Two sample Bayesian acceptance sampling plan

Abstract

This article presents the optimal Bayesian acceptance sampling plan (BASP) for two sample cases. In constructing the BASP, the decision-theoretic approach is used with a specified loss function, and the Bayes decision rule is developed by minimizing the Bayes risk. Both batches of products from the production line are accepted or rejected simultaneously based on the observed sample. Such implementation has a significant advantage in reducing the cost and time by deciding on the acceptance or rejection of both the batches in a single-life testing experiment. The optimal BASP is derived under the assumption of Weibull and exponential lifetimes, although it can be extended for other lifetime distributions also. Some numerical results have been presented to show the performances of the proposed BASP. We have presented the analysis of two data sets; (i) real and (ii) simulated, mainly to show how the proposed method can be used in practice.

Appendices

Appendix

1.1 Calculation of $\phi ({\textbf {W,Z}})$ in Weibull distribution

$$\begin{aligned} \phi ({\textbf {w, z}})= & {} C_1+C_2-C_1 \frac{I_{1}({\textbf {w,z}})}{I({\textbf {w,z}})}-C_2 \frac{I_{2}({\textbf {w,z}})}{I({\textbf {w,z}})} \end{aligned}$$

where

$$\begin{aligned} I({\textbf {w,z}})= & {} \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty } \alpha ^{k} \lambda ^{k_1}_1 \lambda ^{k_2}_2 \prod _{i=1}^{k}w_{i}^{\alpha -1} e^{- (\lambda _1+ \lambda _2)u(\alpha )}\pi (\lambda _1,\lambda _2)\pi (\alpha ) d \lambda _1 d \lambda _2 d \alpha \\= & {} \frac{\Gamma (a_1{+}a_2)\Gamma (a_0{+}k)\Gamma (a_1+k_1)\Gamma (a_2+k_2)b_{0}^{a_0} d^{c}}{\Gamma (a_0)\Gamma (a_1)\Gamma (a_2)\Gamma (a_1+a_2+k)\Gamma (c)}\int _{0}^{\infty } \frac{\alpha ^{k+c-1} \prod _{i=1}^{k}w_{i}^{\alpha -1}e^{-d \alpha }}{(b_0+u(\alpha ))^{a_0+k}} d\alpha ,\\{} & {} I_{1}({\textbf {w,z}}) \\{} & {} = \int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty } e^{-\lambda _1 L_1^{\alpha }} \alpha ^{k} \lambda ^{k_1}_1 \lambda ^{k_2}_2 \prod _{i=1}^{k}w_{i}^{\alpha -1} e^{- (\lambda _1+ \lambda _2)u(\alpha )} \pi (\alpha ,\lambda _1,\lambda _2)d \lambda _1 d \lambda _2 d \alpha \\{} & {} = \sum _{l=0}^{\infty }\frac{(-1)^l}{l!}\int _{0}^{\infty }\int _{0}^{\infty }\int _{0}^{\infty } L_{1}^{l \alpha } \alpha ^{k} \lambda ^{k_1+l}_1 \lambda ^{k_2}_2 \\{} & {} \quad \prod _{i=1}^{k}w_{i}^{\alpha -1} e^{- (\lambda _1+ \lambda _2)u(\alpha )}\pi (\lambda _1,\lambda _2)\pi (\alpha ) d \lambda _1 d \lambda _2 d \alpha \\{} & {} = \sum _{l=0}^{\infty }\frac{(-1)^l}{l!} \frac{\Gamma (a_1+a_2)\Gamma (a_0+k+l)\Gamma (a_1+k_1+l)\Gamma (a_2+k_2)b_{0}^{a_0}d^{c}}{\Gamma (a_0)\Gamma (a_1)\Gamma (a_2)\Gamma (a_1+a_2+k+l)\Gamma (c)}\\{} & {} \quad \int _{0}^{\infty } \frac{\alpha ^{k+c-1} \prod _{i=1}^{k}w_{i}^{\alpha -1} L_{1}^{l\alpha }e^{-d \alpha }}{(b_0+u(\alpha ))^{a_0+k+l}} d\alpha . \end{aligned}$$

Similarly,

$$\begin{aligned}{} & {} I_{2}({\textbf {w,z}}) = \sum _{l=0}^{\infty }\frac{(-1)^l}{l!} \frac{\Gamma (a_1+a_2)\Gamma (a_0+k+l)\Gamma (a_1+k_1)\Gamma (a_2+k_2+l)b_{0}^{a_0}d^c}{\Gamma (a_0)\Gamma (a_1)\Gamma (a_2)\Gamma (a_1+a_2+k+l\Gamma (c))}\\{} & {} \quad \int _{0}^{\infty } \frac{\alpha ^{k+c-1} \prod _{i=1}^{k}w_{i}^{\alpha -1} L_{2}^{l\alpha }e^{-d \alpha }}{(b_0+u(\alpha ))^{a_0+k+l}} d\alpha \end{aligned}$$

