Non-parametric Maximum Likelihood Estimation for Case-Cohort and Nested Case-Control Designs with Competing Risks Data

Abstract

Assuming cause-specific hazards given by Cox’s regression model, we provide non-parametric maximum likelihood estimator (NPMLEs) in the nested case-control or case-cohort design with competing risks data. We propose an iterative algorithm based on self-consistency equations derived from score functions to compute NPMLE and compute the predicted cumulative incidence function with their corresponding confidence intervals and bands. Consistency and asymptotic normality are established, together with a consistent estimator of the asymptotic variance based on the observed profile likelihood. Simulation studies show that the numerical performance of NPMLE approach is satisfactory and compares well with that of weighted partial likelihood. Our method is applied to the Taiwan National Health Insurance Research Database (NHIRD) to analyze the occurrences of liver and lung cancers in type 2 diabetic mellitus patients.

Appendices

Appendix 1: Proof of Lemma 1

$P_{M} l_{2,\left (\boldsymbol {\beta }_{1}^{\left (q\right )},\cdots ,\boldsymbol {\beta }_{K}^{\left (q\right )},\boldsymbol {p},\varLambda _{1}^{\left (q\right )},\cdots ,\varLambda _{K}^{\left (q\right )} \right )} \left [\boldsymbol {h}_{2} \right ]=0$ means $\sum _{i=1}^{M}\left \{I_{\left (o_{i}=1\right )} \left (\frac {\boldsymbol {\mu }_{i}^{T} \boldsymbol {h}_{2} }{\boldsymbol {\mu }_{i}^{T} \boldsymbol {p}} \right )+I_{\left (o_{i}=0\right )}\right .$ $\left .\left (\frac {\boldsymbol {\phi }_{i}^{T} \boldsymbol {h}_{2} }{\boldsymbol {\phi }_{i}^{T} \boldsymbol {p}} \right )\right \}=0$. Here $\boldsymbol {\mu }_{i}=\left (I_{\boldsymbol {Z}_{i}=\boldsymbol {W}_{1} },I_{\boldsymbol {Z}_{i}=\boldsymbol {W}_{2}},\cdots ,I_{\boldsymbol {Z}_{i}=\boldsymbol {W}_{J}} \right )^{T}$and

$$\displaystyle \begin{aligned} \begin{array}{rcl}&\displaystyle &\displaystyle \boldsymbol{\phi}_{i}=\left(\exp \left(-\sum_{k=1}^{K}\varLambda_{k}^{\left(q\right)} \left(X_{i} \right)\exp \left(\boldsymbol{W}_{1}^{T} \boldsymbol{\beta}_{k}^{\left(q\right)} \right) \right),\right.\\ &\displaystyle &\displaystyle \qquad \qquad \left.\cdots,\exp \left(-\sum_{k=1}^{K}\varLambda_{k}^{\left(q\right)} \left(X_{i} \right)\exp \left(\boldsymbol{W}_{J}^{T} \boldsymbol{\beta}_{k}^{\left(q\right)} \right) \right)\right)^{T}.\end{array} \end{aligned} $$

Setting $\boldsymbol {h}_{2}=\left (0,\cdots ,0,1,0,\cdots ,-1\right )^{T}$, having the lth and Jth coordinate equal to 1 and −1 respecting and all the other coordinates being 0, we know

$$\displaystyle \begin{aligned}\sum_{i=1}^{M}\left\{I_{\left(o_{i}=1\right)} \left(\frac{\mu_{i,l}}{\boldsymbol{\mu}_{i}^{T} \boldsymbol{p}} \right){+}I_{\left(o_{i}=0\right)} \left(\frac{\phi_{i,l}}{\boldsymbol{\phi}_{i}^{T} p} \right)\right\}{=}\sum_{i=1}^{M}\left\{I_{\left(o_{i}=1\right)} \left(\frac{\mu_{i,J}}{\boldsymbol{\mu}_{i}^{T} \boldsymbol{p}} \right){+}I_{\left(o_{i}=0\right)} \left(\frac{\phi_{i,J}}{\boldsymbol{\phi}_{i}^{T} \boldsymbol{p}} \right)\right\}. \end{aligned}$$

