Statistical Inference for Type-I Generalized Birnbaum–Saunders Distribution

Abstract

A new generalized Birnbaum–Saunders distribution (Type-I GBS) was presented in Owen (IEEE Trans Reliab 55:475–479, 2006) to model a lifetime of a product under cyclic stresses by using a long memory process on the crack extensions. This highly flexible model includes the original BS distribution as a special case and can be widely applied in fatigue studies. In this article, we present the relevant properties, parameter estimation, and hypothesis testing for the distribution. We explore the traditional maximum likelihood estimation approach, and propose a new inference method for the GBS-I distribution. An extensive simulation study is carried out to assess performance of the methods, and a real data is analyzed where it is shown that the GBS-I model with the proposed method provides an efficient estimation and achieves a better fit than the classic likelihood-based procedure.

Appendices

Appendix A: Fisher Information for the GBS-I

From the expressions in (7)–(9), and the following facts

$$\begin{aligned} \frac{\partial \delta (t)}{\partial \kappa }= & {} -\delta (t)\log t, ~\frac{\partial \delta (t)}{\partial \beta } = -\frac{\varepsilon (t)}{2\beta }, \\ \frac{\partial \varepsilon _1(t)}{\partial \kappa }= & {} -\delta (t)-\varepsilon _1(t)\log t, ~\frac{\partial \varepsilon _1(t)}{\partial \beta } = -\frac{\delta _1(t)}{2\beta }, \\ \frac{\partial \left( \varepsilon (t)\delta (t)\right) }{\partial \kappa }= & {} -2\varepsilon (t)\delta (t)\log t, \frac{\partial \left( \varepsilon (t)\delta (t)\right) }{\partial \beta } = -\frac{1}{2\beta }[\varepsilon ^2(t)+\delta ^2(t)], \\ \frac{\partial \left( \frac{\varepsilon (t)}{\delta _1(t)}\right) }{\partial \kappa }= & {} \frac{\varepsilon ^2(t)}{\delta _1^2(t)}, ~\frac{\partial \left( \frac{\varepsilon (t)}{\delta _1(t)}\right) }{\partial \beta } = -\frac{t^{1-2\kappa }}{\beta \delta _1^2(t)}, \\ \frac{\partial \left( \frac{\varepsilon _1(t)}{\delta _1(t)}\right) }{\partial \kappa }= & {} \frac{\varepsilon (t)\varepsilon _1(t)-\delta (t)\delta _1(t)}{\delta _1^2(t)} = -\frac{2t^{1-2\kappa }}{\delta _1^2(t)}, \\ \frac{\partial \left( \frac{\varepsilon _1(t)}{\delta _1(t)}\right) }{\partial \beta }= & {} \frac{\varepsilon _1^2(t)-\delta _1^2(t)}{2\beta \delta _1^2(t)} = -\frac{2\kappa (1-\kappa )t^{1-2\kappa }}{\beta \delta _1^2(t)}, \end{aligned}$$

the second partial derivatives are given by

$$\begin{aligned} \frac{\partial ^2 \ell }{\partial \alpha ^2}= & {} \frac{n}{\alpha ^2} - \frac{3}{\alpha ^4}\sum _{i=1}^n \varepsilon ^2(t_i), ~\frac{\partial ^2 \ell }{\partial \alpha \partial \kappa } = -\frac{2}{\alpha ^3}\sum _{i=1}^n \varepsilon ^2(t_i)\log t_i, \end{aligned}$$

(30)

$$\begin{aligned} \frac{\partial ^2 \ell }{\partial \alpha \partial \beta }= & {} -\frac{1}{\alpha ^3\beta }\sum _{i=1}^n \varepsilon (t_i)\delta (t_i), ~\frac{\partial ^2 \ell }{\partial \kappa ^2} = -\sum _{i=1}^n \frac{\varepsilon ^2(t_i)}{\delta _1^2(t_i)} -\frac{2}{\alpha ^2}\sum _{i=1}^n \varepsilon ^2(t_i)\log ^2(t_i), \nonumber \\\end{aligned}$$

(31)

$$\begin{aligned} \frac{\partial ^2 \ell }{\partial \kappa \partial \beta }= & {} \frac{1}{\beta }\sum _{i=1}^n \frac{t_i^{1-2\kappa }}{\delta _1^2(t_i)} -\frac{1}{\alpha ^2\beta }\sum _{i=1}^n \varepsilon (t_i)\delta (t_i)\log (t_i), \qquad \end{aligned}$$

