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\begin{document}
\begin{slide}
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\sloppy



\begin{center}

{ON TESTING DEPENDENCE BETWEEN \\

 TIME  TO FAILURE AND CAUSE OF \\

FAILURE 
 VIA CONDITIONAL PROBABILITY}\\

\vspp

Isha Dewan,\\Indian Statistical Institute,\\

New Delhi, India\\
\vspp
J. V. Deshpande, \\ Department of Statistics, University of Pune, \\ Pune  India \\
\vspp
Sangita Kulathinal, \\ Department of Epidemiology and Health Promotion, \\

National Public Health Institute, \\ 

Mannerheimintie 166, \\ 00300 Helsinki, Finland

\end{center}
\end{slide}
\begin{slide}
 
\NI{\bf Introduction}\\

Competing risks data:  $T=$ failure time\\
 $\delta=$ an indicator of the failure type, $\delta \in \{0, 1\}$

Censored data:

Acturial studies:

Planning studies: $T$ could be the time to first employment and $\delta$ indicating the 
employment obtained
\end{slide}

\begin{slide}


\NI{\bf Joint distribution of $T$ and $\delta$}\\

Subsurvival functions:
$$S_i(t) = P[ T \ge t, \delta=i ], \; i=0,1$$

Subdistribution functions:
$$F_i(t) = P[ T \ge t, \delta=i ], \; i=0,1$$

The  survival function of $T$ is given by
$$S(t) = P[ T \ge t] = S_0(t) + S_1(t)$$

The  distribution function of $T$ is given by
$$F(t) = P[ T < t] = F_0(t) + F_1(t)$$

Consider the following two conditional probabilities
\begin{eqnarray*}
\Phi_1(t) & = & pr(\delta = 1 \mid T \ge t)=S_1(t)/S(t) \; \mbox{and}\\
\Phi_0^*(t) & = & pr(\delta = 0 \mid T < t)=F_0(t)/F(t),
\end{eqnarray*}
\end{slide}
\begin{slide}
 $$\Phi_1(t)=\phi, \forall \; t > 0$$ is equivalent to independence of $T$ and $\delta.$ 

 Similarly $\Phi_0^*(t)$ constant is equivalent to independence of $T$ and $\delta.$ 

 $T$ and $\delta$ can be studied separately.

 If $T$ and $\delta$ are independent then 
$$S_i(t)=S(t) P[\delta = i].$$
  Thus the hypothesis of equality of incidence functions, or that of equality of cause-spectific hazard rates reduces to testing whether \linebreak $P[\delta=1]=P[\delta=0]=1/2,$ a Bernoulli trial with success probability half.  The problem from two-dimensions reduces to one-dimensions.
\end{slide}
\begin{slide}

\NI{\bf Few definitions:}\\ 

\begin{defn}
$X_2$ is Right Tail Increasing in $X_1,$ $RTI(X_2 \mid X_1),$ if
$$P[X_2>t_2\mid X_1>t_1]$$
 is increasing in $t_1$ for all $t_2.$
\end{defn}


\begin{defn}
$X_2$ is Left Tail Decreasing in $X_1,$ $LTD(X_2 \mid 
X_1),$ if 
$pr(X_2 \le t_2 \mid X_1 \le t_1)$ is
decreasing in $t_1$ for all $t_2.$
\end{defn}


\begin{defn}
$X_1$ and $X_2$ are Positively Quadrant Dependent, $PQD(X_1, X_2),$ if
$$P[X_1>t_1, X_2>t_2]\ge P[X_1>t_1] P[X_2>t_2], \forall t_1, t_2.$$
\end{defn}

\begin{defn}
Let $K(x_1,x_2)$ be the joint probability density function of $X_1$ and $X_2$. Then $X_1$ and $X_2$ are Totally Positive of Order 2, $TP_2$, if 
 $$f(s_1,t_1) f(s_2, t_2) \ge f(s_2, t_1) f(s_1, t_2)$$
for all $ s_1 < s_2, t_1 < t_2 $.
\end{defn}
\end{slide}
\begin{slide}
$TP_2$ $\Longrightarrow$ $RTI(X_2 \mid X_1)$ $\Longrightarrow$ $PQD(X_1, X_2)$

Similarly

$TP_2$ $\Longrightarrow$ $LTD(X_2 \mid X_1)$ $\Longrightarrow$ $PQD(X_1, X_2)$

But no hierarchy between  $RTI(X_2 \mid X_1)$  and
 $LTD(X_2 \mid X_1)$ 

\NI{\bf Dependence of $T$ and $\delta$ and Monotonicity of $\Phi_1(t)$}\\
\vsp
\noindent (1) Independence of $T$ and $\delta$ is equivalent to
\vsp
(a) $\Phi_1(t)=\phi=pr(\delta = 1),$  for  all   $t>0,$ a constant and

(b) $\Phi_0^*(t)=1-\phi=\phi_0=pr(\delta = 0),$  for all  $t>0,$  a constant.
\end{slide}
\begin{slide}

