Efficient Estimation of the Additive Risks Model for Interval-Censored Data

Abstract

In contrast to the popular Cox model which presents a multiplicative covariate effect specification on the time to event hazards, the semiparametric additive risks model (ARM) offers an attractive additive specification, allowing for direct assessment of the changes or the differences in the hazard function for changing value of the covariates. The ARM is a flexible model, allowing the estimation of both time-independent and time-varying covariates. It has a nonparametric component and a regression component identified by a finite-dimensional parameter. This chapter presents an efficient approach for maximum-likelihood (ML) estimation of the nonparametric and the finite-dimensional components of the model via the Minorize-Maximize (MM) algorithm for case-II interval-censored data. The operating characteristics of our proposed MM approach are assessed via simulation studies, with illustration on a breast cancer dataset via the R package MMIntAdd. It is expected that the proposed computational approach will not only provide scalability to the ML estimation scenario but may also simplify the computational burden of other complex likelihoods or models.

Appendix

We shall use the second part of Lemma 1 from Wang et al. (2022) in proving Theorem 1, and we present this result in the following proposition. The proof of Proposition 1 can be found in Wang et al. (2022).

Proposition 1 (Wang et al., 2022)

For any τ, τ ₀ ≥ 0

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathit{\text{log}}\left\{\frac{1-\mathit{\text{exp}}(-\tau)}{1-\mathit{\text{exp}}(-\tau_0)}\right\}\geq (\tau-\tau_0)A_1(\tau_0)-(\tau-\tau_0)^2A_2(\tau_0)+\mathit{\text{log}}\left(\frac{\tau_0}{\tau}\right)+1-\frac{\tau_0}{\tau}, \end{array} \end{aligned} $$

where A ₁(τ ₀) = exp(−τ ₀)∕{1 −exp(−τ ₀)} and A ₂(τ ₀) = exp(−τ ₀)∕2{1 −exp(−τ ₀)}².

1.1 Proof of Theorem 1

In ℓ ₂(λ, β) and ℓ ₄(λ, β), (λ ₁, …, λ _m)^⊤ are not entangled with β. Therefore, there is no need to develop the minorization functions for them. In the following, we show how to find the minorization functions for ℓ ₁(λ, β) and ℓ ₃(λ, β). Define $u(L_i,X_i)=\sum _{k: t_k\leq L_i}\lambda _{k}+\beta ^{\top } Z_{x_i}(L_i)$, $u(R_i,X_i)=\sum _{k: t_k\leq R_i}\lambda _{k}+\beta ^{\top } Z_{x_i}(R_i)$ and $u(L_i, R_i,X_i)=\sum _{k: L_i<t_k\leq R_i}\lambda _{k}+\beta ^{\top } \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}$. According to our model assumption (1), u(L _i, X _i) > 0, u(R _i, X _i) > 0 and u(L _i, R _i, X _i) > 0 for all i. Now, we can re-write

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ell_1(\lambda,\beta)& =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\mbox{log}[1-\mbox{exp}\{- \sum_{k: t_k\le L_i}\lambda_k-\beta^{\top} Z_{x_i}(L_i)\}]\\ & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\mbox{log}[1-\mbox{exp}\{-u(L_i, X_i)\}]\\ & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\left(\mbox{log}[1-\mbox{exp}\{-u_0(L_i, X_i)\}]+\mbox{log}\left[\frac{1-\mbox{exp}\{-u(L_i, X_i)\}}{1-\mbox{exp}\{-u_0(L_i, X_i)\}}\right]\right). \end{array} \end{aligned} $$

