Point and interval estimation under progressive type-I interval censoring with random removal

Abstract

This work considers point and interval estimation based on data from a life test under progressive type-I interval censoring with random removal. The asymptotic properties of the maximum likelihood estimators (MLEs) are established under appropriate regularity conditions. Asymptotic confidence intervals and $\beta$-content $\gamma$-level tolerance interval are obtained by using the asymptotic normality of MLEs. A simulation study is undertaken to assess the performance of the MLEs, confidence intervals and tolerance interval. Lastly, the minimum sample size required to achieve a desired $\beta$-content $\gamma$-level tolerance interval is determined.

Appendix

Here we provide the proofs of the lemmas and results stated in the main article.

For the sake of convenience, we drop index i in the proof as items are independent and identical. Let $\delta _{1,j}$ and $\delta _{2,j}$ represent the indicator functions that an item fails in jth interval, and is censored at $T_j$ respectively. Note that $\sum _{l=j+1}^k (\delta _{1,l}+\delta _{1,l})=1$ indicates that an item is at risk at $T_j$, thus $P\left( \sum _{l=j+1}^k (\delta _{1,l}+\delta _{1,l})=1\right) =\prod _{l=0}^{j}(1-p_l)(1-q_l)$, where $p_0=q_0=0$. In all the subsequent expressions we use $F(T_j)$ instead of $F(T_j;\varvec{\theta })$ for $j=1,2,\ldots ,k$.

Proof of Lemma 1 (i)

By differentiating the contribution of an item in log-likelihood function (6) with respect to parameter $\theta _u$ for $u=1,2,\ldots ,m$, we have,

(15)

By using (2) and (3), we have $\displaystyle \mathrm {E}[\delta _{1,j}]=\prod \nolimits _{l=0}^{j-1} (1-p_l)(1-q_l)q_j$ and $\displaystyle \mathrm {E}[\delta _{2,j}]=\prod \nolimits _{l=0}^{j-1} (1-p_l)(1-q_l)(1-q_j)p_j$, where $p_0=q_0=0$. Then we get, for $u=1,2,\ldots ,m,$

The last two expressions hold, since $\prod _{l=0}^{j-1}(1-q_l)=1-F(T_{j-1})$ and $p_k=1$. This implies . Hence Lemma 1 (i) holds. $\square$

Proof of Lemma 1 (ii)

On differentiating equation with respect to $p_j$ for $j=1,2,\ldots ,k-1$, we have

Now $\displaystyle \mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{1,l})\right] =P\left( \sum _{l=j+1}^k (\delta _{1,l}+\delta _{1,l})=1\right) =\prod _{l=0}^{j}(1-p_l)(1-q_l)$, for $j=1,\ldots ,k-1$. So,

Hence . $\square$

Proof of Lemma 2 (i)

We have, for $v=1,2,\ldots ,m$,

(16)

where $T_0=0$. By using (2) and (3) and following arguments similar to the ones used for proving Lemma 1 (i), we have,

$$\begin{aligned} \sum _{j=1}^k \mathrm {E} \left[ \frac{\delta _{1,j}}{F(T_j)-F(T_{j-1})} \left( \frac{\partial ^2 F(T_j)}{\partial {\theta _u} \partial {\theta _v}}-\frac{\partial ^2 F(T_{j-1})}{\partial {\theta _u} \partial {\theta _v}} \right) -\frac{\delta _{2,j}}{1-F(T_{j})}\frac{\partial ^2 F(T_j)}{\partial {\theta _u} \partial {\theta _v}}\right] =0. \end{aligned}$$

