Detecting trend change in hazard functions—an L-statistic approach

Abstract

A unified test is proposed to detect trend change in hazard functions. Test statistics based on a weighted integral approach are constructed utilizing a measure of deviation from exponentiality. We exploit L-statistic theory to obtain the exact and asymptotic distributions of our statistics and establish the consistency of the test. Theoretical results of previous works are obtained as special cases. A simulation study shows significant improvement in power depending on appropriate choice of parameters j and k introduced in our work. Finally, applications to real life data sets are presented to illustrate our results.

Appendix

Proof of Proposition 1:

Substituting the limits of integration in (1) we have,

$$\begin{aligned} T_{j,k}(F)= & {} \int _0^{F^{-1}(p)}\int _0^t[r(s)-r(t)][\bar{F}(s)]^{j}[\bar{F}(t)]^{k}ds dt \nonumber \\&+\int _{F^{-1}(p)}^\infty \int _{F^{-1}(p)}^t[r(t)-r(s)][\bar{F}(s)]^{j}[\bar{F}(t)]^{k}ds dt \nonumber \\= & {} I_1 + I_2 \end{aligned}$$

(8)

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

$$\begin{aligned} I_1= & {} \int _0^{F^{-1}(p)}\left( \int _0^t f(s)[\bar{F}(s)]^{j-1}ds\right) [\bar{F}(t)]^{k}dt\\&-\int _0^{F^{-1}(p)}\left( \int _0^t [\bar{F}(s)]^{j}ds\right) f(t)[\bar{F}(t)]^{k-1}dt \\= & {} \frac{1}{j}\int _0^{F^{-1}(p)}\left( 1-[\bar{F}(t)]^{j}\right) [\bar{F}(t)]^{k}dt \\&- \frac{1}{k}\int _0^{F^{-1}(p)}\left( [\bar{F}(s)]^{k}-(1-p)^k\right) [\bar{F}(s)]^{j}ds\\= & {} \int _0^{F^{-1}(p)} \frac{1}{j}\left( (\bar{F}(s))^k - (\bar{F}(s))^{j+k} \right) ds \\&-\int _0^{F^{-1}(p)} \frac{1}{k}\left( (\bar{F}(s))^{j+k} - (1-p)^k (\bar{F}(s))^j\right) ds \\= & {} \int _0^{F^{-1}(p)} \frac{(\bar{F}(s))^k}{j}ds - \left( \frac{j+k}{jk} \right) \int _0^{F^{-1}(p)} (\bar{F}(s))^{j+k} ds\\&+ \frac{(1-p)^k}{k}\int _0^{F^{-1}(p)}(\bar{F}(s))^j ds \end{aligned}$$

and

$$\begin{aligned} I_2= & {} \int _{F^{-1}(p)}^\infty \int _{F^{-1}(p)}^t f(t)[\bar{F}(t)]^{k-1}[\bar{F}(s)]^{j}ds dt \\&- \int _{F^{-1}(p)}^\infty \int _{F^{-1}(p)}^t f(s)[\bar{F}(s)]^{j-1}[\bar{F}(t)]^{k}ds dt \\= & {} \int _{F^{-1}(p)}^\infty [\bar{F}(s)]^{j}\left( -\int _{s}^{\infty } [\bar{F}(t)]^{k-1}d\bar{F}(t) \right) ds \\&- \int _{F^{-1}(p)}^\infty [\bar{F}(t)]^{k}\left( -\int _{F^{-1}(p)}^{t} [\bar{F}(s)]^{j-1}d\bar{F}(s) \right) dt \\= & {} \int _{F^{-1}(p)}^\infty \frac{(\bar{F}(s))^{j+k}}{k} ds - \frac{(1-p)^j}{j}\int _{F^{-1}(p)}^\infty (\bar{F}(s))^k ds\\&+ \int _{F^{-1}(p)}^\infty \frac{(\bar{F}(s))^{j+k}}{j} ds \\= & {} \frac{(j+k)}{jk} \int _{F^{-1}(p)}^\infty (\bar{F}(s))^{j+k} ds - \frac{(1-p)^j}{j}\int _{F^{-1}(p)}^\infty (\bar{F}(s))^k ds \end{aligned}$$

Therefore, using equation (8) and the results of the following integrations for any $l>0$, $\int _0^{F^{-1}(p)}(\bar{F}(s))^l ds = (1-p)^l F^{-1}(p)+ l \int _0^{F^{-1}(p)}s(\bar{F}(s))^{l-1} dF(s)$ and $\int _{F^{-1}(p)}^\infty (\bar{F}(s))^l ds = -(1-p)^l F^{-1}(p)+ l \int _{F^{-1}(p)}^\infty s(\bar{F}(s))^{l-1} dF(s),$ we obtained the final result. $\square$

