in Others edited by
1,187 views
0 votes
0 votes

$K\left(x, x^{\prime}\right)=f(x) g\left(x^{\prime}\right)+f\left(x^{\prime}\right) g(x)$, where $f$ and $g$ are real-valued functions $\left(\mathcal{R}^{D} \rightarrow \mathcal{R}\right)$ is not a valid kernel function. What additional terms would you include to the summation in $K$ to make it a valid kernel?

  1. $f(x)+g(x)$
  2. $f(x) g(x)+f\left(x^{\prime}\right) g\left(x^{\prime}\right)$
  3. $f(x) f\left(x^{\prime}\right)+g(x) g\left(x^{\prime}\right)$
  4. $f\left(x^{\prime}\right)+g\left(x^{\prime}\right)$
in Others edited by
by
1.2k views

1 Answer

0 votes
0 votes
To make $K(x, x')$ a valid kernel function, it needs to satisfy the following conditions:

Symmetry: \( K(x, x') = K(x', x) \) for all inputs \( x \) and \( x' \).

Non-negativity: \( K(x, x) \geq 0 \) for all inputs \( x \).

The original function $K(x, x') = f(x)g(x') + f(x')g(x)$ lacks symmetry. Adding the terms  

$f(x)g(x)$  and  $f(x')g(x')$ addresses this issue:

New function: $K(x, x') = f(x)g(x') + f(x')g(x) + f(x)g(x) + f(x')g(x')$

This new function satisfies both conditions:

Symmetry: It is evident that \( K(x, x') = K(x', x) \) due to the arrangement of terms.

Non-negativity:It's always true that \( f(x)g(x) \geq 0 \) and \( f(x')g(x') \geq 0 \) for

real-valued functions \( f \) and \( g \), ensuring non-negativity.

Therefore, by adding the terms B) $f(x)g(x) + f(x')g(x')$ , we create a valid kernel function.
edited by

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

64.3k questions

77.9k answers

244k comments

80.0k users