\noindent (2) $PQD(\delta, T)$ is equivalent to

(a) $\Phi_1(t) \ge \Phi_1(0)=\phi, \; for \; all \; t>0,$ and 

(b) $\Phi_0^*(t) \ge \Phi_0^*(\infty)=1-\phi, \; for \; all \; t>0.$  

\noindent (3) $RTI(\delta \mid T)$ is equivalent to $\Phi_1(t) \uparrow t.$

\noindent (4) Subsurvival functions $S_i(t)$ being $TP_2$ is equivalent to $\Phi_1(t) \uparrow t.$

\noindent (5) $LTD(\delta \mid T)$ is equivalent to $\Phi_0^*(t) \downarrow t.$

\noindent (6) Subdistribution function 
$F_i(t)$ being $TP_2$ is equivalent to $\Phi_0^*(t) \downarrow t.$



\end{slide}

\begin{slide}
\NI{\bf Examples:}

  In the clinical trials carried out to study the performance of intrauterine device where termination of the device could be due to several reasons it is often of interest to know the chances of termination due to a specific reason given that the device was intact for some specified period.  In such situation, conditional probabilities are of interest. 


\NI{\bf $\Phi_1(t)$ constant} \\
\vsp
1. Bivariate exponential distributions: Block and Basu (1974), 
Farlie-Gumbel-Morgenstern bivariate exponential 
distribution, Gumbel Type A distribution

2. Constant warning-constant inspection model:
 a warning is emitted at time $X-d$ before it fails, where $d<1$ is a constant

3. Hoel data (1972): the data were obtained from a laboratory 
experiment on male mice which had received  a radiation dose of 300r at an age of 5 to 6 
weeks. The death occurred due to cancer ($\delta =1$), or other causes ($\delta =0$).


\end{slide}
\begin{slide}



\NI{\bf $\Phi_1(t)$  increasing}\\


1. Random sign censroing (Cooke {\it et al.}, 1993): lifetime of a unit, $X,$ is censored 
by $Z=X-W\eta,$ 
where $0<W<X$ is the time at which a warning is emitted by the unit before its failure, and 
$\eta$ is a random variable taking values $\{-1, 1\}$ and is independent 
of $X.$ When $W=aX$ for some $0<a<1$ and $X$ is assumed to be exponential, 
it is easy to see that

2. Nair data (1993): the data are times to 
failure ( in millions of operations ) of 37 switches which could fail in two modes (A or B)

\NI{\bf $\Phi_1(t)$ decreasing}\\


 1. Proportional warning-constant inspection model: similar to the constant warning-constant 
inspection
model except that the warning is emitted at time $X/\eta$ if the component 
fails at $X$ and where $\eta>1$ is a constant

\end{slide}

\begin{slide}
\NI {\bf Hazard rate ordering and aging}\\

Let $r_i(t)$ and $h_i(t)$ denote crude and cause-specific hazard rates, 
respectively,$i=0, 1.$ Then
\begin{eqnarray*}
r_i(t) & = & \frac{f_i(t)}{S_i(t)} \\
h_i(t) & = & \frac{f_i(t)}{S(t)}.
\end{eqnarray*}
The overall hazard rate of $T$ is
$$r(t) = f(t)/S(t) = h_1(t)+h_2(t)$$
where $f_i(.), f(.)$ are densities corresponding to $S_i(.)$ and $S(.),$ 
respectively.  Note that, $r_1(t)$ is the hazard rate of  the conditional 
survival function of $T$ given $\delta =1,$ $S_1(t)/\phi.$


\bt
$\Phi_1(t) \uparrow t$ is equivalent to $r_1(t) \le r(t) \le r_0(t).$
\et

This means that given that the failure is due to risk 1, the failure rate of a unit is less than the overall failure rate while given that the failure is due to risk 0, the failure rate is higher than the overall failure.  

\end{slide}
\begin{slide}


Another interesting result stated below connects the monotonicity of 
$\Phi_1(t)$ with the ordering between two survival functions.
\bt
$\Phi_1(t) \uparrow t$ implies that the survival function of $T$ given 
$\delta = 1$ is larger than the survival function of $T$ given $\delta =0,$ 
{\it i.e.,}
    $S_1(t)/\phi \ge S_0(t)/(1-\phi).$
\et

It is important to note that the hazard rates $r_1(t)$ and $r_0(t)$ 
corresponds to the above two distributions.


\bt
$\Phi_1(t) \uparrow t$ is equivalent to $h_1(t) \le r(t)$ and $h_0(t) \ge
(1-\Phi_1(t)) r(t).$
\et

The above theorem implies that

 $h_1(t) \le \frac{\Phi_1(t)}{1-\Phi_1(t)}
 h_0(t).$
\end{slide}
\begin{slide}

It is interesting and also useful to express the cause-specific hazard rate in terms of $\Phi_1(t).$  This enables one to study the aging through the properties of $\Phi_1(t).$

\bt
(a)  $h_1(t) = -\Phi_1'(t) + \Phi_1(t) r(t),$ where $\Phi_1'(t)$ is the first derivative of $\Phi_1(t)$ with respect to $t.$

(b) If $\Phi_1(t)$ is monotone increasing and concave then $h_1(t)$ is an increasing function of $t,$ provided $r(t)$ is IFR.
\et

In case of independent latent lifetimes, the  survival function/hazard rate of $X$ is expressed in terms of $h_1(t).$  If $h_1(t)$ is IFR then $X$ will also have IFR distribution.

