On mean derivative estimation of longitudinal and functional data: from sparse to dense

Abstract

Derivative estimation of the mean of longitudinal and functional data is useful, because it provides a quantitative measure of changes in the mean function that can be used for modeling of the data. We propose a general method for estimation of the derivative of the mean function that allows us to make inference about both longitudinal and functional data regardless of the sparsity of data. The $L^2$ and uniform convergence rates of the local linear estimator for the true derivative of the mean function are derived. Then the optimal weighting scheme under the $L^2$ rate of convergence is obtained. The performance of the proposed method is evaluated by a simulation study, and additionally compared with another existing method. The method is used to analyse a real data set involving children weight growth failure.

Appendix

1.1 Proof of Lemma 1

Proof

Denote the equi-distant partition on [0, 1] by $\chi (\eta _n)$ in which the grid length is $\eta _n \equiv \left( \sum _{i=1}^n N_i \omega _i^2 \log (1/\sum _{i=1}^n N_i \omega _i^2 )\right) ^{2+r}$. Then

$$\begin{aligned} \sup _{t \in [0,1]} \vert S_{r}(t) - ES_{r}(t) \vert \le \sup _{t \in \chi (\eta _n)} \vert S_{r}(t) - ES_{r}(t) \vert +D_1 +D_2,\; \; r=0,1,\ldots ,4, \end{aligned}$$

(13)

where

$$\begin{aligned} D_1 = \sup _{\vert s-t \vert< \eta _n} \vert S_{r}(t) - S_{r}(s) \vert ,\;\; D_2 = \sup _{\vert s-t \vert < \eta _n } \vert ES_{r}(t) - ES_{r}(s) \vert . \end{aligned}$$

For the second term on the right-hand side of Eq. (13), we deduce that

$$\begin{aligned} D_1&\le \sup _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h}\vert K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - t}{h}\right) ^r - K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \vert \\&\le \sup _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \vert \sum _{l=1}^{r}\displaystyle {r\atopwithdelims ()l}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l}\\&\quad -K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \vert \\&\quad +\sup _{\vert s-t \vert < \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h}\vert K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} -s}{h}\right) ^r - K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \vert \\&= B_{1} +B_{2}, \end{aligned}$$

where

$$\begin{aligned} B_{1}= & {} \sup \limits _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h}\vert K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} -s}{h}\right) ^r - K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \vert , \\ B_{2}= & {} \sup _{\vert s-t \vert < \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \vert \sum _{l=1}^{r}\displaystyle {r\atopwithdelims ()l}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l}\\&-K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \vert . \end{aligned}$$

Note that Assumption (A1) implies that $K(.) \le M_K$ for a constant $M_K$, therefore

$$\begin{aligned} B_{1}&\le \sup _{\vert s-t \vert < \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h}\vert \frac{t_{ij} -s}{h}\vert ^{r}M_K\vert \frac{s -t}{h}\vert \le \frac{ \eta _n}{h^{r+2}}M_K\nonumber \\&=o(a_{n}). \end{aligned}$$

(14)

Using the fact that if for some i and j, $\vert \frac{t_{ij}-s}{h}\vert >1$ then $K\big (\frac{t_{ij}-s}{h}\big )=0$, we have

$$\begin{aligned} B_{21}&= \sup _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \left| \sum _{l=1}^{r} \left( \displaystyle {r \atopwithdelims ()l}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l}-\right. \right. \nonumber \\&\quad \left. \left. K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \right) I_{\lbrace \vert \frac{t_{ij}-s}{h}\vert >1 \rbrace } \right| \nonumber \\&\le \sup _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \left| \sum _{l=1}^{r}\displaystyle {r\atopwithdelims ()l} K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l} \right| \nonumber \\&\le \sup _{\vert s-t \vert < \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i}\sum _{l=1}^{r} \frac{1}{h} \displaystyle {r\atopwithdelims ()l}\vert s-t\vert ^{l} \nonumber \\&\le \sum _{l=1}^{r}\displaystyle {r\atopwithdelims ()l}\frac{\eta _n^{l}}{h^{r+1}}\nonumber \\&=o(a_{n}), \end{aligned}$$