Applying Proposition 1 to the second term of the above display with τ = u(L _i, X _i) and τ ₀ = u ₀(L _i, X _i), we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ell_1(\lambda,\beta)& \geq&\displaystyle \sum_{i=1}^n\Delta_{L,i}\bigg( \mbox{log}[1-\mbox{exp}\{-u_0(L_i, X_i)\}]+ \{ u(L_i, X_i)-u_0(L_i, X_i)\} A_1(u_0(L_i, X_i)) \\ & &\displaystyle -\{ u(L_i, X_i)-u_0(L_i, X_i)\}^2 A_2(u_0(L_i, X_i))+ \mbox{log}\left\{ \frac{u_0(L_i, X_i)}{u(L_i, X_i)}\!\right\} +1- \frac{u_0(L_i, X_i)}{u(L_i, X_i)} \! \bigg) \\ & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\Bigg[\{A_1(u_0(L_i,X_i))+2A_2(u_0(L_i,X_i))u_0(L_i,X_i)\}u(L_i,X_i)\\ & &\displaystyle -A_2(u_0(L_i,X_i))u^2(L_i,X_i) \\ & &\displaystyle +\mbox{log}\left\{\frac{u_0(L_i,X_i)}{u(L_i,X_i)}\right\}-\frac{u_0(L_i,X_i)}{u(L_i,X_i)} +C_1(u_0(L_i,X_i))\Bigg]\\ & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\Bigg[\!\{A_1(u_0(L_i,X_i))+2A_2(u_0(L_i,X_i))u_0(L_i,X_i)\}\!\left(\sum_{k: t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)\!\right)\\ & &\displaystyle -A_2(u_0(L_i,X_i))\left(\sum_{k: t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)\right)^2+\mbox{log}\left(\frac{u_0(L_i,X_i)}{\sum_{k: t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)}\right)\\ & &\displaystyle -\left(\frac{u_0(L_i,X_i)}{\sum_{k: t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)}\right)+C_1(u_0(L_i,X_i))\Bigg],{} \end{array} \end{aligned} $$

(7)

where C ₁(u ₀(L _i, X _i)) is the constant term that only depends on u ₀(L _i, X _i), given as $C_1(u_0(L_i,X_i))=\mbox{log}[1-\mbox{exp}\{-u_0(L_i, X_i)\}]-A_1(u_0(L_i,X_i))u_0(L_i,X_i)-A_2(u_0(L_i,X_i))u_0^2(L_i,X_i)+1.$ Next, we look into the following three terms of (7). First,

$$\displaystyle \begin{aligned} \begin{array}{rcl} -\left(\sum_{t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)\right)^2& =&\displaystyle -\left(\sum_{t_k\le L_i}\frac{\lambda_{k0}}{u_0(L_i,X_i)}\frac{u_0(L_i,X_i)}{\lambda_{k0}}\lambda_k\right. \\ & &\displaystyle \left.+\frac{\beta_0^{\top} Z_{x_i}(L_i)}{u_0(L_i,X_i)}\frac{u_0(L_i,X_i)}{\beta_0^{\top} Z_{x_i}(L_i)}\beta^{\top} Z_{x_i}(L_i)\right)^2\\ & \ge&\displaystyle -\bigg\{\sum_{t_k\leq L_i}\frac{u_0(L_i,X_i)}{\lambda_{k0}}\lambda_k^2+\frac{u_0(L_i,X_i)}{\beta_0^{\top} Z_{x_i}(L_i)}(\beta^{\top} Z_{x_i}(L_i))^2\bigg\}, \end{array} \end{aligned} $$

where, the inequality is obtained by applying Jensen’s inequality on the concave function f(x) = −x ² and noting that $\sum _{k: t_k\le L_i}\lambda _{k0}/u_0(L_i,X_i)+ \beta _0^{\top } Z_{x_i}(L_i)/u_0(L_i,X_i)=1$. Second, applying the standard inequality log(x) ≥ 1 − 1∕x for any generic x > 0, we have