Analogous results for reverse hazard rate and $\phi_0^*(t)$.

\end{slide}


\begin{slide}
\NI {\bf Test statistic and its distribution}\\

{\bf General dependence between $T$ and $\delta$}

To test $H_0: T \; \mbox{and} \; \delta$ are independent against the alternative of dependence\\ \\
$ H_0 : \Phi_1(t) \mbox{is a constant}$\\
$ H_1: \Phi_1(t) \mbox{is not a constant}$

A test based on Kendall's $\tau$

Concordance: \\
$  Ti > Tj,\; \delta_i=1, \delta_j=0 $ or \\
$ Ti < Tj,\; \delta_i =0, \delta_j=1$

Discordance:\\
$ Ti > Tj, \;\delta_i =0, \delta_j=1$ or \\
$ Ti < Tj, \;\delta_i =1, \delta_j=0$\\
Define the kernel
\bea
\psi(T_i, \delta_i, T_j, \delta_j)& = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_i > T_j,\; \delta_i =1, \delta_j=0 \nonumber  \\
  & \mbox{or} \;  T_i < T_j, \;\delta_i =0, \delta_j=1 \nonumber \\
-1  & \mbox{if}\; \ T_i > T_j, \;\delta_i =0, \delta_j=1 \nonumber \\
   & \mbox{or} \; T_i < T_j, \; \delta_i =1, \delta_j=0 \nonumber \\
0 & \mbox {otherwise}. \end{array} \right.
\eea

\end{slide}
\begin{slide}
U-statistic is given by
$$ U_k = \frac{1}{{n \choose 2}}\sum _{1 \le i < j \le n}\psi(T_i, \delta_i, T_j, \delta_j) .$$

\ben
E(U_k) &= & 2 P(\delta =1) +   4 \int_0^\ity S(t) dS1(t)
\een

Under $H_0$, $E(U_k) = 0$

Let $R_j$ be the rank of $T_j$. Let $ T_{(1)} < T_{(2)} < T_{(n)} $ be the ordered $T_i's$. 
Let
\bea
W_j & = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_{(j)} \;\mbox{corresponds to} \; \delta =1 \nonumber  \\
  0 & \mbox {otherwise}. \end{array} \right.
\eea

Then $V_k = {{n \choose 2}} U_k$ can be written  as

\bea
V_k &=& \sum_{j=1}^n (2 R_j -n -1) \delta_j \nonumber\\
 &=& \sum_{j=1}^n (2j -n -1) W_j \nonumber\\
 &=& \sum_{j=1}^n a_j W_j 
\eea
where $a_j = 2j -n -1.$

\end{slide}
\begin{slide}
Under $H_0$, $P[W_i=1] = \phi,\; P[W_i=0] = 1 - \phi.$ 
Moment generating function of $V_k$, under $H_0$:
$$ M(t) = \prod_{j=1}^n[ \phi\; e^{t(2j-n-1)} + (1-\phi)].$$

\bt
Under $H_0$,  $\sqrt{n}U_k$ converges in distribution to $N(0, \sigma^2),$ 
where
$$\hat{\sigma^2} = \frac{4}{3}\hat{\phi} (1 - \hat{\phi})$$
as $n \rightarrow \infty,$
\et

A test procedure for testing $H_0$ against $H_1$ is then: reject $H_0$ at $5\%$ 
level of significance if $\mid Z = \sqrt{n} U_k /\hat{\sigma} \mid$ is larger 
than $1.96,$ a cut-off point of standard normal distribution. 
\end{slide}

\begin{slide}

{\bf Testing independence against $PQD(T, \delta)$}

$$H_0 : T \;\;\mbox{and}\;\; \delta \;\;\mbox{are independent}$$

against the alternative 
$$H_2: T \;\; \mbox{and} \;\; \delta \;\;\mbox{are  positive quadrant dependent}.$$


{\bf Test based on $\phi_1(t)$}

$$ H_2: \Phi_1(t) \ge \phi \; \mbox{for all} \; t \;$$

with strict inequality for some $t$.



Consider 
\bea
\Delta_3(S_1, S) &=& \int_0^\infty [S_1(t) - \phi S(t)] dF(t) \nonumber \\
& = &  pr(T_2 > T_1, \delta_2=1) - \phi/2.
\eea
Under $H_0,$ 

$S_1(t)/S(t) = \phi = pr(\delta = 1).$ 
\end{slide}
\begin{slide}


This implies that $\Delta_3(S_1, S) =0.$ 

 Under $H_2,$ $S_1(t) > \phi S(t)$ 

and hence 
$\Delta_3(S_1, S) \ge 0.$

Define the symmetric kernel
\bea
\psi_3(T_i, \delta_i, T_j, \delta_j)& = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \; \;  T_j > T_i,\; \delta_j = 1 \nonumber  \\
  &  \mbox{or if}\; \;  T_i > T_j,\; \delta_i = 1 \nonumber  \\
0 & \mbox {otherwise}. \end{array} \right.
\eea
which is equivalent to
\bea
\psi_3(T_i, \delta_i, T_j, \delta_j)& = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \; \; T_j > T_i,\; \delta_j = 1, \delta_i=0 \nonumber  \\
  &  \mbox{ if} \; \; T_i > T_j,\; \delta_i = 1 ,\delta_j=0 \nonumber  \\
  &  \mbox{ if} \;  \delta_i = \delta_j = 1 \nonumber  \\
0 & \mbox {otherwise}. \end{array} \right.
\eea
Then the U-statistic corresponding to $\Delta_3(S_1, S)$ is given by
$$ U_3 = \frac{1}{{n \choose 2}}\sum _{1 \le i < j \le n}\psi_3(T_i, \delta_i, T_j, \delta_j) .$$



