A New Decision Theoretic Sampling Plan for Type-I and Type-I Hybrid Censored Samples from the Exponential Distribution

Abstract

The study proposes a new decision theoretic sampling plan (DSP) for Type-I and Type-I hybrid censored samples when the lifetimes of individual items are exponentially distributed with a scale parameter. The DSP is based on an estimator of the scale parameter which always exists, unlike the MLE which may not always exist. Using a quadratic loss function and a decision function based on the proposed estimator, a DSP is derived. To obtain the optimum DSP, a finite algorithm is used. Numerical results demonstrate that in terms of the Bayes risk, the optimum DSP is as good as the Bayesian sampling plan (BSP) proposed by Lin et al. (2002) and Liang and Yang (2013). The proposed DSP performs better than the sampling plan of Lam (1994) and Lin et al. (2008a) in terms of Bayes risks. The main advantage of the proposed DSP is that for higher degree polynomial and non-polynomial loss functions, it can be easily obtained as compared to the BSP.

Appendix

1.1 A Proof of Theorem 3.1

Proof.

The Bayes risk of DSP with respect to the loss function (3.1) is given by

$$\begin{array}{@{}rcl@{}} r(n, \tau, \zeta) & = & n(C_s-r_s)+ E(M)r_{s} + \tau C_{\tau} + a_0 + a_1 \mu_1 + a_2 \mu_2 \\ & &+ {\int}_{0}^{\infty}(C_r - a_0 - a_1 \lambda-a_2 \lambda ^2)P(\widehat{\lambda} \geq \zeta)\frac{b^a}{{\Gamma} (a)} \lambda^{a-1}e^{-\lambda b} d\lambda \\ & = & n(C_{s}-r_{s}) + E(M)r_{s} + \tau C_{\tau}+a_0 +a_1 \mu_1 +a_2 \mu_2 \\ & & + \sum\limits_{l = 0}^{2}C_{l}\frac{b^a}{{\Gamma} (a)}{\int}_{0}^{\infty} \lambda^{a+l-1} e^{-\lambda b}P(\widehat{\lambda} \geq \zeta)\ d\lambda, \end{array} $$

(7.1)

where C_l is defined as

$$ C_{l} = \left\{\begin{array}{lllll} C_{r}- a_{l} &\text{if} \quad l = 0,\\ -a_{l} &\text{if} \quad l = 1,2. \end{array}\right. $$

(7.2)

Using Lemma 3.1 in Eq. 7.1 we get

$$\begin{array}{@{}rcl@{}} & & {\int}_{0}^{\infty}\lambda^{a+l-1}e^{-\lambda b}P(\widehat{\lambda} \geq \zeta)\ d\lambda \\ &=& {\int}_{0}^{\infty}\lambda^{a+l-1}e^{-\lambda (b+n\tau)}\ d\lambda \ \textit{I}_{(\zeta= 0)} +\sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^j \\ && \times {\int}_{0}^{\infty} {\int}_{\zeta}^{\frac{1}{\tau_{j,m}}}\lambda^{a+l+m-1}\frac{e^{-\lambda\{b+\frac{m}{y} \}}}{y^2}\big(\frac{1}{y}-\tau_{j,m}\big)^{m-1}dy \ d\lambda \\ &=& \frac{{\Gamma}{(a+l)}}{(b+n\tau)^{(a+l)}} \textit{I}_{(\zeta= 0)} + \sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^j\frac{(m)^m}{{\Gamma} (m)}\\ &&\times{\int}_{\zeta}^{\frac{1}{\tau_{j,m}}}\frac{\big(\frac{1}{y}-\tau_{j,m}\big)^{m-1}{\Gamma}{(a+l+m)}}{y^2\{b+\frac{m}{y}\}^{a+l+m}}dy \\ & & = \frac{{\Gamma}{(a+l)}}{(b+n\tau)^{(a+l)}} \textit{I}_{(\zeta= 0)} + \sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}\frac{(m)^m (-1)^j}{{\Gamma} (m)}\\ &&\times{\int}_{0}^{\frac{1}{\zeta}-\tau_{j,m}}\frac{v^{m-1}{\Gamma}{(a+l+m)}}{\{b+m\tau_{j,m}+mv\}^{a+l+m}}dv. \ \ \ \ \ \ \end{array} $$

