Say I have the following kernel function: $$K(u) = \frac{15}{16}(1-u^2)^2\ 1_{(|u|\leq1)}$$

As I understand, $u$ denotes some absolute distance parameter for each observation $x_i$ from a given point $x$, so: $|x-x_i|$. Then the indicator function above is denoting that the above kernel function is only evaluated for points within the determined band (i.e. within the kernel shape), that is, only for data points whose $u$ distance parameter satisfies: $(|u|\leq1)$.

I have two questions on this:

  1. In $(|u|\leq1)$, why does the inequality have to be equal or greater to 1? Shouldn't any determination of what falls within the boundaries of the kernel shape depend on (or also) the bandwidth parameter ($h$)? e.g. if $h$=0.5 and the point at the center of the x-axis of the kernel shape equals $m$, then the width of the shape equals ($m+h$ and $m-h$), we would then evaluate the kernel function for points that falls within this distance.

  2. Another post touched on the topic of how to introduce the bandwidth parameter into a kernel function. I read it, but I am still not completely clear on the rationale. From what I gathered, the correct way to introduce this parameter, at least for the bi-weight case, is: $K(u) = \frac{15}{16h}(1-(\frac{|u|}{h})^2)^2$. Is this correct, and if yes, why is the $h$ needed both next to the 16 and dividing the $u$?

  1. The kernel function $K(u)$ is the standard kernel, with $h=1$. To obtain the kernel with bandwidth $h$, you write $K_h(u) = \frac{1}{h}K(\frac{u}{h})$.

  2. You need the $\frac{1}{h}$ to make the kernel integrate to 1. Equivalently, you get it by making a subsitution. Start with a standard kernel $K(z)$ and then make a substitution (change of variable) $z = \frac{u}{h}$. The "Jacobian" of the transformation yields the $\frac{1}{h}$.

    This same effect happens with any scale parameter of a density. e.g. it's why there's a $\frac{1}{\sigma}$ out the front of the pdf for a normal distribution with standard deviation $\sigma$.

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  • $\begingroup$ Thanks. For question 1, I'm not clear whether you're addressing what I posted. $\endgroup$ – StatsScared Jun 14 '16 at 18:16
  • $\begingroup$ @StatsScared I explained why the bounds on $u$ in the definition of $K$ don't depend on $h$ ,,, which is that $K(u)$ has $h=1$. $K_h$ has the corresponding restriction (which is in terms of $h$) hold automatically because of how it's defined in terms of $K$. $\endgroup$ – Glen_b Jun 14 '16 at 22:28
  • $\begingroup$ Thank you for elaborating, @Glen_b. I'm still not clear though.. If the bandwidth is h=2.5, for example, would I have to modify the bounds on the $u$? $\endgroup$ – StatsScared Jun 20 '16 at 20:30
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    $\begingroup$ Have you tried evaluating it? If $u=2.4$, say, then $z=u/h = 2.4/2.5 = 0.96$, which is smaller than 1 in absolute value, i.e. within the domain of $K(z)$. Clearly if $u$ is between $-h$ and $h$ then $u/h$ is between $-1$ and $1$ and you can evaluate $K()$ there. $\endgroup$ – Glen_b Jun 21 '16 at 1:12
  • $\begingroup$ Thanks, again. So just to be completely clear, writing $K_(u) = \frac{15}{16}(1-u^2)^2\ 1_{(|u|\leq1)}$ would only be correct if $h=1$. For all other cases, the notation should be the following: $K_h(u) = \frac{15}{16h}(1-u^2)^2\ 1_{(|u|/h\leq1)}$ ? This also makes me wonder, why the change of variable $u=\frac{X_i-x}{h}$ is never explicitly listed in kernel notation.. $\endgroup$ – StatsScared Jun 23 '16 at 16:10

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