Then

$$\displaystyle \begin{aligned}\sum _{i=1}^{M}\left[I_{\left(o_{i}=1\right)} I_{\boldsymbol{Z}_{i}=\boldsymbol{W}_{l} }+I_{\left(o_{i}=0\right)} \alpha _{il} \right] =p_{l} \sum_{i=1}^{M}\left\{I_{\left(o_{i} =1\right)} \left(\frac{I_{\boldsymbol{Z}_{i} =\boldsymbol{W}_{J} } }{p_{J} } \right)+I_{\left(o_{i} =0\right)} \left(\frac{\alpha _{iJ} }{p_{J} } \right)\right\}.\end{aligned}$$

Straight-forward simplification shows that

$$\displaystyle \begin{aligned}M{=}\sum_{l=1}^{J}\sum_{i=1}^{M}\left[I_{\left(o_{i}=1\right)} I_{\boldsymbol{Z}_{i}=\boldsymbol{W}_{l} }+I_{\left(o_{i}=0\right)} \alpha_{il} \right]{=}\sum_{i=1}^{M}\left\{I_{\left(o_{i} =1\right)} \left(\frac{I_{\boldsymbol{Z}_{i}=\boldsymbol{W}_{J}}}{p_{J}} \right){+}I_{\left(o_{i}=0\right)} \left(\frac{\alpha_{iJ}}{p_{J}} \right)\right\},\end{aligned}$$

then we get that

$$\displaystyle \begin{aligned}p_{l}=\frac{1}{M} \sum_{i=1}^{M}\left[I_{\left(o_{i}=1\right)} I_{\boldsymbol{Z}_{i}=\boldsymbol{W}_{l}}+I_{\left(o_{i}=0\right)} \alpha_{il} \right].\end{aligned}$$

This shows that

$$\displaystyle \begin{aligned}\boldsymbol{p}=\left(\frac{1}{M} \boldsymbol{1}^{T} \boldsymbol{\alpha} \left(\boldsymbol{p}\right)\right)^{T},\end{aligned}$$

and the proof is complete.

Making use of Lemma 1, we define

$$\displaystyle \begin{aligned}\boldsymbol{p}^{\left(q+1\right)}=\left(\frac{1}{M} \boldsymbol{1}^{T} \boldsymbol{\alpha} \left(\boldsymbol{p}^{\left(q\right)} \right)\right)^{T}.\end{aligned}$$

Appendix 2: Proof of Lemma 6

For any h ∈ H _p. If we can show that ϕ _θ,h is continuous at θ ₀, then we have, for arbitrary ε > 0, there is some $\delta =\delta \left (\varepsilon ,\boldsymbol {\theta }_{0}\right )>0$ such that $\left \| \phi _{\boldsymbol {\theta },\boldsymbol {h}}-\phi _{\boldsymbol {\theta }_{0},\boldsymbol {h}} \right \| <\varepsilon$ whenever $\left \| \boldsymbol {\theta }-\boldsymbol {\theta }_{0} \right \| <\delta$. Hence, whenever $\left \| \boldsymbol {\theta }-\boldsymbol {\theta }_{0} \right \|<\delta$,

$$\displaystyle \begin{aligned}\left|E\left(\phi_{\boldsymbol{\theta},\boldsymbol{h}}-\phi_{\boldsymbol{\theta}_{0},\boldsymbol{h}} \right)^{2} \right|\le E\left\| \phi_{\boldsymbol{\theta},\boldsymbol{h}}-\phi_{\boldsymbol{\theta}_{0},\boldsymbol{h}} \right\|{}^{2} \le \varepsilon^{2},\end{aligned}$$

which means that $E\left (\phi _{\boldsymbol {\theta },\boldsymbol {h}}-\phi _{\boldsymbol {\theta }_{0},\boldsymbol {h}} \right )^{2} \to 0$ as θ →θ ₀ for any h ∈ H _p. Thus,

$$\displaystyle \begin{aligned}\sup_{\boldsymbol{h}\in H_{p}} E\left(\phi_{\boldsymbol{\theta},\boldsymbol{h}}-\phi_{\boldsymbol{\theta}_{0},\boldsymbol{h}} \right)^{2} \to 0\end{aligned}$$

as θ →θ ₀.