(7.3)

Using C_j,m = b + mτ_j,m in Eq. 7.3, we can write

$$\begin{array}{@{}rcl@{}} &&\sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^{j}\frac{(m)^{m}}{{\Gamma} (m)}\frac{{\Gamma}{(a+l+m)}}{C_{j,m}^{a+l+m}}\\ &&\times{\int}_{0}^{\frac{1}{\zeta}- \tau_{j,m}}\frac{v^{m-1}}{\Big(1+\frac{mv}{C_{j,m}}\Big)^{a+l+m}}dv \\ &=&\sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^{j}\frac{{\Gamma}{(a+l)}}{(C_{j,m})^{a+l}}\frac{{\Gamma}{(a+l+m)}}{{\Gamma} (m){\Gamma}{(a+l)}}\\ &&\times{\int}_{0}^{\frac{m(\frac{1}{\zeta}- \tau_{j,m})}{C_{j,m}}}\frac{z^{m-1}}{(1+z)^{a+l+m}}dz. \end{array} $$

(7.4)

Now taking a transformation z = u/(1 − u), we have

$$\begin{array}{*{20}l} {\int}_{0}^{C^{*}_{j,m}}\frac{z^{m-1}}{(1+z)^{a+l+m}}dz={\int}_{0}^{S^{*}_{j,m}}u^{m-1}(1-u)^{a+l-1}du=B_{S^{*}_{j,m}}(m,a+l), \end{array} $$

where $\displaystyle C^{*}_{j,m}=\frac {m(\frac {1}{\zeta }- \tau _{j,m})}{C_{j,m}}$, $\displaystyle S^{*}_{j,m}=\frac {C^{*}_{j,m}}{1+C^{*}_{j,m}}$, and

$$ B_{x}(\alpha,\beta)={{\int}_{0}^{x}}u^{\alpha -1}(1-u)^{\beta-1}du, \ \ \ \ \ 0\leq x \leq 1, $$

is the incomplete beta function. If the cumulative distribution function of the beta distribution is given by I_x(α, β) = B_x(α, β)/B(α, β), then using Eq. 7.4 the Bayes risk is finally obtained as

$$\begin{array}{@{}rcl@{}} r(n,\tau,\zeta) \!\!&=&\!\! n(C_{s}-r_{s}) + E(M)r_{s} +\tau C_{\tau}+a_{0} +a_{1} \mu_{1} +a_{2} \mu_{2} \\ && + \sum\limits_{l = 0}^{2} C_{l}\frac{b^{a}}{{\Gamma} (a)}\bigg[\frac{{\Gamma}{(a+l)}}{(b+n\tau)^{(a+l)}} \textit{I}_{(\zeta= 0)}\\ &&+\sum\limits_{m = 1}^{n}\sum\limits_{j = 0}^{m} (-1)^{j}\binom{n}{m}\binom{m}{j} \frac{{\Gamma}{(a+l)}}{(C_{j,m})^{a+l}}I_{S^{*}_{j,m}}(m,a+l)\bigg], \end{array} $$

(7.5)

where $E(M) = {\sum }_{m = 1}^{n}{\sum }_{j = 0}^{m}m\binom {n}{m}\binom {m}{j}(-1)^{j}\frac {b^{a}}{(b+(n-m+j)\tau )^{a}}$. □

In general, for higher degree polynomial i.e for k > 2, the Bayes risk can be evaluated in a similar way for Type-I censoring.

1.2 B Proof of Theorem 3.2

Proof.

