Nonparametric Regression

Nonparametric Regression

Given the usual conditional expectation function

\[E[y_i | \mathbf{x_i} = \mathbf{x}] = m(x)\]

we can estimate \(m(x)\) as a traditional linear model or nonparametrically. One approach to nonparametric estimation is to use a binned means estimator. This takes the average \(y\) for some values of \(x_i\) close to \(x\) (assuming \(x_i\) is continuous). In other words, take the average \(y_i\) for all values \(x_i\) where \(|x_i - x| \leq h\) for some small \(h\). Here \(h\) is the bandwidth.

Binned means estimator

\[\begin{equation} \widehat{m}(x) = \frac{\sum_{i=1}^n \mathbb{1}(|x_i - x| \leq h)y_i}{\sum_{i=1}^n \mathbb{1}(|x_i - x|\leq h)} \end{equation}\]

This gives us a step-function because the indicator function makes the weights discontinuous. Alternatively, we can replace the indicator function with a kernel function:

Kernel regression estimator

\[\begin{equation} \widehat{m}_k(x) = \frac{\sum_{i=1}^n K\left(\frac{x_i - x}{h}\right)y_i}{\sum_{i=1}^n K\left(\frac{x_i - x}{h}\right)} \end{equation}\]

Some commonly used kernels:

Gaussian Kernel: \[K(u) = \frac{1}{\sqrt{2 \pi}} exp \left (-\frac{u^2}{2}\right)\]

Epanechnikov Kernel: \[ K(u) = \left\{ \begin{array}{ll} \frac{3}{4\sqrt{5}}\left(1 - \frac{u^2}{5}\right) & \mbox{if $|u| < \sqrt{5}$}\\ 0 & \mbox{otherwise}\end{array} \right. \]

Local Linear approximation

Let’s start with simulating some data using the simstudy packge:

library(tidyverse)
## simulate data using simstudy package
library(simstudy)
set.seed(234)
df <- genData(100, defData(varname = "x", formula = "20;60", dist = 'uniform'))
theta1 = c(0.1, 0.8, 0.6, 0.4, 0.6, 0.9, 0.9)
knots <- c(0.25, 0.5, 0.75)
df <- genSpline(dt = df, newvar = "y",
                predictor = "x",
                theta = theta1,
                knots = knots,
                degree = 3,
                newrange = "90;160",
                noise.var = 64)

Here is a scatterplot with dashed vertical lines showing the knots used to generate the data (quantiles).

## scatter plot of the data
ggplot(data = df, aes(x = x, y = y)) +
  geom_point() +
  geom_vline(xintercept = quantile(df$x, knots), linetype = 'dashed') +
  theme_bw()

#using locpoly 
library(KernSmooth)
## KernSmooth 2.23 loaded
## Copyright M. P. Wand 1997-2009
m_ll <- data.frame(locpoly(df$x, df$y, degree = 1, bandwidth = 3))
# degree = 1 for locally linear
ggplot(data = df, aes(x = x, y = y)) +
  geom_point() +
  geom_vline(xintercept = quantile(df$x, knots), linetype = 'dashed') +
  geom_line(data = m_ll,aes(x = x, y = y),color = "blue") +
  theme_bw()

tricube kernal \[ K(u) = \left\{ \begin{array}{ll} (1-|u|^3)^3 & \mbox{for $|u| < 1$}\\ 0 & \mbox{for $|u|\geq 1$ }\end{array} \right. \]

kernel.function <- function(u){
  tmp <- rep(NA,length(u))
  for(i in 1:length(u)){
    if(abs(u[i])<1){
    k <- (1 - abs(u[i])^3)^3
  }else if(abs(u[i])>=1){
    k <- 0
  }
    tmp[i] <- k
  }
  return(tmp)
}

Decide on evaluation points \(x_0\). Our data is (roughly) from 20 to 60 so let’s use that to define our evaluation points and we will use increments of 20.

# evaluation points
x0 <- seq(20,60,20)
# bandwidth
h <- 10

We have 3 evaluation points, let’s do a loop to calculate the scaled distance from each value of \(x\) to \(x_0\).

# empty container
distances <- matrix(NA, nrow = length(df$x), ncol = 3)
colnames(distances) <- x0
for(i in 1:length(x0)){
  # calc distance and put in column i
  distances[,i] <- df$x - x0[i]
  # divide by bandwidth (this does scaling)
  distances <- distances/h
}

Now apply our kernel function to estimate the weights

distances <- data.frame(distances)
weights <- map_df(distances, kernel.function)

Estimate predicted values at the evaluation points using our weights

# eval point 20
W_20 <- diag(weights$X20)
X <- as.matrix(cbind(rep(1,100), df$x)) #create X matrix with intercept
Y <- as.matrix(df$y)
B_WLS_20 <- solve(t(X)%*%W_20%*%X)%*%(t(X)%*%W_20%*%Y)
# eval point 40
W_40 <- diag(weights$X40)
B_WLS_40 <- solve(t(X)%*%W_40%*%X)%*%(t(X)%*%W_40%*%Y)
# eval point 60
W_60 <- diag(weights$X60)
B_WLS_60 <- solve(t(X)%*%W_60%*%X)%*%(t(X)%*%W_60%*%Y)

