```Econometrics and Free Software by Bruno Rodrigues.
Check out my package that adds logging to R functions, {chronicler}.
Or read my free ebooks, to learn some R and build reproducible analytical pipelines..
You can also watch my youtube channel or find the slides to the talks I've given here.
Buy me a coffee, my kids don't let me sleep.
```

# Dealing with heteroskedasticity; regression with robust standard errors using R

R

First of all, is it heteroskedasticity or heteroscedasticity? According to McCulloch (1985), heteroskedasticity is the proper spelling, because when transliterating Greek words, scientists use the Latin letter k in place of the Greek letter κ (kappa). κ sometimes is transliterated as the Latin letter c, but only when these words entered the English language through French, such as scepter.

Now that this is out of the way, we can get to the meat of this blogpost (foreshadowing pun). A random variable is said to be heteroskedastic, if its variance is not constant. For example, the variability of expenditures may increase with income. Richer families may spend a similar amount on groceries as poorer people, but some rich families will sometimes buy expensive items such as lobster. The variability of expenditures for rich families is thus quite large. However, the expenditures on food of poorer families, who cannot afford lobster, will not vary much. Heteroskedasticity can also appear when data is clustered; for example, variability of expenditures on food may vary from city to city, but is quite constant within a city.

To illustrate this, let’s first load all the packages needed for this blog post:

``````library(robustbase)
library(tidyverse)
library(sandwich)
library(lmtest)
library(modelr)
library(broom)``````

First, let’s load and prepare the data:

``````data("education")

education <- education %>%
rename(residents = X1,
per_capita_income = X2,
young_residents = X3,
per_capita_exp = Y,
state = State) %>%
mutate(region = case_when(
Region == 1 ~ "northeast",
Region == 2 ~ "northcenter",
Region == 3 ~ "south",
Region == 4 ~ "west"
)) %>%
select(-Region)``````

I will be using the `education` data set from the `{robustbase}` package. I renamed some columns and changed the values of the `Region` column. Now, let’s do a scatterplot of per capita expenditures on per capita income:

``````ggplot(education, aes(per_capita_income, per_capita_exp)) +
geom_point() +
theme_dark()``````

It would seem that, as income increases, variability of expenditures increases too. Let’s look at the same plot by `region`:

``````ggplot(education, aes(per_capita_income, per_capita_exp)) +
geom_point() +
facet_wrap(~region) +
theme_dark()``````

I don’t think this shows much; it would seem that observations might be clustered, but there are not enough observations to draw any conclusion from this plot (in any case, drawing conclusions from only plots is dangerous).

Let’s first run a good ol’ linear regression:

``````lmfit <- lm(per_capita_exp ~ region + residents + young_residents + per_capita_income, data = education)

summary(lmfit)``````
``````##
## Call:
## lm(formula = per_capita_exp ~ region + residents + young_residents +
##     per_capita_income, data = education)
##
## Residuals:
##     Min      1Q  Median      3Q     Max
## -77.963 -25.499  -2.214  17.618  89.106
##
## Coefficients:
##                     Estimate Std. Error t value Pr(>|t|)
## (Intercept)       -467.40283  142.57669  -3.278 0.002073 **
## regionnortheast     15.72741   18.16260   0.866 0.391338
## regionsouth          7.08742   17.29950   0.410 0.684068
## regionwest          34.32416   17.49460   1.962 0.056258 .
## residents           -0.03456    0.05319  -0.650 0.519325
## young_residents      1.30146    0.35717   3.644 0.000719 ***
## per_capita_income    0.07204    0.01305   5.520 1.82e-06 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 39.88 on 43 degrees of freedom
## Multiple R-squared:  0.6292, Adjusted R-squared:  0.5774
## F-statistic: 12.16 on 6 and 43 DF,  p-value: 6.025e-08``````

Let’s test for heteroskedasticity using the Breusch-Pagan test that you can find in the `{lmtest}` package:

