Gini index estimation for lifetime data

Abstract

Lifetime data is often right-censored. Recent literature on the Gini index estimation with censored data focuses on independent censoring. However, the censoring mechanism is likely to be dependent censoring in practice. This paper proposes two estimators of the Gini index under independent censoring and covariate-dependent censoring, respectively. The proposed estimators are consistent and asymptotically normal. We also evaluate the performance of our estimators in finite samples through Monte Carlo simulations. Finally, the proposed methods are applied to real data.

Appendix

Proof of Theorem 1

Let $\check{F}_{cn}(t)$, $\check{\mu }$, $\check{\eta }$, and $\check{G}$ be the counterparts, respectively, obtained by substituting $\hat{K}_c(T_i)$ with $K_c(T_i)$ in $F_{cn}(t)$, $\hat{\mu }$, $\hat{\eta }$, and $\hat{G}$. Given Assumptions 1 and 2, we know $\hat{K}_c(t)=K_c(t)+o_p(1)$. Because both $\hat{\mu }$ and $F_{cn}(t)$ are continuous at $K_c(T_i)$, we then have $\hat{\mu }=\check{\mu }+o_p(1)$, and $F_{cn}(t)=\check{F}_{cn}(t)+o_p(1)$. Because $\hat{\eta }$ is continuous at $(K_c(T_i), \check{F}_{cn}(T_i))$, we then have $\hat{\eta }=\check{\eta }+o_p(1)$. Given that $\hat{G}$ is continuous at $(\check{\mu }, \check{\eta })$, we then have $\hat{G}=\check{G}+o_p(1)$. Applying the law of large numbers and the law of iterated expectations to $\check{\mu }$ and $\check{\eta }$ yields $\check{\mu }=\mu +o_p(1)$ and $\check{\eta }=\eta +o_p(1)$. Thus, $\check{G}=G+o_p(1)$ since $\check{G}$ is continuous at $(\mu , \eta )$. Then, we have $\hat{G}=G+o_p(1)$. Therefore, we complete the proof of the first part of Theorem 1.

We take three steps to prove the second part of Theorem 1. The first is to approximate $\hat{\mu }-\mu$ using a sum of i.i.d. variables. The second is to approximate $\hat{\eta }-\eta$ using a sum of i.i.d. variables. Finally, we complete the proof of Theorem 1.

For $\hat{\mu }-\mu$, we have

$$\begin{aligned} \begin{aligned} \hat{\mu }-\mu&=\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i}{\hat{K}_c(T_i)}-\mu \\&=\left[ \frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i}{K_c(T_i)} -\mu \right] +\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c(T_i) -\hat{K}_c(T_i)\right] }{K_c(T_i)\hat{K}_c(T_i)}\\&=.I_{n1}+I_{n2}.\\ \end{aligned} \end{aligned}$$

(11)

Similar to Robins and Rotnitzky (1992), we have

$$\begin{aligned} \frac{\delta _i}{K_c(T_i)}=1-\int _0^\infty \frac{dM_i^c(u)}{K_c(u)}. \end{aligned}$$

(12)

Then, we have

$$\begin{aligned} I_{n1}=\frac{1}{n}\sum \limits _{i=1}^n\tilde{T}_i-\mu -\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{\tilde{T}_i}{K_c(u)}dM_i^c(u) \end{aligned}$$

(13)

According to the martingale representation (Gill 1980), we have

$$\begin{aligned} \begin{aligned} \frac{K_c(t)-\hat{K}_c(t)}{K_c(t)}&=\int _{0}^t\frac{\hat{K}_c(u-)}{K_c(u)} \frac{dM^c(u)}{S(u)}\\&=\int _0^\infty \frac{I(t\ge u)\hat{K}_c(u-)}{K_c(u)}\frac{dM^c(u)}{S(u)}\\&=\int _0^\infty \frac{I(t\ge u)}{nK_c(u)\hat{K}_{\tilde{T}}(u-)}dM^c(u).\\ \end{aligned} \end{aligned}$$

(14)

Substituting (14) into $I_{n2}$ obtains

$$\begin{aligned} \begin{aligned} I_{n2}&=\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _iT_i}{\hat{K}_c(T_i)}\int _0^\infty \frac{I(T_i\ge u)}{nK_c(u)\hat{K}_{\tilde{T}}(u-)}dM_c(u)\\&=\int _0^\infty \left[ \frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _iT_iI(T_i\ge u)}{\hat{K}_c(T_i)\hat{K}_{\tilde{T}}(u-)}\right] \frac{dM^c(u)}{nK_c(u)}\\&=\int _0^\infty \frac{\hat{F}(\tilde{T}, u)}{nK_c(u)}dM^c(u)\\&=\int _0^\infty \frac{F(\tilde{T}, u)}{nK_c(u)}dM^c(u)+o_p(n^{-1/2})\\&=\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{F(\tilde{T}, u)}{K_c(u)}dM_i^c(u)+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(15)