Note that $E(U_3) = 2 \Delta_3(S_1, S) + \phi.$ 

Under $H_0,$ $E(U_3) = \phi.$ 

Under $H_2$ $E(U_3) \ge \phi.$
\end{slide}
\begin{slide}
Note that the statistic $U_3$ has earlier been proposed by  Bagai {\it et al.} (1989) for testing 
the equality of failure rates of two independent competing risks. 
Then, \begin{equation}
{n \choose 2}U_3  =  \sum_{i=1}^n  (R_i -1) \delta_i  =  \sum_{i=1}^n (i-1) W_i.
\end{equation}
Under $H_0$, the moment generating function is given by 
$$ M(t) = \prod_{j=1}^n [ (1 - \phi) + \phi exp\{t(j-1)\}].$$
When $\phi=1/2,$ $M(t)$ is same as that of Wilcoxon signed rank statistics with $n$ replaced by $(n+1)$.
\bt
As $n$ tends to $\infty,$ under $H_0,$ $n^{1/2}\{U_3-E(U_3)\}$ converges in distribution to 
$N(0, \sigma_3^2),$ where ${\sigma_3^2} = (4/3) {\phi}(1  -  {\phi}).$

\et

A consistent estimator of variance is 
$$\hat{\sigma_3^2} = (4/3)\hat{\phi_0} (1 - \hat{\phi_0}).$$
We reject the null hypothesis for large values of $Z=n^{1/2} (U_3-\hat{\phi})/\hat{\sigma_2}.$
\end{slide}
\begin{slide}

{\bf Test based on $\phi_0^*(t)$}

$H_2$ is also equivalent to 
$$ H_2: \Phi_0^*(t) \ge \phi_0 \; \mbox{for all} \; t 
$$
with strict inequality for some $t$.

Exactly  as in the earlier section, we have 
\bt
As $n$ tends to $\infty,$ $n^{1/2}\{U_3^*-E(U_3^*)\}$ converges in distribution to 
$N(0, {\sigma_3^*}^2),$ where 
\begin{equation}
{n \choose 2}U_3^*  =  n(n-1)/2 - \sum_{i=1}^n  (n-i) W_i
\end{equation}
and ${\sigma_3*}^2 = (4/3) {\phi_0}(1  -  {\phi_0}).$
\et
A consistent estimator of variance is
$$\hat{\sigma_3^2} = (4/3)\hat{\phi} (1 - \hat{\phi}).$$

 We reject the null hypothesis for large values of $Z=n^{1/2} (U_3^*-\hat{\phi_0})/\hat{\sigma_3^*}.$ 

Note that
$$U_k = U_3 + U_3^* - 1.$$
\end{slide}
\begin{slide}

{\bf Testing independence against $RTI(\delta \mid T)$}

Here, we consider testing $H_0$ againstthe alternative that $\delta$ is RTI in $T$/. This is equivalent to 
$$H_3: \Phi_1(t) \uparrow t, t > 0.$$

\noindent {\bf A. Test I - $U_1$}

$\Phi_1(t) \uparrow t$ is equivalent to 

$\Phi_1(t_1) \le \Phi_1(t_2),$ 
whenever $t_1 \le t_2.$ 

This gives

 $\delta(t_1, t_2) = S_1(t_2) S(t_1) - S_1(t_1)S(t_2) \ge 0, 
t_1 \le t_2$
 with strict inequality for some $(t_1, t_2).$ 

Define
\begin{eqnarray}
\Delta_1(S_1, S) & = & {\int \int}_{t_1 \le t_2} \delta(t_1, t_2) 
dF_1(t_1) dF_1(t_2)\nonumber \\
& = & \int_0^{\infty} [S_1^{2}(t)-\phi^2/2]S(t) dF_1(t). \nonumber
\end{eqnarray}
Under $H_0,$
 $$S_1(t)/S(t) = \phi,\Delta_1(S_1, S) =0.$$

Under $H_3,$
$$\Delta_1(S_1, S) \ge 0.$$

Define the kernel $\psi^*_1((T_i, \delta_i, T_j, \delta_j, T_k, \delta_k, T_l, \delta_l)= 
\psi^*_1 $ (say)
\bea
\psi^*_1 & = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_k > T_j > T_l > T_i, \nonumber \\  
 & \delta_i = \delta_j = \delta_k = 1, \delta_l = 0 \nonumber  \\
-1  & \mbox{if}\;  T_l > T_j > T_k > T_i, \nonumber \\
 &  \delta_i = \delta_j = \delta_k = 1, \delta_l = 0 \nonumber \\
0 & \mbox {otherwise}. \end{array} \right.
\eea
Then the U-statistic corresponding to $\Delta_1(S_1, S)$ is $U_1$
 $$ \frac{1}{{n \choose 4}}\sum _{1 \le i_1 < i_2 < i_3 < i_4 \le n}
\psi_1(T_{i_1}, \delta_{i_1}, T_{i_2}, \delta_{i_2}, T_{i_3}, \delta_{i_3}, T_{i_4}, \delta_{i_4}), $$

where $\psi_1$ is the symmetric version corresponding to $\psi_1^*$.

