I am trying to understand kernel density estimation and found the graphic below illustrating different kernel functions on Wikipedia. I have no trouble reconciling it with the two statements

  1. "the kernel integrates to one", i.e. $\int K(z)\mathrm{d}z=1$, and
  2. "the kernel is symmetric around zero", i.e. $K(z)=K(-z)$.

But I've several times seen the statement "the kernel has zero mean" and I can't make sense out of it.

Does it mean

  1. $\mathbb{E}[K(z)]=\int K(z)z\mathrm{d}z=0$ or
  2. something else entirely?

Some common kernels

  • 2
    $\begingroup$ The first thing. $\endgroup$
    – Glen_b
    Commented May 16, 2015 at 8:57
  • $\begingroup$ Why this property of zero mean of a kernel is important in kernel density estimation? $\endgroup$ Commented May 17, 2015 at 5:11
  • 1
    $\begingroup$ @xeon Otherwise, the convolution of the kernel with a distribution changes the location. That can be compensated by subtracting the kernel's mean from the result--which is identical to centering the kernel at a zero mean in the first place. So, no importance whatsoever is attached to having a zero mean, but it is useless to have a nonzero mean. $\endgroup$
    – whuber
    Commented May 19, 2015 at 20:36
  • 1
    $\begingroup$ Note that symmetry (property 2) together with the existence of a first moment of $K$ implies that moment must be zero: stats.stackexchange.com/questions/46843. $\endgroup$
    – whuber
    Commented May 19, 2015 at 20:39

1 Answer 1


The first statement is almost, but not entirely correct. $K$ is a density function and what we mean when we refer to the mean of the density function $K$ is $$\int K(z)z\mathrm{d}z.$$ Similarly, the variance of $K$ is $$\int K(z)z^2\mathrm{d}z.$$ So, it doesn't make any sense to write $\mathbb{E}[K(z)]$. Rather, the statement "the kernel has zero mean" should be formally written as $$\int K(z)z\mathrm{d}z=0.$$ (Thanks, Glen_b!)


Your Answer

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

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