``bptest(lmfit)``
``````##
##  studentized Breusch-Pagan test
##
## data:  lmfit
## BP = 17.921, df = 6, p-value = 0.006432``````

This test shows that we can reject the null that the variance of the residuals is constant, thus heteroskedacity is present. To get the correct standard errors, we can use the `vcovHC()` function from the `{sandwich}` package (hence the choice for the header picture of this post):

``````lmfit %>%
vcovHC() %>%
diag() %>%
sqrt()``````
``````##       (Intercept)   regionnortheast       regionsouth        regionwest
##      311.31088691       25.30778221       23.56106307       24.12258706
##         residents   young_residents per_capita_income
##        0.09184368        0.68829667        0.02999882``````

By default `vcovHC()` estimates a heteroskedasticity consistent (HC) variance covariance matrix for the parameters. There are several ways to estimate such a HC matrix, and by default `vcovHC()` estimates the “HC3” one. You can refer to Zeileis (2004) for more details.

We see that the standard errors are much larger than before! The intercept and `regionwest` variables are not statistically significant anymore.

You can achieve the same in one single step:

``coeftest(lmfit, vcov = vcovHC(lmfit))``
``````##
## t test of coefficients:
##
##                      Estimate  Std. Error t value Pr(>|t|)
## (Intercept)       -467.402827  311.310887 -1.5014  0.14056
## regionnortheast     15.727405   25.307782  0.6214  0.53759
## regionsouth          7.087424   23.561063  0.3008  0.76501
## regionwest          34.324157   24.122587  1.4229  0.16198
## residents           -0.034558    0.091844 -0.3763  0.70857
## young_residents      1.301458    0.688297  1.8908  0.06540 .
## per_capita_income    0.072036    0.029999  2.4013  0.02073 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1``````

It’s is also easy to change the estimation method for the variance-covariance matrix:

``coeftest(lmfit, vcov = vcovHC(lmfit, type = "HC0"))``
``````##
## t test of coefficients:
##
##                      Estimate  Std. Error t value  Pr(>|t|)
## (Intercept)       -467.402827  172.577569 -2.7084  0.009666 **
## regionnortheast     15.727405   20.488148  0.7676  0.446899
## regionsouth          7.087424   17.755889  0.3992  0.691752
## regionwest          34.324157   19.308578  1.7777  0.082532 .
## residents           -0.034558    0.054145 -0.6382  0.526703
## young_residents      1.301458    0.387743  3.3565  0.001659 **
## per_capita_income    0.072036    0.016638  4.3296 8.773e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1``````

As I wrote above, by default, the `type` argument is equal to “HC3”.

Another way of dealing with heteroskedasticity is to use the `lmrob()` function from the `{robustbase}` package. This package is quite interesting, and offers quite a lot of functions for robust linear, and nonlinear, regression models. Running a robust linear regression is just the same as with `lm()`:

``````lmrobfit <- lmrob(per_capita_exp ~ region + residents + young_residents + per_capita_income,
data = education)

summary(lmrobfit)``````
``````##
## Call:
## lmrob(formula = per_capita_exp ~ region + residents + young_residents + per_capita_income,
##     data = education)
##  \--> method = "MM"
## Residuals:
##     Min      1Q  Median      3Q     Max
## -57.074 -14.803  -0.853  24.154 174.279
##
## Coefficients:
##                     Estimate Std. Error t value Pr(>|t|)
## (Intercept)       -156.37169  132.73828  -1.178  0.24526
## regionnortheast     20.64576   26.45378   0.780  0.43940
## regionsouth         10.79695   29.42746   0.367  0.71549
## regionwest          45.22589   33.07950   1.367  0.17867
## residents            0.03406    0.04412   0.772  0.44435
## young_residents      0.57896    0.25512   2.269  0.02832 *
## per_capita_income    0.04328    0.01442   3.000  0.00447 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Robust residual standard error: 26.4
## Multiple R-squared:  0.6235, Adjusted R-squared:  0.571
## Convergence in 24 IRWLS iterations
##
## Robustness weights:
##  observation 50 is an outlier with |weight| = 0 ( < 0.002);
##  7 weights are ~= 1. The remaining 42 ones are summarized as
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
## 0.05827 0.85200 0.93870 0.85250 0.98700 0.99790
## Algorithmic parameters:
##        tuning.chi                bb        tuning.psi        refine.tol
##         1.548e+00         5.000e-01         4.685e+00         1.000e-07
##           rel.tol         scale.tol         solve.tol       eps.outlier
##         1.000e-07         1.000e-10         1.000e-07         2.000e-03
##             eps.x warn.limit.reject warn.limit.meanrw
##         1.071e-08         5.000e-01         5.000e-01
##      nResample         max.it       best.r.s       k.fast.s          k.max
##            500             50              2              1            200
##    maxit.scale      trace.lev            mts     compute.rd fast.s.large.n
##            200              0           1000              0           2000
##                   psi           subsampling                   cov
##            "bisquare"         "nonsingular"         ".vcov.avar1"
## compute.outlier.stats
##                  "SM"
## seed : int(0)``````

This however, gives you different estimates than when fitting a linear regression model. The estimates should be the same, only the standard errors should be different. This is because the estimation method is different, and is also robust to outliers (at least that’s my understanding, I haven’t read the theoretical papers behind the package yet).

Finally, it is also possible to bootstrap the standard errors. For this I will use the `bootstrap()` function from the `{modelr}` package:

