Is alpha*RBF a valid kernel, where alpha >= 0 is a parameter?

Question

I wonder if K = alpha*RBF can be a valid kernel satisfying Mercer's condition, where alpha >= 0 is a parameter.

As the parameter alpha >= 0 is a non-negative number and RBF is already a valid kernel, so their composition should also be a valid kernel. However, the Gram matrix thus created, does not have diagonal elements 1 which is generally true for any other kernel's Gram matrix, say, RBF kernel's Gram matrix has all diagonal elements 1.

Answer 1

Yes.

Fact about kernels:

$K$ is a kernel iff for some $\phi : \mathbb{R}^n \to \mathbb{H}$, ($\mathbb{H}$ is an appropriate Hilbert space)

$K(x,y) = \langle\phi(x), \phi(y) \rangle$

From this fact, for $\alpha \geq 0$, $\alpha K(x,y) = \alpha \langle\phi(x), \phi(y) \rangle = \langle \sqrt{\alpha} \phi(x), \sqrt{\alpha} \phi(y) \rangle$

It follows that $K'(x,y) = \alpha K(x,y)$ is a kernel for $\phi'(x) = \sqrt{\alpha}\phi(x)$

A kernel doesn't have to satisfy $K(x,x) = 1$. The simplest example for which $K(x,x) \neq 1$ is linear kernel, $K(x,y) = \langle x,y \rangle$ for $\|x\| \neq 1$

In fact, from the definition of kernel, it follows that $K(x,x) = 1$ if and only if $\|x\|=1$.

Answer 2

Yes. In fact, for any positive-definite kernel $k(x, y)$, $k'(x, y) = \alpha k(x, y)$ for $\alpha \ge 0$ is also positive definite.

Here is one definition of positive-definiteness: for any $m$, any set of points $\{x_i\}_{i=1}^m$, and any corresponding set of weights $w_i$, we have that $$ \sum_{i=1}^n \sum_{j=1}^n w_i k(x_i, x_j) w_j \ge 0 .$$

If this holds for $k$, then it also holds for $k'$, simply because $$ \sum_{i=1}^n \sum_{j=1}^n w_i k'(x_i, x_j) w_j = \alpha \sum_{i=1}^n \sum_{j=1}^n w_i k(x_i, x_j) w_j \ge 0 .$$

For another example of using a bunch of properties like this, check out this answer.

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As mentioned by others already, it is not a general requirement that $k(x, x) = 1$. That's true of Gaussian RBF kernels, but it's not required for kernels in general, and the linear kernel $k(x, y) = x^T y$ is a simple counterexample.