Proof of Theorem 4.1

(a) For the fixed $k_1$ and $k_2$, let us denote $\phi (k_1,k_2,u)=E[g(\lambda _1,\lambda _2)|k_1,k_2,u].$ The joint posterior PDF of $\lambda _1$ and $\lambda _2$ can be obtained as

$$\begin{aligned} \pi (\lambda _1,\lambda _2|k_1,k_2,u)=\frac{\lambda _1^{k_1} \lambda _2^{k_2}e^{-(\lambda _1+\lambda _2)u}\pi (\lambda _1,\lambda _2)}{\int _{0}^{\infty } \int _{0}^{\infty } \lambda _1^{k_1} \lambda _2^{k_2}e^{-(\lambda _1+\lambda _2)u}\pi (\lambda _1,\lambda _2) d\lambda _1 d\lambda _2}. \end{aligned}$$

(19)

Simplifying we obtain $(\lambda _1,\lambda _2)| (k_1,k_2,u) \sim BG(a_0+k,b_0+u,a_1+k_1,a_2+k_2)$ which implies that the Beta-Gamma is a conjugate prior. Therefore it is straight forward that

$$\begin{aligned} \left. \frac{\lambda _1}{\lambda _1+\lambda _2} \right| (k_1,k_2,u)\sim Beta(a_1+k_1,a_2+k_2) \ \text{ and } \ \lambda _1+\lambda _2|(k_1,k_2,u) \sim G(a_0+k,b_0+u). \end{aligned}$$

Let us define $\displaystyle \eta =\frac{\lambda _1}{\lambda _1+\lambda _2}$ and $\displaystyle \lambda = \lambda _1+\lambda _2$ then by the following one-to-one transformation $\lambda _1=\lambda \eta$ and $\lambda _2= \lambda (1-\eta ),$ the joint posterior density function of $\lambda _1$ and $\lambda _2$ is

$$\begin{aligned} \pi (\lambda _1,\lambda _2|k_1,k_2,u)= \frac{1}{\lambda }\times \pi (\eta | k_1,k_2,u) \times \pi (\lambda |k_1,k_2,u) \ \ \forall \lambda _1, \lambda _2>0. \end{aligned}$$

(20)

Note that for a fixed $k_1$ and $k_2,$ the posterior density function of $\eta | (k_1,k_2,u)$ is constant w.r.t. u,  and the posterior density function of $\lambda |(k_1,k_2,u)$ is a function of u. Since the distribution function of $\lambda |(k_1,k_2,u)$ is gamma which belongs to the exponential family, it has Monotone Likelihood Ratio (MLR) property. Then it is straightforward to show that for $u_1 < u_2,$ the likelihood ratio $\frac{\pi (\lambda |k_1,k_2,u_1)}{\pi (\lambda |k_1,k_2,u_2)}$ is strictly increasing in $\lambda .$ Therefore, the posterior distribution of $\lambda |(k_1,k_2,u)$ is stochastically decreasing function of u,  and hence due to (20) the joint posterior distribution of $(\lambda _1,\lambda _2)|(k_1,k_2,u)$ is stochastically decreasing function of u. So, if $g(\lambda _1, \lambda _2)$ is a positive function in $\lambda _1, \lambda _2$ and $u_1<u_2,$ then

$$\begin{aligned} E[g(\lambda _1,\lambda _2)|k_1,k_2,u_1] > E[g(\lambda _1,\lambda _2)|k_1,k_2,u_2], \end{aligned}$$

which implies that $\phi (k_1,k_2,u_1)> \phi (k_1,k_2,u_2).$ Hence $\phi (k_1,k_2,u)$ is strictly decreasing in u for fixed $k_1$ and $k_2.$

(b) From (a), it is straightforward that $\delta _B$ is strictly increasing in u for fixed $k_1$ and $k_2.$

Bayes risk derivation in exponential distribution

The Bayes risk is given by

$$\begin{aligned}{} & {} R_{B}^{\delta _B}(n,k,R_1,\ldots ,R_{k-1})\\{} & {} \quad = n(C_{s}-r_{s_1}-r_{s_2})+E[K_1]r_{s_1}+E[K_2]r_{s_2}+E[W_k]C_{\tau } \\{} & {} \qquad + \frac{\Gamma (a_1+a_2)}{\Gamma (a_0)\Gamma (a_2)\Gamma (a_2)}\sum _{i=0}^{4}c_{i} \frac{\Gamma (a_1+p_i)\Gamma (a_2+q_i)\Gamma (a_0+p_i+q_i)}{\Gamma (a_1+a_2+p_i+q_i)b_{0}^{p_i+q_i}} +R^{\delta _{B}}, \end{aligned}$$