Now, we want to show that ϕ _θ,h is continuous at θ ₀, for any h ∈ H _p. It is clearly that the following functions $\exp \left (\boldsymbol {Z}_{1}^{T} \boldsymbol {\beta }_{k} \right )$, $\exp \left (\boldsymbol {W}_{j}^{T} \boldsymbol {\beta }_{k}\right )$, $\exp \left (-\sum _{k=1}^{2}\varLambda _{k} \left (X_{1} \right )\exp \left (\boldsymbol {W}_{j}^{T} \boldsymbol {\beta }_{k} \right ) \right )$ and $pr\left (\boldsymbol {Z}_{1}=\boldsymbol {W}_{j} |T_{1} \ge X_{1} \right )$ are continuous and we know $\sum _{j=1}^{J}\exp \left (-\sum _{k=1}^{2}\varLambda _{k} \left (X_{1} \right )\exp \left (\boldsymbol {W}_{j}^{T} \boldsymbol {\beta }_{k} \right ) \right )p_{j}$ is bounded and bounded away from zero. Finally, we may see that $\int _{0}^{X_{1}}h\left (t\right )d\varLambda _{k} \left (t\right )$ is continuous at Λ _k0, since

$$\displaystyle \begin{aligned}{\mathop{\sup}\limits_{h\in BV\left[0,\tau \right]}} \int_{0}^{\tau}\left|h\left(t\right)\right|d\left|\varLambda_{k} \left(t\right)-\varLambda_{k0} \left(t\right)\right| \le c\int_{0}^{\tau}d\left|\varLambda_{k} \left(t\right)-\varLambda_{k0} \left(t\right)\right|,\end{aligned}$$

we know $\left \| \varLambda _{k}-\varLambda _{k0} \right \|{ }_{V} \to 0$, which means

$$\displaystyle \begin{aligned}\int_{0}^{\tau}d\left|\varLambda_{k} \left(t\right)-\varLambda_{k0} \left(t\right)\right|=\int_{0}^{\tau}\left|\left(\varLambda_{k} \left(t\right)-\varLambda_{k0} \left(t\right)\right)^{/} \right|dt \to 0.\end{aligned}$$

This completes the proof.

Appendix 3: Σ is Positive Definite and Symmetric

• (i) Σ is positive definite.

• (ii) Σ is symmetric.

Proof of (i)

Note that $\varSigma ^{-1}{=}\left (\sigma _{12}^{-1} \left (\boldsymbol {e}_{1},0,0,0\right ),\sigma _{12}^{-1} \left (\boldsymbol {e}_{2},0,0,0\right ),\cdots ,\sigma _{12}^{-1} \right .$ $\left .\left (\boldsymbol {e}_{D},0,0,0\right )\right )$. Here we want to show Σ ⁻¹ is positive definite. Let $\boldsymbol {h}_{12}=\left (y_{1},y_{2},\cdots ,y_{D} \right )^{T}$ be given. Consider

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \boldsymbol{h}_{12}^{T} \varSigma^{-1} \boldsymbol{h}_{12} \\ &\displaystyle =&\displaystyle y_{1} \sigma_{12}^{-1} \left(\boldsymbol{e}_{1},0,0,0\right)_{1} y_{1}+y_{2} \sigma_{12}^{-1} \left(\boldsymbol{e}_{1},0,0,0\right)_{2} y_{1}+\cdots+y_{D} \sigma_{12}^{-1} \left(\boldsymbol{e}_{1},0,0,0\right)_{D} y_{1} \\ &\displaystyle &\displaystyle + y_{1} \sigma_{12}^{-1} \left(\boldsymbol{e}_{2},0,0,0\right)_{1} y_{2}+y_{2} \sigma_{12}^{-1} \left(\boldsymbol{e}_{2},0,0,0\right)_{2} y_{2}+\cdots+y_{D} \sigma_{12}^{-1} \left(\boldsymbol{e}_{2},0,0,0\right)_{D} y_{2} \\ &\displaystyle &\displaystyle + \cdots \\ &\displaystyle &\displaystyle + y_{1} \sigma_{12}^{-1} \left(\boldsymbol{e}_{D},0,0,0\right)_{1} y_{D}+y_{2} \sigma_{12}^{-1} \left(\boldsymbol{e}_{D},0,0,0\right)_{2} y_{D}+\cdots+y_{D} \sigma_{12}^{-1} \left(\boldsymbol{e}_{D},0,0,0\right)_{D} y_{D} \\ &\displaystyle =&\displaystyle y_{1} \sigma_{12}^{-1} \left(\left(y_{1},y_{2},\cdots,y_{D} \right),0,0,0\right)_{1}+y_{2} \sigma_{12}^{-1} \left(\left(y_{1},y_{2},\cdots,y_{D} \right),0,0,0\right)_{2} \\ &\displaystyle &\displaystyle + \cdots \\ &\displaystyle &\displaystyle + y_{D} \sigma_{12}^{-1} \left(\left(y_{1},y_{2},\cdots,y_{D} \right),0,0,0\right)_{D} \\ &\displaystyle =&\displaystyle \boldsymbol{h}_{12}^{T} \sigma_{12}^{-1} \left(h_{12},0,0,0\right)>0, \end{array} \end{aligned} $$

