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!

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    $\begingroup$ 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
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    $\begingroup$ @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


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


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

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 $r=2$.

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

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