 $E(U_1) = 24 \Delta_1(S_1, S).$ 

Under $H_0, E(U_1) = 0$ 

Under $H_3,$ $E(U_1) \ge 0.$
\end{slide}
\begin{slide}
 Let $T's$ corresponding to $1's$ be called $X's$  [$n_1$] 

those 
corresponding to $0's$ be called $Y's$ [$n_2$]. 

Let $R_{(i)} (S_{(j)})$ be the rank of $X_{(i)} (Y_{(j)})$ be the $ith (jth) $ 
ordered statistic 
in the $X (Y)$ sample in the combined arrangement of $n_1 X's$ and $ n_2 Y's$ (in fact $nT's$). 

Hence
$$ {n \choose 4} U_1 = \sum_{j=1}^{n_2} (S_{(j)}-j) {n_1 + j - S_{(j)} \choose 2}
- \sum_{j=1}^{n_2} { S_{(j)}- j \choose 3}.$$

In terms of $X's$  and 
$Y's$ the above statistic is the same as that proposed by Kochar (1979) for testing equality of failure 
rates, the only difference being that the number of $X's $ and $Y's$ is random.
\bt
As $n$ tends to $\infty,$ under $H_0,$ $n^{1/2}\{U_1-E(U_1)\}$ converges in distribution to 
$N(0, \sigma_1^2),$ where 
${\sigma_1^2} = (96/35){\phi^5}(1 - {\phi}).$
\et
The null hypothesis is rejected for large values of $n^{1/2}U_1/\hat{\sigma_1}$ where 
$\hat{\sigma_1^2} = (96/35)\hat{\phi^5}(1 - \hat{\phi})$.
\end{slide}
\begin{slide}
\noindent {\bf B. Test II - $U_2$}



As mentioned earlier, $H_3$ is equivalent to $S_i(t)$ being $TP_2.$

 Under $TP_2$,
$$S_1(t_2)S_0(t_1)-S_1(t_1)S_0(t_2) > 0, t_1 < t_2.$$
Consider
$$\Delta_2(S_1, S) = \int_{t_1 < t_2} [S_1(t_2) S_0(t_1) - S_1(t_1)S_0(t_2)] $$
$$d[F_1(t_1)F_0(t_2)+F_1(t_2)F_0(t_1)].$$

Under $H_0 \;\;\Delta_2(S_1, S) =0$
Under $H_3 \;\; \Delta_2(S_1, S) \ge 0.$

 Define the kernel $\psi^*_2$ as
\bea
& = & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_k > T_j > T_l > T_i,\; \delta_i = \delta_k = 1, \delta_j = \delta_l = 0 \nonumber  \\
  &  \mbox{if} \;  T_k > T_i > T_l > T_j,\; \delta_i = \delta_k = 1, \delta_j = \delta_l = 0 \nonumber  \\
-1  & \mbox{if}\;  T_l > T_j > T_k > T_i,\; \delta_i = \delta_k = 1, \delta_j = \delta_l = 0 \nonumber \\
   & \mbox{or} \; T_l > T_i > T_k > T_j,\; \delta_i = \delta_k = 1, \delta_j = \delta_l = 0 \nonumber \\
0 & \mbox {otherwise}. \nonumber \end{array} \right.
\eea
\end{slide}
\begin{slide}
The corresponding U-statistic $U_2$ is given by 
$$  \frac{1}{{n \choose 4}}\sum _{1 \le i_1 < i_2 < i_3 < i_4 \le n}\psi_2(T_{i_1}, 
\delta_{i_1}, T_{i_2}, \delta_{i_2}, T_{i_3}, \delta_{i_3}, T_{i_4}, \delta_{i_4}), $$

 $ \psi_2$ is the symmetric version of $ \psi^*_2$.

\bea
E(U_2) & = & 24\Delta_2(S_1, S) \nonumber\\
& = & 24[\phi^2 (1-\phi)^2/4  \nonumber \\
&&     -\phi (1-\phi) \int^\infty_0 S_0(t) dF_1(t)\nonumber \\
&&+ \int_0^\infty S_1(t) S_0^2(t) 
dF_1(t)].\nonumber
\eea
\bea
{n \choose 4} U_2 & = & \sum _{i=1}^{n_1} (n_1 -i) {R_{(i)} -i \choose 2}  \nonumber \\
& - & \sum _{i=1}^{n_1} (n_1 -i) (R_{(i)} -i) (n_2 - R_{(i)} + i)  \nonumber \\
& + & \sum_{j=1}^{n_2} (S_{(j)} -j ) ( n_1 -S_{(j)} + j)(j-1) \nonumber \\
& - & \sum_{j=1}^{n_2} (n_2 - j) {(S_{(j)} - j \choose 2}.\nonumber
\eea
\end{slide}
\begin{slide}
In terms of $X's $ and $Y's$, the above statistic is the same as another one proposed by 
Kochar (1979) to test for equality of failure rates with $n_1$ and $n_2$ fixed.
\bt
As $n$ tends to $\infty,$ under $H_0,$ $n^{1/2}\{U_2-E(U_2)\}$ converges in distribution to $N(0, \sigma_2^2),$ where 
${\sigma_2^2} = (384/35) {\phi^3} ( 1 - {\phi})^3.$
\et
We reject the null hypothesis for large value of $n^{1/2}U_2/\hat{\sigma_2}$ where 
$\hat{\sigma_2^2} = (384/35)\hat{\phi^3} ( 1 - \hat{\phi})^3.$