# predicted values
yhat <- data.frame(y20 = X%*%B_WLS_20,
                   y40 = X%*%B_WLS_40,
                   y60 = X%*%B_WLS_60)
yhat <- pivot_longer(yhat, names_to = "group", names_prefix = "y",
                     values_to="yhat",cols = c(1:3))

df_big <- bind_cols(bind_rows(df,df,df),yhat)
#first eval point
ggplot(data = df_big %>% filter(group==20), aes(x = x, y = y)) +
  geom_point() +
  geom_point(aes(x = x, y = yhat),color="red") +
  theme_bw()

ggplot(data = df_big %>% filter(group==40), aes(x = x, y = y)) +
  geom_point() +
  geom_point(aes(x = x, y = yhat), color="green") +
  theme_bw()

ggplot(data = df_big %>% filter(group==60), aes(x = x, y = y)) +
  geom_point() +
  geom_point(aes(x = x, y = yhat), color="blue") +
  theme_bw()

ggplot(data = df_big, aes(x = x, y = y)) +
  geom_point() +
  geom_point(aes(x = x, y = yhat, color=group)) +
  theme_bw()

Switching gears – let’s do a quick example on how to demean variables using dplyr. We will use the built-in mtcars dataset.

First, let’s demean only one column, the mpg column and make a new variable using mutate():

df <- mtcars %>%
  mutate(mpg_demean = mpg - mean(mpg))

You have to assign this to an object, in this case I’ve stored it as an entirely new data.frame df. I can now use this new variable:

head(df$mpg_demean)
## [1]  0.909375  0.909375  2.709375  1.309375 -1.390625 -1.990625

Here, the demeaned variable is calculated but not stored because I have not assigned this operation to anything. If I try to call the variable mpg_demean later I will get an error because it does not exist.

mtcars %>%
  mutate(mpg_demean2 = mpg - mean(mpg)) %>%
  head()
##    mpg cyl disp  hp drat    wt  qsec vs am gear carb mpg_demean2
## 1 21.0   6  160 110 3.90 2.620 16.46  0  1    4    4    0.909375
## 2 21.0   6  160 110 3.90 2.875 17.02  0  1    4    4    0.909375
## 3 22.8   4  108  93 3.85 2.320 18.61  1  1    4    1    2.709375
## 4 21.4   6  258 110 3.08 3.215 19.44  1  0    3    1    1.309375
## 5 18.7   8  360 175 3.15 3.440 17.02  0  0    3    2   -1.390625
## 6 18.1   6  225 105 2.76 3.460 20.22  1  0    3    1   -1.990625
head(df$mpg_demean2)
## NULL

If I want to de-mean multiple variables at once use mutate_each. Here I am selecting two columns, vs and am, then using mutate_all to demean them both then I’m joining them back to the original data.

df <- df %>%
  select(vs, am) %>%
  mutate_all(list(~. - mean(.))) %>%
  bind_cols(df,.)
## New names:
## * vs -> vs...8
## * am -> am...9
## * vs -> vs...13
## * am -> am...14
head(df)
##    mpg cyl disp  hp drat    wt  qsec vs...8 am...9 gear carb mpg_demean vs...13
## 1 21.0   6  160 110 3.90 2.620 16.46      0      1    4    4   0.909375 -0.4375
## 2 21.0   6  160 110 3.90 2.875 17.02      0      1    4    4   0.909375 -0.4375
## 3 22.8   4  108  93 3.85 2.320 18.61      1      1    4    1   2.709375  0.5625
## 4 21.4   6  258 110 3.08 3.215 19.44      1      0    3    1   1.309375  0.5625
## 5 18.7   8  360 175 3.15 3.440 17.02      0      0    3    2  -1.390625 -0.4375
## 6 18.1   6  225 105 2.76 3.460 20.22      1      0    3    1  -1.990625  0.5625
##    am...14
## 1  0.59375
## 2  0.59375
## 3  0.59375
## 4 -0.40625
## 5 -0.40625
## 6 -0.40625

Using base R you could do something like this:

df$wt_demean <- df$wt - mean(df$wt)
head(df)
##    mpg cyl disp  hp drat    wt  qsec vs...8 am...9 gear carb mpg_demean vs...13
## 1 21.0   6  160 110 3.90 2.620 16.46      0      1    4    4   0.909375 -0.4375
## 2 21.0   6  160 110 3.90 2.875 17.02      0      1    4    4   0.909375 -0.4375
## 3 22.8   4  108  93 3.85 2.320 18.61      1      1    4    1   2.709375  0.5625
## 4 21.4   6  258 110 3.08 3.215 19.44      1      0    3    1   1.309375  0.5625
## 5 18.7   8  360 175 3.15 3.440 17.02      0      0    3    2  -1.390625 -0.4375
## 6 18.1   6  225 105 2.76 3.460 20.22      1      0    3    1  -1.990625  0.5625
##    am...14 wt_demean
## 1  0.59375  -0.59725
## 2  0.59375  -0.34225
## 3  0.59375  -0.89725
## 4 -0.40625  -0.00225
## 5 -0.40625   0.22275
## 6 -0.40625   0.24275

