## 3 Confidence Interval of Damage: Stationary Case
### Case S: Single Stress-Time History  
This is the most common and interesting case in which only a single stress-time history \(x(t)\) of duration \(T\) is available, e.g., from only one measurement. In order to construct the confidence interval on \(E[D(T)]\) as in Case M, a sample of damage values needs to be obtained first. To this end, in a preliminary stage, \(x(t)\) is divided into \(N_{B}\geq 2\) disjoint blocks of equal time length \(T_{B}=T/N_{B}\). The damage of blocks is \(D_{B,j}(T_{B})\), \(j=1,2,\ldots N_{B}\). It is worth emphasizing that since the entire record \(x(t)\) is stationary and its subdividing blocks fully disjoint (not overlapped), the rainflow cycles counted in each block form independent sets, and their amplitudes have the same statistical distribution. As a result, the block damages \(D_{B,j}(T_{B})\), \(j=1,2,\ldots N_{B}\) are independent and identically distributed random variables, with a common value of variance \(Var[D_{B,i}]=\sigma_{D_{B}}^{2}\) and zero covariance, \(Cov\big{(}D_{B,i},D_{B,j}\big{)}=0\) (\(i\
eq j\)). The damage of the undivided record \(x(t)\) is  
\[D(T)\cong\sum_{j=1}^{N_{B}}D_{B,j}(T_{B})=N_{B}\cdot\overline{D}_{B}(T_{B}) \tag{4}\]  
where \(\overline{D}_{B}(T_{B})\) denotes the sample mean of block damage:  
\[\overline{D}_{B}(T_{B})=\frac{1}{N_{B}}\sum_{j=1}^{N_{B}}D_{B,j}(T_{B})\ \ \,\ \ \ \partial_{D_{B}}^{2}=\frac{1}{N_{B}-1}\ \sum_{j=1}^{N_{B}}[D_{B,j}(T_{B})-\overline{D}_{B}(T_{B}) ]^{2} \tag{5}\]  
while \(\hat{\sigma}_{D_{B}}^{2}\) is the sample variance of block damage, to be used shortly. The sign \(\cong\) in the previous equation signposts a small approximation. In fact, the block subdivision determines a small fraction of rainflow cycles to be lost, namely, those cycles formed by peaks and valleys falling in distinct blocks after subdivision. While these cycles would be counted in the undivided record \(x(t)\), after subdivision they are not counted anymore. Compared to the number of cycles in each block, the amount of lost cycles is negligible if the block length \(T_{B}\) is sufficiently long. This condition on \(T_{B}\), in turn, implies that the number of blocks \(N_{B}\) cannot increase indefinitely. As a rule of thumb, the minimum value for \(T_{B}\) must assure a minimum of \(10^{3}\) cycles in each block; this condition ensures that the approximation in Equation (4) is perfectly acceptable [13]. As for Case M, the sample values in Equation (5) allow the \(100(1-\alpha)\%\) confidence interval on \(E[D_{B}(T_{B})]\) (for a single block) to be constructed as [27]  
\[\overline{D}_{B}(T_{B})-t_{\alpha/2,N_{B}-1}\ \frac{\partial_{D_{B}}}{\sqrt{N_{B}}} \leq E[D_{B}(T_{B})]\leq\overline{D}_{B}(T_{B})+t_{\alpha/2,N_{B}-1}\ \frac{\partial_{D_{B}}}{\sqrt{N_{B}}} \tag{6}\]  
Here, \(t_{\alpha/2,N_{B}-1}\) is the upper \(\alpha/2\) percentage point of the \(t\) distribution with \(N_{B}-1\) degrees of freedom; for \(N_{B}>30\), it is \(t_{\alpha/2,N_{B}-1}\ \cong z_{\alpha/2}\). The confidence interval in Equation (6) can be further elaborated. As an intermediate step, take the expected value and variance of Equation (4):  
\[ \begin{split} E[D(T)]&=E\Bigg{[}\sum_{j=1}^{N_{B}}D _{B,j}(T_{B})\Bigg{]}=N_{B}\cdot E[D_{B}(T_{B})]\\ \sigma_{D}^{2}&=Var\Bigg{[}\sum_{j=1}^{N_{B}}D_{B,j }(T_{B})\Bigg{]}=N_{B}\cdot\sigma_{D_{B}}^{2}\end{split}  \tag{7}\]  
where \(\sigma_{D_{B}}^{2}=Var\big{[}D_{B,j}(T_{B})\big{]}\) is the variance of block damage--note that, likewise the expected damage \(E[D(T)]\), also the variances \(\sigma_{D}^{2}\) and \(\sigma_{D_{B}}^{2}\) refer to an infinite population of damage values and then are unknown. In both formulae in Equation (7), the second equal sign takes advantage of the fact, already mentioned above, that the random variables \(D_{B,j}\) are independent and identically distributed, which implies that \(E\big{[}D_{B,j}\big{]}=E[D_{B}(T_{B})]\) and \(Var\big{[}D_{B,j}(T_{B})\big{]}=\sigma_{D_{B}}^{2}\) for any \(j\), and that \(Cov\big{(}D_{B,i},D_{B,j}\big{)}=0\) for \(i\
eq j\).