Exact and asymptotic tests of exponentiality against nonmonotonic mean time to failure type alternatives

Abstract

The performance and effectiveness of an age replacement policy can be assessed by its mean time to failure (MTTF) function. In this paper we propose a class of tests to detect trend change in MTTF function. We develop test statistics utilising a measure of deviation based on a weighted integral approach. We derive the exact and asymptotic distributions of our test statistics exploiting L-statistic theory and also establish the consistency of the test as a consequence of our results. A Monte Carlo study is conducted to evaluate the performance of the proposed test. Finally, we apply our test to some real life data sets for illustrative purposes.

Appendix

Proof of Proposition 2.1

Now substituting the limits of integration in (2) we have

$$\begin{aligned} S_{k,j}(F)&= \int _0^{F^{-1}(p)} \int _0^t\left[ M_F(t)-M_F(s) \right] \left( F(t)\right) ^k\left( F(s)\right) ^j dF(s)dF(t) \nonumber \\&\quad + \int _{F^{-1}(p)}^{\infty }\int _{F^{-1}(p)}^t \left[ M_F(s)-M_F(t) \right] \left( F(t)\right) ^k\left( F(s)\right) ^j dF(s)dF(t) \nonumber \\&= I_1 + I_2 . \end{aligned}$$

(13)

A lengthy and involved calculation using interchange of the order of integration and integration by parts yields

$$\begin{aligned} I_1&= \int _0^{F^{-1}(p)} \int _0^t \left[ \left( F(t)\right) ^{k-1}\left( F(s)\right) ^j\int _0^t \bar{F}(x)d x \right. \nonumber \\&\quad \left. -\left( F(s)\right) ^{j-1}\left( F(t)\right) ^k \int _0^s \bar{F}(x)d x \right] dF(s) dF(t) \nonumber \\&= \int _0^{F^{-1}(p)}\left( F(t)\right) ^{k-1} \int _0^t \bar{F}(x)d x \left( \int _0^t \left( F(s)\right) ^j dF(s) \right) dF(t) \nonumber \\&\quad - \int _0^{F^{-1}(p)} \left( F(t)\right) ^k \left( \int _0^t \left( F(s)\right) ^{j-1} \left( \int _0^s \bar{F}(x)d x \right) dF(s) \right) d F(t) \nonumber \\&= \frac{1}{j+1} \int _0^{F^{-1}(p)} \left( F(t)\right) ^{k+j} \left( \int _0^t \bar{F}(x)d x \right) dF(t) \nonumber \\&\quad - \int _0^{F^{-1}(p)} \left( F(s)\right) ^{j-1} \int _0^s \bar{F}(x)d x \left( \int _s^{F^{-1}(p)} \left( F(t)\right) ^k dF(t) \right) dF(s) \nonumber \\&= \frac{1}{j+1} \int _0^{F^{-1}(p)} \left( F(t)\right) ^{k+j} \left( \int _0^t \bar{F}(x)d x \right) dF(t) \nonumber \\&\quad - \frac{p^{k+1}}{k+1} \int _0^{F^{-1}(p)} \left( F(s)\right) ^{j-1} \left( \int _0^s \bar{F}(x)d x \right) dF(s) \nonumber \\&\quad + \frac{1}{k+1} \int _0^{F^{-1}(p)} \left( F(s)\right) ^{k+j} \left( \int _0^s \bar{F}(x)d x \right) dF(s) \nonumber \\&= \frac{p^{k+j+1}}{(k+j+1)(j+1)}\int _0^{F^{-1}(p)} \bar{F}(t)d t \nonumber \\&\quad - \frac{1}{(k+j+1)(j+1)}\int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{k+j+1}d t \nonumber \\&\quad - \frac{p^{k+j+1}}{j(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)d t + \frac{p^{k+1}}{j(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{j}d t \nonumber \\&\quad + \frac{p^{k+j+1}}{(k+j+1)(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)d t \nonumber \\&\quad - \frac{1}{(k+j+1)(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{k+j+1}d t \nonumber \\&= \left[ \frac{p^{k+j+1}}{(k+j+1)(j+1)} + \frac{p^{k+j+1}}{(k+j+1)(k+1)} - \frac{p^{k+j+1}}{j(k+1)}\right] \int _0^{F^{-1}(p)} \bar{F}(t)d t\nonumber \\&\quad + \frac{p^{k+1}}{j(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{j}d t \nonumber \\&\quad - \left[ \frac{1}{(k+j+1)(j+1)} + \frac{1}{(k+j+1)(k+1)} \right] \int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{k+j+1}d t \end{aligned}$$

