Inference and optimal censoring scheme for progressively Type-II censored competing risks model for generalized Rayleigh distribution

Abstract

This paper considers the statistical inference for the competing risks model from generalized Rayleigh distribution based on progressive Type-II censoring when the parameters of the latent lifetime distributions are different or common. Maximum likelihood estimates are obtained, where the existence of the point estimators are proved, and the confidence intervals are established via the observed Fisher information matrix as well. Bayesian estimates of unknown parameters and reliability characteristics are derived under symmetric and asymmetric loss functions, and Monte Carlo Markov Chain sampling method is used to compute the Bayesian point estimates and the highest posterior density credible intervals. In addition, Bootstrap methods are also considered to obtain bias-corrected point estimates and approximate confidence intervals. Then we carry out hypothesis test using likelihood ratio test statistics. Monte Carlo simulation and a set of real data are presented to assess the performance of our proposed methods. Finally, the optimal censoring scheme issue is studied.

Appendix

Proof of Theorem 3.1:

I. For given $\lambda _k>0$, we have $0<B_{ki}<1$.

(1) When $\alpha _k\rightarrow 0^+$, then $B_{ki}^{\alpha _k}\rightarrow 1^-$ and $1-B_{ki}^{\alpha _k}\rightarrow 0^+$. $\sum _{i=1}^mI(\delta _i=k)\log (B_{ki})<0$ is a constant, and we have

$$\begin{aligned} \lim \limits _{\alpha _k\rightarrow 0^+}\frac{m_k}{\alpha _k}=+\infty , \quad \lim \limits _{\alpha _k\rightarrow 0^+}-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{B_{ki}^{\alpha _k}\log (B_{ki})}{1-B_{ki}^{\alpha _k}}=+\infty , \end{aligned}$$

therefore, $\lim \limits _{\alpha _k\rightarrow 0^+}\frac{\partial l}{\partial \alpha _k}=+\infty$.

(2) When $\alpha _k\rightarrow +\infty$, then $B_{ki}^{\alpha _k}\rightarrow 0^+$, similarly, $\sum _{i=1}^mI(\delta _i=k)\log (B_{ki})<0$ is a constant, and we have

$$\begin{aligned} \lim \limits _{\alpha _k\rightarrow +\infty }\frac{m_k}{\alpha _k}=0, \quad \lim \limits _{\alpha _k\rightarrow +\infty }-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{B_{ki}^{\alpha _k}\log (B_{ki})}{1-B_{ki}^{\alpha _k}}=0, \end{aligned}$$

therefore, $\lim \limits _{\alpha _k\rightarrow +\infty }\frac{\partial l}{\partial \alpha _k}$ is a negtive constant.

We can further get

$$\begin{aligned} \frac{\partial ^2 l}{\partial \alpha _k^2}=-\frac{m_k}{\alpha _k^2}-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{B_{ki}^{\alpha _k}\log ^2(B_{ki})}{(1-B_{ki}^{\alpha _k})^2}<0. \end{aligned}$$

So for given $\lambda _k$, $\frac{\partial l}{\partial \alpha _k}$ is a monotone decreasing function of $\alpha _k$ and the equation $\frac{\partial l}{\partial \alpha _k}=0$ has an unique solution.

II. For given $\alpha _k>0$ and $\alpha _k \ne 1$.

(1) When $\lambda _k\rightarrow +\infty$, then $B_{ki}\rightarrow 1^-$ and $\lim \nolimits _{\lambda _k\rightarrow +\infty }B_{ki}^\prime =2x_i^2 \lim \nolimits _{\lambda _k\rightarrow +\infty } \frac{\lambda _k}{e^{(\lambda _k x_i)^2}}=0^+$,

$$\begin{aligned} \begin{aligned} \frac{\partial l}{\partial \lambda _k}=&2\frac{m_k}{\lambda _k}+\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^\prime }{B_{ki}}\\&-2\lambda _k\left[ \sum _{i=1}^mx_i^2I(\delta _i=k)+\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{\alpha _kB_{ki}^{\alpha _k-1}x_i^2 e^{-(\lambda _k x_i)^2}}{1-B_{ki}^{\alpha _k}}\right] , \end{aligned} \end{aligned}$$

where $\lim \nolimits _{\lambda _k\rightarrow +\infty }2\frac{m_k}{\lambda _k}=0$, $\lim \nolimits _{\lambda _k\rightarrow +\infty }\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^\prime }{B_{ki}}=0$ and

$$\begin{aligned} \begin{aligned}&\lim \limits _{\lambda _k\rightarrow +\infty }\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{\alpha _kB_{ki}^{\alpha _k-1}x_i^2 e^{-(\lambda _k x_i)^2}}{1-B_{ki}^{\alpha _k}}\\&\quad =\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\alpha _k x_i^2 \lim \limits _{\lambda _k\rightarrow +\infty }\frac{ e^{-(\lambda _k x_i)^2}}{1-B_{ki}^{\alpha _k}}=\sum _{i=1}^m(I(\delta _i=3-k)+r_i)x_i^2. \end{aligned} \end{aligned}$$

