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\begin{document}



\centerline{ \Large \bf On Goodness-of-Fit in Accelerated Life Testing}

\vspace{1cm}



\leftline{\scriptsize VILIJANDAS BAGDONAVI\v{C}IUS \hfill
Vilijandas.Bagdonavicius@maf.vu.lt }
\leftline{\scriptsize \it Department of Statistics, Vilnius University, Lithuania}
\bigskip

\leftline{\scriptsize MIKHAIL S. NIKULIN \hfill
M.S.Nikouline@mi2s.u-bordeaux2.fr}

\leftline{\scriptsize \it University Victor Segalen Bordeaux 2, Bordeaux, 33076, France \&}
\leftline{\scriptsize \it  Steklov Mathematical Institute, 27 Fontanka, 191011,
Saint Petersburg, Russia  }
\bigskip

\leftline{\scriptsize {\bf Abstract.} Goodness-of-fit test for the
 generalized Sedyakin's model is proposed  when accelerated experiments are done under
 }
\leftline{\scriptsize  step-stresses.
Various alternatives are considered. Power of the test
against the approaching alternatives is investigated.
 }
\bigskip

\leftline{\scriptsize {\bf Keywords:} Accelerated life testing,
additive accumulation of damages, chi-squared goodness-of-fit, power function, }
 \leftline{\scriptsize  \hspace{1.5cm} proportional hazards, Sedyakin's model, step-stress}
\bigskip

\leftline{\bf AMS Classification:} 60E15, 62F12, 62J99, 62P10
\bigskip

\section{Introduction}
In accelerated life testing (ALT) units are tested at
higher-then-usual levels of stress to induce early failures. The
results are extrapolated to estimate the lifetime distribution at
the design stress using models which relate the lifetime to the
stress. The most general accelerated life model is the generalized Sedyakin's model
(see Bagdonavi\v{c}ius (1978), Nelson (1990), Bagdonavi\v{c}ius and Nikulin (1999). We'll consider a goodness-of-fit
test for this model when the data are obtained from accelerated experiments.

\par To see the possible alternatives to this model
 consider the most important accelerated life models and their relations.

\par Suppose that stresses are deterministic time functions:
$$ x(\cdot)=(x_1(\cdot),...,x_m(\cdot))^T\,:[0,\infty)\to B\in
{R^m}.
$$
We suppose that $x(\cdot)$ are right-continuous. If $x(\cdot)$ is constant in time, we'll write $x$
instead of $x(\cdot)$ in all formulae. \par Denote by
$T_{x(\cdot)}$ the time-to-failure, by
$S_{x(\cdot)}(t)=P\{T_{x(\cdot)}>t\}$ the survival function, by
$$
\alpha_{x(\cdot)}(t) =\lim_{h\downarrow 0}\, \frac{1}{h}\, P
\{T_{x(\cdot)}\in (t,t+h] \mid T_{x(\cdot)}>t\}
$$
the hazard rate and by
$$
A_{x(\cdot)}(t)=\int^t_0
\alpha_{x(\cdot)}(u)du=-ln\{S_{x(\cdot)}(t)\}
$$
the accumulated hazard rate under the stress $x(\cdot)$. We write $x(\cdot)<y(\cdot)$ if
$S_{x(\cdot)}(t)>S_{y(\cdot)}(t)$ for all $t>0$.
\par We formulate the models in terms of the hazard rate $\alpha_{x(\cdot)}(t)$ which for any moment $t$ caracterizes
the risk of failure just after this moment for the units which survived until $t$.

\vspace{0.5cm}

\par {\bf Definition 1}.  {\it The generalized Sedyakin's (GS) model (Bagdonavi\v{c}ius  (1978))
holds on a set of stresses $E$ if there exist a positive on
$E\times{R}^+$ function $g$ such that for all} $x(\cdot) \in {E}$
$$
\alpha_{x(\cdot)}(t) = g \left(x(t),S_{x(\cdot)}(t) \right).
$$
Thus the hazard rate at the moment $t$ is a function of the value of the stress
at this moment and of the probability of
survival until this moment.

Equivalently, the model can be written in the form
$$
\alpha_{x(\cdot)}(t) = g_1 \left(x(t),S_{x(\cdot)}(t) \right)
$$
with $g_1(x,s)=g(x,exp\{-s\})$.

Note that GS model does not state the presence of any connections
between the survival or hazard rate functions under different constant
stresses, because for any hazard rates $\alpha_x(t)$ under constant
stresses $x\in E_0$ we can take $g_1(x,s)=\alpha_x(A_x^{-1}(s))$ and thus the GS model
holds on $E_0$. So
the GS model shows only the influence of time-variability of stresses
on survival.

In terms of hazard rates and accumulated hazards
under time varying and constant in time stresses this model has
the form $$
\alpha_{x(\cdot)}(t)=\alpha_{x(t)}\left(A_{x(t)}^{-1}(A_{x(\cdot)}(t))\right).
$$
\par in accelerated life testing The most used time-varying stresses
in accelerated life testing are the step-stresses:
units are placed on test at an initial
low stress and if they do not fail in a predetermined time $t_1$,
the stress is increased. If they do not fail in a predetermined time $t_2>t_1$,
the stresses is increased once more, and so on.

Consider a set $E_m$ of step-stresses of the form
$$
x(\tau)=\left\{\begin{array}{cc} x_1,& 0 \leq \tau < t_1,\\ x_2,&
t_1 \leq \tau < t_2,\\ \cdots & \cdots \\ x_m,& t_{m-1} \leq \tau
< t_m.\\
\end{array} \right.  \eqno(1)
$$
Set $t_0=0$.
\par If the GS model holds on $E_m$ then
the hazard rate function $\alpha_{x(\cdot)}(t)$ and the survival function $S_{{ x}(\cdot)}(t)$ verify
 the equalities:
$$ \alpha_{{ x}(\cdot)}(t) = \alpha_{{ x}_{i}}(t-t_{i-1}+t_{i-1}^{*}), \quad
\mbox{if}\quad t \in [t_{i-1},t_{i}),
 $$
$$ S_{{ x}(\cdot)}(t) = S_{{ x}_{i}}(t-t_{i-1}+t_{i-1}^{*}), \quad
\mbox{if}\quad t \in [t_{i-1},t_{i}) \: (i=1,2,\ldots ,m), \eqno(2)
 $$ where where $t^{*}_{i}$ can be found by solving the
equations $$ S_{{ x}_{1}}(t_{1}) = S_{{ x}_{2}}(t_{1}^{*}), \ldots
, S_{{ x}_{i}}(t_{i}-t_{i-1} + t^{*}_{i-1}) = S_{{
x}_{i+1}}(t^{*}_{i}) \: (i=1,\ldots ,m-1). \eqno(3)
 $$
%Figure 1

Thus the survival functions under step-stresses are obtained from
the survival functions under constant in time stresses by the {\it
rule of time shift}.

Note also that
$$
 S_{{ x}(\cdot)}(t_i)=S_{x_{i+1}}(t^{*}_{i}).
$$
The moment $t_i$ under the stress ${ x}(\cdot)$ is equivalent to the moment $t^{*}_{i}$ under
the stress $x_{i+1}$.
\vspace{0.5cm}
\par {\bf Definition 2.} {\it
The additive accumulation of damages (AAD) model
(Bagdonavi\v{c}ius (1978))
 holds on}  $E $ {\it if there exists a positive function $r$ on $E$
and a positive on $[0,\infty)$ function $q$  such that for all}
$x(\cdot) \in E $ $$ \alpha_{x(\cdot)}(t) = r
\{x(t)\}\,q\{S_{x(\cdot)}(t)) \}. $$

\par The AAD model implies that under constant in time stresses survival functions have the form
$$ S_x(t)=S_0\{r(x)t\}, $$ where the function $S_0$ does not
depend on $x$, i.e. survival functions under different constant in
time stresses differ only in scale parameters.
\par If the AAD model holds on $E_m$ then the hazard rate function $\alpha_{x(\cdot)}(t)$ and
the survival function $S_{{ x}(\cdot)}(t)$ verify the equalities
(2) with $t^{*}_{i}$ determined by the equalities $$
t^{*}_{i}=\frac{1}{r(x_{i+1})}\sum^{i}_{j=1} r(x_j)(t_j-t_{j-1})
,\quad \mbox{if}\quad t\in[t_{i-1},t_i). $$ Thus the AAD model is
a restriction of the GS model when strict connection between
survival functions under different constant in time stresses are
assumed. The survival functions under step-stresses are obtained
from the survival functions under constant in time functions by
the same rule of time shift as in the case of GS model.

