# What is the order of this kernel function?

Suppose we have kernel function $$K\left(t\right)=\frac{35}{32}\left(1-t^2\right)^3\mathbf{1}\left(|t|\leq 1\right)$$. What is the minimal positive integer $$r$$ such that $$\int K(t)t^rdt\neq0$$?(this minimal positive integer $$r$$ is the order of the kernel.) It's easy to see that $$\int K(t)tdt=0$$, but integration $$\int K(t)t^rdt$$ for $$r>1$$ seems difficult to compute. Any ideas on how to compute them so that we can pin down the order of the kernel? Thanks!

• Because $K$ is a symmetric even function, the integral is strictly positive for all even $r$ and zero for all odd $r.$ In light of this, it looks like your definition of the "order" might be incorrect. Indeed, it conflicts with Wikipedia--check it out and fix your question accordingly.
– whuber
Jul 3, 2020 at 14:17
• @whuber You are right. Thanks! And the problem is solved too, i.e., it's a second order kernel. Jul 3, 2020 at 18:05
• @whuber There seems to be two versions of definitions for this order that differ by 1. Jul 3, 2020 at 18:49

Using the definition of kernel order found in "Choice of Kernel Order in Density Estimation" Peter Hall and J. S. Marron The Annals of Statistics Vol. 16, No. 1 (Mar., 1988), pp. 161-173

https://www.jstor.org/stable/2241429?seq=1

$$\int_{-1}^{1}\frac{35}{32}(1-t^2)^3dt=1$$

With $$r=1$$:

$$\int_{-1}^{1}\frac{35}{32}(1-t^2)^3t^1=0 \quad \text{(odd function)}$$

$$r=2$$ is the smallest integer for which the integral is different from 0:

$$\int_{-1}^{1}\frac{35}{32}(1-t^2)^3t^2=1/9 \quad \text{(even function)}$$

(The integral can be easily calculated expanding the polynomial).

Thus, the order is $$r=2$$.

• Thank you! This is helpful. Jul 3, 2020 at 23:36