3
$\begingroup$

Are there cases where a kernel function, must have 1 as the maximum value ??

The definition of a Kernel can be found in the following link, https://en.wikipedia.org/wiki/Kernel_(statistics)#In_non-parametric_statistics

For example, in Approximate Bayesian Computing (ABC), we approximate the true likelihood of the data with the following integral

$$\int K_{h}(x_{true}-x)p(x|\theta)dx$$

where $x_{true}$ is our observed data, and $p(x|\theta)$ the model from which we sample.

In this case, can we assume that $K_{h}(x_{true}-x)$ is a weight function, so the sum across all $x$ has to be 1, so we can assume that the maximum value of $K_{h}(x_{true}-x)$ is 1?

Because it is known that a scaled kernel, in general, is no constrained to have a maximum value of 1.

But in the case of the ABC approximation, should it have 1 as each maximum value?

$\endgroup$
3
$\begingroup$

In the original ABC version (Tavaré et al., 1999), the (dichotomous) probability of acceptance of a simulation $x(\theta)$ from $p(\cdot|\theta)$ (and hence of the parameter value $\theta$) is $$\mathbb I_{|x_\text{obs}-x(\theta)|<\epsilon}\in\{0,1\}$$ It is a natural generalisation (Fearnhead and Prangle, 2010) to consider a smoother function of the distance like $$K(|x_\text{obs}-x(\theta)|/\epsilon)$$ provided of course that $K(|x_\text{obs}-x(\theta)|/\epsilon)\in(0,1)$. There is no constraint though that $K(\cdot/\epsilon)$ is a probability density and indeed it does not integrate to one (since the rescaled version of a true density $K$ would be $K(\cdot/\epsilon)/\epsilon$ instead). Nonetheless, the name kernel was adopted (or abused!) for $K(\cdot/\epsilon)$ by analogy with non-parametrics (Blum and François, 2010).

The associated ABC posterior then writes as $$\pi(\theta)K(|x_\text{obs}-x(\theta)|/\epsilon)p(x|\theta)\Big/ \int_\Theta\int_{\mathfrak X} \pi(\theta)K(|x_\text{obs}-x(\theta)|/\epsilon)p(x|\theta)\text dx\text d\theta$$ by an accept reject argument, which does not require $K(\cdot/\epsilon)$ to be a probability density.

$\endgroup$
2
  • $\begingroup$ So, If I understand well, I can consider the kernel (if it belongs in (0,1) ) as a smoothed version of the cut-off acceptance probability? $\endgroup$
    – Fiodor1234
    Mar 3 '21 at 11:52
  • $\begingroup$ A smoothed version of the indicator function. $\endgroup$
    – Xi'an
    Mar 3 '21 at 11:53
3
$\begingroup$

The answer is no. You must have that $\int K_h(u)\mathrm{d}u = 1$, but that doesn't mean that $K_h(u) \leq 1$. The same applies to pdfs: a pdf $f$ can take values bigger than $1$, but it's integral $\int_a^b f(x)\mathrm{d}x = P(a \leq X \leq b)$ must be always less than $1$.

You can think of the uniform distribution centered at $0$ with support $(-\epsilon, \epsilon)$ for an example: the pdf (and kernel) is $$ K_\epsilon(x) = \dfrac{1}{2\epsilon}\mathbb{I}(-\epsilon \leq x \leq \epsilon)$$ Just set $\epsilon$ to any small value ( e.g. $\varepsilon\approx 0.001$) and you get that $K_\epsilon$ takes only two values: either $0$ or $500 = 1/0.002$.

TLDR: A kernel or a pdf can take arbitrarily high $y$-values, but the higher the value, the smaller the support.

$\endgroup$
6
  • $\begingroup$ Thank you for your answer. In the case, where you consider the scaled kernel as weight then we have to divide it by each maximum value, and then it would be true that it has max value 1?? $\endgroup$
    – Fiodor1234
    Mar 3 '21 at 9:45
  • $\begingroup$ What are you using this kernel for ? If your goal is to estimate a density (KDE) then you don't need to normalize anything. $\endgroup$
    – ArnoV
    Mar 3 '21 at 9:46
  • $\begingroup$ I'm using it for the Approximate Bayesian Computation. Where you approximate the likelihood with the integral $\int K_{h}(x_{true}-x)p(x|\theta)dx$ $\endgroup$
    – Fiodor1234
    Mar 3 '21 at 9:51
  • $\begingroup$ And maybe that scaled kernel is used as a weight, i.e weights which $x$ is closer to $x_{true}$. $\endgroup$
    – Fiodor1234
    Mar 3 '21 at 9:51
  • $\begingroup$ I'm no ABC expert, maybe you should change your question to include ABC specific terminology and narrow the topic, this way users with ABC experience will answer your question(s) ! $\endgroup$
    – ArnoV
    Mar 3 '21 at 9:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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