Template:Bounds on the Parameters FMB ED

Bounds on the Parameters
For the failure rate $$\hat{\lambda }$$ the upper ($${{\lambda }_{U}}$$) and lower ($${{\lambda }_{L}}$$) bounds are estimated by [30]:


 * $$\begin{align}

& {{\lambda }_{U}}= & \hat{\lambda }\cdot {{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}} \\ & &  \\  & {{\lambda }_{L}}= & \frac{\hat{\lambda }} \end{align}$$

where $${{K}_{\alpha }}$$ is defined by:


 * $$\alpha =\frac{1}{\sqrt{2\pi }}\int_^{\infty }{{e}^{-\tfrac{2}}}dt=1-\Phi ({{K}_{\alpha }})$$

If $$\delta $$ is the confidence level, then $$\alpha =\tfrac{1-\delta }{2}$$ for the two-sided bounds, and $$\alpha =1-\delta $$ for the one-sided bounds. The variance of $$\hat{\lambda },$$ $$Var(\hat{\lambda }),$$ is estimated from the Fisher matrix, as follows:


 * $$Var(\hat{\lambda })={{\left( -\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} \right)}^{-1}}$$

where $$\Lambda $$ is the log-likelihood function of the exponential distribution, described in Appendix for log-likelihood function.

Note that no true MLE solution exists for the case of the two-parameter exponential distribution. The mathematics simply break down while trying to simultaneously solve the partial derivative equations for both the $$\gamma $$ and $$\lambda $$ parameters, resulting in unrealistic conditions. The way around this conundrum involves setting $$\gamma ={{t}_{1}},$$ or the first time-to-failure, and calculating $$\lambda $$ in the regular fashion for this methodology. Weibull++ treats $$\gamma $$ as a constant when computing bounds, i.e. $$Var(\hat{\gamma })=0.$$ (See the discussion in Appendix for log-likelihood function for more information.)