Therefore, $\lim \limits _{\lambda _k\rightarrow +\infty }-2\lambda _k[\cdot ]=-\infty$ such that $\lim \limits _{\lambda _k\rightarrow +\infty }\frac{\partial l}{\partial \lambda _k}=-\infty$.

(2) When $\lambda _k\rightarrow 0^+$, then $B_{ki}\rightarrow 0^+$ and $B_{ki}^\prime \rightarrow 0^+$, based on (3.4), we have

$$\begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}2\frac{m_k}{\lambda _k}=+\infty , \quad \lim \limits _{\lambda _k\rightarrow 0^+}-2\lambda _k\sum _{i=1}^mx_i^2I(\delta _i=k)=0. \end{aligned}$$

(A.1)

For $\lim \limits _{\lambda _k\rightarrow 0^+}\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^\prime }{B_{ki}}$, we can compute it as follows.

$$\begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}\frac{B_{ki}^\prime }{B_{ki}}=\lim \limits _{\lambda _k\rightarrow 0^+}\frac{2x_i^2\lambda _ke^{-(\lambda _k x_i)^2}}{1-e^{-(\lambda _k x_i)^2}}=2x_i^2 \lim \limits _{\lambda _k\rightarrow 0^+} \frac{\lambda _k}{1-e^{-(\lambda _k x_i)^2}}=+\infty . \end{aligned}$$

Hence,

$$\begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^\prime }{B_{ki}}= \left\{ \begin{aligned}&+\infty , \qquad \alpha _k>1\\&-\infty , \qquad 0<\alpha _k<1 \end{aligned} \right. \end{aligned}$$

(A.2)

We have already known $B_{ki}^{\alpha _k}\rightarrow 0^+$ and

$$\begin{aligned} B_{ki}^{\alpha _k-1} \rightarrow \left\{ \begin{aligned} 0^+, \qquad&\alpha _k>1\\ +\infty , \qquad&0<\alpha _k<1 \end{aligned} \right. \end{aligned}$$

hence, when $\alpha _k>1$, $\lim \limits _{\lambda _k\rightarrow 0^+}-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{\alpha _kB_{ki}^{\alpha _k-1}B_{ki}^\prime }{1-B_{ki}^{\alpha _k}}=0$, but when $0<\alpha _k<1$,

$$\begin{aligned}&\lim \limits _{\lambda _k\rightarrow 0^+}-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{\alpha _kB_{ki}^{\alpha _k-1}B_{ki}^\prime }{1-B_{ki}^{\alpha _k}}\\&\quad =-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)2x^2_i\alpha _k \lim \limits _{\lambda _k\rightarrow 0^+}B_{ki}^{\alpha _k-1}\lambda _k. \end{aligned}$$

And we can get

$$\begin{aligned} \begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}B_{ki}^{\alpha _k-1}\lambda _k&=\lim \limits _{\lambda _k\rightarrow 0^+}\frac{\lambda _k}{B_{ki}^{1-\alpha _k}}=\lim \limits _{\lambda _k\rightarrow 0^+} \frac{1}{1-\alpha _k}\frac{B_{ki}^{\alpha _k}}{B_{ki}^\prime }\\&=\lim \limits _{\lambda _k\rightarrow 0^+} \frac{1}{1-\alpha _k}\frac{\alpha _kB_{ki}^{\alpha _k-1}B_{ki}^\prime }{B_{ki}^{\prime \prime }}\\&=\frac{\alpha _k}{1-\alpha _k}\lim \limits _{\lambda _k\rightarrow 0^+}B_{ki}^{\alpha _k-1}\lambda _k \end{aligned} \end{aligned}$$

for arbitrary $0<\alpha _k<1$, here $B_{ki}^{\prime \prime }$ is obtained by deriving $B_{ki}^\prime$ to $\lambda _k$, namely $B_{ki}^{\prime \prime }=2x_i^2e^{-(\lambda _k x_i)^2}(1-2x_i^2\lambda _k^2)$. So $\lim \limits _{\lambda _k\rightarrow 0^+}B_{ki}^{\alpha _k-1}\lambda _k=0$

