Dimension reduction for functional regression with a binary response

Abstract

We propose here a novel functional inverse regression method (i.e., functional surrogate assisted slicing) for functional data with binary responses. Previously developed method (e.g., functional sliced inverse regression) can detect no more than one direction in the functional sufficient dimension reduction subspace. In contrast, the proposed new method can detect multiple directions. The population properties of the proposed method is established. Furthermore, we propose a new method to estimate the functional central space which do not need the inverse of the covariance operator. To practically determine the structure dimension of the functional sufficient dimension reduction subspace, a modified Bayesian information criterion method is proposed. Numerical studies based on both simulated and real data sets are presented.

Appendix

1.1 Proof the Theorem 3.1

Let K be the true value of the dimensionality of $S_{y|x}$. By the law of large number, we have $\hat{\varGamma }_e-\varGamma _e=O_p(1/\sqrt{n})$. And the Theorem 2.7 of Horváth and Kokoszka (2012), we can get

$$\begin{aligned} \hat{\lambda }_i-\lambda _i=O_p(1/\sqrt{n}). \end{aligned}$$

Therefore, If $\hat{K}>K,$

$$\begin{aligned} G(\hat{K})-G(K)=&\left( \sqrt{n}\frac{\sum _{i=1}^{\hat{K}}\hat{\lambda }_i}{\sum _{i=1}^{p}\hat{\lambda }_i}-2\ln (n)\frac{\hat{K}}{p}\right) -\left( \sqrt{n}\frac{\sum _{i=1}^K\hat{\lambda }_i}{\sum _{i=1}^{p}\hat{\lambda }_i}-2\ln (n)\frac{K}{p}\right) \nonumber \\ =&\sqrt{n}\frac{\sum _{i=K+1}^{\hat{K}}\hat{\lambda }_i}{\sum _{i=1}^p\hat{\lambda }_i}+2\ln (n)\frac{K-\hat{K}}{p} \rightarrow \sqrt{n}\frac{\sum _{i=K+1}^{\hat{K}}\lambda _i}{\sum _{i=1}^p\lambda _i}\nonumber \\&\qquad \qquad \qquad \qquad +2\ln (n)\frac{K-\hat{K}}{p}>0. \end{aligned}$$

(9)

If $\hat{K}<K$,

$$\begin{aligned} G(\hat{K})-G(K)=&\left( \sqrt{n}\frac{\sum _{i=1}^{\hat{K}}\hat{\lambda }_i}{\sum _{i=1}^p\hat{\lambda }_i}-2\ln (n)\frac{\hat{K}}{p}\right) -\left( \sqrt{n}\frac{\sum _{i=1}^K\hat{\lambda }_i}{\sum _{i=1}^p\hat{\lambda }_i}-2\ln (n)\frac{K}{p}\right) \nonumber \\&=-\sqrt{n}\frac{\sum _{i=\hat{K}+1}^K\hat{\lambda }_i}{\sum _{i=1}^p\hat{\lambda }_i}+2\ln (n)\frac{K-\hat{K}}{p} \rightarrow 2\ln (n)\frac{K-\hat{K}}{p}>0. \end{aligned}$$

(10)

It follows from (9) and (10) that $\hat{K}\rightarrow K$ in probability.