## 2 Reliability Assessment Method Based on Fusion of Simulation and Test Data  
### Life Reliability Assessment Method for Zero-Failure Data  
#### 2.1.1 Bayes Estimation Method for Failure Probability  
Assume that the distribution function of the product lifetime satisfies \(T\sim F_{\theta}(t),t>0\), \(\theta\in\Theta\). There are S test pieces divided into \(m\) batches for a fixed time truncated test before the life test, and the truncated time \(t_{i}(i=1,\cdots,m)\) satisfies \(t_{1}<t_{2}<\cdots<t_{m}\). None of the \(n_{i}\) specimens in each batch have failed; i.e., the life of all \(n_{i}\) specimens in the \(i\)th batch is \(\succ t_{i}\). The above zero-failure test data is denoted as \((t_{i},n_{i}),i=1,\cdots,m\). Introduce \(s_{i}\) to represent the total number of specimens without failure at \(t_{i}\) time, and \(s_{i}=\sum\limits_{j=i}^{m}n_{j}\). The failure probability is \(p_{i}=F_{\theta}(t_{i})=P(T\leq t_{i}),i=1,\cdots,m\). Step 1: Construct the uninformative prior distribution of failure probability \(p_{i}\). Since the product lifetime test data are zero-failure data, for high-reliability products, when \(n_{m}\) is relatively large, it can be assumed that the probability that the product lifetime is \(<\!t_{m}\) is low. We assume that the probability of failure does not exceed \(q\). Using the a priori information [24]:  
\[p_{m}=P\{t<t_{m}\}<q\leq 0.5 \tag{1}\]  
when \(F_{\theta}(t)\) is a concave function on \(t\), take \(t_{0}=0\); then, we have \(p_{0}\simeq 0\), and,  
\[0<\frac{p_{1}}{t_{1}}<\frac{p_{2}}{t_{2}}<...<\frac{p_{m}}{t_{m}}<\frac{q}{t_{ m}} \tag{2}\]  
From Equation (2), we know that for any \(2\leq i\leq m\), \(p_{i}\leq q\cdot\frac{t_{i}}{t_{m}}\) when we can take the uniform distribution on \(\left[\hat{p}_{i-1},q\cdot\frac{t_{i}}{t_{m}}\right]\) as the prior distribution of \(p_{i}\), that is,  
\[\pi(p_{i})= \begin{cases}\frac{t_{m}}{qt_{i}-t_{m}\hat{p}_{i-1}},&\hat{p}_{i-1}< p_{i}<\frac{qt_{i}}{t_{m}}\\ 0,&\text{other}\end{cases}  \tag{3}\]  
where the value of \(p_{1}\) needs to be estimated in advance. First, we consider a fixed timed truncated test with \(n\) specimens. The test truncated time is \(\tau_{r}\). Suppose that there are \(r\) specimen failures before the truncated time \(\tau\), and \(\tau_{r}\) and \(\tau_{r+1}\) represent the \(r\)th and \((r+1)\)th failure moments, respectively. Then, we have \(\tau_{r}\leq\tau\leq\tau_{r+1}\) and \(F_{\theta}(\tau_{r})\leq F_{\theta}(\tau_{r+1})\), and,  
\[E[F_{\theta}(\tau_{r})]=\frac{r}{n+1},E[F_{\theta}(\tau_{r+1})]=\frac{r+1}{n+1} \tag{4}\]  
Take the estimate of \(F_{\theta}(\tau)\) as \(\hat{F}_{\theta}(\tau)=\frac{r+0.5}{n+1}\), at which time \(\hat{F}_{\theta}(\tau)\) is the summation estimate of \(F_{\theta}(\tau)\), i.e., \(\lim\limits_{n\rightarrow\infty}\hat{F}_{\theta}(\tau)=F_{\theta}(\tau)\). Thus, at the moment of \(t_{1}\), the estimate of \(p_{1}\) can be taken as \(\hat{p}_{1}=\frac{0.5}{s_{1}+1}\) in the case where none of the \(s_{1}\) specimens fails. Step 2: Calculate the likelihood function of failure probability \(p_{i}\). There are \(s_{i}\) specimens that have zero-failure at time \(t_{i}\), and the corresponding likelihood is \(L_{A}(p_{i}|s_{i})=(1-p_{i})^{s_{i}}\). In addition, the test data before \(t_{i}\) are \(\big{(}t_{j},n_{j}\big{)},1\leq j\leq i-1\). For each \(j\), there are \(n_{j}\) specimens without failure at time \(t_{j}\); thus, each of these \(n_{j}\) specimens will fail with probability \(P_{j}=\frac{p_{i}-p_{j}}{1-p_{j}}\) before \(t_{i}\) and with probability \(1-P_{j}=\frac{1-p_{j}}{1-p_{j}}\) after \(t_{i}\).