for h ₁₂ nonzero. This completes the proof.

Proof of (ii)

We want to show that Σ ⁻¹ is symmetric. Fix i. Choose h ^′ such that Let $\boldsymbol {h}_{1}^{\prime }=\sigma _{1}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )$, $\boldsymbol {h}_{2}^{\prime }=\sigma _{2}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )$, $\boldsymbol {h}_{3}^{\prime }=\sigma _{3}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )$, $\boldsymbol {h}_{4}^{\prime }=\sigma _{4}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )$ and $\boldsymbol {h}_{5}^{\prime }=\sigma _{5}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )$. Choose h such that $\boldsymbol {h}_{1}=\sigma _{1}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$, $\boldsymbol {h}_{2}=\sigma _{2}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$, $\boldsymbol {h}_{3}=\sigma _{3}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$, $\boldsymbol {h}_{4}=\sigma _{4}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$ and $\boldsymbol {h}_{5}=\sigma _{5}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$. It remains to check the following case from (8), we can obtain

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \sigma_{1} \left(\boldsymbol{h}\right)^{T} \boldsymbol{h}_{1}^{\prime}+\sigma_{2} \left(\boldsymbol{h}\right)^{T} \boldsymbol{h}_{2}^{\prime}+\sigma_{3} \left(\boldsymbol{h}\right)^{T} \boldsymbol{h}_{3}^{\prime} \\ &\displaystyle &\displaystyle +\int_{0}^{\tau }\sigma_{4} \left(\boldsymbol{h}\right)\left(t\right)\boldsymbol{h}_{4}^{\prime} \left(t\right)d\varLambda_{10} \left(t\right) +\int_{0}^{\tau}\sigma_{5} \left(\boldsymbol{h}\right)\left(t\right)\boldsymbol{h}_{5}^{\prime} \left(t\right)d\varLambda_{20} \left(t\right)\\ &\displaystyle &\displaystyle =\boldsymbol{e}_{k}^{T} \sigma_{12}^{-1} \left(\boldsymbol{e}_{i} ,0,0,0\right).\\ &\displaystyle &\displaystyle \sigma_{1} \left(\boldsymbol{h}^{\prime} \right)^{T} \boldsymbol{h}_{1} +\sigma _{2} \left(\boldsymbol{h}^{\prime} \right)^{T} \boldsymbol{h}_{2} +\sigma_{3} \left(\boldsymbol{h}^{\prime} \right)^{T} \boldsymbol{h}_{3} \\ &\displaystyle &\displaystyle +\int _{0}^{\tau }\sigma _{4} \left(\boldsymbol{h}^{\prime} \right)\left(t\right)\boldsymbol{h}_{4} \left(t\right)d\varLambda _{10} \left(t\right) +\int_{0}^{\tau }\sigma_{5} \left(\boldsymbol{h}^{\prime} \right)\left(t\right)\boldsymbol{h}_{5} \left(t\right)d\varLambda_{20} \left(t\right)\\ &\displaystyle &\displaystyle =\boldsymbol{e}_{i}^{T} \sigma_{12}^{-1} \left(\boldsymbol{e}_{k} ,0,0,0\right). \end{array} \end{aligned} $$

Hence $\boldsymbol {e}_{k}^{T} \sigma _{12}^{-1} \left (\boldsymbol {e}_{i},0,0,0\right )=\boldsymbol {e}_{i}^{T} \sigma _{12}^{-1} \left (\boldsymbol {e}_{k},0,0,0\right )$. This completes the proof.

