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In chapter 3 & 6 of Bishop's Pattern Recognition and Machine Learning, he showed that the equivalent kernel based on eqn (3.62)

$$ k(x,x') = \beta \phi(x)^T (\alpha I + \beta \Phi^T \Phi )^{-1}\phi(x')$$

and the standard kernel based on the feature space mapping eqn (6.10),

$$ k(x,x') = \phi(x)^T \phi(x') = \sum_{j=1}^M \phi_j(x)\phi_j(x') $$

both of which have a sinc like behavior corresponding to Figures 3.11(left) and 6.1(left), see below.

enter image description here

This is quite impossible given that when $x=0$, all $\phi_j(x)=x^j=0$, which implies $\phi(x)^T\phi(x')=0\ \forall x'$. Is my assumption that $\phi_j(x)=x^j$ wrong?

My understanding is that polynomial basis are global functions, so how can one arrive at the conclusion that its kernel have a local behavior, it's not obvious to me at this point.

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    $\begingroup$ This question will not be answerable unless you reproduce the necessary context, such as equations and figures. $\endgroup$
    – Sycorax ♦
    Commented Jul 5, 2018 at 0:10
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    $\begingroup$ Added equations and figures. $\endgroup$
    – David KWH
    Commented Jul 5, 2018 at 15:41
  • $\begingroup$ Thank you! The question is now in the "reopen" queue, where reviewers will voted on whether the question is sufficiently clear to be re-opened. (I have voted to re-open the question.) $\endgroup$
    – Sycorax ♦
    Commented Jul 5, 2018 at 16:06

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I had the same question, and it turns out that the figure is wrong! The corrected one appears on the book errata: Link

enter image description here

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