where

$$\begin{aligned} R^{\delta _B}= & {} E_{\lambda _1,\lambda _2}[(C_r-c_0-c_1\lambda _1 -c_2 \lambda _2 -c_3 \lambda _1^2 -c_4 \lambda _2^2)P(U\le \xi (K_1,K_2))] \\= & {} \sum _{i=0}^{4}C_{i}E_{\lambda _1,\lambda _2}[\lambda _1^{p_i}\lambda _{2}^{q_i}P(U \le \xi (K_1,K_2))] \\= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}}{\Gamma (k)}E_{\lambda _1,\lambda _2}\bigg [\lambda _1^{p_i}\lambda _2^{q_i} \int _{0}^{\xi (r,k-r)}\lambda _1^{k}\lambda _2^{k-r}u^{k-1}e^{-\lambda u}du\bigg ] \\= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}}{\Gamma (k)} \int _{0}^{\xi (r,k-r)}E_{\lambda _1,\lambda _2}\bigg [\bigg (\frac{\lambda _1}{\lambda }\bigg )^{r+p_i}\bigg (1-\frac{\lambda _1}{\lambda }\bigg )^{k-r+q_i} \lambda ^{k+p_i+q_i}e^{-\lambda u}\bigg ] u^{k-1} du \end{aligned}$$

Since $\displaystyle \frac{\lambda _1}{\lambda }$ and $\lambda$ is independent we can write

$$\begin{aligned} R^{\delta _B}= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}}{\Gamma (k)} \int _{0}^{\xi (r,k-r)}E_{\lambda _1,\lambda _2}\bigg [\bigg (\frac{\lambda _1}{\lambda }\bigg )^{r+p_i}\bigg (1-\frac{\lambda _1}{\lambda }\bigg )^{k-r+q_i} \bigg ]\\{} & {} E_{\lambda _1,\lambda _2}\bigg [\lambda ^{k+p_i+q_i}e^{-\lambda u}\bigg ] u^{k-1} du \\= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}\Gamma (k+p_i+q_i+a_0)B(r+p_i+a_1, k-r+q_i+a_2)}{\Gamma (k)\Gamma (a_0)B(a_1,a_2)b_{0}^{-a_0}} \\{} & {} \int _{0}^{\xi (r,k-r)}\frac{ u^{k-1}}{(u+b_0)^{k+p_i+q_i+a_0}} du \\= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}\Gamma (k+p_i+q_i+a_0)B(r+p_i+a_1, k-r+q_i+a_2)}{\Gamma (k)\Gamma (a_0)B(a_1,a_2)b_{0}^{p_i+q_i}} \\{} & {} \int _{0}^{\frac{\xi (r,k-r)}{b_0}}\frac{ u^{k-1}}{(1+u)^{k+p_i+q_i+a_0}} du \\= & {} \sum _{i=0}^{4}\sum _{r=0}^{k}{k \atopwithdelims ()r}\frac{C_{i}\Gamma (p_i+q_i+a_0)B(r{+}p_i+a_1, k{-}r+q_i+a_2)}{\Gamma (a_0)B(a_1,a_2)b_{0}^{p_i+q_i}} I_{S_{r}}(k,p_i{+}q_i{+}a_0), \end{aligned}$$

where

$$\begin{aligned} \int _{0}^{\frac{\xi (r,k-r)}{b_0}}\frac{ u^{k-1}}{(1+u)^{k+p_i+q_i+a_0}} du = \int _{0}^{S_{r}}v^{k-1}(1-v)^{p_i+q_i+a_0-1} dv = B_{S_r}(k,p_i+q_i+a_0), \end{aligned}$$

$S_{r}=\frac{\xi (r,k-r)}{b_0}$and $I_{S_r}(k,p_i+q_i+a_0)=\frac{B_{S_r}(k,p_i+q_i+a_0)}{B(k,p_i+q_i+a_0)},$ and $B(a,b)=\int _{0}^{1}z^{a-1}(1-z)^{b-1}dz=\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}$ is the beta function, and $B_{x}(a,b)=\int _{0}^{x}u^{a -1}(1-u)^{b-1}du, \ \text {where} \ 0\le x \le 1,$ is the incomplete beta function. We denote the cumulative distribution function of beta as, $I_x(a, b)=\frac{B_x(a, b)}{B(a, b)}.$ Then Bayes risk is given as