Note that the Bayes risk can be written as

$$\begin{array}{*{20}l} \begin{array}{lllll} r(n, \tau, \zeta) &=& n(C_{s}-r_{s}) + \tau C_{\tau} + E(M)r_{s}\\&&+ E_{\lambda}\big\{(a_{0} +a_{1}\lambda+ {\ldots} +a_{k} \lambda^{k})P(\widehat{\lambda}< \zeta) + C_{r} P(\widehat{\lambda} \geq \zeta)\big\}. \end{array} \end{array} $$

Now we know that a₀ + a₁λ + … + a_kλ^k ≥ 0 and C_r, the rejection cost, is non negative. Since (n₀, τ₀, ζ₀) is the optimal sampling plan so the corresponding Bayes risk is

$$\begin{array}{*{20}l} r(n_{0}, \tau_{0}, \zeta_{0}) \geq n_{0} (C_{s}-r_{s}) + \tau_{0} C_{\tau}. \end{array} $$

(7.6)

Now when ζ = 0 we reject the batch without sampling and the corresponding Bayes risk is given by r(0, 0, 0) = C_r. When ζ = ∞ we accept the batch without sampling and corresponding Bayes risk is given by r(0, 0, ∞) = a₀ + a₁μ₁ + … + a_kμ_k. Then the optimal Bayes risk is

$$\begin{array}{*{20}l} r(n_{0}, \tau_{0}, \zeta_{0}) \leq min \big\{ r(0,0,0),r(0,0,\infty),r(n, \tau, \zeta^{\prime})\big\}. \end{array} $$

(7.7)

Hence from Eqs. 7.6 and 7.7 we have

$$\begin{array}{*{20}l} n_{0}(C_{s}-r_{s}) + \tau_{0} C_{\tau} \leq min \big\{ r(0,0,0),r(0,0,\infty),r(n, \tau, \zeta^{\prime})\big\}. \end{array} $$

from where it follows that

$$\begin{array}{*{20}l} &n_{0} \leq min \bigg\{ \frac{C_{r}}{C_{s}-r_{s}},\frac{a_{0} + a_{1}\mu_{1} + {\ldots} + a_{k} \mu_{k}}{C_{s}-r_{s}},\frac{r(n, \tau, \zeta^{\prime})}{C_{s}-r_{s}}\bigg\} \\ &\tau_{0} \leq min \bigg\{ \frac{C_{r}}{C_{\tau}},\frac{a_{0} + a_{1}\mu_{1} + {\ldots} + a_{k} \mu_{k}}{C_{\tau}},\frac{r(n, \tau, \zeta^{\prime})}{C_{\tau}}\bigg\}. \end{array} $$

□

1.3 C Proof of Theorem 4.1

Proof.

The Bayes risk of DSP with respect to the loss function (4.1) is given by

$$\begin{array}{@{}rcl@{}} r(n, r,\tau, \zeta) & = & n(C_s-r_{s}) + E(M)r_{s} + E(\tau^{*}) C_{\tau} + a_0 + a_1 \mu_1 + a_2 \mu_2 \\ & & + {\int}_{0}^{\infty}(C_r - a_0 - a_1 \lambda-a_2 \lambda ^2)P(\widehat{\lambda} \geq \zeta)\frac{b^a}{{\Gamma} (a)} \lambda^{a-1}e^{-\lambda b} d\lambda \\ & = & n(C_{s}-r_{s}) + E(M)r_{s} + E(\tau^{*}) C_{\tau}+a_0 +a_1 \mu_1 +a_2 \mu_2 \\ & & + {\sum}_{l = 0}^{2}C_{l}\frac{b^a}{{\Gamma} (a)}{\int}_{0}^{\infty} \lambda^{a+l-1} e^{-\lambda b}P(\widehat{\lambda} \geq \zeta)\ d\lambda \end{array} $$