Appendix 4: $\sqrt {M} \left (\widehat {\boldsymbol {\beta }}_{M}-\boldsymbol {\beta }_{0} \right )$ Has Asymptotic Variance Σ ⁻¹

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle var\left(\sqrt{M} \left(\widehat{\boldsymbol{\beta}}_{M}-\boldsymbol{\beta}_{0} \right)\right) \\ &\displaystyle &\displaystyle =var\left(\varSigma^{-1} \sqrt{M} \left(P_{M}-P_{0} \right)\tilde{l}_{0} \right) \\ &\displaystyle &\displaystyle =\varSigma^{-1} var\left(\left(\sum_{i=1}^{M} \tilde{l}_{0} \left(X_{i},E_{i},\boldsymbol{Z}_{i} \cdot O_{i},O_{i} \right) /\sqrt{M}\right)\right)\varSigma^{-1} \\ &\displaystyle &\displaystyle =\varSigma^{-1} var\left(\tilde{l}_{0} \right)\varSigma^{-1} \\ &\displaystyle &\displaystyle =\varSigma^{-1} E\left(\tilde{l}_{0} \tilde{l}_{0}^{T} \right)\varSigma^{-1}-\varSigma^{-1} \left\{E\left(\tilde{l}_{0} \right)E\left(\tilde{l}_{0} \right)^{T} \right\}\varSigma^{-1} \\ &\displaystyle &\displaystyle =\varSigma^{-1}. \end{array} \end{aligned} $$

Because $E\left (\tilde {l}_{0} \tilde {l}_{0}^{T} \right )=\varSigma$ and

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \boldsymbol{e}_{i}^{T} \varSigma^{-1} \left\{E\left(\tilde{l}_{0} \right)E\left(\tilde{l}_{0} \right)^{T} \right\}\varSigma^{-1} \boldsymbol{e}_{k} \\ &\displaystyle &\displaystyle =\left\{\boldsymbol{e}_{i}^{T} \varSigma^{-1} E\left(\tilde{l}_{0} \right)\left(l_{3,\boldsymbol{\theta}_{0}} \left[\boldsymbol{h}_{3} \right]+l_{4,\boldsymbol{\theta}_{0}} \left[\boldsymbol{h}_{4} \right]+l_{5,\boldsymbol{\theta}_{0} } \left[\boldsymbol{h}_{5} \right]\right)\right\} \\ &\displaystyle &\displaystyle \times \left\{\left(l_{3,\boldsymbol{\theta}_{0} } \left[\boldsymbol{h}_{3} \right]+l_{4,\boldsymbol{\theta}_{0}} \left[\boldsymbol{h}_{4} \right]+l_{5,\boldsymbol{\theta}_{0}} \left[\boldsymbol{h}_{5} \right]\right)^{-1} E\left(\tilde{l}_{0} \right)^{T} \varSigma^{-1} \boldsymbol{e}_{k} \right\} \\ &\displaystyle &\displaystyle =0. \end{array} \end{aligned} $$

This completes the proof.

Appendix 5: Two Results Due to V&W [22]

We quote two important results used in this paper.

Theorem 3.3.1 of V&W [22]

Let Ψ _n and Ψ be random maps and a fixed map, respectively, from Θ into a Banach space such that

$$\displaystyle \begin{aligned}\sqrt{n} \left(\varPsi_{n}-\varPsi \right)\left(\widehat{\theta}_{n} \right)-\sqrt{n} \left(\varPsi_{n}-\varPsi \right)\left(\theta_{0} \right)=o_{p^{*}} \left(1+\sqrt{n} \left\| \widehat{\theta}_{n}-\theta_{0} \right\| \right),\end{aligned}$$

and such that the sequence $\sqrt {n} \left (\varPsi _{n}-\varPsi \right )\left (\theta _{0} \right )$ converges in distribution to a tight random element Z. Let $\theta \to \varPsi \left (\theta \right )$ be Fréchet differentiable at θ ₀ with a continuously invertible derivative $\dot {\varPsi }_{0}$. If $\varPsi \left (\theta _{0} \right )=0$, $\widehat {\theta }_{n}$ satisfies $\varPsi _{n} \left (\widehat {\theta }_{n} \right )=o_{p^{*} } \left (n^{-1/2} \right )$, and converges in outer probability to θ ₀, then

$$\displaystyle \begin{aligned}\sqrt{n} \dot{\varPsi}_{\theta_{0} } \left(\widehat{\theta}_{n}-\theta_{0} \right)=-\sqrt{n} \left(\varPsi_{n}-\varPsi \right)\left(\theta_{0} \right)+o_{p^{*}} \left(1\right).\end{aligned}$$

Consequently, $\sqrt {n} \left (\widehat {\theta }_{n}-\theta _{0} \right )\to -\dot {\varPsi }_{\theta _{0} }^{-1} Z$.

