0
$\begingroup$

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!

$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Jul 3, 2020 at 14:17
  • 1
    $\begingroup$ @whuber You are right. Thanks! And the problem is solved too, i.e., it's a second order kernel. $\endgroup$ Jul 3, 2020 at 18:05
  • $\begingroup$ @whuber There seems to be two versions of definitions for this order that differ by 1. $\endgroup$ Jul 3, 2020 at 18:49

1 Answer 1

0
$\begingroup$

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 $r=2$.

$\endgroup$
1
  • $\begingroup$ Thank you! This is helpful. $\endgroup$ Jul 3, 2020 at 23:36

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.