The GS model seems very natural but it not means that it is universal.
Consider some alternatives to it.
In 1972 Sir D.Cox proposed the {\it proportional hazards} (PH)
model for constant in time covariates (stresses): for all $x\in E_0$
$$
\alpha_{x}(t)=r\{x\}\;\alpha_{0}(t),
$$
where the function $\alpha_{0}$ does not depend on $x$.

For constant stresses the survival functions are from the class of functions
 $\{S_0^{r(x)}(t)\}$, where
$$
S_0(t)=\exp \{-\int_0^t\alpha_0 (u)du\}.
$$
\par In the statistical literature the following {\it formal} generalization of the PH
model is
often used.
\vspace{0.5cm}
\par {\bf Definition 3.} {\it The  proportional hazards
model  holds on the set of time-varying and constant stresses $E$ if for
all} $x(\cdot) \in {E}$ $$
\alpha_{x(\cdot)}(t)=r\{x(t)\}\;\alpha_{0}(t). \eqno(4)
$$
Under the step-stresses the {\it time-shift rule does not take place}:
$$
\alpha_{{ x}(\cdot)}(t) = \alpha_{{ x}_{i}}(t), \quad
\mbox{if}\quad t \in [t_{i-1},t_{i}), \: (i=1,2,\ldots ,m). \eqno(5)
$$
%Figure 2.

The intensity under the stress $x_i$ in the interval
$[t_{i-1},t_{i})$ does not depend on stresses which were used
until the moment $t_{i-1}$. This is not natural when items are
aging. Thus application of the PH model when stresses are not
constant in time must be done with great care.
\par One of possible causes when the time shift property is not satisfied, can be the
presence of unobservable time-varying stresses. For example,
suppose that not only the stress $x(\cdot)$ but also some
non-observable stress $y(\cdot)$ influences the reliability of
items and on the set of stresses of the form
$z(\cdot)=(x(\cdot),y(\cdot))$ the AAD model with exponential
time-to-failure distributions under constant in time stresses
holds: $$ \alpha_{z(\cdot)}(t)=r_1\{x(t)\}\;r_2\{y(t)\},\quad
\alpha_{z}(t)=r_1\{x\}\;r_2\{y)\}. $$ Then
given $y(\cdot)$ the PH model holds on the set of stresses of the
form $x(\cdot)$ (with $\alpha_0(t)=r_2\{y(t)\}$) and
times-to-failure under constant $x$ are not necessary exponential.

Now we shall "repair" the PH model with time-varying stresses to
obtain a restriction of the GS model which coincides with the PH
model for constant stresses and has the time shift property for
step-stresses.
\vspace{0.5cm}
 \par {\bf Definition 4.} {\it The first generalized proportional hazards (GPH1)
model holds on $E$ if for all} $x(\cdot) \in {E}$ $$
\alpha_{x(\cdot)}(t)=r\{x(t)\}\;\alpha_{0}\left(A_0^{-1}
\left(\frac{A_{x(\cdot)}(t)}{r\{x(t)\}}\right)\right), \eqno(6) $$
were the function $\alpha_{0}$ does not depend on $x$ and
$$
A_0(t)=\int_0^t\alpha_0(u)du.
$$
Under constant in time stresses the model (6) implies the PH model: $$
\alpha_{x}(t)=r(x)\; \alpha_0(t). $$ If $x(\cdot)$ is the step-stress (1)
then the hazard rate function $\alpha_{x(\cdot)}(t)$ and the
survival function $S_{{ x}(\cdot)}(t)$ verify
 the equalities (2) with $t^{*}_{i}$ defined by the equalities
$$
t^{*}_{1}=A_0^{-1}\left(\frac{r(x_1)}{r(x_2)}A_0(t_1)\right),
\quad t^{*}_{i}=A_0^{-1}
\left(\frac{r(x_i)}{r(x_{i+1})}A_0(t_i-t_{i-1}+t^{*}_{i-1})\right).
$$

Consider more general then PH models which do not verify the time shift property.

\vspace{0.5cm}
\par {\bf Definition 5.} {\it The second generalized proportional hazards (GPH2)
model ( Bagdonavi\v{c}ius  \& Nikulin (1998)) holds on $E$ if for
all} $x(\cdot) \in {E}$ $$
\alpha_{x(\cdot)}(t)=r\{x(t)\}\;q\{S_{x(\cdot)}(t)\}\;\alpha_{0}(t).
$$
\par The particular cases of the GPH1 model are the PH model ($q(u)\equiv 1$) and the
AAD model ($\alpha_0(t) \equiv \alpha_0=const$).

Suppose that not only the stress $x(\cdot)$ but also some non-observable stress $y(\cdot)$
influences the reliability of items and on the set of stresses of the form
$z(\cdot)=(x(\cdot),y(\cdot))$ the AAD model holds with not necessary exponential time-to-failure
distributions under constant in time stresses:
$$
\alpha_{z(\cdot)}(t)=r_1\{x(t)\}\;r_2\{y(t)\}\;q\{S_{z(\cdot)}(t)\}.
$$
Then given $y(\cdot)$ the GPH2 model holds on the set of stresses of the form
$x(\cdot)$ (with $\alpha_0(t)=r_2(y(t)))$. Thus for time-varying covariates
the GPH2 model is much more natural then the PH model.
If the GPH2 model holds on a set of general step-stresses $E_m$,
then for any $t\in [t_{i-1},t_i)$ $$ S_{x(\cdot)}(t)=
G\left\{(H(S_{x_i}(t))-H(S_{x_i}(t_{i-1})))+
\sum^{i-1}_{j=1}\frac{r(x_j)}{r(x_i)}(H(S_{x_i}(t_j))
-H(S_{x_i}(t_{j-1})))\right\}, $$ where $$ H(u)=\int_u^{1}\frac{dv}{vq(v)},\quad, G=H^{-1}. $$
\vspace{0.5cm}
\par {\bf Definition 6.} {\it The third generalized proportional hazards (GPH3)
model  holds on $E$ if for
all} $x(\cdot) \in {E}$
$$
\alpha_{x(\cdot)}(t)=u\{x(t),S_{x(\cdot)}(t)\}\,\alpha_{0}(t).
$$
The particular cases of the GPH3 model are the GS model
($\alpha_0(t) \equiv \alpha_0=const$) and GPH2 model $(u(x,s)=r(x)\,q(s)).$



 The numerous examples of real data show that taking two constant in time
stresses, say $x_1$ and $x_2$, the ratio
$\alpha_{x_2}(t)/\alpha_{x_1}(t)$ (which is constant under the PH
model), can be {\it increasing or decreasing} in time and even a
{\it cross-effect} of hazard rates can be observed. Such data can
be modeled by sub-models of the GPH2 or more general GPH3 model.