We have

$$\begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}-\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\frac{\alpha _kB_{ki}^{\alpha _k-1}B_{ki}^\prime }{1-B_{ki}^{\alpha _k}}=0. \end{aligned}$$

(A.3)

According to (A.1), (A.2) and (A.3), for $\alpha _k>1$, $\lim \limits _{\lambda _k\rightarrow 0^+}\frac{\partial l}{\partial \lambda _k}=+\infty$. When $0<\alpha _k<1$, by using L’H${\hat{o}}$pital’s rule twice, we have

$$\begin{aligned} \begin{aligned} \lim \limits _{\lambda _k\rightarrow 0^+}\frac{\partial l}{\partial \lambda _k}&=\lim \limits _{\lambda _k\rightarrow 0^+}2\frac{m_k}{\lambda _k}+\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^\prime }{B_{ki}} \\&=\lim \limits _{\lambda _k\rightarrow 0^+}2\sum _{i=1}^{m}I(\delta _i=k)\left[ \frac{1}{\lambda _k}+(\alpha _k-1)\frac{x_i^2\lambda _ke^{-(\lambda _k x_i)^2}}{1-e^{-(\lambda _k x_i)^2}}\right] \\&=\lim \limits _{\lambda _k\rightarrow 0^+}2\sum _{i=1}^{m}I(\delta _i=k) \frac{1-e^{-(\lambda _k x_i)^2}+(\alpha _k-1)x_i^2\lambda _k^2 e^{-(\lambda _k x_i)^2}}{\lambda _k(1-e^{-(\lambda _k x_i)^2})} \\&=+\infty . \end{aligned} \end{aligned}$$

Therefore, the equation $\frac{\partial l}{\partial \lambda _k}=0$ has solutions, namely, the MLE of $\lambda _k$ certainly exists.

$$\begin{aligned} \frac{\partial ^2 l}{\partial \lambda _k^2}= & {} -2\frac{m_k}{\lambda _k^2}-2\sum _{i=1}^mx_i^2I(\delta _i=k)+\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^{\prime \prime }B_{ki}-(B_{ki}^\prime )^2}{B_{ki}^2}\nonumber \\&-\alpha _k\sum _{i=1}^m(I(\delta _i=3-k)+r_i) \nonumber \\&\cdot \frac{[(\alpha _k-1)B_{ki}^{\alpha _k-2}(B_{ki}^\prime )^2+B_{ki}^{\alpha _k-1}B_{ki}^{\prime \prime }](1-B_{ki}^{\alpha _k})+(B_{ki}^{\alpha _k-1}B_{ki}^{\prime })^2\alpha _k}{(1-B_{ki}^{\alpha _k})^2} \end{aligned}$$

(A.4)

$$\begin{aligned}= & {} -2\frac{m_k}{\lambda _k^2}-2\sum _{i=1}^mx_i^2I(\delta _i=k)+\sum _{i=1}^m(\alpha _k-1)I(\delta _i=k)\frac{B_{ki}^{\prime \prime }B_{ki}-(B_{ki}^\prime )^2}{B_{ki}^2} \nonumber \\&-\alpha _k\sum _{i=1}^m(I(\delta _i=3-k)+r_i)\nonumber \\&\frac{B_{ki}^{\alpha _k-2}(1-B_{ki}^{\alpha _k})(B_{ki}^{\prime \prime }B_{ki}-(B_{ki}^\prime )^2)+\alpha _kB_{ki}^{\alpha _k-2}(B_{ki}^\prime )^2}{(1-B_{ki}^{\alpha _k})^2}\nonumber \\ \end{aligned}$$

(A.5)

Using the inequality $e^{-t}>1-2t$ for $t>0$, we have

$$\begin{aligned} B_{ki}^{\prime \prime }B_{ki}-(B_{ki}^\prime )^2=2x_i^2e^{-(\lambda _k x_i)^2}(1-2x_i^2\lambda _k^2-e^{-(\lambda _k x_i)^2})<0. \end{aligned}$$

When $\alpha _k>1$, it is evident that $\frac{\partial ^2 l}{\partial \lambda _k^2}<0$ can be derived from (A.4), and MLE of $\lambda _k$($k=1,2$) is unique in this case. $\square$