(7.8)

where C_l is defined as earlier. Let $\zeta ^{*}=max\{\frac {1}{n\tau },\zeta \}$, where ζ > 0 and

$$\begin{array}{@{}rcl@{}} R_{l,j,m} & = & {\int}_{0}^{\infty}{\int}_{\zeta^{*}}^{\infty}\lambda^{a+l-1}\frac{e^{-\lambda \{b+\tau(n-m+j)\}}}{y^2}\pi \Big(\frac{1}{y}-\tau_{j,m}; m, m\lambda \Big)dy \ d\lambda \\ & = & \frac{(m)^m}{{\Gamma}(m)}{\int}_{0}^{\infty}{\int}_{\zeta^{*}}^{\frac{1}{\tau_{j,m}}}\lambda^{a+l+m-1}\frac{e^{-\lambda\{b+\frac{m}{y} \}}}{y^2}\big(\frac{1}{y}-\tau_{j,m}\big)^{m-1}dy \ d\lambda \\ & = & \frac{(m)^m}{{\Gamma}(m)}{\int}_{\zeta^{*}}^{\frac{1}{\tau_{j,m}}}\frac{\big(\frac{1}{y}-\tau_{j,m}\big)^{m-1}{\Gamma}{(a+l+m)}}{y^2\{b+\frac{m}{y}\}^{a+l+m}}dy \\ & = & \frac{(m)^m}{{\Gamma}(m)}{\int}_{0}^{\frac{1}{\zeta^{*}}-\tau_{j,m}}\frac{v^{m-1}{\Gamma}{(a+l+m)}}{\{b+m\tau_{j,m}+mv\}^{a+l+m}}dv \\ & = &\frac{(m)^m}{{\Gamma}(m)}\frac{{\Gamma}{(a+l+m)}}{C_{j,m}^{a+l+m}}{\int}_{0}^{\frac{1}{\zeta^{*}}- \tau_{j,m}}\frac{v^{m-1}}{\Big(1+\frac{mv}{C_{j,m}}\Big)^{a+l+m}}dv \\ & = & \frac{{\Gamma}{(a+l)}}{(C_{j,m})^{a+l}}\frac{{\Gamma}{(a+l+m)}}{{\Gamma} (m){\Gamma}{(a+l)}}{\int}_{0}^{\frac{m(\frac{1}{\zeta^{*}}- \tau_{j,m})}{C_{j,m}}}\frac{z^{m-1}}{(1+z)^{a+l+m}}dz, \end{array} $$

where C_{j, m} = b + mτ_{j, m}. Now taking a transformation z = u/(1 − u), we have

$$\begin{array}{*{20}l} {\int}_{0}^{C^{*}_{j,m}}\frac{z^{m-1}}{(1+z)^{a+l+m}}dz={\int}_{0}^{S^{*}_{j,m}}u^{m-1}(1-u)^{a+l-1}du=B_{S^{*}_{j,m}}(m,a+l), \end{array} $$

where $\displaystyle C^{*}_{j,m}=\frac {m(\frac {1}{\zeta ^{*}}- \tau _{j,m})}{C_{j,m}}$ and $\displaystyle S^{*}_{j,m}=\frac {C^{*}_{j,m}}{1+C^{*}_{j,m}}$. Using B_x(α, β) and I_x(α, β) defined earlier, we obtain the expression

$$\begin{array}{@{}rcl@{}} R_{l,j,m}= \frac{{\Gamma}{(a+l)}}{(C_{j,m})^{a+l}} I_{S^*_{j,m}}(m,a+l). \end{array} $$

(7.9)

