Is kernel density in kernel density estimation derived or defined? If defined, why is it defined this way, if derived, how to derive it? In particular, why $h^d$ and not $h$ in the multivariate case, where kernel density is defined as $$\frac{1}{nh^d}\sum_{i=1}^n K(\frac{x-x_i}{h})$$
-
2$\begingroup$ Could you please clarify what you mean by "derive" and "define"? Your second question is answered by considering that the integral of the kernel density must be $1$. $\endgroup$ – whuber♦ Sep 26 '17 at 17:01
-
$\begingroup$ Derive meaning it can be gotten from a base formula and expand to become this. Define meaning someone thinks it is a good idea for density to look this way so it cannot be explained. $\endgroup$ – user10024395 Sep 27 '17 at 1:40
-
$\begingroup$ @user2675516 what "base formula"? Given such definition, all of the statistics, probability theory and mathematics are "defined" since they all are based on some definitions, assumptions and abstract models... $\endgroup$ – Tim♦ Mar 1 '18 at 8:26
$\begingroup$
$\endgroup$
You can "derive" it from the empirical distribution function $$ F_n(x) = \frac{1}{n} \sum_{i=1}^{n}\mathbf{1}_{\{X_i \leq x\}}. $$ Since the density function at a point x of its support is just defined as the derivative of the cumulative distribution function, a straightforward estimator of the density function is $$ \hat{f}_{n}(x) = \frac{F_n(x + h_n) - F_n(x - h_n)}{2h_n} = \frac{1}{2nh_n}\mathbf{1}_{\{x-h_n \leq X_i \leq x+h_n\}}. $$ This kernel estimator would give uniform weight $\frac{1}{2}$ to each observation in the window $(x-h_n,x+h_n)$. This motivates the estimation of the nonparametric density with a smoothed kernel function: $$ \hat{f}_n(x) = \frac{1}{nh_n}\sum_{i=1}^{n} K \left( \frac{X_i-x}{h_n} \right). $$