An alternative to the GS model under step-stresses can be obtained by taking into
account the influence of switch-off's of stresses on reliability of items.
Switch-off's can imply failures of items. Suppose that an item is observed under
 the stress (1) and after the switch-off at the
moment $t_i$ from the stress $x_i$ to the stress $x_{i+1}$ the survival
function has a jump:
$$
S_{x(\cdot)}(t_i)=S_{x(\cdot)}(t_i-)\,\delta_i;
$$
here $\delta_i$ is the probability for an item not to fail because of the switch-off at
the moment $t_i$. In this case the GS model for step-stresses can be
modified as follows:
$$
S_{x(\cdot)}(t)=S_{x_i}(t-t_{i-1}+t_{i-1)}^{**}), \eqno(7)
$$
where
$$
t_{1}^{**}=S_{x_2}^{-1}\{S_{x_1}(t_1)\,\delta_1\},\quad t_{i}^{**}=
S_{x_{i+1}}^{-1}\{S_{x_i}(t_i-t_{i-1}+t_{i-1)}^{**})\,\delta_i\}. \eqno(8)
$$
Thus the time shift is modified by the jumps.

\par Goodness-of-fit tests for the PH model are numerous (see, for example,  Wei (1984), Lin (1991),
Lin, Wei and Ying (1993) ). Tests for the AAD model are considered in
Bagdonavi\v{c}ius (1978,
1990).
Tests for the GPH models are considered in Bagdonavi\v{c}ius and
Nikulin (2000). A test for the GS model in the case of a special
experiment with simple step-stress is considered in
Bagdonavi\v{c}ius and
Nikoulina (1997).
\par We consider a test for the GS model from the data of accelerated experiments with
multistage step-stresses. Power of the test against several important alternatives is
investigated.

\section{Test statistic for the GS model}





Suppose that a group of $n_{0}$ units is tested under the
step-stress (1) and $m$ groups of $n_1,\cdots,n_m$
items are tested under constant in time
stresses $x_{1}\cdots,x_{m}$ ($x_{1}<\cdots<x_{m}$), respectively.
The units are observed time $t_m$ given for the experiment.



%Suppose that a group of $n_{0}$ items is tested under the
%step-stress (1) and $k$ groups of $n_1,\cdots,n_k$
%items are tested under constant in time
%stresses $x_{(1)}\cdots,x_{(k)}$, respectively.
%Here $x_{(1)}<\cdots<x_{(k)}$ ($2\leq k\leq m$) are the different ordered values of constant
%stresses from the set of stresses
% $\{x_{1},\cdots,x_{m}\}$ ,where $x_i$ are given in (1).
% We write $x<y$ if $S_x(t)> S_y(t)$
%for all $t>0$.
%In accelerated life testing usually  $m=k\geq 2$ and $x_i=x_{(i)}$,
%i.e. $x_{1}<\cdots<x_{k}$.


%If possible influence of the
%switch-off's
%on the reliability is considered then it is sufficient to take two alternating values
%$x_{(1)}$ and $x_{(2)}$, i.e. $x_{2i-1}=x_{(1)}$, $x_{2i}=x_{(2)}$.

\par The idea of goodness-of-fit is based on comparing two estimators $\hat A_{x(\cdot)}^{(1)}$
and $\hat A_{x(\cdot)}^{(2)}$ of
the accumulated hazard rate $A_{x(\cdot)}$.
One estimator can be obtained from the experiment under the step-stress (1) and another
from the experiments under the stresses $x_{1},\cdots,x_{m}$ by using the equalities
(2) and (3).

Denote by $N_{i}(t)$ and $Y_{i}(t)$ the numbers of observed failures in the interval $[0,t]$ and the
number of items at risk just prior the moment $t$, respectively, for the group
of items tested under
the stress $x_{i}$ and $N(t)$, $Y(t)$ the analogous numbers for the group of items tested under the
stress $x(\cdot)$.
\par Set
$$
\alpha_{i}=\alpha_{x_i},\quad \alpha=\alpha_{x(\cdot)},\quad A_{i}=A_{x_i}
,\quad  A=A_{x(\cdot)}
\quad (i=1,...,m).
$$
If the GS model holds on $E=\{x_{1},\cdots,x_{m}, x(\cdot)\}$, then the accumulated hazards $A$ can be written in
terms of accumulated hazards $A_{i}$ (cf. (2) and (3)):
$$
A(t)=
A_{i+1}(t-t_{i}+t_{i}^*),\; t \in [t_{i},t_{i+1})\; (i=0,...,m-1), \eqno(9)
$$
where
$$
t_0^*=0,\quad t_1^*=A_{2}^{-1}(A_{1}(t_1)),\cdots,
$$
$$
t_{i+1}^*=A_{i+2}^{-1}(A_{i+1}(t_{i+1}-t_i+t_i^*))\quad (i=0,...,m-2). \eqno(10)
$$
 The first estimator $\hat A^{(1)}$ of the accumulated hazard $A$ can be the well known Nelson -Aalen estimator obtained
from the experiment under the step-stress (1):
$$
\hat A^{(1)}(t)=\int_0^t\frac{dN(v)}{Y(v)}.
$$
The second is suggested by
the GS model (formulae (9) and (10)) and is obtained from the experiments under
constant stresses:
$$
\hat A^{(2)}(t)=\hat A_{i+1}(t-t_{i}+\hat t_{i}^*),\; t \in [t_{i},t_{i+1}),
\; (i=0,...,m-1),
$$
where
$$
\hat t_0^*=0,\quad \hat t_1^*=\hat A_{2}^{-1}(\hat A_{1}(t_1)),\;\cdots,\;
\hat t_{i+1}^*=\hat A_{i+2}^{-1}(\hat A_{i+1}(t_{i+1}-t_i+\hat
t_i^*)), \quad (i=0,...,m-2),
$$
$$
\hat A_{i}^{-1}(s)=\inf \{u:\hat A_{i}(u)\geq s\},
$$
$$
\hat A_{i}(t)=\int_0^t\frac{dN_{i}(v)}{Y_{i}(v)}\quad (i=1,...,m).
$$
The test is based on the statistic
$$
T_n=\int_0^{t_m} K(v)\, d\{\hat A^{(1)}(t)-\hat A^{(2)}(t)\}, \eqno(11)
$$
where $K$ is the weight function.

We'll consider the weight functions of the type: for $v\in [t_i,t_i+\Delta t_i)$
$$
K(v)=\frac{1}{\sqrt{n}}\;
\frac{Y(v)Y_{i+1}(v-t_i+\hat t_i^*)}{Y(v)+Y_{i+1}(v-t_i+\hat t_i^*)}\;
g\left(\frac{Y(v)+Y_{i+1}(v-t_i+\hat t_i^*)}{n}\right),
$$
where $g$ is nonnegative bounded continuous function with bounded variation on $[0,1]$ and
$n=\sum_{i=0}^{m}n_i$.
\par Take notice that the properties of this statistic are different from the properties
of the logrank-type statistics
used testing hypothesis of the equality of survival functions. The properties of $T_n$ would
be similar if $t_i^*$ would be known. The problem is that the points
$t_i^*$
are unknown and are estimated.
Thus when seeking
the limit distribution of the statistic
$T_n$ we must take in mind that the estimators $\hat A_{i}$ are approaching $A_{i}$ with
the same rate as the estimators $\hat t_i^*$ are approaching $t_i^*$.

%The minimal lenghts of experiments under $x_1,\cdots,x_m$ which are needed for the statistic $T$ and
%the estimator $\hat A_2(t),t\in [0,t_m]$ to be defined
%are
%$$
%t_1,\hat t_1^*+t_2-t_1,\cdots,\hat t_{m-1}^*+t_m-t_{m-1},
%$$
%respectively.

%We'll consider the case when the lenghts of the experiments under the constant stresses $x_i$ are $t_i$.
The condition $x_1<\cdots<x_m$ implies that
$$
{\bf P}\{T_n \quad \mbox{is defined}\}\rightarrow 1\quad \mbox{as}\quad  n_i\rightarrow \infty .
$$


\section{Asymptotic distribution of the test statistic}

To find the asymptotic distribution of the test statistic, consider at first
the asymptotic distribution of the estimators $\hat t_i^*$.
Denote by $\stackrel{\cal D}{\rightarrow}$ the convergence in distribution.
\vspace{0.5cm}
\par {\bf Assumptions A}.

a) The hazard rates $\alpha_i$ are positive and continuous on $(0,\infty)$;

b) $A_i(t)<\infty$ for all $t<0$;

c)
  $n\rightarrow \infty$,\quad  $n_i/n\rightarrow l_i$,\quad
$l_i \in (0,1)$.