Using Lemma 4.1 in Eq. 7.8 and by Eq. 7.9 we get

$$\begin{array}{@{}rcl@{}} & & {\int}_{0}^{\infty}\lambda^{a+l-1}e^{-\lambda b}P(\widehat{\lambda} \geq \zeta)\ d\lambda \\ & =& {\int}_{0}^{\infty}\lambda^{a+l-1}e^{-\lambda (b+n\tau)}\ d\lambda \ \textit{I}_{(\zeta= 0)} +\sum\limits_{m = 1}^{r-1}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^j \\ && \times {\int}_{0}^{\infty}{\int}_{\zeta^{*}}^{\infty}\lambda^{a+l-1}\frac{e^{-\lambda \{b+\tau(n-m+j)\}}}{y^2} \pi \Big(\frac{1}{y}-\tau_{j,m}; m, m\lambda \Big)dy \ d\lambda \\ & & + {\int}_{0}^{\infty}{\int}_{\zeta^{*}}^{\infty}\lambda^{a+l-1}\frac{e^{-\lambda b}}{y^2} \pi \Big(\frac{1}{y}; r, r\lambda \Big)dy \ d\lambda + \sum\limits_{k = 1}^{r}\binom{n}{r}\binom{r-1}{k-1}\\ &&\times(-1)^k \frac{r}{(n-r+k)} \\ & & \times{\int}_{0}^{\infty}{\int}_{\zeta^{*}}^{\infty}\lambda^{a+l-1}\frac{e^{-\lambda \{b+\tau(n-r+k)\}}}{y^2} \pi \Big(\frac{1}{y}-\tau_{k,r}; r, r\lambda \Big)dy \ d\lambda \\ & =& \frac{{\Gamma}{(a+l)}}{(b+n\tau)^{(a+l)}} \textit{I}_{(\zeta= 0)}+\sum\limits_{m = 1}^{r-1}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^j R_{l,j,m} +R_{l,r-n,r} \\ & & +\sum\limits_{k = 1}^{r}\binom{n}{r}\binom{r-1}{k-1}(-1)^k \frac{r}{(n-r+k)} R_{l,k,r}. \end{array} $$

Thus Bayes risk of DSP under Type-I hybrid censoring is given by

$$\begin{array}{@{}rcl@{}} r(n, r,\tau, \zeta) & = & n(C_{s}-r_{s}) + E(M)r_{s} + E(\tau^{*}) C_{\tau}+a_0 +a_1 \mu_1 +a_2 \mu_2 \\ & & + \sum\limits_{l = 0}^{2}C_{l}\frac{b^a}{{\Gamma} (a)}\bigg\{\frac{{\Gamma}{(a+l)}}{(b+n\tau)^{(a+l)}} \textit{I}_{(\zeta= 0)}\\ &&+\sum\limits_{m = 1}^{r-1}\sum\limits_{j = 0}^{m}\binom{n}{m}\binom{m}{j}(-1)^j R_{l,j,m} +R_{l,r-n,r} \\ & & +\sum\limits_{k = 1}^{r}\binom{n}{r}\binom{r-1}{k-1}(-1)^k \frac{r}{(n-r+k)} R_{l,k,r}\bigg\}, \end{array} $$

(7.10)

where

$$\begin{array}{@{}rcl@{}} E(M)& = &\sum\limits_{m = 1}^{r-1}\sum\limits_{j = 0}^{m}m\binom{n}{m}\binom{m}{j}(-1)^{j}\frac{b^a}{(b+(n-m+j)\tau)^a} \\ & & \hspace{2.5cm}+\sum\limits_{k=r}^{n}\sum\limits_{j = 0}^{k}r\binom{n}{k}\binom{k}{j}(-1)^j\frac{b^a}{(b+(n-k+j)\tau)^a} \\ E(\tau^{*})& = & r\binom{n}{r}\sum\limits_{j = 0}^{r-1}\binom{r-1}{j}(-1)^{r-1-j}\\ &&\times\bigg\{\frac{b}{(n-j)^2(a-1)}- \frac{tb^a}{(n-j)((n-j)\tau+b)^a} \\ & & -\frac{b^a}{(n-j)^2(a-1)((n-j)\tau+b)^{a-1}}\bigg\} \\ &&+ \sum\limits_{k=r}^{n}{\sum}_{j = 0}^{k}\tau\binom{n}{k}\binom{k}{j}(-1)^j\frac{b^a}{(b+(n-k+j)\tau)^a}. \end{array} $$

For computation of E(M) and E(τ^∗) see Liang and Yang (2013). □

In general, for higher degree polynomial, i.e., for k > 2, the Bayes risk can be evaluated in a similar way for Type-I hybrid censoring.