(14)

and

$$\begin{aligned} I_2&= \int _{F^{-1}(p)}^{\infty } \int _{F^{-1}(p)}^t \left[ \left( F(s)\right) ^{j-1}\left( F(t)\right) ^k\int _0^s \bar{F}(x)d x\right. \nonumber \\&\quad \left. -\left( F(s)\right) ^{j}\left( F(t)\right) ^{k-1} \int _0^t \bar{F}(x)d x \right] dF(s)dF(t) \nonumber \\&= \int _{F^{-1}(p)}^{\infty } \int _s^{\infty } \left( F(s)\right) ^{j-1}\left( F(t)\right) ^k \left( \int _0^s \bar{F}(x)d x \right) dF(t) dF(s) \nonumber \\&\quad - \int _{F^{-1}(p)}^{\infty } \int _{F^{-1}(p)}^t \left( F(s)\right) ^{j}\left( F(t)\right) ^{k-1} \left( \int _0^t \bar{F}(x)d x \right) dF(s)dF(t) \nonumber \\&= \int _{F^{-1}(p)}^{\infty } \left( F(s)\right) ^{j-1} \left( \int _0^t \bar{F}(x)d x \right) \nonumber \\&\quad \left( \int _s^{\infty } \left( F(t)\right) ^k dF(t) \right) dF(s) \nonumber \\&\quad - \int _{F^{-1}(p)}^{\infty } \left( F(t)\right) ^{k-1} \left( \int _0^t \bar{F}(x)d x \right) \left( \int _{F^{-1}(p)}^t \right. \nonumber \\&\quad \left. \left( F(s)\right) ^{j} dF(s) \right) dF(t) \nonumber \\&= \frac{1}{k+1} \int _{F^{-1}(p)}^{\infty } \left( F(s)\right) ^{j-1} \left( \int _0^s \bar{F}(x)d x \right) dF(s) \nonumber \\&\quad - \frac{1}{k+1} \int _{F^{-1}(p)}^{\infty } \left( F(s)\right) ^{k+j} \left( \int _0^s \bar{F}(x)d x \right) dF(s) \nonumber \\&\quad - \frac{1}{j+1} \int _{F^{-1}(p)}^{\infty } \left( F(t)\right) ^{k+j} \left( \int _0^t \bar{F}(x)d x \right) dF(t) \nonumber \\&\quad + \frac{p^{j+1}}{j+1} \int _{F^{-1}(p)}^{\infty } \left( F(t)\right) ^{k-1} \left( \int _0^t \bar{F}(x)d x \right) dF(t) \nonumber \\&\quad = \left[ \frac{1}{j(k+1)}-\frac{1}{(k+1)(k+j+1)}- \frac{1}{(j+1)(k+j+1)}\right. \nonumber \\&\quad \left. + \frac{p^{j+1}}{k(j+1)} \right] \int _0^{\infty } \bar{F}(t)d t \nonumber \\&\quad + \left[ \frac{p^{k+j+1}}{(k+1)(k+j+1)}+\frac{p^{k+j+1}}{(j+1)(k+j+1)}-\frac{p^{k+j+1}}{k(j+1)}\right. \nonumber \\&\quad \left. -\frac{p^j}{j(k+1)} \right] \int _0^{F^{-1}(p)}\bar{F}(t)d t \nonumber \\&\quad + \left[ \frac{1}{(k+1)(k+j+1)} + \frac{1}{(j+1)(k+j+1)} \right] \int _{F^{-1}(p)}^{\infty }\bar{F}(t) \left( F(t)\right) ^{k+j+1}d t \nonumber \\&\quad - \frac{p^{j+1}}{k(j+1)} \int _{F^{-1}(p)}^{\infty }\bar{F}(t) \left( F(t)\right) ^{k}d t - \frac{1}{j(k+1)}\int _{F^{-1}(p)}^{\infty }\bar{F}(t) \left( F(t)\right) ^{j}d t . \end{aligned}$$