Thus for $u,v=1,2,\ldots ,m$, we have

$$\begin{aligned} \displaystyle \mathrm {E}&\left[ -\frac{\partial ^ 2 {\mathcal {l}}^{I^*} (\varvec{\theta }) }{\partial {\theta _u} \partial {\theta _v}}\right] \nonumber \\&\quad = \sum _{j=1}^k \frac{\mathrm {E}[\delta _{1,j}]}{\left( F(T_j)-F(T_{j-1})\right) ^2} \left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) \left( \frac{\partial F(T_j)}{\partial \theta _v}-\frac{\partial F(T_{j-1})}{\partial \theta _v}\right) \nonumber \\&\qquad +\sum _{j=1}^k \frac{\mathrm {E}[\delta _{2,j}]}{\left( 1-F(T_{j})\right) ^2} \frac{\partial F(T_j)}{\partial \theta _u}\frac{\partial F(T_j)}{\partial \theta _v} \nonumber \\&\quad = \sum _{j=1}^k \frac{\prod _{l=0}^{j-1}(1-p_l)(1-q_l)}{\left( 1-F(T_{j-1})\right) ^2} \left[ \frac{1}{q_j} \left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) \left( \frac{\partial F(T_j)}{\partial \theta _v}-\frac{\partial F(T_{j-1})}{\partial \theta _v}\right) \right. \nonumber \\&\qquad \left. +\; \frac{1}{1-q_j}\frac{\partial F(T_j)}{\partial \theta _u}\frac{\partial F(T_j)}{\partial \theta _v} \right] . \end{aligned}$$

(17)

Next by using (15), we have

$$\begin{aligned}&\displaystyle \frac{\partial {\mathcal {l}}^{I^*} (\varvec{\theta })}{\partial \theta _u} \frac{\partial {\mathcal {l}}^{I^*} (\varvec{\theta })}{\partial \theta _v}\\&\quad = \left[ \sum _{j=1}^k \frac{1}{1-F(T_{j-1})} \left[ \frac{\delta _{1,{j}}}{q_j} \left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) -\frac{\delta _{2,{j}}}{1- q_j}\frac{\partial F(T_j)}{\partial \theta _u} \right] \right] \\&\qquad \times \left[ \sum _{s=1}^k \frac{1}{1-F(T_{s-1})} \left[ \frac{\delta _{1,{s}}}{q_s} \left( \frac{\partial F(T_s)}{\partial \theta _v}-\frac{\partial F(T_{s-1})}{\partial \theta _v}\right) -\frac{\delta _{2,{s}}}{1- q_s}\frac{\partial F(T_s)}{\partial \theta _v} \right] \right] \end{aligned}$$

Since out of the 2k indicator variables, $\delta _{1,1},\delta _{1,2},\ldots , \delta _{1,k}, \delta _{2,1}, \delta _{2,2},\ldots ,\delta _{2,k}$, exactly one can take value 1 and others are zero, we observe that when $j\ne s$, $\displaystyle P\left( \delta _{1,j}\delta _{1,s}=1\right) = P\left( \delta _{2,j}\delta _{2,s}=1\right) =0.$ Also $P\left( \delta _{1,j}\delta _{2,s}=1\right) =0$ for all j and s. As a result, we have, for $j\ne s$,

$$\begin{aligned} \mathrm {E}\left[ \delta _{1,j}\delta _{1,s}\right] =\mathrm {E}\left[ \delta _{2,j} \delta _{2,s}\right] =0 \end{aligned}$$

(18)

and when $j=s$, we get

$$\begin{aligned} \mathrm {E}\left[ \delta _{1,j}^2\right] =P(\delta _{1,j}=1), \mathrm {E}\left[ \delta _{2,j}^2\right] =P(\delta _{2,j}=1) \text{ and } \mathrm {E}\left[ \delta _{1,j}\delta _{2,j}\right] =0. \end{aligned}$$

(19)

