2
$\begingroup$

This question is related to my previous question Bias for kernel density estimator (periodic case)

A kernel $K(x)$ is of the order $p$ if $$\int_{-\infty}^{\infty}K(x)x^{j}=\delta_{0,j}\ j=0,...p-1$$ $$\int_{-\infty}^{\infty}K(x)x^{p}\neq0\ $$

Does it mean that for the kernel with period 1 the definition of the order of the kernel is $$\int_{0}^{1}K(x)Min(x,1-x)^{j}=\delta_{0,j}\ j=0,...p-1$$ $$\int_{0}^{1}K(x)Min(x,1-x)^{p}\neq0\ $$

$\endgroup$
  • $\begingroup$ Could you supply a reference for this definition of order of a kernel? $\endgroup$ – whuber Dec 6 '10 at 22:01
  • $\begingroup$ for example Tsybakov "Introduction to nonparametric estimation ", definition 1.3 $\endgroup$ – Katja Dec 6 '10 at 22:15
  • $\begingroup$ books.google.com/… $\endgroup$ – Katja Dec 6 '10 at 22:16
2
$\begingroup$

I think the correct analog of this definition in the periodic case is that coefficients $1$ through $p-1$ of the Fourier Series for $K$ all vanish.

The purpose of the definition of order is to obtain estimates of the bias of the kernel estimator. When $K$ "kills" powers $1$ through $p-1$ of $x$, then the bias will be approximately of order $h^p$ for a bandwidth $h$. This is proven in Tsybakov's Proposition 1.2 by expanding the pdf in a power series: multiplication by $K$ kills off the terms through order $p-1$, leaving the Taylor error term of order $p$; elementary estimates of that integral finish the job.

The analog of a power series for periodic functions is the Fourier Series. The analog is a perfect one: we can think of a periodic function as being defined on the unit circle in the complex plane. It has a complex coordinate $q = e^{i x}$ (where now the period is $2\pi$ rather than $1$, but that's inconsequential). Expanding $K(q)$ in a power series expresses it as a sum of powers of $q$. However, from

$$q^j = (e^{i x})^j = e^{i x j} = \cos(j x) + i \sin(j x)$$

we see that this expansion is just the Fourier Series (both the sine and cosine terms). Consequently you should be able to emulate the proof of Proposition 1.2 with very little change at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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