Penalized spline estimation for panel count data model with time-varying coefficients

Abstract

We consider a panel count data model with both time-varying and time-invariant coefficients. We estimate the baseline function and the time-varying coefficients using penalized splines based on the pseudolikelihood method. We evaluate the performance of three efficient Newton–Rapshon-based algorithms and another adaptive barrier algorithm. We propose a novel cross-validated score to select the smoothing parameters and deduce an easy-to-compute approximation to the score. Extensive simulations are conducted to compare the four algorithms, to compare the proposed penalized spline estimation with regression spline estimation and kernel estimation, and to assess the inference performance and robustness of the penalized spline estimation. Finally, we illustrate our method by using a data set from a childhood wheezing study.

The derivation of the approximated CVL score

Observe that $\mathrm{max}_{\gamma \in \varvec{\varTheta } } l^{\lambda }(\gamma )$ can be reformulated as the optimization problem with inequality constraints:

$$\begin{aligned} \min (-l^{\lambda }(\gamma )) \qquad \mathrm{subject \ to \ } {\left\{ \begin{array}{ll} \ g_{1}(\gamma )=\eta _{1,0}-\eta _{2,0}\le 0 \\ \ g_{2}(\gamma )=\eta _{2,0}-\eta _{3,0}\le 0 \\ \quad \qquad \qquad \vdots \\ \ g_{q_{n}-1}(\gamma )=\eta _{q_{n}-1,0}-\eta _{q_{n},0}\le 0 \end{array}\right. } \end{aligned}$$

(8)

By the Karush-Kuhn-Tucher (KKT) condition, if $b^{\lambda }$ is the solution to the optimization problem (8), then there exist $u_{i}^{*}\ge 0, i=1,2,\ldots ,q_{n}-1$, such that $\nabla _{\gamma } l^{\lambda }(b^{\lambda })=\sum _{i=1}^{q_{n}-1} u_{i}^{*}\nabla _{\gamma } g_{i}(b^{\lambda })$. Define $f(\gamma ) = -l^{\lambda }(\gamma )+\sum _{i=1}^{q_{n}-1} u_{i}^{*} g_{i}(\gamma )$ and ${\hat{\gamma }} = \mathrm{argmin}f(\gamma )$, by the concavity of $l^{\lambda }(\gamma )$ and the sufficient condition for KKT problems, we have

$$\begin{aligned} b^{\lambda }=\mathop {\mathrm{argmax}}\limits _{\gamma \in \varvec{\varTheta } } l^{\lambda }(\gamma ) \Leftrightarrow b^{\lambda }=\mathrm{argmin}f(\gamma ) = {\hat{\gamma }}. \end{aligned}$$

Similarily, there exist $w_{k}^{*}\ge 0, k=1,2,\ldots ,q_{n}-1$, such that $\nabla _{\gamma } l^{\lambda }_{(-(i,j))}(b^{\lambda }_{(-(i,j))})=\sum _{k=1}^{q_{n}-1} w_{k}^{*}\nabla _{\gamma } g_{k}(b^{\lambda }_{(-(i,j))})$, and define $f_{(-(i,j))}(\gamma ) = -l^{\lambda }_{(-(i,j))}(\gamma )+\sum _{k=1}^{q_{n}-1} w_{k}^{*} g_{k}(\gamma )$ and ${\hat{\gamma }} _{(-(i,j))} = \mathrm{argmin}f_{(-(i,j))}(\gamma )$, then we have

$$\begin{aligned} b^{\lambda }_{(-(i,j))}=\mathop {\mathrm{argmax}}\limits _{ \gamma \in \varvec{\varTheta } } l^{\lambda }_{(-(i,j))}(\gamma ) \Leftrightarrow b^{\lambda }_{(-(i,j))}=\mathrm{argmin}f_{(-(i,j))}(\gamma ) = {\hat{\gamma }}_{(-(i,j))}. \end{aligned}$$

