Penalized Relative Error Estimation of a Partially Functional Linear Multiplicative Model

Abstract

Functional data become increasingly popular with the rapid technological development in data collection and storage. In this study, we consider both scalar and functional predictors for a positive scalar response under the partially linear multiplicative model. A loss function based on the relative errors is adopted, which provides a useful alternative to the classic methods such as the least squares. Penalization is used to detect the true structure of the model. The proposed method can not only identify the significant scalar variables but also select the basis functions (on which the functional variable is projected) that contribute the response. Both estimation and selection consistency properties are rigorously established. Simulation is conducted to investigate the finite sample performance of the proposed method. We analyze the Tecator data to demonstrate application of the proposed method.

Appendix

Denote δ = (θ ^⊤, β ^⊤)^⊤ and $\hat {\delta }=(\hat {\theta }^{\top }, \hat {\beta }^{\top } )^{\top }$. To facilitate proof of the theorems, we introduce Lemmas 1–3 where proof of Lemmas 1 and 2 can be found in [7] and proof of Lemma 3 can be found in [9].

Lemma 1

Under (C1), (C2), and (C5), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle n^{-1}\sum_{i=1}^{n}(\hat{\eta}_{ij}\hat{\eta}_{ik}-\eta_{ij}\eta_{ik})=O_{p}(j^{\alpha/2+1}n^{-1/2} +k^{\alpha/2}n^{-1/2}),\\ &\displaystyle n^{-1}\sum_{i=1}^{n}(\hat{\eta}_{ij}-\eta_{ij})z_{ik}=O_{p}(j^{\alpha/2+1}n^{-1/2} ). \end{array} \end{aligned} $$

Lemma 2

Under (C1)–(C5), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} n^{-1}\sum_{i=1}^{n}\sum_{j=K+1}^{\infty}\eta_{ij}\beta_{j}z_{il}=O_{p}(n^{-1/2}). \end{array} \end{aligned} $$

Lemma 3

Under (C1) and (C2), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle |\hat{\eta}_{ij}-\eta_{ij}|=O_{p}(K^{\alpha+1}/\sqrt{n}),\\ &\displaystyle \|\hat{\phi}_{j}-\phi_{j}\|=O_{p}(K^{2\alpha+2}/n). \end{array} \end{aligned} $$

Proof of Theorem 1

Let $\alpha _{n}=\sqrt {K/n}$. We first show that $\|\hat {\delta }-\delta \|=O_{p}(\alpha _{n})$. Following [4], it is sufficient to show that, for any given ε > 0, there exists a large constant C, such that

$$\displaystyle \begin{aligned} \begin{array}{rcl} P\{ \inf_{\|u\|=C}Q_{n}(\delta+\alpha_{n}u))>Q_{n}(\delta)) \}=1-\varepsilon, \end{array} \end{aligned} $$

where δ _C = {δ ^∗ = δ + α _n u, ∥u∥ = C} for C > 0. This implies that, with probability 1 − ε, there exists a minimum in the ball δ _C. Hence, the consistency of δ is established.

By some simplifications, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \psi_{n}(u)&\displaystyle =&\displaystyle Q_{n}(\delta+\alpha_{n}u))-Q_{n}(\delta))\\ &\displaystyle =&\displaystyle \sum_{i=1}^{n}\{|1-Y_{i}^{-1}\exp\{\hat{W}_{i}^{\top}(\delta+\alpha_{n}u)\}| - |1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)|\}\\ &\displaystyle &\displaystyle + \sum_{i=1}^{n}\{|1-Y_{i}\exp\{-\hat{W}_{i}^{\top}(\delta+\alpha_{n}u)\}|]- |1-Y_{i}\exp\{-\hat{W}_{i}^{\top}\delta\}|\}\\ &\displaystyle &\displaystyle +n\sum_{l=1}^{q}\{\xi_{l}|\theta_{l}+u_{l}\alpha_{n}|-\xi_{l}|\theta_{l}|\} +n\sum_{j=1}^{K}\{ \lambda_{j}|\beta_{j}+u_{j+q}\alpha_{n}|- \lambda_{j}|\beta_{j}|\} \\ &\displaystyle \equiv&\displaystyle I_{1}+I_{2}+I_{3}+I_{4}. \end{array} \end{aligned} $$