Then, substituting (13) and (15) into (11) obtains

$$\begin{aligned} \begin{aligned} \hat{\mu }-\mu =&\;\frac{1}{n}\sum \limits _{i=1}^n\tilde{T}_i-\mu -\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{\tilde{T}_i}{K_c(u)}dM_i^c(u)\\&\quad +\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{F(\tilde{T}, u)}{K_c(u)}dM_i^c(u)+o_p(n^{-1/2}). \end{aligned} \end{aligned}$$

(16)

Let $h_{ij}=I(\tilde{T}_j\le \tilde{T}_i)\tilde{T}_i+I(\tilde{T}_i\le \tilde{T}_j) \tilde{T}_j$. Then, for $\hat{\eta }-\eta$, we have

$$\begin{aligned} \begin{aligned} \hat{\eta }-\eta&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{\delta _i\delta _jh_{ij}}{\hat{K}_c(T_i)\hat{K}_c(T_j)}-\eta \\&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n\frac{\delta _i\delta _jh_{ij}}{K_c(T_i)K_c(T_j)}-\eta \\&\quad +\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{\delta _i\delta _j\left[ K_c(T_i)K_c(T_j)-\hat{K}_c(T_i)\hat{K}_c(T_j)\right] h_{ij}}{K_c(T_i)K_c(T_j)\hat{K}_c(T_i)\hat{K}_c(T_j)}\\&=.J_{n1}+J_{n2}.\\ \end{aligned} \end{aligned}$$

(17)

$J_{n1}$ is a U-statistic. Then, we have the following by the projection method of U-statistics

$$\begin{aligned} \begin{aligned} J_{n1}&=\frac{2}{n}\sum \limits _{i=1}^n\left[ \frac{\delta _i}{K_c(T_i)}E \left( \left. \frac{\delta _jh_{ij}}{K_c(T_j)}\right| i\right) -\eta \right] +o_p(n^{-1/2})\\&=\frac{2}{n}\sum \limits _{i=1}^n\left[ \frac{\delta _i}{K_c(T_i)}E(h_{ij}|i) -\eta \right] +o_p(n^{-1/2})\\&=\frac{2}{n}\sum \limits _{i=1}^n\left[ \frac{\delta _ih_\eta (\tilde{T}_i)}{K_c(T_i)}-\eta \right] +o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(18)

Substituting (12) into (18) yields

$$\begin{aligned} J_{n1}=\frac{2}{n}\sum \limits _{i=1}^n[h_\eta (\tilde{T}_i)-\mu ] -\frac{2}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{h_\eta (\tilde{T}_i)}{K_c(u)}dM_i^c(u)+o_p(n^{-1/2}). \end{aligned}$$

(19)

We decompose $J_{n2}$ into the following three components:

$$\begin{aligned} \begin{aligned} J_{n2}&=\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{\delta _i\delta _j\left[ K_c(T_i)-\hat{K}_c(T_i)\right] \left[ K_c(T_j) -\hat{K}_c(T_j)\right] h_{ij}}{K_c(T_i)K_c(T_j)\hat{K}_c(T_i)\hat{K}_c(T_j)}\\&\quad +\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{\delta _i\delta _jh_{ij}\left[ K_c(T_i)-\hat{K}_c(T_i)\right] }{K_c(T_i)K_c(T_j)\hat{K}_c(T_i)}\\&\quad +\frac{1}{n^2}\sum \limits _{i=1}^n\sum \limits _{j=1}^n \frac{\delta _i\delta _jh_{ij}\left[ K_c(T_j)-\hat{K}_c(T_j)\right] }{K_c(T_j)K_c(T_i)\hat{K}_c(T_j)}\\&=.J_{n2,1}+J_{n2,2}+J_{n2,3}.\\ \end{aligned} \end{aligned}$$

(20)