(15)

Now using (13), (14) and (15), we get

$$\begin{aligned} S_{k,j}(F)&= \left[ \frac{1}{j(k+1)}-\frac{1}{(k+1)(k+j+1)}- \frac{1}{(j+1)(k+j+1)}\right. \nonumber \\&\quad \left. + \frac{p^{j+1}}{k(j+1)} \right] \int _0^{\infty } \bar{F}(t)d t \nonumber \\&\quad + \left[ \frac{2p^{k+j+1}}{(k+1)(k+j+1)}+\frac{2p^{k+j+1}}{(j+1)(k+j+1)}-\frac{p^{k+j+1}}{k(j+1)}-\frac{p^{k+j+1}}{j(k+1)}\right. \nonumber \\&\quad \left. -\frac{p^j}{j(k+1)} \right] \int _0^{F^{-1}(p)}\bar{F}(t)d t \nonumber \\&\quad - \left[ \frac{1}{(k+j+1)(j+1)} + \frac{1}{(k+j+1)(k+1)} \right] \int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^{k+j+1}d t \nonumber \\&\quad + \left[ \frac{1}{(k+j+1)(j+1)} + \frac{1}{(k+j+1)(k+1)} \right] \int _{F^{-1}(p)}^{\infty } \bar{F}(t)\left( F(t)\right) ^{k+j+1}d t \nonumber \\&\quad +\frac{p^{k+1}}{j(k+1)}\int _0^{F^{-1}(p)} \bar{F}(t)\left( F(t)\right) ^j d t \nonumber \\&\quad - \frac{p^{j+1}}{k(j+1)} \int _{F^{-1}(p)}^{\infty }\bar{F}(t) \left( F(t)\right) ^{k}d t \nonumber \\&\quad - \frac{1}{j(k+1)}\int _{F^{-1}(p)}^{\infty }\bar{F}(t) \left( F(t)\right) ^{j}d t . \end{aligned}$$

(16)

$\square$

Proof of Proposition 2.2

Direct but somewhat lengthy calculations yield

$$\begin{aligned} \int _{F_n^{-1}(p)}^{\infty } \bar{F}_n(t) \left( F_n(t)\right) ^{\eta }d t&= \sum _{i=m+1}^n \left( 1- \frac{i-1}{n}\right) \left( \frac{i-1}{n}\right) ^{\eta } \left[ X_{(i)}-X_{(i-1)} \right] \nonumber \\&= \sum _{i=m+1}^n \left[ \left( 1- \frac{i-1}{n}\right) \left( \frac{i-1}{n}\right) ^{\eta } - \left( 1- \frac{i}{n}\right) \left( \frac{i}{n}\right) ^{\eta } \right] \nonumber \\&\quad - \left( 1- \frac{m}{n}\right) \left( \frac{m}{n}\right) ^{\eta } X_{(m)} X_{(i)} , \end{aligned}$$

(17)

$$\begin{aligned} \int _0^{F_n^{-1}(p)} \bar{F}_n(t) \left( F_n(t)\right) ^{\zeta }d t =&\sum _{i=1}^m \left( 1- \frac{i-1}{n}\right) \left( \frac{i-1}{n}\right) ^{\zeta } \left[ X_{(i)}-X_{(i-1)} \right] \nonumber \\&= \sum _{i=1}^{m-1} \left\{ \left( 1- \frac{i-1}{n}\right) \left( \frac{i-1}{n}\right) ^{\zeta } - \left( 1- \frac{i}{n}\right) \left( \frac{i}{n}\right) ^{\zeta } \right\} \Bigg ] X_{(i)} \nonumber \\&\quad +\left( 1- \frac{m-1}{n}\right) \left( \frac{m-1}{n}\right) ^{\zeta } X_{(m)} , \end{aligned}$$

(18)

and

$$\begin{aligned} \int _0^{F_n^{-1}(p)} \bar{F}(t)d t= \sum _{i=1}^m \left( 1-\frac{i-1}{n}\right) \left[ X_{(i)}-X_{(i-1)} \right] =\frac{1}{n}\sum _{i=1}^{m-1}X_{(i)}+ \left( 1-\frac{m-1}{n}\right) X_{(m)} .\nonumber \\ \end{aligned}$$