\vspace{0.5cm}


\par {\bf Lemma}. {\it Suppose that Assumptions A hold. Then}
$$
\sqrt{n}(\hat{t}_{j}^*-t_{j}^*)\stackrel{\cal D}{\rightarrow}a_j \sum^j_{l=1}d_{jl}
\{U_{l}(t^*_{l-1}+\Delta t_{l-1})
-U_{l+1}(t_l^*)\}, \eqno(12)
$$
{\it where
$$
d_{jl}=\prod^{j-1}_{s=l}c_s, \quad l=1,...,j-1; \quad d_{jj}=1,
$$
$$
a_{j}=\frac{1}{\alpha_{j+1}(t_{j}^*)},
\quad c_{s}=\frac{\alpha_{s+1}(t_{s}^* + \Delta t_{s})}{\alpha_{s+1}(t_{s}^*)},
$$
$U_{1},\cdots,U_{m}$ and $U$ are independent Gaussian martingales with $U_{i}(0)=U(0)=0$ and
%$$
%{\bf \cov}(U_{i}(s_1),U_{i}(s_2))=\frac{1}{l_{j_i}}\;
%\frac{1-S_i(s_1\wedge s_2)}{S_i(s_1\wedge s_2)},\quad
%{\bf \cov}(U(s_1),U(s_2))=\frac{1}{l_{0}}\;
%\frac{1-S(s_1\wedge s_2)}{S(s_1\wedge s_2)},
%$$
$$
{\bf \cov}(U_{i}(s_1),U_{i}(s_2))=\frac{1}{l_{i}}\;
\frac{1-S_i(s_1\wedge s_2)}{S_i(s_1\wedge s_2)}:=\sigma^2_i(s_1\wedge s_2),
$$
$$
{\bf \cov}(U(s_1),U(s_2))=\frac{1}{l_{0}}\;
\frac{1-S(s_1\wedge s_2)}{S(s_1\wedge s_2)}:=\sigma^2(s_1\wedge s_2)
$$
%where $j_i=j$ if $x_i=x_{(j)}$, $S_i=\exp\{-A_i\}$, $S=\exp\{-A\}$.
with}  $S_i=\exp\{-A_i\}$, $S=\exp\{-A\}$.

% Note that if $j_{i_1},\cdots,j_{i_l}$ ($l\leq k$) are different, then
%  $U_{i_1},\cdots,U_{i_l}$ are independent .

 Proof. Under Assumptions A for any $t\in (0,t_m)$
the estimators $\hat A_{{i}}$ and $\hat A^{(1)}$
are uniformly consistent on $[0,t]$, and
$$
\sqrt{n}(\hat{A}_{{i}}-A_{{i}})  \stackrel{\cal D}{\rightarrow}
U_{i},\quad
\sqrt{n}(\hat{A}^{(1)}-A)  \stackrel{\cal D}{\rightarrow} U \eqno(13)
$$
on $D[0,t]$, the space of cadlag functions on $[0,t]$ with Skorokhod metric.
\par We prove (12) by recurrence. If $i=1$ then
$$
\sqrt{n}(\hat{t}_{1}^*-t_{1}^*)=\sqrt{n}(\hat{A}_{2}^{-1}(\hat{A}_{1}(t_1))-A_{2}^{-1}
(\hat{A}_{1}(t_1)))+
\sqrt{n}(A_{2}^{-1}(\hat{A}_{1}(t_1))-A_{2}^{-1}(A_{1}(t_1))). \eqno(14)
$$
\par For any $0<s_1<s_2<\infty$ (see Andersen et al (1993))
$$
\sqrt{n}(\hat{A}_{2}^{-1}-A_{2}^{-1})  \stackrel{\cal D}{\rightarrow} U_2^* \eqno(15)
$$
on $D[s_1,s_2]$, where
$$
U_2^*(s)=-\frac{e^{-s}U_2(A_2^{-1}(s))}{p_2(A_2^{-1}(s))}
$$
and $p_i$ is the density of $T_{x_i}$.
Note that
$$
U_2^*(A_1(t_1))=-\frac{U_2(t_1^*)}{\alpha_{2}(t_1^*)}. \eqno(16)
$$
Consistence of the estimator $\hat{A}_{1}(t_1)$, the convergence (15) and the
formula (16) imply that
$$
\sqrt{n}\{\hat{A}_{2}^{-1}(\hat{A}_{1}(t_1))-A_{2}^{-1}(\hat{A}_{1}(t_1))\}
 \stackrel{\cal D}{\rightarrow}
-\frac{U_2(t_1^*)}{\alpha_{2}(t_1^*)}=-a_1U_{2}(t_1^*). \eqno(17)
$$
Using the delta method and the convergence (13), we obtain
$$
\sqrt{n}\{A_{2}^{-1}(\hat{A}_{1}(t_1))-A_{2}^{-1}(A_{1}(t_1))\}
\stackrel{\cal D}{\rightarrow}
\frac{1}{\alpha_{2}(t_1^*)}U_1(t_1)=a_1U_{1}(t^*_{0}+\Delta t_{0}), \eqno(18)
$$
Thus (14), (17) and (18) imply that
$$
\sqrt{n}(\hat{t}_{1}^*-t_{1}^*)
\stackrel{\cal D}{\rightarrow} a_{1}d_{11}\{U_{1}(t^*_{0}+\Delta t_{0})-U_{2}(t_1^*)\}.
$$
Suppose that (12) holds for $i=j$. Then similarly as in the case $i=1$ we have
$$
\sqrt{n}(\hat{t}_{j+1}^*-t_{j+1}^*)=\sqrt{n}\{\hat{A}_{j+2}^{-1}(\hat{A}_{j+1}
(\hat{t}_{j}^*+\Delta t_{j}))-
A_{j+2}^{-1}(A_{j+1}(t_{j}^* + \Delta t_{j}))\}=
$$
$$
\sqrt{n}\{\hat{A}_{j+2}^{-1}(\hat{A}_{j+1}(\hat{t}_{j}^* +\Delta t_{j}))-
A_{j+2}^{-1}(\hat{A}_{j+1}(\hat{t}_{j}^* +\Delta t_{j}))\}+
$$
$$
\sqrt{n}\{A_{j+2}^{-1}(\hat{A}_{j+1}(\hat{t}_{j}^* +\Delta t_{j}))-
A_{j+2}^{-1}(A_{j+1}(\hat t_{j}^* +\Delta t_{j}))\}=
$$
$$
\sqrt{n}\{A_{j+2}^{-1}({A}_{j+1}(\hat{t}_{j}^* +\Delta t_{j}))-
A_{j+2}^{-1}(A_{j+1}( t_{j}^* +\Delta t_{j}))\}=
$$
$$
a_{j+1}\{U_{j+1}(t^*_{j}+\Delta t_{j})
-U_{j+2}(t_{j+1}^*)\}+a_{j+1}\frac{c_{j}}{a_{j}}\sqrt{n}(\hat{t}_{j}^*-t_{j}^*)+\Delta_n,
$$
where $\Delta_n\stackrel{P}{\rightarrow}0$ as $n\to 0$. The last formula and the assumption of recurrency
imply that
$$
\sqrt{n}(\hat{t}_{j+1}^*-t_{j+1}^*)  \stackrel{\cal D}{\rightarrow}
$$
$$
a_{j+1}\{U_{j+1}(t^*_{j}+\Delta t_{j})
-U_{j+2}(t_{j+1}^*)\}
$$
$$
+
a_{j+1}\frac{c_{j}}{a_{j}}a_{j}\left\{\sum^{j-1}_{l=1}\prod^{j-1}_{s=l}c_s
\{U_{l}(t^*_{l-1}+\Delta t_{l-1})
-U_{l+1}(t_{l}^*)\}+U_{j}(t^*_{j-1}+\Delta t_{j-1})
-U_{j+1}(t_{j}^*)\right\}=
$$
$$
a_{j+1}\left\{U_{j+1}(t^*_{j}+\Delta t_{j})
-U_{j+2}(t_{j+1}^*)+\sum^{j}_{l=1}\prod^{j}_{s=l}c_s \{U_{l}(t^*_{l-1}+\Delta t_{l-1})
-U_{l+1}(t_{l}^*)\}\right\}=
$$
$$
a_{j+1}\sum^{j+1}_{l=1}d_{j+1,l}\{U_{l}(t^*_{l-1}+\Delta t_{l-1})
-U_{l+1}(t_{l}^*)\}.
$$
The proof is complete.