Tests proposed in this section will help in discriminating between the  constant or proportional 
warning-constant inspection 
 and random sign censoring  models and also to 
determine whether the corresponding mode of failure becomes more likely with increasing age.

\end{slide}
\begin{slide}

{\bf Testing independence against $LTD(\delta \mid T)$}

Here, we consider testing $H_0$ against the alternative $H_4$ that $\delta$ is LTD in $T$ where $H_4$ can equivalently be stated as 
$$H_4: \Phi_0^*(t) \downarrow t, t > 0.$$

\noindent {\bf A. Test I - $U_1^*$}

$\Phi_0^*(t) \downarrow t$ is equivalent to 

$\Phi_0^*(t_1) \ge \Phi_0^*(t_2),$ whenever $t_1 \le t_2.$ 

$\delta(t_1, t_2) = F_0(t_1) F(t_2) - F_0(t_2)F(t_1) \ge 0, $

$t_1 \le t_2$
with strict inequality 
for some $(t_1, t_2).$

 Define
\begin{eqnarray}
\Delta_1(F_0, F) & = & {\int \int}_{t_1 \le t_2} \delta(t_1, t_2) dF_0(t_1) dF_0(t_2)\nonumber \\
& = & \int_0^{\infty} [F_0^2(t)-\phi_0^2/2]F(t) dF_0(t). \nonumber
\end{eqnarray}
\end{slide}
\begin{slide}

Under $H_0,$

 $F_0(t)/F(t) = \phi_0,\;\;\Delta_1(F_0, F) =0.$  

Under $H_4,$ $\Delta_1(F_0, F) \ge 0.$

Define the kernel $\psi^*_1$
\bea
&  & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_k < T_j < T_l < T_i,\; \delta_i = \delta_j = \delta_k = 0, \delta_l = 1 \nonumber  \\
-1  & \mbox{if}\;  T_l < T_j < T_k < T_i,\; \delta_i = \delta_j = \delta_k = 0, \delta_l = 1 \nonumber \\
0 & \mbox {otherwise}. \end{array} \right.
\eea

Then $U_1^*$, the corresponding U-statistic is  
$$  \frac{1}{{n \choose 4}}\sum _{1 \le i_1 < i_2 < i_3 < i_4 \le n}\psi_1(T_{i_1}, 
\delta_{i_1}, T_{i_2}, \delta_{i_2}, T_{i_3}, \delta_{i_3}, T_{i_4}, \delta_{i_4}),$$

where $\psi_1$ is the symmetric version corresponding to $\psi_1^*$.
\end{slide}
\begin{slide}
 $E(U_1^*) = 24 \Delta_1(F_0, F).$

 Under $H_0, \;\;E(U_1^*) = 0$ 

Under $H_4,$ $E(U_1^*) \ge 0.$

A rank representation of $U_1^*$ is
$$ {n \choose 4} U_1^* = \sum_{j=1}^{n_1} {R_{(j)}-j \choose 2} (n_2 + j - R_{(j)}) $$
$$- \sum_{j=1}^{n_1} { n_2-R_{(j)}+ j \choose 3}.$$

\bt
As $n$ tends to $\infty,$ under $H_0,$ $n^{1/2}\{U_1^*-E(U_1^*)\}$ converges in distribution to $N(0, {\sigma_1^*}^2),$ 
where 
${{\sigma_1^*}^2} = (96/35) {\phi_0^5}(1 - {\phi_0}) =(96/35){\phi}(1 - {\phi})^5.$
\et

We reject the null hypothesis for large values of $n^{1/2}U_1^*/\hat{\sigma_1^*},$ where 
$\hat{{\sigma_1^*}^2} = (96/35)\hat{\phi_0^5}(1 - \hat{\phi_0}) =(96/35)\hat{\phi}(1 - \hat{\phi})^5.$

\end{slide}
\begin{slide}

\noindent {\bf B. Test II - $U_2^*$}

Test for  $H_0$ against $H_4$ using the $TP_2$ 
property of the subdistribution functions of $(T, \delta).$

 $H_4$ is equivalent to $F_i(t)$ being $TP_2.$ Under $TP_2$,
 $$F_1(t_2)F_0(t_1)-F_1(t_1)F_0(t_2) > 0$$
 for $t_1 < t_2.$
 Consider
$$\Delta_2(F_0, F) = \int_{t_1 < t_2} [F_1(t_2) F_0(t_1) F_1(t_1)F_0(t_2)]
$$
$$[dF_1(t_1)dF_0(t_2)+dF_1(t_2)dF_0(t_1)].$$