Given that $\frac{K_c(T_i)-\hat{K}_c(T_i)}{K_c(T_i)}=O_p(n^{-1/2})$ and $\frac{K_c(T_j)-\hat{K}_c(T_j)}{K_c(T_j)}=O_p(n^{-1/2})$ (see Gill 1980), we can prove $J_{n2,1}=o_p(n^{-1/2})$. We approximate $J_{n2,2}$ as follows:

$$\begin{aligned} \begin{aligned} J_{n2,2}&=\frac{1}{n}\sum \limits _{i=1}^nE\left[ \left. \frac{\delta _jh_{ij}}{K_c(T_j)}\right| i\right] \frac{\delta _i\left[ K_c(T_i)-\hat{K}_c(T_i)\right] }{K_c(T_i)\hat{K}_c(T_i)}+o_p(n^{-1/2})\\&=\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _ih_\eta (\tilde{T}_i)\left[ K_c(T_i) -\hat{K}_c(T_i)\right] }{K_c(T_i)\hat{K}_c(T_i)}+o_p(n^{-1/2})\\&=\int _0^\infty \frac{1}{n}\sum \limits _{i=1}^n \frac{\delta _ih_\eta (\tilde{T}_i)I(T_i\ge u)}{\hat{K}_c(T_i) \hat{K}_{\tilde{T}}(u)}\frac{dM^c(u)}{nK_c(u)}+o_p(n^{-1/2})\\&=\int _0^\infty \frac{\hat{F}(h_\eta , u)}{nK_c(u)}dM^c(u)+o_p(n^{-1/2})\\&=\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{F(h_\eta , u)}{K_c(u)}dM_i^c(u)+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(21)

$J_{n2,3}=J_{n2,2}$. Therefore, we have

$$\begin{aligned} \begin{aligned} \hat{\eta }-\eta =&\,\,\frac{2}{n}\sum \limits _{i=1}^n[h_\eta (\tilde{T}_i)-\eta ] -\frac{2}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{h_\eta (\tilde{T}_i)}{K_c(u)}dM_i^c(u)\\&+\frac{2}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{F(h_\eta , u)}{K_c(u)}dM_i^c(u)+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(22)

Central limit theory implies that $\hat{\mu }-\mu =O_p(n^{-1/2})$ and $\hat{\eta }-\eta =O_p(n^{-1/2})$. Then, taking the Taylor series expansion of $\hat{G}$ at $(\eta ,\mu )$ yields that

$$\begin{aligned} \begin{aligned} \hat{G}&=\frac{\eta }{\mu }-1+\frac{\hat{\eta }-\eta }{\mu } -\frac{\eta }{\mu ^2}(\hat{\mu }-\mu )+o_p(n^{-1/2})\\&=G+\frac{1}{\mu }[(\hat{\eta }-\eta )-(G+1)(\hat{\mu }-\mu )]+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(23)

Combining (16), (22) and (23), we obtain

$$\begin{aligned} \sqrt{n}(\hat{G}-G)= & {} \frac{1}{\mu }\left\{ \frac{2}{\sqrt{n}} \sum \limits _{i=1}^n[h_\eta (\tilde{T}_i)-\eta ]-\frac{2}{\sqrt{n}} \sum \limits _{i=1}^n\int _0^\infty \frac{h_\eta (\tilde{T}_i)}{K_c(u)}dM_i^c(u)\right. \nonumber \\&+\;\frac{2}{\sqrt{n}}\sum \limits _{i=1}^n\int _0^\infty \frac{F(h_\eta , u)}{K_c(u)}dM_i^c(u)\nonumber \\&-(G+1)\left[ \frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\tilde{T}_i-\mu -\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\int _0^\infty \frac{\tilde{T}_i}{K_c(u)}dM_i^c(u)\right. \nonumber \\&\left. \left. +\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\int _0^\infty \frac{F(\tilde{T}, u)}{K_c(u)}dM_i^c(u)\right] \right\} +o_p(1). \end{aligned}$$

(24)

Applying the central limit theory (Fleming and Harrington 1991) to (24), we can prove the second part of Theorem 1.$\square$

Proof of Theorem 2

Note that

$$\begin{aligned} \begin{aligned}&E\left( \int _0^\infty \frac{F(h_\eta , t)h_{\eta }(\tilde{T}_i)}{K_c(t)^2}\lambda _c(t)S_i(t)dt\right) \\&\quad =E\left[ \int _0^\infty \frac{E\left( F(h_\eta , t)h_{\eta } (\tilde{T}_i)I(\tilde{T}_i>t)\right) I(C_i>t)}{K_c(t)^2}\lambda _c(t)dt\right] \\&\quad =E\left[ \int _0^\infty \frac{F(h_{\eta }F(h_\eta , t),t)K_{\tilde{T}_i} (t)I(C_i>t)}{K_c(t)^2}\lambda _c(t)dt\right] \\&\quad =E\left( \int _0^\infty \frac{F(h_{\eta }F(h_\eta , t),t)}{K_c(t)^2}S_i(t)\lambda _c(t)dt\right) \\ \end{aligned} \end{aligned}$$