$$\begin{aligned}{} & {} R_{B}^{\delta _B}(n,k,R_1,\ldots ,R_{k-1}) \\{} & {} \hspace{1.5cm} =n(C_{s}-r_{s_1}-r_{s_2})+E[K_1]r_{s_1}+E[K_2]r_{s_2}+E[W_k]C_{\tau } \\{} & {} \hspace{2cm}+ \frac{\Gamma (a_1+a_2)}{\Gamma (a_0)\Gamma (a_2)\Gamma (a_2)}\sum _{i=0}^{4}c_{i} \frac{\Gamma (a_1+p_i)\Gamma (a_2+q_i)\Gamma (a_0+p_i+q_i)}{\Gamma (a_1+a_2+p_i+q_i)b_{0}^{p_i+q_i}}\\{} & {} \hspace{2cm} + \sum _{i=0}^{4}\sum _{r=0}^{k} {k \atopwithdelims ()r}\frac{C_{i}\Gamma (p_i+q_i+a_0)B(r+p_i+a_1, k-r+q_i+a_2)}{\Gamma (a_0)B(a_0,a_1)b_0^{p_i+q_i}}\\{} & {} \quad I_{S_r}(k,p_i+q_i+a_0), \end{aligned}$$

where $K_1(K_2)$ follows a Binomial distribution with parameters k and $\frac{\lambda _1}{\lambda } (\frac{\lambda _2}{\lambda }),$ therefore

$$\begin{aligned} E[K_1]= & {} E_{(\lambda _1, \lambda _2)}E_{({\textbf {W,Z}})|(\lambda _1, \lambda _2)}[K_1] = E_{(\lambda _1, \lambda _2)} \Big [ k \frac{\lambda _1}{\lambda } \Big ] = k \frac{B(a_1 + 1, a_2)}{B(a_1, a_2)} \end{aligned}$$

similarly,$\displaystyle \ E(K_2) = k \frac{B(a_1, a_2 +1)}{B(a_1, a_2)}$, and from Mondal and Kundu (2019a) for all $i=1, \ldots , k,$ $W_i = \sum _{s=1}^{i}V_s$, where $V_s \sim Exp(\frac{1}{E_s})$ independently and $E_s= \lambda (n-\sum _{j=1}^{s-1}(R_j+1)),$ using this the $E[W_k]$ is given as

$$\begin{aligned} E[W_k]= & {} b_0 \frac{\Gamma (a_0-1)}{\Gamma (a_0)} \times \sum _{s=1}^{k} \frac{1}{(n-\sum _{j=1}^{s-1}(R_j+1)) } \quad \end{aligned}$$

Therefore, $E(W_k)$ exists only if $a_0>1.$

1.1 Proof of Theorem 3.1

Let $(n_B, k_B, R_{1_B}, \ldots , R_{k_{B}-1}, \delta _B)$ be the optimal BASP, then minimum Bayes risk is given by $R_{B}^{\delta _B}(n_B, k_B, R_{1_B}, \ldots , R_{k_{B}-1})$. Let $(0,0,0, \ldots , 0,\delta _{B}=0)$ denote the sampling plan, when we reject the batch without sampling with Bayes risk $R_{B}^{\delta _{B}=0}(0,0, 0, \ldots , 0)= C_r$ and $(0,0, 0, \ldots , 0,\delta _{B}=1)$ denote the sampling plan when we accept the batch without sampling with Bayes risk $R_{B}^{\delta _{B}=1}(0,0,0, \ldots , 0)= E_{\alpha , \lambda _1, \lambda _2}[g(\alpha , \lambda _1,\lambda _2)]$. Then the Bayes risk of optimal BASP is

$$\begin{aligned} R_{B}^{\delta _{B}}(n_B, k_B, R_{1_B}, \ldots , R_{k_{B}-1})\le \min \big \{E_{\alpha , \lambda _1, \lambda _2}[g(\alpha , \lambda _1,\lambda _2)],C_r\big \}. \end{aligned}$$

(21)

From (5) it is clear that

$$\begin{aligned} R_{B}^{\delta _B}(n_B, k_B, R_{1_B}, \ldots , R_{k_{B}-1})\ge n_B(C_s-r_{s_1}-r_{s_2}). \end{aligned}$$

(22)

Now from (22) and (21), it follows that

$$\begin{aligned} n_B(C_s-r_{s_1}-r_{s_2})\le \min \big \{E_{\alpha ,\lambda _1, \lambda _2}[g(\alpha , \lambda _1,\lambda _2)],C_r\big \}, \end{aligned}$$

which implies that

$$\begin{aligned} n_{B}\le \min \bigg \{\frac{E_{\alpha , \lambda _1, \lambda _2}[g(\alpha , \lambda _1,\lambda _2)]}{C_s-r_{s_1}-r_{s_2}},\frac{C_{r}}{C_s-r_{s_1}-r_{s_2}}\bigg \}, \end{aligned}$$

and $0\le k_{B}\le n_{B}.$