Consider the limit distribution of the statistic $T_n$. Note that
$$
\frac{K(v)}{\sqrt{n}}\stackrel{\bf P}\rightarrow k(v)=\frac{l_0l_{i+1}}{l_0+l_{i+1}}S(v)\;g\left((l_0+l_{i+1})S(v)\right).
$$
\vfill
\eject
Set
$$
e_j=a_j\{k(t_j)\; \alpha_{j+1}(t_j^*)-k(t_j+\Delta t_j)\; \alpha_{j+1}(t_j^*+\Delta t_j)
$$
$$
+\int_{t_{j}^*}^{t_{j}^*+\Delta
t_{j}}\alpha_{i+1}(t_i^*+\Delta t_i)(v) d\, k(v+t_{j}- t_{j}^*) \quad (j=1,...,m-1),
$$
and $f_0=f_{m}=0$,
$f_i=\sum_{j=i}^{m-1}e_jd_{ji}$, $(i=1,...,m-1)$.
\vspace{0.5cm}
\par {\bf Theorem 1}. {\it Under Assumptions A}
 $$ T_n
\stackrel{\cal D}\rightarrow
\int^{t_m}_0k(v)d\, U(v)+\sum_{i=0}^{m-1}\{f_{i+1}U_{i+1}(t_{i}^*+\Delta
t_{i})- f_{i}U_{i+1}(t_{i}^*)
$$
$$
-\int_{t_{i}^*}^{t_{i}^*+\Delta
t_{i}}k(v+t_{i}- t_{i}^*)d\, U_{i+1}(v)\}. \eqno(19)
$$
\vspace{0.5cm}
\par Proof. Write the statistic (11)
in the form $$ T_n=\int_0^{t_m}K(v)\, d\{\hat
A^{(1)}(t)-A(t)\} $$ $$ +
\sum_{i=1}^{m-1}\int_{t_{i}^*
}^{\hat{t}_{i}^*}K(v+t_i-\hat{t}_{i}^*)\,d\hat A_{i+1}(v)-
\sum_{i=1}^{m-1}\int_{t_{i}^*+\Delta t_i }^{\hat{t}_{i}^*+\Delta
t_i} K(v+t_i-\hat{t}_{i}^*)\,d\hat A_{i+1}(v) $$ $$
-\sum_{i=1}^{m-1}\int_{t_{i}^* }^{t_{i}^*+\Delta t_i}
\{K(v+t_i-\hat{t}_{i}^*)-K(v+t_i-{t}_{i}^*)\}\,d\hat A_{i+1}(v) $$ $$
-\sum_{i=0}^{m-1}\int_{t_{i}^* }^{t_{i}^*+\Delta t_i}
K(v+t_i-{t}_{i}^*)\,d\{\hat A_{i+1}(v)-A_{i+1}(v)\}. $$
It is easy to show
that under Assumptions A
$$ \int_{0 }^{t_m} K(v)\, d\{\hat
A^{(1)}(t)-A(t)\}=\int_0^{t_m}
k(v)\,d\, U(v)+o_p(1), $$ $$ \int_{t_{i}^*
}^{\hat{t}_{i}^*}K(v+t_i-\hat{t}_{i}^*)\,d\hat A_{i+1}(v)=k(t_i)\, \alpha_{i+1}(t_i^*)\, \sqrt
{n} (\hat{t}_{i}^*-{t}_{i}^*)+o_p(1), $$
$$
\int_{t_{i}^*+\Delta t_i }^{\hat{t}_{i}^*+\Delta t_i}
K(v+t_i-\hat{t}_{i}^*)\,d\,\hat A_{i+1}(v)=k(t_i+\Delta t_i)\,
\alpha_{i+1}(t_i^*+\Delta t_i)\,\sqrt {n}
(\hat{t}_{i}^*-{t}_{i}^*)+o_p(1), $$$$ \int_{t_{i}^*
}^{t_{i}^*+\Delta t_i}
\{K(v+t_i-\hat{t}_{i}^*)-K(v+t_i-{t}_{i}^*)\}\,d\,\hat A_{i+1}(v)=
-\int_{t_{i}^*}^{{t}_{i}^*+\Delta t_i}\alpha_{i+1}(v)\,d\, k(v+t_i-t_i^*)\;
\sqrt {n} (\hat{t}_{i}^*-{t}_{i}^*)+o_p(1), $$
 $$ \int_{t_{i}^*
}^{t_{i}^*+\Delta t_i} K(v+t_i-{t}_{i}^*)\,d\{\hat
A_{i+1}(v)-A_{i+1}(v)\}=\int_{t_{i}^*}^{{t}_{i}^*+\Delta t_i}
k(v+t_i-t_i^*)\,d\,U_{i+1}(v)+o_p(1), $$
where $o_p(1)\stackrel{\bf P}\rightarrow 0$ as $n\rightarrow \infty$.
So the statistic $T_n$ can be written in the form:
$$
T_n =
\int_0^{t_m}
k(v)d U(v)+\sum_{i=1}^{m-1}k(t_i) \alpha_{i+1}(t_i^*) \sqrt{n}(\hat{t}_i^*-t_i^*)- $$
$$ \sum_{i=1}^{m-1}k(t_i+\Delta
t_i)\,\alpha_{i+1}(t_i^*+\Delta t_i)\,\sqrt{n}(\hat{t}_i^*-t_i^*)+
$$
$$
\sum_{i=1}^{m-1}\int_{t_i^*}^{t_i^*+\Delta
t_i}\,
\alpha_{i+1}(v)\,d\, k(v+t_i-t_i^*)\sqrt{n}(\hat{t}_i^*-t_i^*)-
\sum_{i=0}^{m-1}\int_{t_i^*}^{t_i^*+\Delta
t_i} k(v+t_i-t_i^*)\,d U_{i+1}(v)+o_p(1). \eqno(20)
$$
The lemma implies that
$$
\sum_{i=1}^{m-1}\{k(t_i) \alpha_{i+1}(t_i^*)-k(t_i+\Delta
t_i)\alpha_{i+1}(t_i^*+\Delta t_i)+\int_{t_i^*}^{t_i^*+\Delta t_i}\alpha_{i+1}(v)d\,k(v+t_i-t_i^*)\}
\sqrt{n}(\hat{t}_i^*-t_i^*)= $$ $$
\sum_{i=1}^{m-1}e_i\sum_{l=1}^id_{il}\{U_{l}(t_{l-1}^*+\Delta
t_{l-1})-U_{l+1}(t_l^*)\}+o_p(1)
$$
$$
=
\sum_{l=1}^{m-1}\left(\sum_{i=l}^{m-1}e_id_{il}\right)\{U_{l}(t_{l-1}^*+\Delta
t_{l-1})-U_{l+1}(t_l^*)\}+o_p(1) $$ $$
=\sum_{l=1}^{m-1}f_l\{U_{l}(t_{l-1}^*+\Delta
t_{l-1})-U_{l+1}(t_l^*)\}+o_p(1)
$$
$$
= \sum_{i=0}^{m-2}f_{i+1}U_{i+1}(t_{i}^*+\Delta
t_i)-\sum_{i=1}^{m-1}f_i U_{i+1}(t_i^*)+o_p(1). \eqno(19)
 $$
The formulae (20) and (21) imply the result of the theorem.