(25)

Based on (16), the martingale central limit theory implies that

$$\begin{aligned} \sqrt{n}(\hat{\mu }-\mu )\rightarrow ^d N(0,\sigma _\mu ^2), \end{aligned}$$

(26)

where

$$\begin{aligned} \begin{aligned} \sigma _\mu ^2=&\;var(\tilde{T}_i)+E\left( \int _0^\infty \frac{F(h_\eta , t)^2}{K_c(t)^2}\lambda _c(t)S_i(t)dt\right) \\&+E\left( \int _0^\infty \frac{h_{\eta }(\tilde{T}_i)^2}{K_c(t)^2} \lambda _c(t)S_i(t)dt\right) \\&-2E\left( \int _0^\infty \frac{F(h_\eta , t)h_{\eta }(\tilde{T}_i)}{K_c(t)^2}\lambda _c(t)S_i(t)dt\right) .\\ \end{aligned} \end{aligned}$$

Then, $\hat{\mu }=\mu +o_p(1)$. Based on (25), we have

$$\begin{aligned} \begin{aligned} \hat{{\varOmega }}_4&=-\frac{8}{\mu ^2}E\left( \int _0^\infty \frac{F(h_{\eta }F(h_\eta , t),t)}{K_c(t)^2}S_i(t)\lambda _c(t)dt\right) +o_p(1)\\&=-\frac{8}{\mu ^2}E\left( \int _0^\infty \frac{F(h_\eta , t)h_{\eta }(\tilde{T}_i)}{K_c(t)^2}\lambda _c(t)S_i(t)dt\right) +o_p(1)\\&={\varOmega }_4+o_p(1). \end{aligned} \end{aligned}$$

(27)

Similarly, we can prove that the other 10 elements of $\hat{{\varOmega }}$ converge in probability to the corresponding ones of ${\varOmega }$. Therefore, we complete the proof of Theorem 2.$\square$

Proof of Theorem 3

Given Assumptions 1’ and 2’, the consistency of Aalen’s estimator implies $\hat{K}_c^{z_i}(t)=K_c^{z_i}(t)+o_p(1)$. Then, similar to the proof of the first part of Theorem 1, we can complete the proof of the first part of Theorem 3. We omit details here for the sake of conciseness.

We now prove the second part of Theorem 3. Given covariate-dependent censoring, along the lines similar to (11) and (13), we have

$$\begin{aligned} \begin{aligned} \hat{\mu }_z-\mu =&\;\frac{1}{n}\sum \limits _{i=1}^n\tilde{T}_i-\mu -\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{\tilde{T}_i}{K_c^{z_i}(u)}dM_i^c(u)\\&+\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)\hat{K}_c^{z_i}(T_i)}.\\ \end{aligned} \end{aligned}$$

(28)

Note that

$$\begin{aligned} \begin{aligned}&\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)\hat{K}_c(T_i)}-\frac{1}{n} \sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)K_c^{z_i}(T_i)}\\&\quad =\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] ^2}{K_c^{z_i}(T_i)K_c^{z_i}(T_i) \hat{K}_c^{z_i}(T_i)}=o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(29)

Based on (28) and (29), we have

$$\begin{aligned} \begin{aligned} \hat{\mu }_z-\mu =&\;\frac{1}{n}\sum \limits _{i=1}^n\tilde{T}_i-\mu -\frac{1}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{\tilde{T}_i}{K_c^{z_i}(u)}dM_i^c(u)\\&+\frac{1}{n}\sum \limits _{i=1}^n\frac{\delta _i\tilde{T}_i\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)K_c^{z_i}(T_i)}+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(30)

For covariate-dependent censoring, along lines similar to (21) and (22), we have

$$\begin{aligned} \begin{aligned} \hat{\eta }_z-\eta =&\;\frac{2}{n}\sum \limits _{i=1}^n[h_\eta (\tilde{T}_i)-\eta ] -\frac{2}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{h_\eta (\tilde{T}_i)}{K_c^{z_i}(u)}dM_i^c(u)\\&+\frac{2}{n}\sum \limits _{i=1}^n\frac{\delta _ih_\eta (\tilde{T}_i) \left[ K_c^{z_i}(T_i)-\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)\hat{K}_c^{z_i}(T_i)} +o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(31)