``````resamples <- 100

boot_education <- education %>%
modelr::bootstrap(resamples)``````

Let’s take a look at the `boot_education` object:

``boot_education``
``````## # A tibble: 100 x 2
##    strap      .id
##    <list>     <chr>
##  1 <resample> 001
##  2 <resample> 002
##  3 <resample> 003
##  4 <resample> 004
##  5 <resample> 005
##  6 <resample> 006
##  7 <resample> 007
##  8 <resample> 008
##  9 <resample> 009
## 10 <resample> 010
## # … with 90 more rows``````

The column `strap` contains resamples of the original data. I will run my linear regression from before on each of the resamples:

``````(
boot_lin_reg <- boot_education %>%
mutate(regressions =
map(strap,
~lm(per_capita_exp ~ region + residents +
young_residents + per_capita_income,
data = .)))
)``````
``````## # A tibble: 100 x 3
##    strap      .id   regressions
##    <list>     <chr> <list>
##  1 <resample> 001   <lm>
##  2 <resample> 002   <lm>
##  3 <resample> 003   <lm>
##  4 <resample> 004   <lm>
##  5 <resample> 005   <lm>
##  6 <resample> 006   <lm>
##  7 <resample> 007   <lm>
##  8 <resample> 008   <lm>
##  9 <resample> 009   <lm>
## 10 <resample> 010   <lm>
## # … with 90 more rows``````

I have added a new column called `regressions` which contains the linear regressions on each bootstrapped sample. Now, I will create a list of tidied regression results:

``````(
tidied <- boot_lin_reg %>%
mutate(tidy_lm =
map(regressions, broom::tidy))
)``````
``````## # A tibble: 100 x 4
##    strap      .id   regressions tidy_lm
##    <list>     <chr> <list>      <list>
##  1 <resample> 001   <lm>        <tibble [7 × 5]>
##  2 <resample> 002   <lm>        <tibble [7 × 5]>
##  3 <resample> 003   <lm>        <tibble [7 × 5]>
##  4 <resample> 004   <lm>        <tibble [7 × 5]>
##  5 <resample> 005   <lm>        <tibble [7 × 5]>
##  6 <resample> 006   <lm>        <tibble [7 × 5]>
##  7 <resample> 007   <lm>        <tibble [7 × 5]>
##  8 <resample> 008   <lm>        <tibble [7 × 5]>
##  9 <resample> 009   <lm>        <tibble [7 × 5]>
## 10 <resample> 010   <lm>        <tibble [7 × 5]>
## # … with 90 more rows``````

`broom::tidy()` creates a data frame of the regression results. Let’s look at one of these:

``tidied\$tidy_lm[[1]]``
``````## # A tibble: 7 x 5
##   term              estimate std.error statistic  p.value
##   <chr>                <dbl>     <dbl>     <dbl>    <dbl>
## 1 (Intercept)       -571.     109.        -5.22  4.92e- 6
## 2 regionnortheast    -48.0     17.2       -2.80  7.71e- 3
## 3 regionsouth        -21.3     15.1       -1.41  1.66e- 1
## 4 regionwest           1.88    13.9        0.135 8.93e- 1
## 5 residents           -0.134    0.0608    -2.21  3.28e- 2
## 6 young_residents      1.50     0.308      4.89  1.47e- 5
## 7 per_capita_income    0.100    0.0125     8.06  3.85e-10``````