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mbox{log}\left(\frac{u_0(L_i,X_i)}{\sum_{t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)}\right)\geq 1-\frac{\sum_{t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)}{u_0(L_i,X_i)}, \end{array} \end{aligned} $$

and third,

$$\displaystyle \begin{aligned} \begin{array}{rcl} -\frac{u_0(L_i,X_i)}{\sum_{t_k\leq L_i}\lambda_k+\beta^{\top} Z_{x_i}(L_i)}& =&\displaystyle -u_0(L_i,X_i)\bigg\{\sum_{t_k\le L_i}\frac{\lambda_{k0}}{u_0(L_i,X_i)}\frac{u_0(L_i,X_i)}{\lambda_{k0}}\lambda_k\\ & &\displaystyle +\frac{\beta_0^{\top} Z_{x_i}(L_i)}{u_0(L_i,X_i)}\frac{u_0(L_i,X_i)}{\beta_0^{\top} Z_{x_i}(L_i)}\beta^{\top} Z_{x_i}(L_i)\bigg\}^{-1}\\ & \ge &\displaystyle - \bigg[ \sum_{t_k\le L_i} \frac{\lambda_{k0}^2}{u_0(L_i,X_i)}\lambda_k^{-1}+\frac{\{\beta_0^{\top} Z_{x_i}(L_i)\}^2}{u_0(L_i,X_i)}\{\beta^{\top} Z_{x_i}(L_i)\}^{-1}\bigg], \end{array} \end{aligned} $$

where, the last inequality is obtained by applying Jensen’s inequality on the concave function f(x) = −1∕x, and noting that $\sum _{k: t_k\le L_i}\lambda _{k0}/u_0(L_i,X_i)+ \beta _0^{\top } Z_{x_i}(L_i)/u_0(L_i,X_i)=1$. Then, applying the last three inequalities in (7), we obtain $\ell _1(\lambda ,\beta )\ge \ell _{1,\dagger }(\lambda ,\beta |\lambda _0,\beta _0)\equiv \sum _{k=1}^m\mathcal {M}_{1,1,k}(\lambda _k|\lambda _0,\beta _0)+\mathcal {M}_{1,2}(\beta |\lambda _0,\beta _0)+\mathcal {M}_{1,3}(\lambda _0,\beta _0)$, where for k = 1, …, m,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{M}_{1,1,k}(\lambda_k|\lambda_0,\beta_0) & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\Bigg[\{A_1(u_0(L_i,X_i))+2A_2(u_0(L_i,X_i))u_0(L_i,X_i)\}\lambda_k\\ & &\displaystyle -A_2(u_0(L_i,X_i))\left\{\frac{u_0(L_i,X_i)}{\lambda_{k0}}\right\}\lambda_k^2-\frac{\lambda_k}{u_0(L_i,X_i)} \\ & &\displaystyle -\frac{\lambda_{k0}^2}{u_0(L_i,X_i)}\lambda_k^{-1}\Bigg]I(t_k\leq L_i), \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{M}_{1,2}(\beta|\lambda_0,\beta_0) & =&\displaystyle \sum_{i=1}^n\Delta_{L,i}\Bigg[\{A_1(u_0(L_i,X_i))+2A_2(u_0(L_i,X_i))u_0(L_i,X_i)\}\beta^{\top} Z_{x_i}(L_i)\\ & &\displaystyle -A_2(u_0(L_i,X_i))\frac{u_0(L_i,X_i)}{\beta_0^{\top} Z_{x_i}(L_i)}\{\beta^{\top} Z_{x_i}(L_i)\}^2-\frac{\beta^{\top} Z_{x_i}(L_i)}{u_0(L_i,X_i)}\\ & &\displaystyle -\frac{\{\beta_0^{\top} Z_{x_i}(L_i)\}^2}{u_0(L_i,X_i)}\{\beta^{\top} Z_{x_i}(L_i)\}^{-1}\Bigg], \end{array} \end{aligned} $$

and $\mathcal {M}_{1,3}(\lambda _0,\beta _0)=\sum _{i=1}^n\Delta _{L,i}\{\mbox{log}[1-\mbox{exp}\{-u_0(L_i, X_i)\}]-A_1(u_0(L_i,X_i)) u_0(L_i,X_i)-A_2(u_0(L_i,X_i)) u_0^2(L_i,X_i)+1\}$. Next, consider finding the minorization function for ℓ ₃(λ, β). Here, we use the same techniques as finding the minorization function for ℓ ₁(λ, β). Note,