If I want to demean by groups. Here I will use the cylinder variable to group with.

df <- df %>%
  group_by(cyl) %>%
  mutate(mpg_mean_cyl = mpg - mean(mpg))
head(df[,c("mpg","cyl","mpg_mean_cyl")])
## # A tibble: 6 x 3
## # Groups:   cyl [3]
##     mpg   cyl mpg_mean_cyl
##   <dbl> <dbl>        <dbl>
## 1  21       6         1.26
## 2  21       6         1.26
## 3  22.8     4        -3.86
## 4  21.4     6         1.66
## 5  18.7     8         3.6 
## 6  18.1     6        -1.64

Here is an example of some code that runs but may not be doing what we want. What are some differences in this code?

df %>%
  group_by(cyl) %>%
  mutate(new_mpg = mean(mpg, na.rm=TRUE)) %>%
  select(mpg,new_mpg)
## Adding missing grouping variables: `cyl`
## # A tibble: 32 x 3
## # Groups:   cyl [3]
##      cyl   mpg new_mpg
##    <dbl> <dbl>   <dbl>
##  1     6  21      19.7
##  2     6  21      19.7
##  3     4  22.8    26.7
##  4     6  21.4    19.7
##  5     8  18.7    15.1
##  6     6  18.1    19.7
##  7     8  14.3    15.1
##  8     4  24.4    26.7
##  9     4  22.8    26.7
## 10     6  19.2    19.7
## # … with 22 more rows

F-test

We use the F-test to look at multiple hypotheses at once.

fit <- lm(mpg ~ cyl + vs + am + gear + carb, data = mtcars)
summary(fit)
## 
## Call:
## lm(formula = mpg ~ cyl + vs + am + gear + carb, data = mtcars)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -5.5403 -1.1582  0.2528  1.2787  5.5597 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)   
## (Intercept)  25.1591     7.7570   3.243  0.00323 **
## cyl          -1.2239     0.7510  -1.630  0.11521   
## vs            0.8784     1.9957   0.440  0.66347   
## am            3.5989     1.8694   1.925  0.06522 . 
## gear          1.2516     1.4730   0.850  0.40323   
## carb         -1.4071     0.5368  -2.621  0.01444 * 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 2.808 on 26 degrees of freedom
## Multiple R-squared:  0.8179, Adjusted R-squared:  0.7829 
## F-statistic: 23.36 on 5 and 26 DF,  p-value: 7.418e-09

We could do an F-test on the following hypothesis:

\[H_0:\beta_{cyl} = \beta_{vs} = 0\]

This is different than testing the individual hypotheses: \[H_0^{(1)}:\beta_{cyl}=0\] \[H_0^{(2)}:\beta_{vs}=0\]

The F-test will compare the two models: 1) model with cyl and vs and 2) model without cyl and vs.

\[F_0 = (\frac{SSR_{restricted} - SSR_{unrestricted})/q}{SSR_{unrestricted}/(n-k-1)}\]

Where \(\text{SSR}_{unrestricted}\) is the residual sum of squares for the full model. \(\text{SSR}_{restricted}\) is the residual sum of squares for the null model (restricted). \(q\) is how many coefficients are omitted.

full <- lm(mpg ~ cyl + vs + am + gear + carb, data = mtcars)
restrict <- lm(mpg ~ am + gear + carb, data = mtcars)
n <- nrow(mtcars)
k <- 5
q <- 2
SSR_unres <- sum((summary(full)$residuals)^2)
SSR_res <- sum((summary(restrict)$residuals)^2)
SSR_unres
## [1] 205.0521
SSR_res
## [1] 265.9296
F <- ((SSR_res - SSR_unres)/q)/(SSR_unres/(n-k-1))
F
## [1] 3.859546
1 - pf(F,df1 = 2, df2 = n-k-1)
## [1] 0.03406177

Using linearHypothesis() function from the car package:

library(car)
## The following object is masked from 'package:dplyr':
## 
##     recode
linearHypothesis(full, c("cyl = 0", "vs = 0"))
## Linear hypothesis test
## 
## Hypothesis:
## cyl = 0
## vs = 0
## 
## Model 1: restricted model
## Model 2: mpg ~ cyl + vs + am + gear + carb
## 
##   Res.Df    RSS Df Sum of Sq      F  Pr(>F)  
## 1     28 265.93                              
## 2     26 205.05  2    60.878 3.8595 0.03406 *
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Given the p-value we can reject the null hypothesis. This suggests that while individually, the coefficients on cyl and vs were not statistically significant at the 95% level, including them both improves model fit because we reject the null that they are jointly equal to 0.

Avatar
Elisha Cohen
PhD Student in Political Science

I am a PhD student in political science at Emory University. I study political methodology with applications focused on gender, race, and inequality in the United States.