[R] Query on ksmooth(): Theoretical step function vs. plotted diagonal segments

[Rui Barradas]

Às 14:00 de 19/03/2026, 송상은 escreveu:
> Dear R-help members,
> 
> Hello,
> I am studying kernel regression and experimenting with the ksmooth()
> function in R.
> 
> When using a box kernel, the kernel function is an indicator function
> (weight = 1 inside the bandwidth and 0 outside). Based on this definition,
> I expected the Nadaraya–Watson estimator to produce a step-like function:
> the estimate should remain constant while the set of included points is
> unchanged, and then jump when a point enters or leaves the bandwidth window.
> 
> However, when I run the following code using the cars dataset:
> 
> par(mfrow=c(1,1))
> with(cars, {
> plot(speed, dist)
> lines(ksmooth(speed, dist, "normal", bandwidth = 2),
> col = "blue", lwd = 3)
> lines(ksmooth(speed, dist, "box", bandwidth = 2),
> col = "darkorange", lwd = 3)
> })
> legend("topleft", c("Normal Kernel with h=2", "Box Kernel with h=2"),
> lwd = c(2,2),
> col = c("blue","darkorange"), cex = 2)
> 
> the curve produced by the box kernel (dark orange) appears to contain
> diagonal line segments rather than the step-like shape I expected. I have
> attached the resulting plot for reference.
> 
> My understanding is that the theoretical estimator should behave like a
> step function because the kernel weights are either 0 or 1. Therefore, I
> was wondering whether the diagonal segments arise from how ksmooth()
> evaluates the estimator on a grid of x values and then connects those
> points with straight lines for plotting, or if there is another
> implementation detail that explains this behavior.
> 
> Could you please clarify whether this is expected behavior?
> 
> Thank you very much for your time.
> 
> Best regards,
> Sangeun Song
> 
> 
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Hello,

Use type = "s", see help("plot.default") for the possible values of 
argument type.
You have a link to that help page in the documentation of line().

?plot.default
?line

In the code below, I only use it in the 2nd line() call.



# if it's the 1st time you call par(), this does nothing
# (it changes a graphics device from rows*cols == 1*1 to 1*1)
# anyway, save the value so that you can later reset the default
old_par <- par(mfrow = c(1, 1))
# see what mfrow was before
old_par
#> $mfrow
#> [1] 1 1

with(cars, {
   plot(speed, dist)
   lines(ksmooth(speed, dist, "normal", bandwidth = 2),
         col = "blue", lwd = 3)
   lines(ksmooth(speed, dist, "box", bandwidth = 2),
         type = "s",
         col = "darkorange", lwd = 3)
})
legend("topleft", c("Normal Kernel with h=2", "Box Kernel with h=2"),
        lwd = c(2,2),
        col = c("blue","darkorange"), cex = 2)

# reset the default par values
par(old_par)



Hope this helps,

Rui Barradas