%Consider the two before mentioned plans of experiments:
%\par 1) The stress has the form (1) with $x_1<\cdots x_m$. Thus
%$m=k$ and $x_{(i)}=x_i$.
%\par 2) The stress (1) has the alternative form:
% $$
%x(\tau)=\left\{\begin{array}{cc} x_1,& t_{2i} \leq \tau < t_{2i+1},\\ x_2,&
%t_{2i+1} \leq \tau < t_{2i+2}.\\
%\end{array} \right.
%\quad (i=0,1,\cdots,r-1)
%$$
%Thus $m=2r$ and $x_{2i}=x_1$, $x_{2i+1}=x_2$.

\vspace{0.5cm}

\par {\bf Corollary}. Under the assumptions of the theorem
$$ T_n\stackrel{\cal D}\rightarrow N(0, \sigma^2_T),
$$
with
$$
 \sigma^2_T
%=\Delta_1 \quad \mbox{(first plan)},
% $$
%$$
% \sigma^2_T=\Delta_1+\Delta_2 \quad \mbox{(second plan)}, \eqno(20)
% $$
% and
%$$\Delta_1
=\int^{t_m}_0k^2(v)\,d\,\sigma^2(v) +
\sum_{i=0}^{m-1}\{f^2_{i+1}\sigma^2_{i+1}(t_{i}^*+\Delta t_{i})+
f_{i}(f_{i}-2f_{i+1})\sigma^2_{i+1}(t_{i}^*)-
$$
$$
2f_{i+1}\int_{t_{i}^*}^{t_{i}^*+\Delta
t_{i}} k(v+t_{i}-t_{i}^*)\,d\,\sigma^2_{i+1}(v)+
\int_{t_{i}^*}^{t_{i}^*+\Delta
t_{i}}k^2(v+t_{i}-t_{i}^*)\,d\,\sigma^2_{i+1}(v)\}. \eqno(22)
 $$
If $m=2$ then
%  $$
%  \Delta_2= 2{\bf 1}_{\{r \geq
%2\}}\sum_{i=0}^{2r-3}g_i\{f_{i+1}\sigma^2_{i+1}(t_{i}^*+\Delta
%t_{i})-f_{i}\sigma^2_{i+1}(t_{i}^*)- \int_{t_{i}^*}^{t_{i}^*+\Delta
%t_{i}} k(v+t_{i}-t_{i}^*)\,d\sigma^2_{i+1}(v)\}, $$
%where for $r\geq 2$ $$
%g_i=\sum_{l=1}^{r-1-[i/2]}(f_{i+2l+1}-f_{i+2l}), \quad
%(i=0,...,2r-3). $$ If $r=1$, then
%for both plans
$$ \sigma^2_T=
\int^{t_m}_0k^2(v)\,d \,\sigma^2(v)+f_1^2(\sigma_1^2(t_1)+
\sigma_2^2(t_1^*))-2f_1\int^{t_1}_0k(v)
\,d \,\sigma_1^2(v)+
$$
$$
\int^{t_1}_0k^2(v)\,d \,\sigma_1^2(v)+\int_{t_1^*}^\infty
k^2(v+t_1-t_1^*)\,d \,\sigma_2^2(v). $$
\vspace{0.5cm}
\par {\bf Remark}. The variance $\sigma^2_T$ can
be consistently estimated by the statistic
%$ \hat \sigma^2_T$,
%where
$$
 \hat \sigma^2_T
%=\hat\Delta_1 \quad \mbox{(first plan)},
% $$
%$$
%\hat \sigma^2_T=\hat \Delta_1+\hat \Delta_2 \quad \mbox{(second plan)},
% $$
% and
%$$
%\hat\Delta_1
=\int^{t_m}_0\hat{k}^2(v)d\hat{\sigma}^2(v)
+\sum_{i=0}^{m-1}\{\hat{f}^2_{i+1}\hat{\sigma}^2_i(\hat{t}_i^*+
\Delta
t_i)+\hat{f}_{i}(\hat{f}_{i}-2\hat{f}_{i+1})\hat{\sigma}^2_i(\hat{t}_i^*)-
$$
$$
2\hat{f}_{i+1}\int_{\hat{t}_{i}^*}^{\hat{t}_{i}^*+\Delta
t_{i}}\hat{k}(v+t_{i}-\hat{t}_{i}^*)d\,\hat{\sigma}^2_i(v)
+\int_{\hat t_{i}^*}^{\hat t_{i}^*+\Delta
t_{i}}\hat k^2(v+t_{i}-\hat t_{i}^*)\,d\,\hat\sigma^2_{i+1}(v)\},
%$$
%\hat\Delta_2= 2{\bf
%1}_{\{r\geq
%2\}}\sum_{i=0}^{2r-3}\hat{g}_i\{\hat{f}_{i+1}\hat{\sigma}^2_i(\hat{t}_{i}^*+\Delta
%t_{i})-
%\hat{f}_{i}\hat{\sigma}^2_i(\hat{t}_{i}^*)-\int_{\hat{t}_{i}^*}^{\hat{t}_{i}^*+\Delta
%t_{i}} \hat{k}(v+t_{i}-\hat{t}_{i}^*)d\hat{\sigma}^2_i(v)\},
$$
where
$$
\hat{k}(v)=K(v)/\sqrt{n},\quad
\hat{\sigma}^2(v)=\frac{n}{n_0}\left(\frac{1}{\hat{S}(v)}-1\right),\quad
\hat{\sigma}^2_i(v)=\frac{n}{n_i}\left(\frac{1}{\hat{S}_i(v)}-1\right),
$$
$\hat S$ and $\hat S_i$ are the empirical survival functions,
$$
%\hat{g}_i=\sum_{l=1}^{r-1-[i/2]}(\hat{f}_{i+2l+1}-\hat{f}_{i+2l}),\quad
 \hat{f}_0=\hat{f}_{m}=0, \quad
\hat{f}_i=\sum_{s=i}^{2m-1}\hat{e}_s\hat{d}_{si} \quad (i=1,\cdots,m-1),
$$
$$
\hat{e}_s=\hat{a}_s\left(\hat{k}(t_s)-\hat{k}(t_s+\Delta
t_s)-\hat{k}(t_s)\,\hat{\alpha}_{s+1}(\hat{t}_{s}^*)+\hat{k}(t_s+\Delta
t_s)\,\hat{\alpha}_{s+1}(\hat{t}_{s}^*+\Delta
t_{s})\right.
$$
$$
\left.-\int_{\hat{t}_{s}^*}^{\hat{t}_{s}^*+\Delta
t_{s}}\hat{k}(v+t_{s}-\hat{t}_{s}^*) d\,\hat A_{s+1}(v)\right),
$$
$$
\hat{d}_{si}=\prod_{l=i}^{s-1}\hat{c}_l,\quad i=1,...,s-1,\quad
\hat{d}_{ss}=1,
$$
$$
\hat{c}_l=\frac{\hat{\alpha}_{l+1}(\hat{t}_l^*+\Delta
t_l)}{\hat{\alpha}_{l+1}(\hat{t}_l^*)}, \quad
\hat{a}_s=\frac{1}{\hat{\alpha}_{s+1}(\hat{t}_s^*)} ,
$$
and $\hat{\alpha}_{s+1}(\hat{t}_s^*)$, $\hat{\alpha}_{s+1}(\hat{t}_{s}^*+\Delta
t_{s})$ are kernel estimators:
$$
\hat{\alpha}_{s+1}(\hat{t}_s^*)=\frac{1}{b}\int^{t_m}_0 Ker\left(\frac{\hat{t}_i^*-u}{b}\right)d\hat{A}_{i+1}(u),
$$
$$
\hat{\alpha}_{s+1}(\hat{t}_{s}^*+\Delta
t_{s})=\frac{1}{b}\int^{t_m}_0 Ker\left(\frac{\hat{t}_i^*+\Delta t_s-u}{b}\right)d\hat{A}_{i+1}(u);
$$
here $Ker$ is some kernel function.
\section{The test}

The hypothesis
$$
H_0:  \mbox{{\it GS model holds on}}\; E=\{x_{1},\cdots,x_{m},x(\cdot)\}
$$
is rejected with the approximative significance level $\alpha$, if
$$
\left(\frac{T}{\hat\sigma_T}\right)^2>\chi^2_{1-{\alpha}}(1),
$$
where $\chi^2_{1-{\alpha}}(1)$ is the $(1-\alpha)$-quantile
of the chi-square distribution with one degree of freedom.



