Template:Likelihood Ratio Confidence Bounds

As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for θ1 and θ2 that satisfy:

$$ -2\cdot \text{ln}\left( \frac{L(\theta _{1},\theta _{2})}{L(\hat{\theta }_{1}, \hat{\theta }_{2})}\right) =\chi _{\alpha ;1}^{2} EQNREF lratio2 $$

This equation can be rewritten as:

$$ L(\theta _{1},\theta _{2})=L(\hat{\theta }_{1},\hat{\theta } _{2})\cdot e^{\frac{-\chi _{\alpha ;1}^{2}}{2}} EQNREF lratio3 $$

For complete data, the likelihood function for the Weibull distribution is given by:

$$ L(\beta ,\eta )=\prod_{i=1}^{N}f(x_{i};\beta ,\eta )=\prod_{i=1}^{N}\frac{ \beta }{\eta }\cdot \left( \frac{x_{i}}{\eta }\right) ^{\beta -1}\cdot e^{-\left( \frac{x_{i}}{\eta }\right) ^{\beta }} $$

For a given value of α, values for β and η can be found which represent the maximum and minimum values that satisfy Eqn. (\ref {lratio3}). These represent the confidence bounds for the parameters at a confidence level δ, where α = δ for two-sided bounds and α = 2δ − 1 for one-sided.

Similarly, the bounds on time and reliability can be found by substituting the Weibull reliability equation into the likelihood function so that it is in terms of β and time or reliability, as discussed in Chapter 5. The likelihood ratio equation used to solve for bounds on time (Type 1) is:

$$ L(\beta ,t)=\prod_{i=1}^{N}\frac{\beta }{\left( \frac{t}{(-\text{ln}(R))^{ \frac{1}{\beta }}}\right) }\cdot \left( \frac{x_{i}}{\left( \frac{t}{(-\text{ ln}(R))^{\frac{1}{\beta }}}\right) }\right) ^{\beta -1}\cdot \text{exp}\left[ -\left( \frac{x_{i}}{\left( \frac{t}{(-\text{ln}(R))^{\frac{1}{\beta }}} \right) }\right) ^{\beta }\right] $$

The likelihood ratio equation used to solve for bounds on reliability (Type 2) is:

$$ L(\beta ,R)=\prod_{i=1}^{N}\frac{\beta }{\left( \frac{t}{(-\text{ln}(R))^{ \frac{1}{\beta }}}\right) }\cdot \left( \frac{x_{i}}{\left( \frac{t}{(-\text{ ln}(R))^{\frac{1}{\beta }}}\right) }\right) ^{\beta -1}\cdot \text{exp}\left[ -\left( \frac{x_{i}}{\left( \frac{t}{(-\text{ln}(R))^{\frac{1}{\beta }}} \right) }\right) ^{\beta }\right] $$