(15)

where $I_{\lbrace .\rbrace }$ is the indicator function. Otherwise, if for some i and j, $\vert \frac{t_{ij}-s}{h} \vert \le 1$, we can show that

$$\begin{aligned} B_{22}&= \sup _{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \left| \sum _{l=1}^{r} \left( \displaystyle {r \atopwithdelims ()l}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l} \right. \right. \nonumber \\&\quad \left. \left. -K\left( \frac{t_{ij}-s}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^r \right) I_{\lbrace \vert \frac{t_{ij}-s}{h}\vert \le 1 \rbrace } \right| \nonumber \\&\le sup_{\vert s-t \vert< \eta _n} \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h} \vert \sum _{l=1}^{r}\displaystyle {r\atopwithdelims ()l}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - s}{h}\right) ^{r-l}\left( \frac{s-t}{h}\right) ^{l}\nonumber \\&\quad -K\left( \frac{t_{ij}-s}{h}\right) \vert \nonumber \\&\le \frac{M_K}{h}sup_{\vert s-t \vert < \eta _n}\bigg \vert \frac{s-t}{h}\bigg \vert \nonumber \\&=o(a_{n}). \end{aligned}$$

(16)

Hence $B_{2}=B_{21}+B_{22}=o_{p}(a_{n})$ follows directly from (15) and (16). For $D_2$ in (13), Jensen’s inequality implies that $D_2=o(a_{n})$. For the first term on the right-hand side of Eq. (13), define

$$\begin{aligned} v_{ij}=\left( h\frac{\omega _{i}^{2} \log \left( \frac{1}{\sum _{i=1}^{n}N_{i}\omega _{i}^{2}}\right) }{\sum _{i=1}^{n}N_{i}\omega _{i}^{2}}\right) ^{\frac{1}{2}}\frac{1}{h}K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij}-t}{h}\right) ^{r} , \end{aligned}$$

then we can write for any $M >0$,

$$\begin{aligned}&P(S_{r}(t)-E(S_{r}(t))>Ma_{n})\\&\quad =P\left( \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} \frac{1}{h}\left( K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - t}{h}\right) ^{r}\right. \right. \\&\qquad \left. \left. - E\left( K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij} - t}{h}\right) ^{r}\right) \right)>Ma_{n}\right) \\&\quad \le P\left( \sum _{i=1}^n \sum _{j=1}^{N_i}(v_{ij}-E(v_{ij}))>M log\left( \frac{1}{\sum _{i=1}^{n}N_{i}\omega _{i}^{2}}\right) \right) . \end{aligned}$$

By Markov inequality and independence of the $v_{ij}$, we deduce that

$$\begin{aligned} P(S_{r}(t)-E(S_{r}(t))>Ma_{n})&\le \frac{\prod _{i=1}^{n}\prod _{j=1}^{N_{i}}E(e^{v_{ij}-E(v_{ij})})}{\left( \frac{1}{\sum _{i=1}^{n}N_{i}\omega _{i}^{2}}\right) ^M}\\&\le \left( \sum _{i=1}^{n}N_{i}\omega _{i}^{2}\right) ^{M-C} , \end{aligned}$$

for some constant C. Therefore, for a large M we have

$$\begin{aligned} P(\sup _{t \in [0,1]} \vert S_{r}(t) - ES_{r}(t) \vert&\le \sup _{t \in \chi (\eta _n)} \vert S_{r}(t) - ES_{r}(t) \vert >Ma_{n})\nonumber \\&\le 2 (\sum _{i=1}^{n}N_{i}\omega _{i}^{2})^{M-C-r-2}log\left( \frac{1}{\sum _{i=1}^{n}N_{i}\omega _{i}^{2}}\right) ^{-(r+2)}\nonumber \\&=o(a_{n}) ,\; \; r=0,1,\ldots ,4. \end{aligned}$$

(17)

The proof is complete by combining (13)–(17). $\square$

1.2 Proof of Theorem 3.1

Proof

Combining Eq. (9) and Assumptions (A1)–(A3), we conclude that

$$\begin{aligned} \widehat{\mu ^{\prime }}(t)-\mu ^{\prime }(t) =\frac{\frac{f(t)}{h\mu ^2_k}R_1 - f^{\prime } (t) R_0}{f^2 (t)-h^2 f^{\prime 2}(t)\mu _k^2}+ O_p({h^2} + h a_n) . \end{aligned}$$

(18)

From Assumptions (A1) and (A2), we have $f^2 (t)>0$ and $h^2 f^{\prime 2}(t)\mu _k^2=O(h^2)$. Therefore, under Assumption (A5), $f^2 (t)-h^2 f^{\prime 2}(t)\mu _k^2$ tends to a positive constant that is bounded away from zero with probability approaching one as $h \rightarrow 0$, so we have

