit:

$\widehat{\beta} = (X'X + \lambda I_p)^{-1}X'Y$

where $$\lambda \in \mathbb{R}$$ is an hyper-parameter and $$I_p$$ is the identity matrix of dimension $$p$$ ($$p$$ is the number of explanatory variables). This formula above is the closed form solution to the following optimisation program:

$\sum_{i=1}^n \left(y_i - \sum_{j=1}^px_{ij}\beta_j\right)^2$

such that:

$\sum_{j=1}^p(\beta_j)^2 < c$

for any strictly positive $$c$$.

The glmnet() function from the {glmnet} package can be used for ridge regression, by setting the alpha argument to 0 (setting it to 1 would do LASSO, and setting it to a number between 0 and 1 would do elasticnet). But in order to compare linear regression and ridge regression, let me first divide the data into a training set and a testing set:

index <- 1:nrow(Housing)

set.seed(12345)
train_index <- sample(index, round(0.90*nrow(Housing)), replace = FALSE)

test_index <- setdiff(index, train_index)

train_x <- Housing[train_index, ] %>%
select(-price)

train_y <- Housing[train_index, ] %>%
pull(price)

test_x <- Housing[test_index, ] %>%
select(-price)

test_y <- Housing[test_index, ] %>%
pull(price)

I do the train/test split this way, because glmnet() requires a design matrix as input, and not a formula. Design matrices can be created using the model.matrix() function:

library("glmnet")

train_matrix <- model.matrix(train_y ~ ., data = train_x)

test_matrix <- model.matrix(test_y ~ ., data = test_x)

Let’s now run a linear regression, by setting the penalty to 0:

model_lm_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 0)

The model above provides the same result as a linear regression, because I set lambda to 0. Let’s compare the coefficients between the two:

coef(model_lm_ridge)
## 13 x 1 sparse Matrix of class "dgCMatrix"
##                       s0
## (Intercept) -2667.542863
## (Intercept)     .
## lotsize         3.397596
## bedrooms     2081.087654
## bathrms     13294.192823
## stories      6400.454580
## drivewayyes  6530.644895
## recroomyes   5389.856794
## fullbaseyes  4899.099463
## gashwyes    12575.611265
## aircoyes    13078.144146
## garagepl     4155.249461
## prefareayes 10260.781753

and now the coefficients of the linear regression (because I provide a design matrix, I have to use lm.fit() instead of lm() which requires a formula, not a matrix.)

coef(lm.fit(x = train_matrix, y = train_y))
##  (Intercept)      lotsize     bedrooms      bathrms      stories  drivewayyes
## -2667.052098     3.397629  2081.344118 13293.707725  6400.416730  6529.972544
##   recroomyes  fullbaseyes     gashwyes     aircoyes     garagepl  prefareayes
##  5388.871137  4899.024787 12575.970220 13077.988867  4155.269629 10261.056772

as you can see, the coefficients are the same. Let’s compute the RMSE for the unpenalized linear regression:

preds_lm <- predict(model_lm_ridge, test_matrix)

rmse_lm <- sqrt(mean(preds_lm - test_y)^2)

The RMSE for the linear unpenalized regression is equal to 1731.5553157.

Let’s now run a ridge regression, with lambda equal to 100, and see if the RMSE is smaller:

model_ridge <- glmnet(y = train_y, x = train_matrix, alpha = 0, lambda = 100)

and let’s compute the RMSE again:

preds <- predict(model_ridge, test_matrix)

rmse <- sqrt(mean(preds - test_y)^2)

The RMSE for the linear penalized regression is equal to 1726.7632312, which is smaller than before. But which value of lambda gives smallest RMSE? To find out, one must run model over a grid of lambda values and pick the model with lowest RMSE. This procedure is available in the cv.glmnet() function, which picks the best value for lambda:

best_model <- cv.glmnet(train_matrix, train_y)
# lambda that minimises the MSE
best_model$lambda.min ##  61.42681 According to cv.glmnet() the best value for lambda is 61.4268056. In the next section, we will implement cross validation ourselves, in order to find the hyper-parameters of a random forest. ## 6.9 Training, validating, and testing models Cross-validation is an important procedure which is used to compare models but also to tune the hyper-parameters of a model. In this section, we are going to use several packages from the {tidymodels} collection of packages, namely {recipes}, {rsample} and {parsnip} to train a random forest the tidy way. I will also use {mlrMBO} to tune the hyper-parameters of the random forest. ### 6.9.1 Set up Let’s load the needed packages: library("tidyverse") library("recipes") library("rsample") library("parsnip") library("yardstick") library("brotools") library("mlbench") Load the data which is included in the {mlrbench} package: data("BostonHousing2") I will train a random forest to predict the housing prices, which is the cmedv column: head(BostonHousing2) ## town tract lon lat medv cmedv crim zn indus chas nox ## 1 Nahant 2011 -70.9550 42.2550 24.0 24.0 0.00632 18 2.31 0 0.538 ## 2 Swampscott 2021 -70.9500 42.2875 21.6 21.6 0.02731 0 7.07 0 0.469 ## 3 Swampscott 2022 -70.9360 42.2830 34.7 34.7 0.02729 0 7.07 0 0.469 ## 4 Marblehead 2031 -70.9280 42.2930 33.4 33.4 0.03237 0 2.18 0 0.458 ## 5 Marblehead 2032 -70.9220 42.2980 36.2 36.2 0.06905 0 2.18 0 0.458 ## 6 Marblehead 2033 -70.9165 42.3040 28.7 28.7 0.02985 0 2.18 0 0.458 ## rm age dis rad tax ptratio b lstat ## 1 6.575 65.2 4.0900 1 296 15.3 396.90 4.98 ## 2 6.421 78.9 4.9671 2 242 17.8 396.90 9.14 ## 3 7.185 61.1 4.9671 2 242 17.8 392.83 4.03 ## 4 6.998 45.8 6.0622 3 222 18.7 394.63 2.94 ## 5 7.147 54.2 6.0622 3 222 18.7 396.90 5.33 ## 6 6.430 58.7 6.0622 3 222 18.7 394.12 5.21 Only keep relevant columns: boston <- BostonHousing2 %>% select(-medv, -tract, -lon, -lat) %>% rename(price = cmedv) I remove tract, lat and lon because the information contained in the column town is enough. To train and evaluate the model’s performance, I split the data in two. One data set, called the training set, will be further split into two down below. I won’t touch the second data set, the test set, until the very end, to finally assess the model’s performance. train_test_split <- initial_split(boston, prop = 0.9) housing_train <- training(train_test_split) housing_test <- testing(train_test_split) initial_split(), training() and testing() are functions from the {rsample} package. I will train a random forest on the training data, but the question, is which random forest? Because random forests have several hyper-parameters, and as explained in the intro these hyper-parameters cannot be directly learned from the data, which one should we choose? We could train 6 random forests for instance and compare their performance, but why only 6? Why not 16? In order to find the right hyper-parameters, the practitioner can use values from the literature that seemed to have worked well (like is done in Macro-econometrics) or you can further split the train set into two, create a grid of hyperparameter, train the model on one part of the data for all values of the grid, and compare the predictions of the models on the second part of the data. You then stick with the model that performed the best, for example, the model with lowest RMSE. The thing is, you can’t estimate the true value of the RMSE with only one value. It’s like if you wanted to estimate the height of the population by drawing one single observation from the population. You need a bit more observations. To approach the true value of the RMSE for a give set of hyperparameters, instead of doing one split, let’s do 30. Then we compute the average RMSE, which implies training 30 models for each combination of the values of the hyperparameters. First, let’s split the training data again, using the mc_cv() function from {rsample} package. This function implements Monte Carlo cross-validation: validation_data <- mc_cv(housing_train, prop = 0.9, times = 30) What does validation_data look like? validation_data ## # Monte Carlo cross-validation (0.9/0.1) with 30 resamples ## # A tibble: 30 × 2 ## splits id ## <list> <chr> ## 1 <split [409/46]> Resample01 ## 2 <split [409/46]> Resample02 ## 3 <split [409/46]> Resample03 ## 4 <split [409/46]> Resample04 ## 5 <split [409/46]> Resample05 ## 6 <split [409/46]> Resample06 ## 7 <split [409/46]> Resample07 ## 8 <split [409/46]> Resample08 ## 9 <split [409/46]> Resample09 ## 10 <split [409/46]> Resample10 ## # … with 20 more rows Let’s look further down: validation_data$splits[]
## <Analysis/Assess/Total>
## <409/46/455>

The first value is the number of rows of the first set, the second value of the second, and the third was the original amount of values in the training data, before splitting again.

How should we call these two new data sets? The author of {rsample}, Max Kuhn, talks about the analysis and the assessment sets, and I’m going to use this terminology as well.