\section{Consistency and the power of the test against approaching alternatives}



%Consider the first plan of experiments and find
Find the power of the
test against the following alternatives:
$$
H_1:\mbox{{\it PH model with specified
non-exponential time-to-failure} }
$$
$$
\quad \quad \mbox{{\it distributions under
constant stresses}}
$$
Under $H_1$
$$
\hat {A}^{(1)}(v) \stackrel{\bf P}{\rightarrow}{A}_*^{(1)}(v)=A_i(v),\;v\in [t_{i},t_{i+1}) \;
(i=0,\cdots,m-1),
$$
$$
\hat {A}^{(2)}(v) \stackrel{\bf P}{\rightarrow} A^{(2)}(v)=A_i(v-t_{i}+t_{i}^*),\;v\in [t_{i},t_{i+1}) \;
(i=0,\cdots,m-1),
$$
and
$$
 \frac{1}{\sqrt{n}}K(v)\stackrel{\bf P}{\rightarrow}k^*(v),
$$
where for $v\in [t_{i},t_{i+1})$
$$
k^*(v)=\frac{l_0l_{i+1}S_*^{(1)}(v)S_i(v-t_i+t_i^*)}{l_0 S_*^{(1)}(v)+l_{i+1} S_i(v-t_i+t_i^*)}
g\left(l_0 S_*^{(1)}(v)+l_{i+1} S_i(v-t_i+t_i^*)\right),
$$
and $S_*^{(1)}(v)=exp\{-A_*^{(1)}(v)\}$.
Convergence is uniform on $[0,t_m]$.


\vspace{0.5cm}


\par {\bf Proposition 1}. {\it Assume that Assumptions A hold under $H_1$ and
$$
\Delta^* =\int_0^{t_m} k^*(v)\, d\{ A^{(1)}(v)-
A(v)\}\neq 0.
$$
Then the test is consistent against $H_1$.}

\vspace{0.5cm}
Proof. Write the test statistic in the form
$$
T_n=\int_0^{\infty} K(v)\, d\{\hat A^{(1)}(v)- A_*^{(1)}(v)\}-
\int_0^{\infty} K(v)\, d\{\hat A^{(2)}(v)-A^{(2)}(v)\}+
$$
$$
\int_0^{\infty} K(v)\, d\{ A_*^{(1)}(v)-A^{(2)}(v)\}=T_{1n}+T_{2n}+T_{3n}. \eqno(23)
$$
Analogously as in the case when seeking the limit distribution of
the statistic $T_n$ under the hypothesis $H_0$, we obtain that under
$H_1$
$$
T_{1n}+T_{2n}\stackrel{\cal D}\rightarrow N(0, {\sigma_T^*}^2),
$$
where ${\sigma_T^*}^2$ has the same form (22) with only difference
that $k$ is replaced by $k^*$ and $\sigma^2(t)$ is replaced by
$$
(\sigma^{(1)})^2(t)
=\frac{1}{l_0}\left(\frac{1}{S^{(1)}(t)}-1\right).
$$
Under $H_1$ we have
$$
\hat \sigma^2_T\stackrel{\bf P}{\rightarrow}{\sigma_T^*}^2 \eqno(24)
$$
and
$$
\frac{T_{1n}+T_{2n}}{\hat \sigma_T}\stackrel{\cal D}\rightarrow N(0,1). \eqno(25)
$$
The third member in (23) can be written in the form
$$
T_{3n}=\sum_{i=1}^{m-1}\int_{t_i}^{t_{i+1}} K(v)\,
\{\alpha_{i+1}(v)-\alpha_{i+1}(v-t_i+t^*_i)\}\,d v. \eqno(26)
$$
The assumptions of the proposition and the equalities (23)-(26) imply that under $H_1$
$$
\frac{1}{\sqrt{n}}T_{3n} \stackrel{\bf P}{\rightarrow} \Delta^*,
\quad \frac{T_n}{\hat \sigma_T}\stackrel{\bf P}{\rightarrow} \infty .
$$
Thus under $H_1$, see for example, Greenwood and Nikulin (1996),
$$
{\bf P}\left\{\left(\frac{T}{\hat\sigma_T}\right)^2>\chi^2_{1-{\alpha}}(1)\right\}
\rightarrow 1.
$$

The proposition is proved.
\vspace{0.3cm}
\par {\bf Remark}. {\it If $\alpha_i$ are increasing
(decreasing) then the test is consistent against $H_1$}.
\vspace{0.3cm}
Proof. We'll show now by recurrence that $t_i>t_i^*$ for all $i$. Really, the
inequalities $x_1<\cdots<x_m$ imply that
$$
S_1(t_1^*)>S_2(t_1^*)=S_1(t),
$$
which give $t_1>t_1^*$. If we assume that $t_{i-1}>t_{i-1}^*$ then
$$
S_{i+1}(t_i^*)=S_i(t_i-t_{i-1}+t_{i-1}^*)>S_i(t_i-t_{i-1}+t_{i-1})=S_i(t_i)>S_{i+1}(t_i),
$$
which imply $t_i>t_i^*$. If $\alpha_i$ are increasing
(decreasing) then $\Delta^*>0$ ($\Delta^*<0$) under $H_1$. The proposition implies
the consistency of the test.

\par Consider the sequence of approaching alternatives
$$
H_n:\;PH \;\mbox{{\it with}}\;
\alpha_i(t)=\left(\frac{t}{\theta_i}\right)^{\frac{\varepsilon_i}{\sqrt
{n}}}
$$
with fixed $\varepsilon_i>0$ ($i=1,\cdots,m$).
Then
$$
T_{3n} \stackrel{\bf P}{\rightarrow}\mu=- \sum_{i=1}^{m-1}\varepsilon_i\int_{t_i}^{t_{i+1}} k^*(v)\,
\ln (1+\frac{t_i^*-t_i}{v})d v>0
$$
and
$$
 \frac{T_n}{\hat \sigma_T}\stackrel{\cal D}{\rightarrow}N(a,1), \quad
  \left(\frac{T}{\hat \sigma_T}\right)^2\stackrel{\cal
  D}{\rightarrow}\chi^2(1,a),
$$
where $a=\mu/\sigma_T^*$ and $\chi^2(1,a)$ denotes the chi-square
distribution with one degree of freedom and the non-centrality
parameter $a$ (or a random variable having such distribution).
\par The power function of the test is approximated by
the function
$$
\beta=\lim_{n\to \infty}{\bf P}\left\{
\left(\frac{T}{\hat\sigma_T}\right)^2>\chi^2_{1-{\alpha}}(1)\mid H_n\right\}=
{\bf P}\left\{\chi^2(1,a)
>\chi^2_{1-{\alpha}}(1)\right\}. \eqno(27)
$$