Thus by using (18) and (19), we get

$$\begin{aligned}&\displaystyle \mathrm {E}\left[ \frac{\partial {\mathcal {l}}^{I^*} (\varvec{\theta })}{\partial \theta _u} \frac{\partial {\mathcal {l}}^{I^*} (\varvec{\theta })}{\partial \theta _v}\right] \nonumber \\&\quad = \sum _{j=1}^k \frac{1}{\left( 1-F(T_{j-1})\right) ^2} \left[ \frac{\mathrm {E}[\delta _{1,j}^2]}{q_j^2}\left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) \left( \frac{\partial F(T_j)}{\partial \theta _v}-\frac{\partial F(T_{j-1})}{\partial \theta _v}\right) \right. \nonumber \\&\qquad \left. - \;\frac{\mathrm {E}[\delta _{1,j}\delta _{2,j} ]}{q_j(1-q_j)} \left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) \frac{\partial F(T_j)}{\partial \theta _v}\right. \nonumber \\&\qquad \left. - \;\frac{\mathrm {E}[\delta _{1,j}\delta _{2,j} ]}{q_j(1-q_j)} \left( \frac{\partial F(T_j)}{\partial \theta _v}-\frac{\partial F(T_{j-1})}{\partial \theta _v}\right) \frac{\partial F(T_j)}{\partial \theta _u}+ \frac{\mathrm {E}[\delta _{2,j}^2]}{(1-q_j)^2}\frac{\partial F(T_j)}{\partial \theta _u} \frac{\partial F(T_j)}{\partial \theta _v}\right] . \nonumber \\&\quad = \sum _{j=1}^k \frac{\prod _{l=0}^{j-1}(1-p_l)(1-q_l)}{\left( 1-F(T_{j-1})\right) ^2} \left[ \frac{1}{q_j}\left( \frac{\partial F(T_j)}{\partial \theta _u}-\frac{\partial F(T_{j-1})}{\partial \theta _u}\right) \left( \frac{\partial F(T_j)}{\partial \theta _v}-\frac{\partial F(T_{j-1})}{\partial \theta _v}\right) \right. \nonumber \\&\qquad \left. + \;\frac{p_j}{(1-q_j)}\frac{\partial F(T_j)}{\partial \theta _u} \frac{\partial F(T_j)}{\partial \theta _v}\right] . \end{aligned}$$

(20)

Thus, using Eqs. (17) and (20), part (i) of Lemma 2 is proved.

Proof of Lemma 2 (ii)

On differentiating (15) with respect to $p_s$ for $s=1,2,\ldots ,k-1$, we have . Hence for $u=1,2,\ldots ,m$ and $s=1,2,\ldots ,k-1$, we get that

(21)

Next we consider the LHS of part (ii) of Lemma 2. For $s=1,2,\ldots ,k-1$, we have

(22)

Hence by using (21) and (22), we prove that Lemma 2 (ii) holds.

Proof of Lemma 2 (ii)

From Eq. (16), it is obvious that

and ,    for $j \ne s$.

Next we consider its LHS,

By using (16), we have

Note that by using (18) and (19), we have, $\mathrm {E}\left[ \delta _{1,j} \delta _{2,s} \right] =0$ for all j and s, and $\mathrm {E}\left[ \delta _{2,j} \delta _{2,s} \right] =\mathrm {E}\left[ \delta _{2,j}\right]$ for $j=s$ and 0 otherwise. Also we have

$$\begin{aligned} \begin{aligned} \mathrm {E}\left[ \delta _{1,j} \sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right]&=\left\{ \begin{array}{cc} \mathrm {E}\left[ \delta _{1,j}\right] &{} \text{ when } j>s \\ 0&{} \text{ otherwise }. \end{array} \right. \\ \mathrm {E}\left[ \delta _{2,j} \sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right]&=\left\{ \begin{array}{cc} \mathrm {E}\left[ \delta _{2,j}\right] &{} \text{ when } j>s\\ 0&{} \text{ otherwise }. \end{array} \right. \\ \end{aligned} \end{aligned}$$

(23)

