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I have the Epanechnikov kernel that looks like this:

$K(x) = \frac{3}{4} * (1 - x^2)$

I have added $\sigma$ to the equation so that I can adjust the width, very much like the gaussian kernel.

It looks like this:

$K(x) = \frac{3}{4} * \left(1 - \left(\frac{x}{\sigma^2}\right)^2\right)$

Can someone verify that the above equation is correct?

1 Answers 1

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A kernel is suppose to have a unit integral. You should think of it as a differential:

$K(x) = \frac{3}{4}(1-x^2) dx.$

(Of course it is understood the values are $0$ in the complement of the interval $[-1,1]$.)

To change the radius from $1$ to $\sigma$, re-express $x$ as $u / \sigma$:

$K(u) = \frac{3}{4}(1 - (u/\sigma)^2) d(u/\sigma) = \frac{3}{4\sigma}(1 - (u/\sigma)^2).$

(Now it is understood $K$ is $0$ in the complement of the interval $[-\sigma, \sigma]$.)

Geometrically this is clear: uniformly stretching the kernel out so it is supported on $[-\sigma, \sigma]$ would multiply its integral by $\sigma$, so you have to rescale its values by $1/\sigma$ to compensate.