Under $H_0\;\; \Delta_2(F_0, F) =0,$

Under $H_4,\;\; \Delta_2(F_0, F) \ge 0.$

Define the kernel $\psi^*_2$
\bea
& & \left \{ \begin{array}{ll}
1 &  \mbox{if} \;  T_k < T_j < T_l < T_i,\; \delta_i = \delta_k = 0, \delta_j = \delta_l = 1 \nonumber  \\
  &  \mbox{if} \;  T_k < T_i < T_l < T_j,\; \delta_i = \delta_k = 0, \delta_j = \delta_l = 1 \nonumber  \\
-1  & \mbox{if}\;  T_l < T_j < T_k < T_i,\; \delta_i = \delta_k = 0, \delta_j = \delta_l = 1 \nonumber \\
   & \mbox{or} \; T_l < T_i < T_k < T_j,\; \delta_i = \delta_k = 0, \delta_j = \delta_l = 1 \nonumber \\
0 & \mbox {otherwise}. \end{array} \right.
\eea
\end{slide}
\begin{slide}
The corresponding  U-statistic $U_2^*$ is given by 
$$  \frac{1}{{n \choose 4}}\sum _{1 \le i_1 < i_2 < i_3 < i_4 \le n}\psi_2(T_{i_1}, 
\delta_{i_1}, T_{i_2}, \delta_{i_2}, T_{i_3}, \delta_{i_3}, T_{i_4}, \delta_{i_4}),$$

 $ \psi_2$ is the symmetric version of $ \psi^*_2$.
\bea
E(U_2^*) & = & \Delta_2(F_0, F) \nonumber \\
& = & \phi_0^2 (1-\phi_0)^2/4 - \phi_0 (1-\phi_0) \int^\infty_0 F_1(t) dF_0(t) \nonumber \\
& & + \int_0^\infty F_0(t) F_1^2(t) dF_0(t).\nonumber
\eea
\bea
{n \choose 4} U_2^* & = & \sum _{i=1}^{n_1} (n_1 -i) {R_{(i)} -i \choose 2} \nonumber \\
& + & \sum _{i=1}^{n_1} (n_1 -i) (R_{(i)} -i) (n_2 - R_{(i)} + i) \nonumber \\
& - & \sum_{j=1}^{n_2} (S_{(j)} -j ) ( n_1 -S_{(j)} + j)(j-1) \nonumber \\
& - & \sum_{j=1}^{n_2} (n_2 - j) {(S_{(j)} - j \choose 2}.\nonumber
\eea
\end{slide}
\begin{slide}
\bt
As $n$ tends to $\infty,$ under $H_0,$ $n^{1/2}\{U_2^*-E(U_2^*)\}$ converges in distribution to $N(0, {\sigma_2^*}^2),$  
where 
${{\sigma_2^*}^2} = (384/35) {\phi_0^3} ( 1 - {\phi_0})^3 = 
 (384/35){\phi^3} ( 1 - {\phi})^3.$
\et

We reject the null hypothesis for large values of $n^{1/2}U_2^*/\hat{\sigma_2^*},$ where 
$\hat{{\sigma_2^*}^2} = (384/35)\hat{\phi_0^3} ( 1 - \hat{\phi_0})^3 = 
 (384/35)\hat{\phi^3} ( 1 - \hat{\phi})^3.$ 

\end{slide}
\begin{slide}

{\bf Asymptotic relative efficiency}

To compare alternative tests proposed in 
this paper 
for testing $H_0$ against $H_2,$ $H_0$ against $H_3$ and $H_0$ against $H_4,$ we compute 
asymptotic relative efficiency of the tests within a semiparametric family of distributions proposed 
in Deshpande (1990). The semiparametric family considered here is 

$F_1(t)  =  p F^ a(t),$

$F_0(t)  =  F(t) - p F^ a(t),$
where $1 \le a \le 2,$ $0 \le p \le 0.5$ and $F(t)$ is a proper distribution function.

$\phi = p$ 

$$\Phi_1(t) = \frac{p(1-F^a(t))}{1-F(t)}$$
which is an increasing function of $t.$ 

$$\Phi_0^*(t)=1-p F^{a-1}(t)$$
which is a decreasing function of $t.$ 
\end{slide}
\begin{slide}
$H_0$ corresponds 
to $a=1,$ and other alternative hypotheses 
correspond to $1 < a \le 2.$

 By the limiting theorem of U-statistics, all the U-statistics proposed here 
have asymptotic normal distribution under both null and the alternative hypothesis. 

The asymptotic relative efficiency of test $U_1$ with respect to test 
$U_2$ is then defined as 

$eff(U_1, U_2) = e(U_2)/e(U_1)$

 $e(U) = \mu'^2(1)/var(U \mid H_0)$

 $\mu'(1)$ is the derivative of expected value of $U$ with respect to $a$ evaluated at $a=1,$ and $var(U \mid H_0)$ is the asymptotic 
variance of $n^{1/2}U$ under $H_0.$

 Tests $U_1$ and $U_2$ are equally efficient and the same is true for 
tests $U_1^*$ and $U_2^*.$ Tests $U_3$ and $U_3^*$ are equally efficient but the general test $U_k$ is 
four times more efficient compared to these tests. This indicates the superiority of $U_k$ as it is 
consistent for the alternative $H_2.$

For this particular family of distributions, the other alternative tests are equally efficient. But this need not be 
true in general.