(6)

For I ₁, by applying the identity in [6],

$$\displaystyle \begin{aligned} \begin{array}{rcl} |x-y|-|x|=-y[I(x>0)-I(x<0)]+2\int_{0}^{y}[I(x\leq s)-I(x\leq0)]ds, \end{array} \end{aligned} $$

which is valid for x ≠ 0, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{1}&\displaystyle =&\displaystyle -\sum_{i=1}^{n}w_{1i}[I(1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)>0) - I(1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)<0)]\\ &\displaystyle &\displaystyle +2\sum_{i=1}^{n}\int_{0}^{w_{1i}}[I(1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)\leq s)-I(1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)\leq0)]ds\\ &\displaystyle \equiv &\displaystyle I_{11}+I_{12}, \end{array} \end{aligned} $$

where

$$\displaystyle \begin{aligned} \begin{array}{rcl} w_{1i}=Y_{i}^{-1}\{\exp(\hat{W}_{i}^{\top}(\delta+\alpha_{n}u)) -\exp(\hat{W}_{i}^{\top}\delta)\}. \end{array} \end{aligned} $$

By Taylor expansion, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{11} &\displaystyle =&\displaystyle \sum_{i=1}^{n}\alpha_{n}u\hat{W}_{i}^{\top}Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta) sgn(1- Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta))\\ &\displaystyle &\displaystyle +\alpha_{n}^{2}u^{\top}\{\sum_{i=1}^{n}\hat{W}_{i}^{\top}\hat{W}_{i}Y_{i}^{-1}\exp(\xi_{i}^{[1]}) sgn(1- Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta))\}u/2\\ &\displaystyle \equiv &\displaystyle I_{111}+I_{112}, \end{array} \end{aligned} $$

where $\xi _{i}^{[1]}$ lies between $\hat {W}_{i}^{\top }(\delta +\alpha _{n}u)$ and $\hat {W}_{i}^{\top }\delta$.

For I ₁₁₁, by Lemma 2, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{111}&\displaystyle =&\displaystyle \sum_{i=1}^{n}\alpha_{n}u\hat{W}_{i}^{\top}\frac{1}{\varepsilon_{i}} sgn(1- Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta))\\ &\displaystyle &\displaystyle +\sum_{i=1}^{n}\alpha_{n}^{2}u^{\top}\frac{1}{\varepsilon_{i}}\hat{W}_{i}^{\top} \hat{W}_{i}u sgn(1- Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta))+o_{p}(1)\\ &\displaystyle \equiv &\displaystyle I_{1111}+I_{1112}+o_{p}(1). \end{array} \end{aligned} $$

It follows directly from Lemma 1 and (C5) that I ₁₁₁₁ = O _p(1)∥α _n u∥. Moreover, under (C5), $I_{1112}=\alpha _{n}^{2}u^{\top }D_{2}u/2$ and $I_{112}=\alpha _{n}^{2}u^{\top }D_{2}u/2+o_{p}(1)\|\alpha _{n}u\|{ }^{2}$.

Hence,

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{11}=O_{p}(1)\|\alpha_{n}u\|+\alpha_{n}^{2}u^{\top}Du/2+o_{p}(1)\|u\|. \end{array} \end{aligned} $$

For I ₁₂, denote $c_{1i} = \exp (\hat {W}_{i}^{\top }\hat {\delta }-W_{i}^{\top }\delta -H)$, $c_{2i} = \exp (\hat {W}_{i}^{\top }\delta -W_{i}^{\top }\delta -H)$ and τ = sε _i, then

