0
$\begingroup$

Suppose we have feature transformation $\Phi(x) = [1, x_1, x_2, x_1x_2]$. Now we want to find the kernel corresponding to $\Phi$.

What I have done is using kernel decomposition, we have: $$ K(x, y) = \Phi(x) .\Phi(y)\\ K(x, y) = 1.1 + x_1y_1 + x_2y_2 + x_1x_2y_1y_2 \\ K(x, y) = 1 + \sum_{i=1}^{N}x_iy_i + \prod_{i=1}^{N}x_iy_i $$

$x, y$ are in $\Bbb{R}^2$

Is that kernel valid?

$\endgroup$

1 Answer 1

0
$\begingroup$

Yes, it is valid.

Kernel is just an inner product in the feature space. In fact, here you have explicitly started from the feature transformation and we understand the feature transformation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.