\end{slide}
\begin{slide}

\NI{\bf Illustrations}\\

In case of Nair's data, the value of $Z$ is $2.70$ and hence the hypothesis of 
independence of the failure time and the type of failure is rejected. The 
nonlinearity of the plot in Figure 1 supports this conclusion.

In case of Hoel's data, the value of $Z$ is $1.86$ and the Figure 2 shows a flat 
curve. Hence, the  time of death and the cause of death are independent.

The following tables gives the value of the test statistics and the conclusions reached for the two data sets. 
\end{slide}
\begin{slide}
\begin{center}
Table 1: Values of the test statistics for Nair's data (1993)
\begin{tabular}{|l|l|l|l|l|}
\hline
U-statistics & Expectation & Variance & $Z$ & Conclusion \\
\hline
$U_k=0.26$ & 0 & 0.33 & 2.70 & Reject $H_0$ \\
$U_1=0.04$ & 0 & 0.03 &  1.45 & Accept $H_0$ \\
$U_2=0.15$ & 0 & 0.17 & 2.26 & Reject $H_0$ \\
$U_1^*=0.06$ & 0 & 0.06 & 2.29 & Reject $H_0$ \\
$U_2^*=0.15$ & 0 & 0.17 & 2.18 &  Reject $H_0$ \\
$U_3=0.59$ & 0.46 & 0.33 & 1.35 & Accept $H_0$ \\
$U_3^*=0.67$ & 0.54 & 0.33 & 1.35 & Accept $H_0$ \\
\hline
\end{tabular}
\end{center}

\vsp
\vsp

\begin{center}
Table 2: Values of the test statistics for Hoel's data (1972)
\begin{tabular}{|l|l|l|l|l|}
\hline
U-statistics & Expectation & Variance & $Z$ & Conclusion \\
\hline
$U_k=0.11$ & 0 & 0.32 & 1.86 & Accept $H_0$ \\
$U_1=0.04$ & 0 & 0.09 &  1.50 & Accept $H_0$ \\
$U_2=0.06$ & 0 & 0.15 & 1.63 & Accept $H_0$ \\
$U_1^*=0.01$ & 0 & 0.02 & 1.14 & Accept $H_0$\\
$U_2^*=0.05$ & 0 & 0.15 & 1.38 &  Accept $H_0$ \\
$U_3=0.66$ & 0.61 & 0.32 & 0.93 & Accept $H_0$ \\
$U_3^*=0.45$ & 0.39 & 0.32 & 0.53 & Accept $H_0$ \\
\hline
\end{tabular}
\end{center}

\end{slide}
\begin{slide}

\noindent {\large \bf {References}}



Aras, G. and Deshpande, J. V. (1992). Statistical analysis of dependent competing 
risks. {\it Statistics and Decisions} {\bf 10}, 323-336.

Bagai, I., Deshpande, J. V. and Kochar, S. C. (1989). Distribution-free tests for the 
stochastic ordering alternatives under the competing risks model. {\it Biometrika} {\bf 76}, 75-81. 

Block, H. W. and Basu, A. P. (1974). A continuous bivariate exponential distribution. 
{\it J. Amer. Stat. Assoc.} {\bf 69}, 1031-1037. 

Cooke, R. M., Bedford, T., Meilijson, I. and Meester, L. (1993). Design of 
reliability data bases for aerospace applications. {\it Reports of the faculty of Technical 
Mathematics and Informatics no. 93-110, Delft}.

Deshpande, J. V. (1990). A test for bivariate symmetry of dependent competing risks. 
{\it Biometrical Journal} {\bf 32}, 736-746.

\end{slide}
\begin{slide}


Dykstra, R., Kochar, S. and Robertson, T. (1996). Testing whether one risk progresses faster than 
the other in a competing risks problem. {\it Statistics and Decisions} {\bf 14}, 209-222.

Hoel, D. G. (1972). A representation of mortality data by competing risks. {\it Biometrics} 
{\bf 28}, 475-488.

Kochar, S. C. (1979). Distribution-free comparison of
two probability distributions with reference to 
their hazard rates. {\it Biometrika} {\bf 66}, 437-442.

Nair, V. N. (1993). Bounds for reliability estimation under dependent censoring. 
{\it International Stat. Review} {\bf 61}, 169-182.

Prentice, R. L., Kalbfleisch, J. D., Peterson, A. V., Fluornoy, N., Farewell, V. S. 
and Breslow, N. E. (1978). The analysis of failure time in the presence of competing risks. 
{\it Biometrics} {\bf 34}, 541-554.

Sengupta, D. and Deshpande, J. V. (1994). Some results on the relative ageing of two life distributions. 
{\it J. Appl. Prob.} {\bf 31}, 991-1003.


\end{slide}





\end{document}