Now, in order to continue I need to pre-process the data. I will do this in three steps. The first and the second steps are used to center and scale the numeric variables and the third step converts character and factor variables to dummy variables. This is needed because I will train a random forest, which cannot handle factor variables directly. Let’s define a recipe to do that, and start by pre-processing the testing set. I write a wrapper function around the recipe, because I will need to apply this recipe to various data sets:

simple_recipe <- function(dataset){
recipe(price ~ ., data = dataset) %>%
step_center(all_numeric()) %>%
step_scale(all_numeric()) %>%
step_dummy(all_nominal())
}

We have not learned yet about writing functions, and will do so in the next chapter. However, for now, you only need to know that you can write your own functions, and that these functions can take any arguments you need. In the case of the above function, which we called simple_recipe(), we only need one argument, which is a dataset, and which we called dataset.

Once the recipe is defined, I can use the prep() function, which estimates the parameters from the data which are needed to process the data. For example, for centering, prep() estimates the mean which will then be subtracted from the variables. With bake() the estimates are then applied on the data:

testing_rec <- prep(simple_recipe(housing_test), testing = housing_test)

test_data <- bake(testing_rec, new_data = housing_test)

It is important to split the data before using prep() and bake(), because if not, you will use observations from the test set in the prep() step, and thus introduce knowledge from the test set into the training data. This is called data leakage, and must be avoided. This is why it is necessary to first split the training data into an analysis and an assessment set, and then also pre-process these sets separately. However, the validation_data object cannot now be used with recipe(), because it is not a dataframe. No worries, I simply need to write a function that extracts the analysis and assessment sets from the validation_data object, applies the pre-processing, trains the model, and returns the RMSE. This will be a big function, at the center of the analysis.

But before that, let’s run a simple linear regression, as a benchmark. For the linear regression, I will not use any CV, so let’s pre-process the training set:

trainlm_rec <- prep(simple_recipe(housing_train), testing = housing_train)

trainlm_data <- bake(trainlm_rec, new_data = housing_train)

linreg_model <- lm(price ~ ., data = trainlm_data)

broom::augment(linreg_model, newdata = test_data) %>%
yardstick::rmse(price, .fitted)
## Warning in predict.lm(x, newdata = newdata, na.action = na.pass, ...):
## prediction from a rank-deficient fit may be misleading
## # A tibble: 1 × 3
##   .metric .estimator .estimate
##   <chr>   <chr>          <dbl>
## 1 rmse    standard       0.439

broom::augment() adds the predictions to the test_data in a new column, .fitted. I won’t use this trick with the random forest, because there is no augment() method for random forests from the {ranger} package which I’ll use. I’ll add the predictions to the data myself.

Ok, now let’s go back to the random forest and write the big function:

my_rf <- function(mtry, trees, split, id){

analysis_set <- analysis(split)

analysis_prep <- prep(simple_recipe(analysis_set), training = analysis_set)

analysis_processed <- bake(analysis_prep, new_data = analysis_set)

model <- rand_forest(mode = "regression", mtry = mtry, trees = trees) %>%
set_engine("ranger", importance = 'impurity') %>%
fit(price ~ ., data = analysis_processed)

assessment_set <- assessment(split)

assessment_prep <- prep(simple_recipe(assessment_set), testing = assessment_set)

assessment_processed <- bake(assessment_prep, new_data = assessment_set)