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ell_3(\lambda,\beta) & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\mbox{log}\left(1-\mbox{exp}\left[-\sum_{k: L_i< t_k\leq R_i}\lambda_k-\beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}\right]\right)\\ & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\mbox{log}[1-\mbox{exp}\{-u(L_i,R_i, X_i)\}]\\ & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\left(\mbox{log}[1-\mbox{exp}\{-u_0(L_i,R_i, X_i)\}]+\mbox{log}\left[\frac{1-\mbox{exp}\{-u(L_i,R_i, X_i)\}}{1-\mbox{exp}\{-u_0(L_i,R_i, X_i)\}}\right]\right). \end{array} \end{aligned} $$

Now applying Proposition 1 to the second term of the above display with τ = u(L _i, R _i, X _i) and τ ₀ = u ₀(L _i, R _i, X _i), we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ell_3(\lambda,\beta)& \geq&\displaystyle \sum_{i=1}^n\Delta_{I,i}\bigg( \mbox{log}[1-\mbox{exp}\{-u_0(L_i,R_i X_i)\}] \\ & &\displaystyle + \{ u(L_i,R_i, X_i)-u_0(L_i,R_i, X_i)\} A_1(u_0(L_i,R_i, X_i)) \\ & &\displaystyle -\{ u(L_i,R_i, X_i)-u_0(L_i,R_i, X_i)\}^2 A_2(u_0(L_i,R_i, X_i))\\ & &\displaystyle + \mbox{log}\left\{ \frac{u_0(L_i,R_i, X_i)}{u(L_i,R_i, X_i)}\right\} +1 - \frac{u_0(L_i,R_i, X_i)}{u(L_i,R_i, X_i)} \bigg) \\ & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\Bigg[\{A_1(u_0(L_i,R_i,X_i))+2A_2(u_0(L_i,R_i,X_i))u_0(L_i,R_i,X_i)\}u(L_i,R_i,X_i)\\ & &\displaystyle -A_2(u_0(L_i,R_i,X_i))u^2(L_i,R_i,X_i) \\ & &\displaystyle +\mbox{log}\left\{\frac{u_0(L_i,R_i,X_i)}{u(L_i,R_i,X_i)}\right\}-\frac{u_0(L_i,R_i,X_i)}{u(L_i,R_i,X_i)} +C_1(u_0(L_i,R_i,X_i))\Bigg]\\ & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\Bigg[\{A_1(u_0(L_i,R_i,X_i))\\ & &\displaystyle +2A_2(u_0(L_i,R_i,X_i))u_0(L_i,R_i,X_i)\}\left(\sum_{k: L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\right)\\ & &\displaystyle -A_2(u_0(L_i,R_i,X_i))\left(\sum_{k: L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\right)^2\\ & &\displaystyle +\mbox{log}\left(\frac{u_0(L_i,R_i,X_i)}{ \sum_{k: L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))} \right)\\ & &\displaystyle -\left(\frac{u_0(L_i,R_i,X_i)}{\sum_{k: L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}\right)+C_1(u_0(L_i,R_i,X_i))\Bigg]{} \end{array} \end{aligned} $$

