Empirical Bayes inference for the Weibull model

Abstract

In this study, the theory of statistical kernel density estimation has been applied for deriving non-parametric kernel prior to the empirical Bayes which frees the Bayesian inference from subjectivity that has worried some statisticians. For comparing the empirical Bayes based on the kernel prior with the fully Bayes based on the informative prior, the mean square error and the mean percentage error for the Weibull model parameters are studied based on these approaches under both symmetric and asymmetric loss functions, via Monte Carlo simulations. The results are quite favorable to the empirical Bayes that provides better estimates and outperforms the fully Bayes for different sample sizes and several values of the true parameters. Finally, a numerical example is given to demonstrate the efficiency of the empirical Bayes.

Appendix

For the scale parameter, $\beta >0$, we can derive the kernel function estimator as follows:

$$\begin{aligned} \hat{{g}}(\beta )&= \frac{1}{nh_1 }\sum _{i=1}^n K\left( \frac{\beta -\beta _i }{h_1 }\right) \\&= \frac{1}{nh_1 }\sum _{i=1}^n K\left( \frac{\beta /\hat{\beta }-\beta _i /\hat{{\beta }}}{h_1 /\hat{{\beta }}}\right) \\&= \frac{1}{nh_1 }\sum _{i=1}^n K\left( \frac{e^{-\ln \left( \hat{{\beta }}/\beta \right) ^{\hat{{\alpha }}}/\hat{{\alpha }}}-e^{-\ln \left( \hat{{\beta }}/\beta _i \right) ^{\hat{{\alpha }}}/\hat{{\alpha }}}}{h_1 /\hat{{\beta }}}\right) \\&= \frac{1}{nh_1 }\sum _{i=1}^n K\left( \frac{e^{-w/\hat{{\alpha }}}-e^{-w/\hat{{\alpha }}}}{h_1 /\hat{{\beta }}}\right) =\frac{1}{nh_1 }\sum _{i=1}^n K\left( \frac{v-v_i }{h_1 /\hat{{\beta }}}\right) \\&= \frac{1}{\hat{{\beta }}nH_1 }\sum _{i=1}^n {K\left( \frac{v-v_i }{H_1 }\right) } =\hat{{g}}(v)/\hat{{\beta }}. \end{aligned}$$

Here $Q=(\hat{{\beta }}/\beta )^{\hat{{\alpha }}}$ is pivotal, thus $W=\ln Q$ and $v=e^{-w/\hat{{\alpha }}}$ are pivotal quantities and $H_1 =h_1 /\hat{{\beta }}$.

Similarly for the shape parameter, $\alpha >0$, we can derive the kernel density estimator as:

$$\begin{aligned} \hat{{g}}(\alpha )&= \frac{1}{nh_2 }\sum _{i=1}^n K\left( \frac{\alpha -\alpha _i }{h_2 }\right) \\&= \frac{1}{n\hat{{\alpha }}h_2 /\hat{{\alpha }}}\sum _{i=1}^n {K\left( \frac{\alpha /\hat{{\alpha }}-\alpha _i /\hat{{\alpha }}}{h_2 /\hat{{\alpha }}}\right) }\\&= \frac{1}{n\hat{{\alpha }}H_2 }\sum _{i=1}^n {K\left( \frac{z-z_i }{H_2 }\right) } =\hat{{g}}(z)/\hat{{\alpha }}, \end{aligned}$$

where $Z=\alpha /\hat{{\alpha }}$ and $H_2 =h_2 /\hat{{\alpha }}$.