3

I came across a paper and it states that a Kronecker (Dirac) delta function is a valid kernel by defining the kernel as below:

$k(x,z)=\boldsymbol{v}_x^T \cdot \boldsymbol{v}_z = \displaystyle\sum_{i=1}^{m} \boldsymbol{v}_x(i) \cdot \boldsymbol{v}_z(i) = 1 - δ(x, z)$.

Is this correct?

Thanks for any helpful reply

Jasper
  • 31
  • 2
  • This would help if you could provide a definition of v and m! The intermediate steps in the equality are not helpful as they are – RUser4512 Sep 05 '18 at 08:34
  • Sorry for lacking the definitions, v is a real-valued vector and m denotes the dimension of v, number of its components. Nothing special, very elementary concepts. – Jasper Sep 05 '18 at 09:06
  • Since "kernel" means a great variety of different things in mathematics (and statistics), please explain what definition you are working with. – whuber Sep 05 '18 at 11:54
  • "kernel" refers to the one employed in Gaussian process, sth must be positive semi-definite – Jasper Sep 06 '18 at 00:24

1 Answers1

3

Yes, this is a kernel. There are various ways to show it. You can use the following result :

The limit of a sequence of kernels is a kernel

If $\kappa_1, \kappa_2, \dots$ are kernels, and $\kappa(x, y) := \lim_{n \to \infty} \kappa_n(x, y)$ exists for all $x, y$, then $\kappa$ is a kernel.

And apply it to a gaussian kernel whose $\sigma$ tends to infinity.

However, given the simplicity of this kernel, the best way is probably to go use the definition of a kernel and prove that $\delta$ respects the properties of a kernel.

RUser4512
  • 10,217
  • Thank you for your lighting response. The sum in the equation should refer to a simple expansion of dot products involving two real-valued vectors. It may not be necessary to introduce kernel limits, but thank you for your help. – Jasper Sep 05 '18 at 09:11