# Is kernel density in kernel density estimation derived or defined?

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 $$\frac{1}{nh^d}\sum_{i=1}^n K(\frac{x-x_i}{h})$$

• 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 $1$. – whuber Sep 26 '17 at 17:01
• 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. – user10024395 Sep 27 '17 at 1:40
• @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... – Tim Mar 1 '18 at 8:26

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: $$\hat{f}_n(x) = \frac{1}{nh_n}\sum_{i=1}^{n} K \left( \frac{X_i-x}{h_n} \right).$$