In the case of independent and identically distributed observations, the theorem may be applied with $\varPsi _{n} \left (\theta \right )h=P_{n} \phi _{\theta ,h}$ and $\varPsi \left (\theta \right )h=P\phi _{\theta ,h}$ for given measurable functions ϕ _θ,h, indexed by Θ and arbitrary index set H. In this case, $\sqrt {n} \left (\varPsi _{n}-\varPsi \right )\left (\theta \right )=\left \{G_{n} \phi _{\theta ,h}:h\in H\right \}$ is the empirical process indexed by the classes of functions $\left \{\phi _{\theta ,h}:h\in H\right \}$.

Lemma 3.3.5 of V&W [22]

Suppose that the class of functions

$$\displaystyle \begin{aligned}\left\{\phi_{\theta,h}-\phi_{\theta_{0},h}:\left\| \theta-\theta_{0} \right\| <\delta,h\in H\right\}\end{aligned}$$

is P-Donsker for some δ > 0 and

$$\displaystyle \begin{aligned}{\mathop{\sup}\limits_{h\in H}} P\left(\phi_{\theta,h}-\phi_{\theta_{0},h} \right)^{2} \to 0\end{aligned}$$

as θ → θ ₀.

If $\widehat {\theta }_{n}$ converges in outer probability to θ ₀, then

$$\displaystyle \begin{aligned}\left\| G_{n} \left(\phi_{\widehat{\theta}_{n},h}-\phi_{\theta_{0},h} \right)\right\|{}_{H}=o_{p^{*}} \left(1+\sqrt{n} \left\| \widehat{\theta}_{n}-\theta_{0} \right\| \right).\end{aligned}$$

Appendix 6: The Conditional Distribution of Z Given Y

In modeling the conditional distribution of Z given Y when Y is of discrete type, denote the stratified levels of Y as $SY=\left (1,\cdots ,S\right )$ and describe Y by using dummy variable according to stratified levels.

Assume that there are J _s distinct values among the observed covariates under Y -stratum s and they are denoted by $\left (\boldsymbol {W}_{s1},\boldsymbol {W}_{s2},\cdots ,\boldsymbol {W}_{sJ_{s}} \right )$. Let 0 ≤ p _sj ≤ 1 and $\sum _{j=1}^{J_{s}}p_{sj}=1$ be given. Then the conditional distribution $f_{e}\left (\boldsymbol {Z}|SY \right )$ is given by

$$\displaystyle \begin{aligned}f_{e} \left(\boldsymbol{Z}|SY\right)=\prod_{s=1}^{S}\left[\sum_{j=1}^{J_{s} }p_{sj} I_{\boldsymbol{Z}=\boldsymbol{W}_{sj}} \right]^{I_{\left(SY=s\right)}},\end{aligned}$$

and the corresponding marginal survival function G of T is given by

$$\displaystyle \begin{aligned}G_{e} \left(t|SY\right)=\prod_{s=1}^{S}\left[\sum_{j=1}^{J_{s}}\left(\exp \left(-\sum_{k=1}^{K}\varLambda_{k} \left(t\right)\exp \left(\boldsymbol{Y}^{T} \boldsymbol{\eta}_{k}+\boldsymbol{W}_{sj}^{T} \boldsymbol{\beta}_{k} \right) \right)\right)p_{sj} \right]^{I_{\left(SY=s\right)}}.\end{aligned}$$

This approach to modeling the covariate distribution is also taken by Scheike and Juul [16]. The proposed NPMLE then applies with the conditional covariate distribution and the marginal survival function of T.

When Y is of continuous type, we can apply kernel smoothing techniques to obtain the estimate for the conditional density of Z given Y . But this is beyond the scope of this paper and deserves further research.