Which is kernel similar gaussian kernel?

Question

I must find a kernel that statisfies as follows:

In the my reference paper, the author suggest gaussian kernel that is

The purpose of that kernel is that it will take a weight for each points around center point (mean value), ànd that weighted value decreases drastically to zero as points go away from center point

The gaussian kernel is very good to approximate that properties. But selectec the kernel size is very challenge task. Could you suggest to me the other kernel that can satisfy above properties And I have a ball kernel that defined

Which is different between ball kernel and gaussian kernel? It is similar the average kernel,right?

Answer 1

Take any decreasing function of the norm such that the integral in (3) converges, and use the value of the integral as a normalization factor.

The simplest is the box function, $f(||u||)=1$ for $||u||\le\sigma$ and $0$ elsewhere. The normalization factor will be the volume of the hypersphere $\frac{\pi^{n/2}\sigma^n}{\Gamma(n/2+1)}$.

If you convolve this kernel with itself several times, you will obtain smoother and smoother kernels of finite support, that still fulfill the requirements.

As a goodie, these kernels are separable, just like the Gaussian.

UPDATE:

Caution: the box filter is separable only in the case of a cube, not a sphere.

Answer 2

You can use a standard symmetric mollification kernel. Such a function is compactly supported but this is in agreement with your conditions.

http://en.wikipedia.org/wiki/Mollifier

More precisely: Take

$$ \varphi: x\mapsto C \exp\left(-\frac{1}{1 - x^2}\right) $$ ($x\in (-1,1)$ and $\varphi=0$ else) where $C$ is chosen such that $$ \frac{1}{C}=\int_{\mathbb R} \exp\left(-\frac{1}{1 - x^2}\right)\mathrm dx. $$

EDIT: Observe that every function $f_\lambda (x)=\frac{1}{\lambda}\varphi\left(\frac{x}{\lambda}\right)$ for $\lambda >0$ has the desired properties.