(32)

$$\begin{aligned} \frac{\partial ^2 \ell }{\partial \beta ^2}= & {} \frac{n}{2\beta ^2}-\frac{\kappa ^2}{\beta }\sum _{i=1}^n \frac{t_i^{-2\kappa }}{\delta _1^2(t_i)} -\frac{1}{\alpha ^2\beta ^3}\sum _{i=1}^n t_i^{2(1-\kappa )}. \end{aligned}$$

(33)

So the Fisher’s information matrix is

$$\begin{aligned} I(\kappa ,\alpha ,\beta )=\left( \begin{array}{ccc} u_{\kappa \kappa } &{} u_{\kappa \alpha } &{} u_{\kappa \beta } \\ u_{\alpha \kappa } &{} u_{\alpha \alpha } &{} u_{\alpha \beta } \\ u_{\beta \kappa } &{} u_{\beta \alpha } &{} u_{\beta \beta } \end{array}\right) , \end{aligned}$$

(34)

where the elements are the expectations of the second partial derivatives in (30)–(33). Unfortunately, due to the complexity, there are no theoretical forms for the elements except $u_{\alpha \alpha } = -E(\partial ^2 \ell /\partial \alpha ^2) = 2n/\alpha ^2$ by using the fact that $Z=\varepsilon (T)/\alpha \sim N(0,1)$. The observed $I(\hat{\kappa },\hat{\alpha },\hat{\beta })$ is evaluated at MLEs $\hat{\kappa },\hat{\alpha },\hat{\beta }$.

Appendix B: Proof of Uniqueness of $\hat{\kappa }_N$

First we show the function in the Eq. (20)

$$\begin{aligned} G(\kappa )=\frac{\sum _{i=1}^n T_i^{1-\kappa }}{\sum _{i=1}^n T_i^{-\kappa }} \end{aligned}$$

is decreasing for $\kappa > 0$. Since

$$\begin{aligned} G'(\kappa ) = \frac{\left( \sum _{i=1}^n T_i^{1-\kappa }\right) \left( \sum _{i=1}^n T_i^{-\kappa }\log T_i\right) - \left( \sum _{i=1}^n T_i^{1-\kappa }\log T_i\right) \left( \sum _{i=1} T_i^{-\kappa }\right) }{\left( \sum _{i=1}^n T_i^{-\kappa }\right) ^2}. \end{aligned}$$

Let $g(\kappa )$ be the numerate function of $G'(\kappa )$. By switching the index of i and j in the summation, we have

$$\begin{aligned} 2g(\kappa )= & {} 2\left[ \sum _{i=1}^n\sum _{j=1}^n T_i^{1-\kappa }T_j^{-\kappa }\log T_j - \sum _{i=1}^n\sum _{j=1}^n T_i^{1-\kappa }T_j^{-\kappa }\log T_i \right] \nonumber \\= & {} \sum _{i=1}^n\sum _{j=1}^n T_i^{1-\kappa }T_j^{-\kappa }\log T_j + \sum _{i=1}^n\sum _{j=1}^n T_j^{1-\kappa }T_i^{-\kappa }\log T_i \\- & {} \sum _{i=1}^n\sum _{j=1}^n T_i^{1-\kappa }T_j^{-\kappa }\log T_i - \sum _{i=1}^n\sum _{j=1}^n T_j^{1-\kappa }T_i^{-\kappa }\log T_j \\= & {} \sum _{i=1}^n\sum _{j=1}^n T_i^{1-\kappa }T_j^{-\kappa }(\log T_j-\log T_i) + \sum _{i=1}^n\sum _{j=1}^n T_j^{1-\kappa }T_i^{-\kappa }(\log T_i-\log T_j) \nonumber \\= & {} - \sum _{i=1}^n\sum _{j=1}^n (T_i T_j)^{-\kappa }(T_i-T_j)(\log T_i-\log T_j). \end{aligned}$$

Since $(T_i-T_j)$ and $(\log T_i - \log T_j)$ have the same sign for any values of $T_i$ and $T_j$, it leads to $(T_i-T_j)(\log T_i - \log T_j) > 0$. Hence the function $g(\kappa ) < 0$, and so $G'(\kappa ) < 0$. Additionally, $\lim _{\kappa \rightarrow 0} G(\kappa ) = \sum _{i=1}^n T_i/n$ and $\lim _{\kappa \rightarrow 1} G(\kappa ) = n(\sum _{i=1}^n T_i^{-1})^{-1}$. It is known that $n(\sum _{i=1}^n T_i^{-1})^{-1} \le (\prod _{i=1}^n T_i)^{1/n} \le (\sum _{i=1}^n T_i)/n$ for any positive values of $T_i's$, and we conclude that there is a unique solution of $\hat{\kappa }_N$ in Eq. (20). Figure 5 shows the plot of $G(\kappa )$ for a sample data $T_1,\ldots ,T_n$.