$$\begin{aligned} E&\int { \left( \frac{\frac{f(t)}{h\mu ^2_k} R_1 - f^{\prime } (t) R_0}{f^2 (t)-h^2 f^{\prime 2}(t)\mu _k^2} \right) ^2 }dt \\&= E \int \left[ \left( \frac{ f(t) }{h^2 \mu ^2_k( f^2 (t) - h^2 f^{\prime 2} (t) \mu _k^2 )} \right) \sum _{i=1}^{n} \omega _i \sum _{j=1}^{N_i} K\left( \frac{t_{ij}-t}{h}\right) \left( \frac{t_{ij}-t}{h} \right) \delta _{ij} \right] ^2 dt \\&\quad +E \int \left[ \left( \frac{ f^{\prime } (t) }{h( f^2 (t) - h^2 f^{\prime 2} (t) \mu _k^2 )} \right) \sum _{i=1}^{n} \omega _i \sum _{j=1}^{N_i} K(\frac{t_{ij}-t}{h}) \delta _{ij} \right] ^2 dt \\&\le \left( \frac{M_1}{h^4} + \frac{M_2}{h^2}\right) \sum _{i=1}^{n} \omega _i^2 N_i \int E \left[ K^2 \left( \frac{s-t}{h}\right) \left( \gamma (s,s)+\sigma ^2\right) \right] ^2 dt\\&\quad + \left( \frac{M_1}{h^4} + \frac{M_2}{h^2}\right) \sum _{i=1}^{n} \omega _i^2 N_i(N_i-1) \int E K\left( \frac{s_1-t}{h}\right) K \left( \frac{s_2-t}{h}\right) \gamma (s_1,s_2) dt \\&\le \left( \frac{M_3}{h^4} + \frac{M_4}{h^2}\right) \sum _{i=1}^{n} \omega _i^2 N_i h + \sum _{i=1}^{n} \omega _i^2 N_i(N_i-1)h^2 , \end{aligned}$$

where $M_{1}$, $M_2$, $M_{3}$, and $M_4$ are some constants. For any $M>0$ by Markov inequality,

$$\begin{aligned}&P\left( \Vert \frac{\frac{f(t)}{h\mu ^2_k} R_1 - f^{\prime 2} (t) R_0}{f^2 (t)-h^2 f^{\prime 2}(t)\mu _k^2} \Vert _2 > M b_n \right) \nonumber \\&\quad \le \frac{ \left( \frac{M_3}{h^4} + \frac{M_4}{h^2}\right) \sum _{i=1}^{n} \omega _i^2 N_i h + \sum _{i=1}^{n} \omega _i^2 N_i(N_i-1)h^2 }{M^2 b_n^2} . \end{aligned}$$

(19)

Therefore, by Assumption (A5), the right-hand side of (19) tends to zero as $M \rightarrow \infty$. Combining (18) and (19) completes the proof. $\square$

1.3 Proof of Theorem 3.2

Proof

We can write

$$\begin{aligned}&\frac{f(t)}{h \mu _k^2} R_1 -f^{\prime } (t) R_0 \\&\quad = \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} K_h (t_{ij} -t) \delta _{ij} \left( -f^{\prime }(t) + \frac{f(t)}{h\mu _k^2}\left( \frac{t_{ij} -t}{h}\right) \right) . \end{aligned}$$

By Lemma 5 in Zhang and Wang (2016), we deduce that

$$\begin{aligned}&\sup _{t\in [0,1] }\vert \sum _{i=1}^n \omega _i \sum _{j=1}^{N_i} K_h (t_{ij} -t) \delta _{ij} \vert \\&\quad =O_p \left( \frac{ \log {n } \left[ \sum _{i=1}^n N_i \omega _i^2 h + \sum _{i=1}^n N_i (N_i -1)\omega _i^2 h^2 \right] }{h} \right) . \end{aligned}$$

Using the fact that $f^{\prime }(t)$, f(t) and $\mu _k^2$ are bounded, and $K_h(t_{ij}-t) =0$ if $\vert t_{ij} -t \vert > h$, we have

$$\begin{aligned} \sup _{t \in [0,1]} \vert \widehat{\mu ^{\prime }} (t) - \mu ^{\prime }(t) \vert = O_p \left( \frac{ \log {n } \left[ \sum _{i=1}^n N_i \omega _i^2 h + \sum _{i=1}^n N_i (N_i -1)\omega _i^2 h^2 \right] }{h^2} + h^2\right) . \end{aligned}$$

$\square$

1.4 Proof of Theorem 3.3

Proof

For a fixed bandwidth h, minimizing the $L^2$ rate is equivalent to the unique minimizer of $b_n$ defined in Theorem 3.1. Applying the method of Lagrange multipliers under the condition $\sum _{i=1}^n N_i \omega _i =1$, we deduce that

$$\begin{aligned} \omega _i^{opt} = \frac{1/h^3 + (N_i -1)/h^2}{\sum _{i=1}^n N_i( 1/h^3 + (N_i -1)/h^2)} . \end{aligned}$$

$\square$