tibble::tibble("id" = id,
"truth" = assessment_processed$price, "prediction" = unlist(predict(model, new_data = assessment_processed))) } The rand_forest() function is available in the {parsnip} package. This package provides an unified interface to a lot of other machine learning packages. This means that instead of having to learn the syntax of range() and randomForest() and, and… you can simply use the rand_forest() function and change the engine argument to the one you want (ranger, randomForest, etc). Let’s try this function: results_example <- map2_df(.x = validation_data$splits,
.y = validation_data$id, ~my_rf(mtry = 3, trees = 200, split = .x, id = .y)) head(results_example) ## # A tibble: 6 × 3 ## id truth prediction ## <chr> <dbl> <dbl> ## 1 Resample01 -0.328 -0.0274 ## 2 Resample01 1.06 0.686 ## 3 Resample01 1.04 0.726 ## 4 Resample01 -0.418 -0.0190 ## 5 Resample01 0.909 0.642 ## 6 Resample01 0.0926 -0.134 I can now compute the RMSE when mtry = 3 and trees = 200: results_example %>% group_by(id) %>% yardstick::rmse(truth, prediction) %>% summarise(mean_rmse = mean(.estimate)) %>% pull ##  0.6305034 The random forest has already lower RMSE than the linear regression. The goal now is to lower this RMSE by tuning the mtry and trees hyperparameters. For this, I will use Bayesian Optimization methods implemented in the {mlrMBO} package. ### 6.9.2 Bayesian hyperparameter optimization I will re-use the code from above, and define a function that does everything from pre-processing to returning the metric I want to minimize by tuning the hyperparameters, the RMSE: tuning <- function(param, validation_data){ mtry <- param trees <- param results <- purrr::map2_df(.x = validation_data$splits,
.y = validation_data$id, ~my_rf(mtry = mtry, trees = trees, split = .x, id = .y)) results %>% group_by(id) %>% yardstick::rmse(truth, prediction) %>% summarise(mean_rmse = mean(.estimate)) %>% pull } This is exactly the code from before, but it now returns the RMSE. Let’s try the function with the values from before: tuning(c(3, 200), validation_data) ##  0.6319843 I now follow the code that can be found in the arxiv paper to run the optimization. A simpler model, called the surrogate model, is used to look for promising points and to evaluate the value of the function at these points. This seems somewhat similar (in spirit) to the Indirect Inference method as described in Gourieroux, Monfort, Renault. If you don’t really get what follows, no worries, it is not really important as such. The idea is simply to look for hyper-parameters in an efficient way, and bayesian optimisation provides this efficient way. However, you could use another method, for example a grid search. This would not change anything to the general approach. So I will not spend too much time explaining what is going on below, as you can read the details in the paper cited above as well as the package’s documentation. The focus here is not on this particular method, but rather showing you how you can use various packages to solve a data science problem. Let’s first load the package and create the function to optimize: library("mlrMBO") fn <- makeSingleObjectiveFunction(name = "tuning", fn = tuning, par.set = makeParamSet(makeIntegerParam("x1", lower = 3, upper = 8), makeIntegerParam("x2", lower = 100, upper = 500))) This function is based on the function I defined before. The parameters to optimize are also defined as are their bounds. I will look for mtry between the values of 3 and 8, and trees between 50 and 500. We still need to define some other objects before continuing: # Create initial random Latin Hypercube Design of 10 points library(lhs)# for randomLHS des <- generateDesign(n = 5L * 2L, getParamSet(fn), fun = randomLHS) Then we choose the surrogate model, a random forest too: # Specify kriging model with standard error estimation surrogate <- makeLearner("regr.ranger", predict.type = "se", keep.inbag = TRUE) Here I define some options: # Set general controls ctrl <- makeMBOControl() ctrl <- setMBOControlTermination(ctrl, iters = 10L) ctrl <- setMBOControlInfill(ctrl, crit = makeMBOInfillCritEI()) And this is the optimization part: # Start optimization result <- mbo(fn, des, surrogate, ctrl, more.args = list("validation_data" = validation_data)) result ## Recommended parameters: ## x1=8; x2=314 ## Objective: y = 0.484 ## ## Optimization path ## 10 + 10 entries in total, displaying last 10 (or less): ## x1 x2 y dob eol error.message exec.time ei error.model ## 11 8 283 0.4855415 1 NA <NA> 7.353 -3.276847e-04 <NA> ## 12 8 284 0.4852047 2 NA <NA> 7.321 -3.283713e-04 <NA> ## 13 8 314 0.4839817 3 NA <NA> 7.703 -3.828517e-04 <NA> ## 14 8 312 0.4841398 4 NA <NA> 7.633 -2.829713e-04 <NA> ## 15 8 318 0.4841066 5 NA <NA> 7.692 -2.668354e-04 <NA> ## 16 8 314 0.4845221 6 NA <NA> 7.574 -1.382333e-04 <NA> ## 17 8 321 0.4843018 7 NA <NA> 7.693 -3.828924e-05 <NA> ## 18 8 318 0.4868457 8 NA <NA> 7.696 -8.692828e-07 <NA> ## 19 8 310 0.4862687 9 NA <NA> 7.594 -1.061185e-07 <NA> ## 20 8 313 0.4878694 10 NA <NA> 7.628 -5.153015e-07 <NA> ## train.time prop.type propose.time se mean ## 11 0.011 infill_ei 0.450 0.0143886864 0.5075765 ## 12 0.011 infill_ei 0.427 0.0090265872 0.4971003 ## 13 0.012 infill_ei 0.443 0.0062693960 0.4916927 ## 14 0.012 infill_ei 0.435 0.0037308971 0.4878950 ## 15 0.012 infill_ei 0.737 0.0024446891 0.4860699 ## 16 0.013 infill_ei 0.442 0.0012713838 0.4850705 ## 17 0.012 infill_ei 0.444 0.0006371109 0.4847248 ## 18 0.013 infill_ei 0.467 0.0002106381 0.4844576 ## 19 0.014 infill_ei 0.435 0.0002182254 0.4846214 ## 20 0.013 infill_ei 0.748 0.0002971160 0.4847383 So the recommended parameters are 8 for mtry and 314 for trees. The user can access these recommended parameters with result$x$x1 and result$x$x2. The value of the RMSE is lower than before, and equals 0.4839817. It can be accessed with result$y. Let’s now train the random forest on the training data with this values. First, I pre-process the training data