(19)

Using (17), (18), (19) and $\int _0^{\infty } \bar{F}_n(t)d t:= \frac{1}{n}\sum _{i=1}^n X_{i} =\frac{1}{n} \sum _{i=1}^n X_{(i)}$, we get our desired result. $\square$

Proof of Proposition 4.1

Integrating by parts we derive the following formulae. The details are omitted for the sake of brevity. For $\gamma ,\, \delta >0,$ we have

$$\begin{aligned}&\int _{F^{-1}(p)}^{\infty } \bar{F}(t) \left( F(t)\right) ^{\delta }d t \nonumber \\&=-(1-p)p^{\delta }F^{-1}(p)-\int _{F^{-1}(p)}^{\infty }x\left[ \delta -(1+\delta )F(x)\right] \left[ F(x) \right] ^{\delta -1} d F(x) \end{aligned}$$

(20)

and

$$\begin{aligned}&\int _0^{F^{-1}(p)} \bar{F}(t) \left( F(t)\right) ^{\gamma }d t \nonumber \\&= (1-p)p^{\gamma }F^{-1}(p)-\int _0^{F^{-1}(p)} x\left[ \gamma -(1+\gamma )F(x)\right] \left[ F(x) \right] ^{\gamma -1} d F(x) . \end{aligned}$$

(21)

Now the proof easily follows using (20) and (21). $\square$ $\square$

Proof of Theorem 4.2

We apply Theorem 4.1 to obtain asymptotic normality of $S_{k, j}(F_n)$. From (10), it is easy to verify that $J_{k,\,j}$ is bounded and continuous a.e. Lebesgue and a.e. $F^{-1}$. Thus, to prove this theorem it is now enough to show that $\sigma ^2(J_{k, j},F) < \infty$. Let $S=\left\{ (x,y): 0\le x, y <\infty \right\}$, $S_1=\left\{ (x,y): 0\le y \le x <\infty \right\}$, $S_2=\left\{ (x,y): 0\le x \le y <\infty \right\}$, $B^\prime =\left\{ x: F(x) \right.$ is a discontinuity point of $\left. J \right\}$, $B_1=\left\{ (x,y): F(x) \right.$ and F(y) are discontinuity points of J in the region $\left. S_1\right\}$ and $B_2=\left\{ (x,y): F(x) \right.$ and F(y) are discontinuity points of J in the region $\left. S_2\right\}$. Since $B^\prime$ is a Lebesgue-null set (using the fact that J is bounded and continuous a.e. $F^{-1}$), it follows that

$$\begin{aligned}&\iint \limits _{S-B^\prime \times B^\prime } J_{k, j}(F(x))J_{k, j}(F(y))\left[ F(\min (x,y))-F(x)F(y)\right] d x d y \nonumber \\&\quad \le |J|^2 \iint \limits _{S-B^\prime \times B^\prime } \left[ F(\min (x,y))-F(x)F(y)\right] d x d y \nonumber \\&\quad = |J|^2 \iint \limits _{S_1-B_1} F(y) {\bar{F}}(x)d y d x + |J|^2 \iint \limits _{S_2-B_2} F(x) {\bar{F}}(y)d x d y \nonumber \\&\quad \le 2 |J|^2 \int _0^{\infty } x {\bar{F}}(x)d x = |J|^2 \int _0^{\infty }x^2 d F(x) <\infty \end{aligned}$$

and

$$\begin{aligned}&\int \limits _{[0,\infty )-B^\prime } J_{k, j}(F(x))\left[ \min (p,F(x))-p F(x)\right] d x \\&\quad \le |J| \int \limits _{[0,\infty )-B^\prime } \left[ \min (p,F(x))-p F(x)\right] d x \\&\quad \le |J| (1-p) F^{-1}(p) + p|J| \int _0^\infty {\bar{F}}(x)d x = |J| (1-p) F^{-1}(p) + p|J| \mu <\infty . \end{aligned}$$

Thus $\sigma ^2(J_{k, j},F) < \infty$. $\square$