One-sided cross-validation for nonsmooth density functions

Abstract

One-sided cross-validation (OSCV) is a bandwidth selection method initially introduced by Hart and Yi (J Am Stat Assoc 93(442):620–631, 1998) in the context of smooth regression functions. Martínez-Miranda et al. (in Gregoriou (ed) Operational risk towards basel III: best practices and issues in modeling, management and regulation, Wiley, Hoboken, 2009) developed a version of OSCV for smooth density functions. This article extends the method for nonsmooth densities. It also introduces the fully robust OSCV modification that produces consistent OSCV bandwidths for both smooth and nonsmooth cases. Practical implementations of the OSCV method for smooth and nonsmooth densities are discussed. One of the considered cross-validation kernels has potential for improving the OSCV method’s performance in the regression context.

Appendix

1.1 Notation

For an arbitrary function g, define the following functionals:

$$\begin{aligned} \begin{array}{l} \displaystyle {D_g(z)=\int _{-\infty }^z g(u)\,du},\\ \displaystyle {G_g(z)=\int _{-\infty }^z ug(u)\,du,\qquad z\in {\mathbb {R}}}. \end{array} \end{aligned}$$

Based on $D_g$ and $G_g$ we define

$$\begin{aligned} B(g)=\int _0^\infty \left\{ z\bigl (1-D_g(z)\bigr ) +G_g(z)\right\} ^2\,dz+\int _0^\infty \left\{ zD_g(-z)+G_g(-z)\right\} ^2\,dz.\nonumber \\ \end{aligned}$$

(12)

The theoretical results in this article involve B(K) and B(L) for the two-sided and one-sided kernels K and L, respectively. In the case when K is symmetrical, one may derive from (12) that

$$\begin{aligned} B(K)=2\int _0^\infty \left\{ z\bigl (1-D_K(z)\bigr )+G_K(z)\right\} ^2\,dz. \end{aligned}$$

By taking into account that L is supported on $[0,\infty )$, it appears that

$$\begin{aligned} B(L)=\int _0^\infty \left\{ z\bigl (1-D_L(z)\bigr )+G_L(z)\right\} ^2\,dz. \end{aligned}$$

1.2 Identity of the left-sided and right-sided OSCV functions in the case of a symmetrical generating kernel H

As we noted above, $K_R(u)=K_L(-u)$ in the case when H is symmetrical.

Theorem

In the case when H is symmetrical, $\text{ OSCV }_{K_L}(b)=\text{ OSCV }_{K_R}(b)$ for any $b>0$. This implies ${{\hat{h}}}_{OSCV,K_L}={{\hat{h}}}_{OSCV,K_R}$.

Proof

For any $b>0$, consider

$$\begin{aligned} R({{\hat{f}}}_{b,K_L})=\frac{1}{n^2b^2}\sum _{i=1}^n\sum _{j=1}^n\int _{-\infty }^\infty K_L\left( \frac{x-X_i}{b}\right) K_L\left( \frac{x-X_j}{b}\right) \,dx. \end{aligned}$$

After the change of variables $u=\displaystyle {\frac{x-\frac{X_i+X_j}{2}}{b}}$, we get

$$\begin{aligned} R({{\hat{f}}}_{b,K_L})= & {} \frac{1}{n^2b}\sum _{i=1}^n\sum _{j=1}^n\int _{-\infty }^\infty K_L\left( u-\frac{X_i-X_j}{2b}\right) K_L\left( u+\frac{X_i-X_j}{2b}\right) \,du\\= & {} \frac{1}{n^2b}\sum _{i=1}^n\sum _{j=1}^n\int _{-\infty }^\infty K_L\left( -z-\frac{X_i-X_j}{2b}\right) K_L\left( -z+\frac{X_i-X_j}{2b}\right) \,dz\\= & {} \frac{1}{n^2b}\sum _{i=1}^n\sum _{j=1}^n\int _{-\infty }^\infty K_R\left( z+\frac{X_i-X_j}{2b}\right) K_L \left( z-\frac{X_i-X_j}{2b}\right) \,dz\\= & {} R({{\hat{f}}}_{b,K_R}). \end{aligned}$$

Next, consider

$$\begin{aligned} \sum _{i=1}^n{{\hat{f}}}_{b,K_L}^{-i}(X_i)= & {} \frac{1}{(n-1)b}\sum _{i=1}^n\sum _{j=1}^n K_L\left( \frac{X_i-X_j}{b}\right) -\frac{n}{(n-1)b}K_L(0)\\= & {} \frac{1}{2(n-1)b}\sum _{i=1}^n\sum _{j=1}^n K_L\left( -\frac{|X_i-X_j|}{b}\right) -\frac{n}{2(n-1)b}K_L(0)\\= & {} \frac{1}{2(n-1)b}\sum _{i=1}^n\sum _{j=1}^n K_R\left( \frac{|X_i-X_j|}{b}\right) -\frac{n}{2(n-1)b} K_R(0)\\= & {} \sum _{i=1}^n{{\hat{f}}}_{b,K_R}^{-i}(X_i). \end{aligned}$$

This finishes the proof that $\text{ OSCV }_{K_L}(b)=\text{ OSCV }_{K_R}(b)$ for any $b>0$. This implies equality ${{\hat{h}}}_{OSCV,K_L}={{\hat{h}}}_{OSCV,K_R}$. $\square$

The above proof can be easily adjusted for the case of a slightly differently defined OSCV functions used in Mammen et al. (2011, 2014).