Similar to the argument of (29), we have

$$\begin{aligned} \begin{aligned}&\frac{2}{n}\sum \limits _{i=1}^n\frac{\delta _ih_\eta (\tilde{T}_i)\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)\hat{K}_c^{z_i}(T_i)}\\&\quad =\frac{2}{n}\sum \limits _{i=1}^n\frac{\delta _ih_\eta (\tilde{T}_i)\left[ K_c^{z_i}(T_i) -\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i) K_c^{z_i}(T_i)}+o_p(n^{-1/2}). \end{aligned} \end{aligned}$$

(32)

Then, we have

$$\begin{aligned} \begin{aligned} \hat{\eta }_z-\eta =&\frac{2}{n}\sum \limits _{i=1}^n[h_\eta (\tilde{T}_i)-\eta ] -\frac{2}{n}\sum \limits _{i=1}^n\int _0^\infty \frac{h_\eta (\tilde{T}_i)}{K_c^{z_i}(u)}dM_i^c(u)\\&+\frac{2}{n}\sum \limits _{i=1}^n\frac{\delta _ih_\eta (\tilde{T}_i) \left[ K_c^{z_i}(T_i)-\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i) K_c^{z_i}(T_i)}+o_p(n^{-1/2}).\\ \end{aligned} \end{aligned}$$

(33)

Similar to the argument of (23), we have

$$\begin{aligned} \hat{G}_z=G+\frac{1}{\mu }[(\hat{\eta }_z-\eta )-(G+1)(\hat{\mu }_z-\mu )]+o_p(n^{-1/2}). \end{aligned}$$

(34)

Based on (31), (33) and (34), we have

$$\begin{aligned} \begin{aligned} \sqrt{n}(\hat{G}_z-G)=&\;\frac{1}{\mu }\left\{ \frac{1}{\sqrt{n}} \sum \limits _{i=1}^n[2(h_\eta (\tilde{T}_i)-\eta )-(G+1)(\tilde{T}_i-\mu )]\right. \\&\quad -\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\int _0^\infty \frac{2h_\eta (\tilde{T}_i)-(G+1)\tilde{T}_i}{K_c(u)}dM_i^c(u)\\&\quad \left. +\;\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\frac{\delta _i\left[ 2h_\eta (\tilde{T}_i) -(G+1)\tilde{T}_i\right] \left[ K_c^{z_i}(T_i)-\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)K_c^{z_i}(T_i)}\right\} \\&+o_p(1). \end{aligned} \end{aligned}$$

(35)

Let $Q_i=2h_\eta (\tilde{T}_i)-(G+1)\tilde{T}_i$. Similar to 7.4.7 of Andersen et al. (1993), we have

$$\begin{aligned}&\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\frac{\delta _i\left[ 2h_\eta (\tilde{T}_i) -(G+1)\tilde{T}_i\right] \left[ K_c^{z_i}(T_i)-\hat{K}_c^{z_i}(T_i)\right] }{K_c^{z_i}(T_i)K_c^{z_i}(T_i)}\nonumber \\&\quad =\frac{1}{\sqrt{n}}\sum \limits _{i=1}^n\frac{\delta _iQ_i}{ K_c^{z_i}(T_i)} \int _0^{\tilde{T}_i-}z_i^T A^{-1}(u)\tilde{z}(u)d\mathbb {M}^c(u)\nonumber \\&\quad =\frac{1}{\sqrt{n}}\int _0^\infty \frac{1}{n}\sum \limits _{i=1}^n \frac{\delta _iQ_i}{K_c^{z_i}(T_i)}I(T_i>u)z_i^T(n^{-1}A)^{-1}(u) \tilde{z}(u)d\mathbb {M}^c(u)\nonumber \\&\quad =\frac{1}{\sqrt{n}}\int _0^\infty \gamma ^T(u)(\mathbf{a}(u))^{-1} \tilde{z}(u)d\mathbb {M}^c(u)+o_p(1), \end{aligned}$$

(36)

where

$$\begin{aligned} \gamma ^T(u)=\mathop {plim}_{n\rightarrow \infty }\frac{1}{n}\sum \limits _{i=1}^n \frac{\delta _i\left[ 2h_\eta (\tilde{T}_i)-(G+1)\tilde{T}_i\right] }{K_c^{z_i}(T_i)}I(T_i>u)z_i^T. \end{aligned}$$

We can obtain the martingale representation of $\sqrt{n} (\hat{G}_z-G)$ by substituting (36) into (35). Then, we can obtain the asymptotic normality of $\sqrt{n} (\hat{G}_z-G)$ using Rebolledos central limit theorem (see Theorem II.5.1 and proposition II.4.1 in Andersen et al. (1993)).$\square$