# Order of the kernel for periodic case

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\$$

• Could you supply a reference for this definition of order of a kernel?
– whuber
Dec 6, 2010 at 22:01
• for example Tsybakov "Introduction to nonparametric estimation ", definition 1.3 Dec 6, 2010 at 22:15
• books.google.com/… Dec 6, 2010 at 22:16

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