training_rec <- prep(simple_recipe(housing_train), testing = housing_train)

train_data <- bake(training_rec, new_data = housing_train)

Let’s now train our final model and predict the prices:

final_model <- rand_forest(mode = "regression", mtry = result$x$x1, trees = result$x$x2) %>%
set_engine("ranger", importance = 'impurity') %>%
fit(price ~ ., data = train_data)

price_predict <- predict(final_model, new_data = select(test_data, -price))

Let’s transform the data back and compare the predicted prices to the true ones visually:

cbind(price_predict * sd(housing_train$price) + mean(housing_train$price),
housing_test$price) ## .pred housing_test$price
## 1  16.76938               13.5
## 2  27.59510               30.8
## 3  23.14952               24.7
## 4  21.92390               21.2
## 5  21.35030               20.0
## 6  23.15809               22.9
## 7  23.00947               23.9
## 8  25.74268               26.6
## 9  24.13122               22.6
## 10 34.97671               43.8
## 11 19.30543               18.8
## 12 18.09146               15.7
## 13 18.82922               19.2
## 14 18.63397               13.3
## 15 19.14438               14.0
## 16 17.05549               15.6
## 17 23.79491               27.0
## 18 20.30125               17.4
## 19 22.99200               23.6
## 20 32.77092               33.3
## 21 31.66258               34.6
## 22 28.79583               34.9
## 23 39.02755               50.0
## 24 23.53336               21.7
## 25 24.66551               24.3
## 26 24.91737               24.0
## 27 25.11847               25.1
## 28 24.42518               23.7
## 29 24.59139               23.7
## 30 24.91760               26.2
## 31 38.73875               43.5
## 32 29.71848               35.1
## 33 36.89490               46.0
## 34 24.04041               26.4
## 35 20.91349               20.3
## 36 21.18602               23.1
## 37 22.57069               22.2
## 38 25.21751               23.9
## 39 28.55841               50.0
## 40 14.38216                7.2
## 41 12.76573                8.5
## 42 11.78237                9.5
## 43 13.29279               13.4
## 44 14.95076               16.4
## 45 15.79182               19.1
## 46 18.26510               19.6
## 47 14.84985               13.3
## 48 16.01508               16.7
## 49 24.09930               25.0
## 50 20.75357               21.8
## 51 19.49487               19.7

Let’s now compute the RMSE:

tibble::tibble("truth" = test_data\$price,
"prediction" = unlist(price_predict)) %>%
yardstick::rmse(truth, prediction)
## # A tibble: 1 × 3
##   .metric .estimator .estimate
##   <chr>   <chr>          <dbl>
## 1 rmse    standard       0.425

As I mentioned above, all the part about looking for hyper-parameters could be changed to something else. The general approach though remains what I have described, and can be applied for any models that have hyper-parameters.

### References

Breiman, Leo. 2001. “Statistical Modeling: The Two Cultures (with Comments and a Rejoinder by the Author).” Statist. Sci. 16 (3): 199–231. https://doi.org/10.1214/ss/1009213726.

1. This package gets installed with R, no need to add it↩︎

2. The formula in the example is shown for illustration purposes.↩︎

3. data must only contain numeric values, and n is the number of clusters.↩︎

4. data must only contain numeric values, or a formula can be provided.↩︎

5. Surv(y_time, y_status) creates a survival object, where y_time is the time to event y_status. It is possible to create more complex survival objects depending on exactly which data you are dealing with.↩︎