This format is easier to handle than the standard `lm()` output:

``tidied\$regressions[[1]]``
``````##
## Call:
## lm(formula = per_capita_exp ~ region + residents + young_residents +
##     per_capita_income, data = .)
##
## Coefficients:
##       (Intercept)    regionnortheast        regionsouth
##         -571.0568           -48.0018           -21.3019
##        regionwest          residents    young_residents
##            1.8808            -0.1341             1.5042
## per_capita_income
##            0.1005``````

Now that I have all these regression results, I can compute any statistic I need. But first, let’s transform the data even further:

``````list_mods <- tidied %>%
pull(tidy_lm)``````

`list_mods` is a list of the `tidy_lm` data frames. I now add an index and bind the rows together (by using `map2_df()` instead of `map2()`):

``````mods_df <- map2_df(list_mods,
seq(1, resamples),
~mutate(.x, resample = .y))``````

Let’s take a look at the final object:

``head(mods_df, 25)``
``````## # A tibble: 25 x 6
##    term              estimate std.error statistic  p.value resample
##    <chr>                <dbl>     <dbl>     <dbl>    <dbl>    <int>
##  1 (Intercept)       -571.     109.        -5.22  4.92e- 6        1
##  2 regionnortheast    -48.0     17.2       -2.80  7.71e- 3        1
##  3 regionsouth        -21.3     15.1       -1.41  1.66e- 1        1
##  4 regionwest           1.88    13.9        0.135 8.93e- 1        1
##  5 residents           -0.134    0.0608    -2.21  3.28e- 2        1
##  6 young_residents      1.50     0.308      4.89  1.47e- 5        1
##  7 per_capita_income    0.100    0.0125     8.06  3.85e-10        1
##  8 (Intercept)        -97.2    145.        -0.672 5.05e- 1        2
##  9 regionnortheast     -1.48    10.8       -0.136 8.92e- 1        2
## 10 regionsouth         12.5     11.4        1.09  2.82e- 1        2
## # … with 15 more rows``````

Now this is a very useful format, because I now can group by the `term` column and compute any statistics I need, in the present case the standard deviation:

``````(
r.std.error <- mods_df %>%
group_by(term) %>%
summarise(r.std.error = sd(estimate))
)``````
``````## # A tibble: 7 x 2
##   term              r.std.error
##   <chr>                   <dbl>
## 1 (Intercept)          220.
## 2 per_capita_income      0.0197
## 3 regionnortheast       24.5
## 4 regionsouth           21.1
## 5 regionwest            22.7
## 6 residents              0.0607
## 7 young_residents        0.498``````

We can append this column to the linear regression model result:

``````lmfit %>%
broom::tidy() %>%
full_join(r.std.error) %>%
select(term, estimate, std.error, r.std.error)``````
``## Joining, by = "term"``
``````## # A tibble: 7 x 4
##   term               estimate std.error r.std.error
##   <chr>                 <dbl>     <dbl>       <dbl>
## 1 (Intercept)       -467.      143.        220.
## 2 regionnortheast     15.7      18.2        24.5
## 3 regionsouth          7.09     17.3        21.1
## 4 regionwest          34.3      17.5        22.7
## 5 residents           -0.0346    0.0532      0.0607
## 6 young_residents      1.30      0.357       0.498
## 7 per_capita_income    0.0720    0.0131      0.0197``````

As you see, using the whole bootstrapping procedure is longer than simply using either one of the first two methods. However, this procedure is very flexible and can thus be adapted to a very large range of situations. Either way, in the case of heteroskedasticity, you can see that results vary a lot depending on the procedure you use, so I would advise to use them all as robustness tests and discuss the differences.

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