(8)

where, C ₁(u ₀(L _i, R _i, X _i)) is the constant term that only depends on u ₀(L _i, R _i, X _i), given by $C_1(u_0(L_i,R_i,X_i))=\mbox{log}[1-\mbox{exp}\{-u_0(L_i,R_i, X_i)\}]-A_1(u_0(L_i,R_i,X_i)) u_0(L_i,R_i,X_i)-A_2(u_0(L_i,R_i,X_i)) u_0^2(L_i,R_i,X_i)+1.$ Similarly, we have the following three inequalities,

$$\displaystyle \begin{aligned} \begin{array}{rcl} & -&\displaystyle \left(\sum_{L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\right)^2\\ & =&\displaystyle -\left(\sum_{L_i<t_k\le R_i}\frac{\lambda_{k0}}{u_0(L_i,R_i,X_i)}\frac{u_0(L_i,R_i,X_i)}{\lambda_{k0}}\lambda_k\right.\\ & &\displaystyle \left.+\frac{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}{u_0(L_i,R_i,X_i)}\frac{u_0(L_i,R_i,X_i)}{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\right)^2\\ & \ge&\displaystyle -\bigg\{\sum_{L_i<t_k\leq R_i}\frac{u_0(L_i,R_i,X_i)}{\lambda_{k0}}\lambda_k^2+\frac{u_0(L_i,R_i,X_i)}{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}(\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i)))^2\bigg\}, \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mbox{log}\left(\!\frac{u_0(L_i,R_i,X_i)}{\sum_{L_i<t_k\leq R_i}\!\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}\!\right)\geq\! 1-\frac{\sum_{L_i<t_k\leq R_i}\!\lambda_k +\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}{u_0(L_i,R_i,X_i)}, \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} & -&\displaystyle \frac{u_0(L_i,R_i,X_i)}{\sum_{L_i<t_k\leq R_i}\lambda_k+\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}\\ & =&\displaystyle -u_0(L_i,R_i,X_i)\bigg\{\sum_{L_i<t_k\leq R_i}\frac{\lambda_{k0}}{u_0(L_i,R_i,X_i)}\frac{u_0(L_i,R_i,X_i)}{\lambda_{k0}}\lambda_k\\ & &\displaystyle +\frac{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}{u_0(L_i,R_i,X_i)}\frac{u_0(L_i,R_i,X_i)}{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))}\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\bigg\}^{-1}\\ & \ge &\displaystyle - \bigg[ \sum_{L_i<t_k\le R_i} \frac{\lambda_{k0}^2}{u_0(L_i,R_i,X_i)}\lambda_k^{-1}+\frac{\{\beta_0^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\}^2}{u_0(L_i,R_i,X_i)}\{\beta^{\top} (Z_{x_i}(R_i)-Z_{x_i}(L_i))\}^{-1}\bigg], \end{array} \end{aligned} $$

where, the first and the third inequalities are obtained by applying Jensen’s inequality on the concave function f(x) = −x ² and f(x) = −1∕x, respectively, and the second inequality is obtained by applying the standard inequality log(x) ≥ 1 − 1∕x. Applying the above two inequalities in (8), we obtain $\ell _3(\lambda ,\beta )\ge \ell _{3,\dagger }(\lambda ,\beta |\lambda _0,\beta _0)\equiv \sum _{k=1}^m\mathcal {M}_{3,1,k}(\lambda _k|\lambda _0,\beta _0)+\mathcal {M}_{3,2}(\beta |\lambda _0,\beta _0)+\mathcal {M}_{3,3}(\lambda _0,\beta _0)$, where

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{M}_{3,1,k}(\lambda_k|\lambda_0,\beta_0) & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\Bigg[\{A_1(u_0(L_i,R_i,X_i))+2A_2(u_0(L_i,R_i,X_i))u_0(L_i,R_i,X_i)\}\lambda_k\\ & &\displaystyle -A_2(u_0(L_i,R_i,X_i))\left\{\frac{u_0(L_i,R_i,X_i)}{\lambda_{k0}}\right\}\lambda_k^2\\ & &\displaystyle -\frac{\lambda_k}{u_0(L_i,R_i,X_i)}-\frac{\lambda_{k0}^2}{u_0(L_i,R_i,X_i)}\lambda_k^{-1}\Bigg]I(L_i< t_k\le R_i),\!\!\quad k=1,\ldots,m, \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{M}_{3,2}(\beta|\lambda_0,\beta_0) & =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\Bigg(\{A_1(u_0(L_i,R_i,X_i))+2A_2(u_0(L_i,R_i,X_i))u_0(L_i,R_i,X_i)\}\\ & &\displaystyle \times \beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\} \\ & &\displaystyle -A_2(u_0(L_i,R_i,X_i))\frac{u_0(L_i,R_i,X_i)[\beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}]^2}{\beta_0^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}}\\ & &\displaystyle -\frac{\beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}}{u_0(L_i,R_i,X_i)} -\frac{[\beta_0^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}]^2}{u_0(L_i,R_i,X_i) \beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\} } \Bigg), \end{array} \end{aligned} $$