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{12}&\displaystyle =&\displaystyle 2\sum_{i=1}^{n}\int_{0}^{w_{1i}}[I(1-\varepsilon_{i}^{-1}c_{2i}\leq s)-I(1-\varepsilon_{i}^{-1}c_{2i}\leq0)]ds\\ &\displaystyle =&\displaystyle 2\sum_{i=1}^{n}\int_{0}^{c_{1i}-c_{2i}}\varepsilon_{i}^{-1}[I(\varepsilon_{i}\leq c_{2i}+\delta) -I(\varepsilon_{i}\leq c_{2i})]d\delta\\ &\displaystyle =&\displaystyle 2\sum_{i=1}^{n}\int_{0}^{c_{1i}-c_{2i}}E_{\varepsilon|X}\{\varepsilon_{i}^{-1}[I(\varepsilon_{i} \leq c_{2i}+\tau)-I(\varepsilon_{i}\leq c_{2i})]\}d\tau+o_{p}(1)\\ &\displaystyle =&\displaystyle 2\sum_{i=1}^{n}\int_{0}^{c_{1i}-c_{2i}}E_{\varepsilon|X}\{[I(\varepsilon_{i}\leq c_{2i}+\tau) -I(\varepsilon_{i}\leq c_{2i})]\}d\delta\\ &\displaystyle &\displaystyle +2\sum_{i=1}^{n}\int_{0}^{c_{1i}-c_{2i}}E_{\varepsilon|X}\{(\varepsilon_{i}^{-1}-1) [I(\varepsilon_{i}\leq c_{2i}+\tau)-I(\varepsilon_{i}\leq c_{2i})]\}d\delta+o_{p}(1)\\ &\displaystyle =&\displaystyle \alpha_{n}^{2}u^{\top}\sum_{i=1}^{n}\hat{W}_{i}\hat{W}_{i}^{\top}u[1+o_{p}(1)]\\ &\displaystyle =&\displaystyle \alpha_{n}^{2}u^{\top}D_{2}u+o_{p}(1)\|\alpha_{n}u\|{}^{2}. \end{array} \end{aligned} $$

Therefore, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{1}=O_{p}(1)\|\alpha_{n}u\|+\alpha_{n}^{2}u^{\top}D_{2}u/2+o_{p}(1)\|u\|. \end{array} \end{aligned} $$

(7)

Similarly, it can be shown that

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{2}=O_{p}(1)\|\alpha_{n}u\|+\alpha_{n}^{2}u^{\top}D_{1}u/2+o_{p}(1)\|u\|. \end{array} \end{aligned} $$

(8)

Adopting the approach in [6], we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle |\theta_{l}+u_{l}\alpha_{n}|-|\theta_{l}| \rightarrow \alpha_{n}\{u_{l}sgn(\theta_{l})I(\theta_{l}\neq 0)+|u_{l}|I(\theta_{l}= 0)\},\\ &\displaystyle |\beta_{j}+u_{j+q}\alpha_{n}|-|\beta_{j}| \rightarrow \alpha_{n}\{u_{j+q}sgn(\beta_{j})I(\beta_{j}\neq 0)+|u_{j+q}|I(\beta_{j}= 0)\}. \end{array} \end{aligned} $$

According to (C7), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{3}>O(1)\|u\| \end{array} \end{aligned} $$

(9)

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{4}>O(1)\|u\|. \end{array} \end{aligned} $$

(10)

Combining (6)–(10), for a sufficiently large C, we have ψ _n(u) > 0. Therefore, $\|\hat {\delta }-\delta \|=O_{p}(\sqrt {K/n})$.

