What sort of kernel density estimator does one use to avoid boundary bias?

Consider the task of estimating the density $f_0(x)$ with bounded support and where the probability mass is not decreasing or going to zero as the boundary is approached. To simplify matters assume that the bound(s) of the density is known.

To focus ideas consider as an example the uniform distribution:

Given a sample size $N$ of iid draws $\mathcal U(0,1)$ one could think of applying the kernel density estimator

$$\hat f(y) = \frac{1}{ns}\sum_i K\left( \frac{x_i-y}{s} \right)$$

with a normal kernel and some smoothing parameter $s$. To illustrate the boundary bias consider (implemented in the software R: A Language and Environment for Statistical Computing):

N <- 10000
x <- runif(N)
s <- .045

M <- 100
y <- seq(0,1,length.out=M)
out <- rep(0,M)
for (i in 1:M)
        weights <- dnorm((x-y[i])/s)
        out[i] <- mean(weights)/s

which generates the following plotenter image description here

clearly the approach has a problem capturing the true value of the density function $f_0(x)$ at $x$ close to the boundary.

The logspline method works better but is certainly not without some boundary bias

N <- 10000
x <- runif(N)
m <- logspline(x,lbound=0,ubound=1,knots=seq(0,1,length.out=21))

enter image description here

  • $\begingroup$ In principle the density could be estimated on logit scale and back transformed as density on logit scale / $[x (1 - x)]$ but in practice even if all values are neither 0 nor 1 this doesn't work well for me. Another answer is to reflect at boundaries any probability mass smoothed outside $[0. 1]$. My favourite answer is to use a quantile plot instead! $\endgroup$
    – Nick Cox
    Feb 20, 2020 at 18:29
  • $\begingroup$ As R isn't yet universal, no harm is done by calling this R code (assuming that is so). I speak a different language. $\endgroup$
    – Nick Cox
    Feb 20, 2020 at 18:30
  • 1
    $\begingroup$ My prejudice is that an unbounded kernel is wrong in principle for such a distribution, even though the impact can be made minute by suitable choice of bandwidth. $\endgroup$
    – Nick Cox
    Feb 20, 2020 at 18:32
  • 2
    $\begingroup$ I believe the answer by G. Simpson to a closely related question at stats.stackexchange.com/a/71291/919 generalizes directly to this case. It splines the log density. $\endgroup$
    – whuber
    Feb 20, 2020 at 18:37
  • 1
    $\begingroup$ A key issue is whether you know the bounds of the domain or need to estimate them from the data. In the example you seem to know them beforehand: should we assume that? $\endgroup$
    – whuber
    Feb 20, 2020 at 22:46

2 Answers 2


If you know the boundaries, then one approach mentioned in Silverman's great little book (Density Estimation for Statistics and Data Analysis) is the "reflection technique". One simply reflects the data about the boundary (or boundaries). (This is what @NickCox mentioned in his comment.)

# Generate numbers from a uniform distribution
  N <- 10000
  x <- runif(N)

# Reflect the data at the two boundaries
  xReflected <- c(-x, x, 2-x)

# Construct density estimate
  d <- density(xReflected, from=0, to=1)
  plot(d$x, 3*d$y, ylab="Probability density", xlab="x", ylim=c(0,1.1), las=1)

Estimate of density

Note that in this case we end up with 3 times the number of data points so we need to multiply by 3 the density that comes out of the density function.

Below is an animated display of 100 simulations (as above) but with the true density and the two estimated densities (one from the original data and one from the reflected data). That there is bias near the boundaries is pretty clear when using density with just the original data.

Animation of density estimates

  • 1
    $\begingroup$ '+1' Seems to be working fine. I read that the reflection methods was particularly efficient when $f'(x)$ the derivative of the density was equal to zero when evaluated at the boundary. This is offcourse the case for the uniform distribution. Could be interesting to see how this methods fares with for example the exponential distribution as data generating (I expect the logspline may do better in that case). $\endgroup$ Feb 21, 2020 at 16:35
  • 1
    $\begingroup$ The logspline approach does indeed work better for estimating the density at the lower boundary of zero for the exponential distribution. But it will depend on the application of the results as to if there is any "practical" difference. One computational issue (which should not be the most important issue) is that the reflection approach is more likely to be readily implemented in a variety of statistics packages as not all have logspline functions available. $\endgroup$
    – JimB
    Feb 21, 2020 at 17:24

I do not know if it is interesting (given the original quesiton and the answers it already received) but, I would like to suggest an alternative method. It could maybe be useful to somebody in the future as well (I hope at least) :-).

If you worry about bounday effects of your density smoothing method I would suggest to use P-splines (see Eilers and Marx, 1991 - the authors specifically talk about boundary bias in density smoothing in par 8). Quoting Eilers and Marx,

the P-spline density smoother is not troubled by boundary effects, as for instance kernel smoothers are.

In general, P-splines combine B-splines and finite difference penalties. The density smoothing problem is a special case of GLM. So we just need to parameterize our smoothing problem accordingly.

To answer the original question I will consider data grouped in a histogram fashion. I will indicate with $y_{i}$ the count (but the reasoning can be adapted to the density case as well) of observations falling in the bin/bar $u_{i}$. To smooth these data I will use the following ingredients:

  • the smoother: Whittaker smoother (special case of P-splines, the bases is the identity matrix)
  • first order difference penalty
  • IWLS algorithm to maximize my penalized likelihood (eq 36 in the reference)
    $$ L = \sum_{i} y_{i} \log \mu_{i} - \sum_{i} \mu_{i} - \lambda \sum_{i} (\Delta^{(1)} \eta_{i})^{2} $$ with $\mu_{i} = \exp(\eta_{i})$.

The results are produced by the code below for a fixed value of $\lambda$ (I left some comments to make it easier to read I hope). As you will notice form the results, the $\lambda$ parameter regulates the smoothness of the final estimates. For a very high $\lambda$ we obtaine a pretty flat line.


# Simulate data
N = 10000
x = runif(N)

# Construct histograms
his = hist(x, breaks = 50, plot = F)
X = his$counts
u = his$mids

# Prepare basis (I-mat) and penalty (1st difference)
B = diag(length(X))
D1 = diff(B, diff = 1)
lambda = 1e6 # fixed but can be selected (e.g. AIC)
P = lambda * t(D1) %*% D1

# Smooth
tol = 1e-8
eta = log(X + 1)
for (it in 1:20) 
  mu = exp(eta)
  z = X - mu + mu * eta
  a = solve(t(B) %*% (c(mu) * B) + P, t(B) %*% z)
  etnew = B %*% a
  de = max(abs(etnew - eta))
  cat('Crit', it, de, '\n')
  if(de < tol) break
  eta = etnew
# Plot
plot(u, exp(eta), ylim = c(0, max(X)), type = 'l', col = 2)
lines(u, X, type = 'h')

To conclude, I hope my suggestion is clear enough and replies (at least partially) the original question.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.