Proof of Proposition 2:

Using a plug-in estimator in (1) (viz. empirical distribution function $\hat{F}_n$ for the unknown life distribution function F), we have

$$\begin{aligned} T_{j,k}(\hat{F}_n)= & {} \int _0^\infty sJ_{j,k}(\hat{F}_n(s))d\hat{F}_n(s)\\&-\left\{ (1-p)^j\left( \dfrac{1}{j}+\dfrac{1}{k} \right) -\dfrac{1}{j} \right\} (1-p)^k \hat{F}^{-1}_n(p)\\= & {} \sum _{i=1}^n \int _{X_{(i-1)}}^{X_{(i)}}sJ_{j,k}(\hat{F}_n(s))d\hat{F}_n(s)\\&-\left\{ (1-p)^j\left( \dfrac{1}{j}+\dfrac{1}{k} \right) -\dfrac{1}{j} \right\} (1-p)^k \hat{F}^{-1}_n(p) \end{aligned}$$

which on simplification yields the result. $\square$

In Theorem 4 the expression for variance of $T_{j,k}^*(\hat{F}_n)$ under $H_0$ is given as follows: for $j \ne \frac{1}{2}$,  $k > \frac{1}{2}$,

$$\begin{aligned} \sigma ^2(J_{j,k},F_0)= & {} \dfrac{1 + 2k ( k-1) + j ( 2 k-1)}{k^2 (2 k-1) (j + 2 k-1) ( 2 j + 2 k-1)} - \dfrac{2 (j-1) (1 - p)^k}{k^2 ( j + k-1) ( 2 j + k-1)} \nonumber \\&+ \dfrac{\{j^2+k^2(2j-1)\} }{ j^2 k^2(2j-1)} (1 - p)^{2 k} - \dfrac{2k(1-p)^{2k-1}}{j^2(2k-1)} \nonumber \\&+ \dfrac{2(k-1)(1-p)^{j+2k-1}}{j^2(j+k-1)(j+2k-1)}+ \dfrac{p (1-p)^{2k-1}}{j^2}\nonumber \\&- \dfrac{( j-1 - 2 j^3) k^2 + ( j-1) ( 2 j-3) k^3+2 k^4(2 j-1) }{j^2 k ^2 (2j-1)(2j+k-1)(2k-1)(j+2k-1)}(1 - p)^{2j+2k-1}\nonumber \\&- \dfrac{j^2(j-1)(2j-1)+j^2 k(j(5-4j)-1)}{j^2 k ^2 (2j-1)(2j+k-1)(2k-1)(j+2k-1)}(1-p)^{2j+2k-1} \end{aligned}$$

(9)

and for $j = \frac{1}{2}$,   $k > \frac{1}{2}$,

$$\begin{aligned} \sigma ^2(J_{\frac{1}{2},k},F_0)= & {} \dfrac{1-2k+4k^2}{2k^3(1-6k+8k^2)} + \dfrac{2(-1+4k-3k^2)+12k^3(4k+1)}{k^3(1-6k+8k^2)} (1-p)^k \nonumber \\&+\dfrac{3(32k^4+1-2k)p(1-p)^{2k}}{2k^2(1-6k+8k^2)} + \left\{ \dfrac{2-11k}{k^3(1-6k+8k^2)}\right. \nonumber \\&\left. - \dfrac{\log (1-p)}{k^2}\right\} (1-p)^{2k} \nonumber \\&+ \dfrac{3(1-p)^{(k+\frac{1}{2})}}{k^2(1-6k+8k^2)}\nonumber \\&+ \dfrac{4(4k+17)-2(10k+41)p-6(2k-3)p^2}{1-6k+8k^2}(1-p)^{(2k-1)} \nonumber \\&- \dfrac{12(4k+3)(1-p)^{k+\frac{1}{2}}}{(1-6k+8k^2)} \nonumber \\&+ \dfrac{4\left\{ 7(k+2)p -(4k+11) \right\} }{(1-6k+8k^2)}(1-p)^{(2k-\frac{3}{2})}\nonumber \\&- \dfrac{12(k+1)p^2 (1-p)^{\left( 2k-\frac{3}{2}\right) }}{(1-6k+8k^2)} \nonumber \\&+ \dfrac{ 6 - 3p(16k^3+1) }{k(1-6k+8k^2)} (1-p)^{\left( 2k-\frac{1}{2}\right) } \end{aligned}$$

(10)