and

$$\displaystyle \begin{aligned} \begin{array}{rcl} \mathcal{M}_{3,3}(\lambda_0,\beta_0)& =&\displaystyle \sum_{i=1}^n\Delta_{I,i}\Bigg[\mbox{log}\left\{1-\mbox{exp}\left(-\left[\sum_{L_i< t_k\leq R_i}\lambda_k+\beta^{\top} \{Z_{x_i}(R_i)-Z_{x_i}(L_i)\}\right]\right)\right\}\\ & &\displaystyle -A_1(u_0(L_i,R_i,X_i))u_0(L_i,R_i,X_i)-A_2(u_0(L_i,R_i,X_i))u_0^2(L_i,R_i,X_i)+1\Bigg]. \end{array} \end{aligned} $$

Finally, we obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} \ell(\lambda, \beta)& =&\displaystyle \ell_1(\lambda, \beta)+ \ell_2(\lambda, \beta)+\ell_3(\lambda, \beta)\\ & \geq&\displaystyle \ell_{\dagger}(\lambda,\beta|\lambda_0,\beta_0)\\ & \equiv&\displaystyle \ell_{1,\dagger}(\lambda,\beta|\lambda_0,\beta_0)+\ell_2(\lambda, \beta)+\ell_{3,\dagger}(\lambda,\beta|\lambda_0,\beta_0)\\ & =&\displaystyle \sum_{k=1}^m\mathcal{M}_{1,1,k}(\lambda_k|\lambda_0,\beta_0)+\mathcal{M}_{1,2}(\beta|\lambda_0,\beta_0)+\mathcal{M}_{1,3}(\lambda_0,\beta_0)+\ell_2(\lambda, \beta)\\ & &\displaystyle +\sum_{k=1}^m\mathcal{M}_{3,1,k}(\lambda_k|\lambda_0,\beta_0)+\mathcal{M}_{3,2}(\beta|\lambda_0,\beta_0)+\mathcal{M}_{3,3}(\lambda_0,\beta_0)\\ & \equiv &\displaystyle \sum_{k=1}^m\mathcal{M}_{1,k}(\lambda_k|\lambda_0,\beta_0)+\mathcal{M}_2(\beta|\lambda_0,\beta_0)+\mathcal{M}_3(\lambda_0,\beta_0), \end{array} \end{aligned} $$

where $\mathcal {M}_{1,k}(\lambda _k|\lambda _0,\beta _0) = \mathcal {M}_{1,1,k}(\lambda _k|\lambda _0,\beta _0)+ \mathcal {M}_{3,1,k}(\lambda _k|\lambda _0,\beta _0)- \lambda _k\sum ^n_{i=1} \Delta _{I, i} I(t_k\leq L_i)$, $\mathcal {M}_{2}(\beta |\lambda _0,\beta _0) = \mathcal {M}_{1, 2}(\beta |\lambda _0,\beta _0)+ \mathcal {M}_{3, 2}(\beta |\lambda _0,\beta _0)- \sum _{i=1}^n\Delta _{I,i}\beta ^{\top } Z_{x_i} (L_i)$, and $\mathcal {M}_3(\lambda _0,\beta _0) = \mathcal {M}_{1, 3}(\lambda _0,\beta _0) +\mathcal {M}_{3, 3}(\lambda _0,\beta _0)$.