A Transformation for the Analysis of Unimodal Hazard Rate Lifetimes Data

Abstract

The family of distributions introduced by [34] is the best known, best understood, most extensively investigated, and commonly employed model used for lifetimes data analysis. A variety of software packages are available to simplify its use. Yet, as is well known, the model is appropriate only when the hazard rate is monotone. However, as suggested in an overview by [23], the software packages may be usefully employed by transforming data when exploratory tools such as TTT transform or nonparametric estimates indicate unimodal, bathtub or J-shaped hazard rates, which are also commonly encountered in practice. Mudholkar et al. [22] discussed the details of one such transformation relevant for the bathtub case. In this paper, specifics of another transformation which is appropriate when data exploration indicates a unimodal hazard rate is discussed. The details of parameter estimation and hypothesis testing are considered in conjunction with earlier alternatives and illustrated using examples from the fields of biological extremes and finance.

Appendix A

Proof of Theorem 1. The hazard function is given by:

$$\begin{aligned} h(x)=\frac{\alpha }{\sigma }\left( \frac{1}{\sigma }\right) \frac{x^{2(\alpha -1)}}{(x+\theta )^{\alpha +1}}(x^2+2\theta x). \end{aligned}$$

(5.18)

Since logarithm is a monotone function, we may consider $\log h(x)$ instead of h(x). We have

$$\begin{aligned} \log h(x)=(\alpha -1)\log \left( \frac{x^2}{x+\theta }\right) +\log (x^2+2\theta x)-2\log (x+\theta ) + C, \end{aligned}$$

(5.19)

where C is independent of x. Then

$$\begin{aligned} \ell _{h(x)}'=\frac{d}{dx}\left( \log h(x)\right)= & {} \frac{(\alpha -1)(x+2\theta )}{x(x+\theta )}+\frac{2\theta ^2}{x(x+\theta )(x+2\theta )}\nonumber \\= & {} \frac{1}{x(x+\theta )}\left\{ (\alpha -1)(x+2\theta )+\frac{2\theta ^2}{x+2\theta }\right\} . \end{aligned}$$

(5.20)

• (a) For $\alpha \ge 1$, the sign of (5.20) is positive for all positive values of x, hence $\log h(x)$ is increasing and hence h(x) is also increasing and (a) is proved.

• (b) $1/2<\alpha <1$: The proof of (b) is more involved, and requires additional machinery. It amounts to proving, roughly speaking that, the hazard function h(x) is unimodal if and only if (5.20) has opposite signs at zero and $\infty$ and has one and only one zero over $0<x<\infty$. For critical point, the derivative of $\log h(x)$ is equal to zero, this means (5.20) is equal to zero and hence

  $$\begin{aligned} \left\{ (\alpha -1)(x+2\theta )+\frac{2\theta ^2}{x+2\theta }\right\} =0. \end{aligned}$$

  (5.21)

  Simplifying (5.21) gives the quadratic equation

  $$\begin{aligned}&(\alpha -1)(x+2\theta )^2+2\theta ^2=0\nonumber \\\Leftrightarrow & {} x^2+4\theta x +4\theta ^2+\frac{2\theta ^2}{(\alpha -1)}=0\nonumber \\\Leftrightarrow & {} x^2+4\theta x +4\theta ^2-\frac{2\theta ^2}{(1-\alpha )}=0. \end{aligned}$$

  (5.22)

  Equation (5.22) is solved by taking only the positive part of the solution, since x is positive, we then obtain

  $$\begin{aligned} x^{*}=\theta \left\{ -2+\sqrt{2/(1-\alpha )}\right\} , \end{aligned}$$

  (5.23)

  hence the derivative of $\log h(x)$ has only one solution $x^{*}$.

  For $\log h(x)$ to have an interior extreme point over $0<x<\infty$, the critical point $x^{*}$ has to be positive, and hence $-2+\sqrt{2/(1-\alpha )}$ should be greater than zero, this implies that $1/2<\alpha <1$. Thus when $1/2<\alpha <1$, the derivative of $\log h(x)$ has only one zero and hence $\log h(x)$ has a turning point and therefore h(x) has a turning point. Next, we show that the turning point is maximum by showing that the derivative of $\log h(x)$ has a positive sign at zero and a negative sign at $\infty$. If we let x approaches 0, then we have

  $$\begin{aligned} \lim _{x\rightarrow 0}\left\{ \frac{1}{x(x+\theta )}\left[ (\alpha -1)(x+2\theta )+\frac{2\theta ^2}{x+2\theta }\right] \right\}= & {} \frac{1}{0}\left[ (\alpha -1)(2\theta )+\theta \right] \nonumber \\= & {} \frac{\theta }{0}\left[ 2(\alpha -1)+1\right] \nonumber \\= & {} +\infty . \end{aligned}$$

  (5.24)

  We let x goes to $\infty$ in the next step.

  $$\begin{aligned} \ell _{h(x)}'= & {} \lim _{x\rightarrow \infty }\left\{ \frac{1}{x(x+\theta )}\left[ (\alpha -1)(x+2\theta )+\frac{2\theta ^2}{x+2\theta }\right] \right\} \nonumber \\= & {} \lim _{x\rightarrow \infty }\left\{ \frac{1}{x(x+\theta )}\left[ (\alpha -1)(x+2\theta )\right] \right\} ,\nonumber \\= & {} \lim _{x\rightarrow \infty } \left\{ \frac{(\alpha -1)}{2x}\right\} ,\,\,\text {by L'Hospital's Rule} \nonumber \\= & {} -(0)\,\,\mathrm{since}\,\,\alpha <1. \end{aligned}$$

  (5.25)

  We have shown that the derivative of $\log h(x)$ has only one zero and also changes sign from ($+$) to (−) over $0<x<\infty$ when $1/2<\alpha <1$. This shows that $\log h(x)$ has a maximum turning point and hence h(x) is unimodal.

• (c) $0<\alpha \le 1/2$, the proof of (c) is straight forward, for $0<\alpha \le 1/2$, (5.20) has no zero over $0<x<\infty$ and hence no turning point and therefore $\log h(x)$ is monotone. Following the steps in (5.24) and (5.25), we see that the derivative of $\log h(x)$ does not change signs and (5.20) is negative in both cases. Thus $\log h(x)$ is decreasing and hence h(x) is a decreasing hazard function.