We also note that an item at risk at $T_j$, must be at risk at $T_s$. Thus it is easy to deduce that

$$\begin{aligned} \begin{aligned} \mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l}) \sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right]&=\left\{ \begin{array}{cc} \mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\right] &{} \text{ when } j>s\\ \mathrm {E}\left[ \sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right] &{} \text{ when } s>j \\ \mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\right] &{} \text{ when } j=s\\ \end{array} \right. . \end{aligned}\nonumber \\ \end{aligned}$$

(24)

and

By using (23) and (24), for $j>s$, we have

$$\begin{aligned}- & {} \frac{\mathrm {E}\left[ \delta _{2,j}\sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right] }{p_j(1-p_s)}+ \frac{\mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right] }{(1-p_j)(1-p_s)}\\&=\frac{\prod _{l=0}^{j-1}(1-p_l)(1-q_l)}{1-p_s} \left[ -\frac{(1-q_j)p_j}{p_j}+\frac{(1-q_j)(1-p_j)}{1-p_j}\right] =0. \end{aligned}$$

Similarly for $j<s$,

$$\begin{aligned} \displaystyle -\frac{\mathrm {E}\left[ \delta _{2,s}\sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\right] }{p_s(1-p_j)}+ \frac{\mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\sum _{l=s+1}^k (\delta _{1,l}+\delta _{2,l})\right] }{(1-p_j)(1-p_s)}=0. \end{aligned}$$

Lastly, for $j=s$

$$\begin{aligned} \frac{\mathrm {E}\left[ \delta _{2,j}^2\right] }{p_j^2}+\frac{\mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})^2\right] }{(1-p_j)^2}= & {} \frac{\mathrm {E}\left[ \delta _{2,j}\right] }{p_j^2}+\frac{\mathrm {E}\left[ \sum _{l=j+1}^k (\delta _{1,l}+\delta _{2,l})\right] }{(1-p_j)^2}\\= & {} \prod _{l=0}^{j-1}(1-p_l)(1-q_l)\frac{(1-q_j)}{p_j(1-p_j)}. \end{aligned}$$

Hence we have

Hence the proof of Lemma 2 is complete. $\square$

Proof of Result 1

Let $\nabla _{\varvec{\theta }}F(T_j)=\left( \frac{\partial F(T_j)}{\partial \theta _1}, \frac{\partial F(T_j)}{\partial \theta _2}, \ldots , \frac{\partial F(T_j)}{\partial \theta _m}\right) ^T$ for $j=1,2,\ldots ,k$. By using Lemma 2 and the assumption that items are independent and identically distributed, we have

where $\mathrm {E}[N_j]=n \prod _{l=0}^{j-1}(1-p_l)(1-q_l)$ and $p_0=q_0=0$.

Similarly for $j,s=1,2,\ldots ,k-1$, we have

where ,    for $j=1,2,\ldots ,k-1$.

Hence Result 1 is proved.

Next we prove the lemmas 3 to 5 necessary to establish the consistency and asymptotic properties of the MLE of $\varvec{\theta }$.

Proof of Lemma 3

By (1), we have $\mathrm {E}[N_j]=n\prod _{l=0}^{j-1}(1-p_l)(1-q_l)$. Thus, $\mathrm {E}\left[ \frac{N_j}{n}\right] =\prod _{l=0}^{j-1}(1-p_l)(1-q_l) =b_j$, say. It is clear that $b_j < \infty$. Now, it is enough to show that the variance of $N_j/n$ tends to zero as $n \rightarrow \infty$. We have

$$\begin{aligned} \text{ Var }\left( \frac{N_j}{n}\right) =\mathrm {E}\left[ \left( \frac{N_j}{n}\right) ^2\right] -\mathrm {E}^2\left[ \frac{N_j}{n}\right]= & {} \frac{1}{n} \sum _{l=0}^{j-1} \prod _{l'=1}^{l} (1-p_{l'})(1-q_{l'}) (1-p_l)q_l. \end{aligned}$$

Thus, $\text{ Var }\left( \frac{N_j}{n}\right) \rightarrow 0$, when $n \rightarrow \infty$. Hence the proof.