A first-order Taylor approximation at ${\hat{\gamma }}$ yields

$$\begin{aligned} \nabla _{\gamma } f_{(-(i,j))} \big ({\hat{\gamma }}_{(-(i,j))}\big ) \approx \nabla _{\gamma } f_{(-(i,j))} \big ({\hat{\gamma }} \big ) + \nabla ^{2}_{\gamma } f_{(-(i,j))} \big ({\hat{\gamma }} \big )\cdot \big ({\hat{\gamma }}_{(-(i,j))}-{\hat{\gamma }}\big ). \end{aligned}$$

Define $f_{(i,j)}(\gamma ) = f(\gamma )-f_{(-(i,j))}(\gamma ) \approx l_{(-(i,j))}(\gamma )-l(\gamma )=-l_{(i,j)}(\gamma )$, then we have

$$\begin{aligned} 0 \approx - \nabla _{\gamma } f_{(i,j)} ({\hat{\gamma }} ) +\big (\nabla ^{2}_{\gamma } f({\hat{\gamma }} )-\nabla ^{2}_{\gamma } f_{(i,j)} ({\hat{\gamma }} ) \big ) \cdot \big ({\hat{\gamma }}_{(-(i,j))}-{\hat{\gamma }}\big ). \end{aligned}$$

(9)

The computation of the second derivatives can consume a lot of computer time and/or storage. Omitting this term $\nabla ^{2}_{\gamma } f_{(i,j)} ({\hat{\gamma }} )$ in (9) leads to

$$\begin{aligned} {\hat{\gamma }}_{(-(i,j))}&\approx {\hat{\gamma }}+ \big ( \nabla ^{2}_{\gamma } f({\hat{\gamma }} )\big )^{-1}\cdot \nabla _{\gamma } f_{(i,j)} ({\hat{\gamma }} ) \\&\approx {\hat{\gamma }}+\big ( \nabla ^{2}_{\gamma } l^{\lambda }({\hat{\gamma }} ) \big )^{-1}\cdot \nabla _{\gamma } l_{(i,j)} ({\hat{\gamma }} ). \end{aligned}$$

Then we have

$$\begin{aligned} l_{(i,j)} \big (b^{\lambda }_{(-(i,j))}\big )&\approx l_{(i,j)}\big [ b^{\lambda }+\big ( \nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \big )^{-1}\cdot \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \big ] \\&\approx l_{(i,j)}( b^{\lambda })+\bigl ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \bigr )^{T} \cdot \bigl [ \big ( \nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \big )^{-1}\cdot \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \bigr ] \\&= l_{(i,j)}( b^{\lambda })+\mathrm{tr}\bigl [ \bigl ( \nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \bigr )^{-1}\cdot \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \cdot \bigl ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } )\bigr )^{T} \bigr ]. \end{aligned}$$

Therefore,

$$\begin{aligned} CVL(\lambda ) \approx l(b^{\lambda })+ \mathrm{tr}\bigg \{ \bigl (\nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \big )^{-1}\cdot \bigg [ \sum \limits _{i=1}^n\sum \limits _{j=1}^{ K_{i}} \Big ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \cdot \bigl ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \bigr )^{T} \Big ) \bigg ] \bigg \}. \end{aligned}$$

In view of the fact that

$$\begin{aligned} \begin{aligned}&E \bigg [ \mathrm{tr}\bigg \{ \bigl (\nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \big )^{-1}\cdot \bigg [ \sum \limits _{i=1}^n\sum \limits _{j=1}^{ K_{i}} \Big ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \cdot \bigl ( \nabla _{\gamma } l_{(i,j)} (b^{\lambda } ) \bigr )^{T} \Big ) \bigg ] \bigg \} \bigg ] \\&\quad = \mathrm{tr}\Bigl [ \bigl (\nabla ^{2}_{\gamma } l^{\lambda }(b^{\lambda } ) \big ) ^{-1} \cdot \bigl (- \nabla ^{2}_{\gamma } l(b^{\lambda } ) \bigl )\Bigr ]. \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm{CVL}(\lambda ) \approx l(b^{\lambda })-\mathrm{tr}\Big [\big (\nabla ^{2}_\gamma l^{\lambda }(b^{\lambda })\big )^{-1}\cdot \nabla ^{2}_\gamma l(b^{\lambda })\Big ]. \end{aligned}$$