Next we show $\|\hat {\gamma }(t)-\gamma (t)\|=o_{p}(1)$. In fact,

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|\hat{\gamma}_{1}(t)-\gamma_{1}(t)\| \leq \sum_{j=1}^{K}\|\hat{\beta}_{j}\hat{\phi}_{j}(t)-\beta_{j}\phi_{j}(t)\|+\| \sum_{j>K}\beta_{j}\phi_{j}(t)\|\\ &\displaystyle \leq &\displaystyle \sum_{j=1}^{K}[|\hat{\beta}_{j}-\beta_{j}|\|\phi_{j}(s)\|+|\hat{\beta}_{j}-\beta_{j}|\|\hat{\phi}_{j}(s)-\phi_{j}(s)\| +|\beta_{j}|\|\hat{\phi}_{j}(s)-\phi_{j}(s)\|]\\ &\displaystyle &\displaystyle +\| \sum_{j>K}\beta_{j}\phi_{j}(s)\|\\ &\displaystyle \equiv&\displaystyle F_{1}+F_{2}. \end{array} \end{aligned} $$

By result of the first part and Lemma 3, we can obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} F_{1}=O_{p}(\frac{K^{2\alpha+3}}{n}). \end{array} \end{aligned} $$

By the square integrable property of γ(t), we have F ₂ = o _p(1). From the above results and condition (C4), we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle \|\hat{\gamma}(t)-\gamma(t)\| =o_{p}(1). \end{array} \end{aligned} $$

This completes the proof of Theorem 1. □

Proof of Theorem 2

For j ∈ A and k ∈ B, the consistency result in Theorem 1 indicates that $\hat {\theta }\rightarrow \theta$ and $\hat {\beta }\rightarrow \beta$ in probability. Therefore, $P(j\in \hat {A}_{n})\rightarrow 1$ and $P(k\in \hat {B}_{n})\rightarrow 1$. Then it suffices to show that $P ( j\in \hat {A}_{n} ) \rightarrow 0$ for ∀j∉A and $P ( k\in \hat {B}_{n} ) \rightarrow 0$ for ∀k∉B.

For $\forall j\in \hat {A}_{n}$ and $\forall k\in \hat {B}_{n}$, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} 0=\frac{\partial \psi_{n}(u)}{\partial u}&\displaystyle =&\displaystyle \sum_{i=1}^{n}\{|1-Y_{i}^{-1}\exp\{\hat{W}_{i}^{\top}(\delta)\}| - |1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)|\}\hat{W}_{ijk}\\ &\displaystyle &\displaystyle + \sum_{i=1}^{n}\{|1-Y_{i}\exp\{-\hat{W}_{i}^{\top}(\delta)\}|]- |1-Y_{i}\exp\{-\hat{W}_{i}^{\top}\delta\}|\}\hat{W}_{ijk}\\ &\displaystyle &\displaystyle +\sum_{i=1}^{n}\{|1-Y_{i}^{-1}\exp\{\hat{W}_{i}^{\top}(\delta)\}| - |1-Y_{i}^{-1}\exp(\hat{W}_{i}^{\top}\delta)|\}\hat{W}_{ijk}\hat{W}_{i}^{\top}u\\ &\displaystyle &\displaystyle + \sum_{i=1}^{n}\{|1-Y_{i}\exp\{-\hat{W}_{i}^{\top}(\delta)\}|]- |1-Y_{i}\exp\{-\hat{W}_{i}^{\top}\delta\}|\}\hat{W}_{ijk}\hat{W}_{i}^{\top}u\\ &\displaystyle &\displaystyle +n\alpha_{n}\xi_{j}sgn(\theta_{j})+n\alpha_{n}\lambda_{k}sgn(\beta_{k})\\ &\displaystyle \equiv&\displaystyle N_{1}+N_{2}+N_{3}+N_{4}+N_{5}+N_{6}. \end{array} \end{aligned} $$

By similar arguments with Theorem 1, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} N_{k}=O_{p}(1), k=1,\cdots,4. \end{array} \end{aligned} $$

According to (C8), we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} N_{k}\rightarrow \infty, k=5,6. \end{array} \end{aligned} $$

Consequently, we must have $P(j\in \hat {A}_{n})\rightarrow 0$ and $P(k\in \hat {B}_{n})\rightarrow 0$ for ∀j∉A and ∀k∉B. This completes the proof Theorem 2. □