Proof of Lemma 4 (i)

From Result 1, the $(u,v)^{th}$ element of is given by

For $u=1,2,\ldots ,m$, we define

$$\begin{aligned} \zeta _1= & {} \max _{1\le j\le k} \sup _{\varvec{\theta } \in \Theta } \frac{1}{F(T_j)-F(T_{j-1})} . \end{aligned}$$

(25)

$$\begin{aligned} \zeta _2= & {} \max _{1\le j\le k} \sup _{\varvec{\theta } \in \Theta } \frac{1}{1-F(T_j)} . \end{aligned}$$

(26)

$$\begin{aligned} A_u= & {} \max _{1\le j\le k} \sup _{\varvec{\theta } \in \Theta } \left| \frac{\partial F(T_j)}{\partial \theta _u}\right| . \end{aligned}$$

(27)

Then

Also, for fixed n, $\mathrm {E} [N_j] <n$. Lastly, for all $\varvec{\theta } \in \Theta$, by using regularity condition V(a), we have $0<\frac{1}{q_j(1-q_j)}<\infty$ and number of inspections k is finite. Hence, .

Proof of Lemma 4 (ii)

Note that is the variance-covariance matrix of the score vector $\varvec{\theta }$, it is a symmetric and non-negative definite. Consider a vector ${\mathbf {a}}(\varvec{\theta })\ne 0$. Then we have

Note that $0<q_j<1$ and $\mathrm {E} [N_j]>0$ for $j=1,2,\ldots ,k$. Also, by the regularity condition V(b), ${\mathbf {a}}(\varvec{\theta })^T \nabla _{\varvec{\theta }}\left( F(T_j)-F(T_{j-1})\right) \ne 0$ and ${\mathbf {a}}(\varvec{\theta })^T\nabla _{\varvec{\theta }}F(T_j) \ne 0$. This implies for ${\mathbf {a}}(\varvec{\theta })\ne 0$. Hence is a positive definite matrix for all $\varvec{\theta } \in \Theta _0$.

Proof of Lemma 5

For $u,v,w=1,2,\ldots ,m$, we define

$$\begin{aligned} A_{uv}= & {} \max _{1\le j\le k} \sup _{\varvec{\theta } \in \Theta } \left| \frac{\partial ^2 F(T_j)}{\partial \theta _u \partial \theta _v} \right| . \end{aligned}$$

(28)

$$\begin{aligned} A_{uvw}= & {} \max _{1\le j\le k} \sup _{\varvec{\theta } \in \Theta } \left| \frac{\partial ^3 F(T_j)}{\partial \theta _u \partial \theta _v \partial \theta _w}\right| . \end{aligned}$$

(29)

By using Eqs. (25)–(27), (28) and (29), it can be easily shown that

Now

where

Proof of Lemma 6 (i)

We show that for $j,j'=1,2,\ldots ,k-1$. Note that for $j\ne j'$ and when $j'=j$. Hence each element of the Fisher Information matrix is finite for all $\left( \varvec{\theta }, \varvec{p}\right) \in \Theta ^*$.

Proof of Lemma 6 (ii)

We prove that is a positive definite matrix. A diagonal matrix is positive definite if and only if all its element are positive. The jth component of is for $j=1,2,\ldots ,k-1$. Hence is positive definite.

Proof of Lemma 7

Given that the regularity condition (VI) holds, we define

$$\begin{aligned} p_{1_j}^*= & {} \underset{p_j \in \Theta _{p_j} }{sup} \frac{1}{p_j} \text{ and } p_{2_j}^* =\underset{p_j \in \Theta _{p_j}}{sup}\frac{1}{1-p_j}. \end{aligned}$$

(30)

By using and equation (30), we have

Now

where

Hence the proof of the lemma is complete. $\square$