%Consider the second plan of experiments and
Find the power of the
test against the following alternatives:
$$
H_2:\mbox{{\it the model (7) with specified time-to-failure}}
$$
$$
\mbox{{\it distributions under
constant stresses}}
$$


Under $H_2$
$$
\hat {A}^{(1)}(v) \stackrel{\bf P}{\rightarrow}{A}_{**}^{(1)}(v)=A_i(v-t_{i}+t_{i}^{**}),\;v\in [t_{i},t_{i+1}) \;
(i=0,\cdots,m-1),
$$
$$
\hat {A}^{(2)}(v) \stackrel{\bf P}{\rightarrow} A^{(2)}(v)=A_i(v-t_{i}+t_{i}^*),\;v\in [t_{i},t_{i+1}) \;
(i=0,\cdots,m-1),
$$
and
$$
 \frac{1}{\sqrt{n}}K(v)\stackrel{\bf P}{\rightarrow}k^{**}(v),
$$
where for $v\in [t_{i},t_{i+1})$
$$
k^{**}(v)=\frac{l_0l_{i+1}S_{**}^{(1)}(v)S_i(v-t_i+t_i^*)}{l_0 S_{**}^{(1)}(v)+l_{i+1} S_i(v-t_i+t_i^*)}
g\left(l_0 S_{**}^{(1)}(v)+l_{i+1} S_i(v-t_i+t_i^*)\right),
$$
and $S_{**}^{(1)}(v)=exp\{-A_{**}^{(1)}(v)\}$.
Convergence is uniform on $[0,t_m]$.




\vspace{0.5cm}


\par {\bf Proposition 2}.  {\it Assume that Assumptions A hold under $H_2$ and
$$
\Delta^{**} =\int_0^{\infty} k^{**}(v)\, d\{ A^{(1)}(t)-
A(t)\}\neq 0.
$$
Then the test is consistent against $H_2$.}

\vspace{0.3cm}

Proof. Write the test statistic in the form (23). Analogously as in the case when seeking the limit distribution of
the statistic $T_n$ under the hypothesis $H_0$, we obtain that under
$H_2$
$$
T_{1n}+T_{2n}\stackrel{\cal D}\rightarrow N(0, {\sigma_T^{**}}^2),
$$
where ${\sigma_T^*}^2$ has the same form (20) with only difference
that $k$ is replaced by $k^{**}$ and $\sigma^2(t)$ is replaced by
$$
(\sigma^{(1)})^2(t)
=\frac{1}{l_1}\left(\frac{1}{S^{(1)}(t)}-1\right).
$$
The third member in (23) can be written in the form
$$
T_{3n}=\sum_{i=1}^{m-1}\int_{t_i}^{t_{i+1}} K(v)\,
\{\alpha_{i+1}(v-t_i+t^{**}_i)-\alpha_{i+1}(v-t_i+t^*_i)\}d v.
$$
The assumptions of the proposition and the the last equality imply that
$$
\frac{1}{\sqrt{n}}T_{3n} \stackrel{\bf P}{\rightarrow} \Delta^{**},
\quad \frac{T_n}{\hat \sigma_T}\stackrel{\bf P}{\rightarrow} \infty .
$$
The proposition is proved.
\vspace{0.3cm}
\par {\bf Remark}. {\it If $\alpha_i$ are increasing
(decreasing) then the test is consistent against $H_2$}.
\vspace{0.3cm}


Proof. We'll show now by recurrence that $t^{**}_i>t_i^*$. Really,
the inequalities $x_1<\cdots<x_m$ imply that
$$
S_2(t^{**}_1)=S_1(t_1)\,\delta_1<S_1(t_1)=S_2(t_1^*),
$$
which give $t^{**}_1>t_1^*$. If we assume that $t^{**}_{i-1}>t_{i-1}^*$ then
$$
S_{i+1}(t_i^{**})=S_i(t_i-t_{i-1}+t_{i-1}^{**})\,\delta_i<S_i(t_i-t_{i-1}+t^{**}_{i-1})
<S_i(t_i-t_{i-1}+t^{*}_{i-1})=S_{i+1}(t^{*}_i),
$$
which imply $t^{**}_i>t_i^*$. If $\alpha_i$ are increasing
(decreasing) then $\Delta>0$ ($\Delta<0$) under $H_2$. The proposition 2 implies the consistency
of the test.
\par Consider the sequence of approaching alternatives
$$
H_n:\mbox{{\it the model (7) with specified time-to-failure distributions}}
$$
$$
\quad \quad \quad \mbox{{\it  under
constant stresses}}\;
\mbox{and}\;\delta_i=1-\frac{\varepsilon_i}{\sqrt{n}}.
$$
Find the limit of $\sqrt{n}(t^{**}_i-t^{*}_i)$ by recurrence. If
$i=1$, then
$$
\sqrt{n}(t^{**}_1-t^{*}_1)=\sqrt{n}\left\{S_2^{-1}
\left(S_1(t_1)(1-\frac{\varepsilon_1}{\sqrt{n}})\right)-
S_2^{-1}\left(S_1(t_1)\right)\right\}\rightarrow
-\frac{\varepsilon_1}{\alpha_2(t^{*}_1)}=-a_1\varepsilon_1
$$
Suppose that
$$
\sqrt{n}(t^{**}_i-t^{*}_i)\longrightarrow
a_i\sum^i_{j=1}d_{ij}\varepsilon_j, \eqno(28)
$$
where $a_i,d_{ij}$ are defined in the formulation of the lemma.
Then
$$
\sqrt{n}(t^{**}_{i+1}-t^{*}_{i+1})=\sqrt{n}(S_{i+2}^{-1}
\left(S_{i+1}(t_{i+1}-t_i+t_i^{**})(1-\frac{\varepsilon_{i+1}}{\sqrt{n}})\right)-
S_{i+2}^{-1}(S_{i+1}(t_{i+1}-t_i+t_i^{*}))
$$
$$=
\sqrt{n}\frac{1}{p_{i+2}(t_{i+1}^*)}\left\{p_{i+1}(t_{i+1}-t_{i}+t_{i}^*)(t^{**}_i-t^{*}_i)-
S_{i+1}(t_{i+1}-t_{i}+t_{i}^*)\frac{\varepsilon_{i+1}}{\sqrt{n}}\right\}+o(1)=
$$
$$
\frac{1}{p_{i+2}(t_{i+1}^*)}\left\{-p_{i+1}(t_{i+1}-t_{i}
+t_{i}^*)a_i\sum^i_{j=1}d_{ij}\varepsilon_j -
S_{i+1}(t_{i+1}-t_{i}+t_{i}^*)\varepsilon_{i+1}\right\}+o(1)=
a_{i+1}\sum^{i+1}_{j=1}d_{i+1,j}\varepsilon_j+o(1).
$$
We note $p_i$ the densities of $T_{x_i}$.
Thus the convergence (28) holds for all i. It implies that
$$
T_{3n} \stackrel{\bf P}{\rightarrow}\mu=-\sum_{i=1}^{m-1}a_i\sum^i_{j=1}d_{ij}\,\varepsilon_j\int_{t_i}^{t_{i+1}} k(v)\,
d\, \alpha_{i+1}(v-t_i+t^{*}_i)
$$
and
$$
 \frac{T_n}{\hat \sigma_T}\stackrel{\cal D}{\rightarrow}N(a,1), \quad
  \left(\frac{T_n}{\hat \sigma_T}\right)^2\stackrel{\cal
  D}{\rightarrow}\chi^2(1,\mid a \mid ),
$$
where $a=\mu/\sigma_T^{**}$.
\par The parameter $\mu$ is positive (negative) if the functions
$\alpha_i$ are convex (concave).

\par The power function of the test is approximated by
the function (27) with $a=\mu/\sigma_T^{**}$.



\vskip 1.0cm


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\bigskip
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\end{document}
