—
2:

 

"Hm

I.
CHIGAN STATE UNIVERSITY UBRARlES

J llllllll\“llllllllillMW \\ “Hill

3 1293 00895 2594

 

 

 

This is to certify that the
dissertation entitled

Weak Convergence of
Kaplan-Meier Process in L -Space

presented by

Mangalam Vasudaven

has been accepted towards fulfillment
of the requirements for

Ph.D. degree in StatlStiCS

I"

'7 C
Major professor

[hue Auqust 9, 1990

MSU is an Affirmative Action/Equal Opportunity Institution 012771

 

————— .. _ _ ____

 

LIBRARY
Michigan State
University

'\ _ r

 

 

PLACE IN RETURN BOX to removeth checkout from your record.
TO AVOID FINES return on or before date due.

DATE DUE DATE DUE DATE DUE

__J
[jgm
#7

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

l—T—z

MSU Is An Affirmative Action/Equal Opportunity Institution
cmmmapn

 

 

 

 

WEAK CONVERGENCE or STANDARDIZED KAPLAN-MEIER
PROCESS m I.2 SPACE

By

Mangalam Vasndaven

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY
Dewrtment of Statistics and Probability
1990

‘3

/

/<,{ a)" (ILL.

ABSTRACT

WEAK CONVERGENCE OF STANDARDIZED KAPLAN—MEIER
PROCESS IN I.2 SPACE

By

Mangalam Vasudaven

Let )(l,...,Xn be i.i.d. with a distribution function F and Yl,...,Yn be
i.i.d. with an unknown censoring distribution function G with Xi’s and
Yi’s independent. In the random censorship model one observes

Zi = Xi A Yi and 6i = 1(Xi g Yi), 1 g i g 11. Such models arise in
clinical trials and survival analysis.

This thesis discusses the weak convergence of some standardized and
rescaled versions Of the Kaplan — Meier Process to a Gaussian process in
L2(A, p) where A is the support set of F and u is a a—finite measure
defined on A. Applications of this result to the goodness-Of-fit test
pertaining to F are also discussed. In addition, certain tests Of
Ho: F = F0, F0 a known distribution function, based on L2- norms Of a
suitably scaled Kaplan - Meier Process with respect to certain random

measures, are shown to be asymptotically distribution free.

TO my parents and my wife Sushama

iii

ACKNOWLEDGEMENTS

I would like to express my appreciation to my thesis adviser Professor
Hira Lal Koul for his support and encouragement during the preparation of
this dissertation, especially for the patience with which he explained statistical
concepts to me. I would also like to thank Professors Roy Erickson, Habib
Salehi and Joseph Gardiner for serving on my guidance committee. My
special thanks go to Professor Erickson for numerous mathematical discussions
which enhanced my insight into Probability Theory.

Thanks are also due to Loretta Ferguson for her technical assistance in
typing this manuscript.

Finally I express my deep appreciation to my family; to my parents
whose help and encouragement I could not have done without, and to my wife
Sushama whose total support of me in this project has been constant even in

the most trying circumstances.

iv

TABLE OF CONTENTS

Links Page
0. Introduction ................... 1
1. Notation, Assumptions and Preliminaries . . . 5
2. Convergence of WI. in L2[0, r], r < 7H . . . . 12
3. Uniform Boundedness of Rn/(F/C) ....... 19
4. Convergence of Wn in L2[0, TH] ........ 27
5. More General Cases ................ 33
6. Asymptotically Distribution—free Tests . . . . 40
References ...................... 56

CHAPTER 0
INTRODUCTION

A common feature of many survival studies is that the time of
occurrence of the event of interest, called a death, may be prevented for an
item of the sample by previous occurrence of some other event, called a loss.
In other words, a censorship may be present due to which the units under
study may not be completely observable. Typically this is the case in a
clinical trial where patients under treatment cannot be followed up due to
withdrawals from study. For example, in medical follow-up studies to
determine the distribution Of the survival times after an Operation, contact
with some of the patients may be lost before their death and others may die
from causes it is desired to exclude from consideration. Similarly, observation
of the life of a vacuum tube may be ended by breakage Of the tube or a need
to use the test facilities for some other purposes. In both examples,
incomplete Observation may also result from a need to get a report out within
a reasonable time. The losses may be either accidental or by design, the
latter resulting from a decision to terminate certain Observations. There are
various types of censorships. For an excellent overview of different types Of
censorship, see Chapter 3 of Gill (1980).

In the random right censorship model (random censorship model, for
short) considered here, for each item, the only data available are the minimum
of the survival time and the time Of loss, and whether or not censoring is
present. Kaplan and Meier (1958) suggested the Product-Limit Estimator
(PLE) Fn of the true survival distribution of the lifetime when there is
random censorship in the data. While the usual empirical distribution function

(EDF) assigns mass l/n to each of the Observations, the PLE redistributes

2

the mass Of censored Observations equally among the Observations to the right,
giving zero mass to all the censored Observations. More precisely, first, one
arranges the Observations in the ascending order and assigns mass l/n to
each Observation, whether or not uncensored. Next, starting from the smallest
Observation, one locates the first censored observation, and redistributes its
mass equally among the observations to its right. Then one moves on to the
next censored Observation and redistributes its new mass equally among the
Observations to its right. This process is continued for all the censored
Observations. Note that, if the largest observation is censored, the total mass
will be less than 1, thereby making the PLE a defective distribution.

This thesis studies the standardized error Wn Of a version of the PLE
as a process with paths in the space of square—integrable functions. It is
shown that this process converges weakly to a Gaussian Process in the space
mentioned above, where the integration is with respect to a measure belonging
to a large class Of a—finite measures. Also proved are the weak convergence
of some rescaled versions of the error process in the aforementioned space.
These results are useful in developing goodness-Of-fit tests pertaining to the
survival distribution F and/or the censoring distribution G. It is
conceivable that they may also be useful in minimum distance estimation
problems.

In the uncensored case Anderson and Darling (1952) considered a
Cramér-von Mises type statistic for the goodness-Of—fit problem of testing
Ho: F = Fo vs 11,: F # F0, where F0 is a completely specified continuous
d.f.. The statistic was Obtained by squaring the standardized E.D.F. and
integrating it with respect to the a—finite measure dFo/(Fo(1—Fo)). The
asymptotic distribution of this statistic was shown to be that of the integral Of

the square of a rescaled Brownian Bridge, where the scaling function is the

3

standard deviation function of the Brownian Bridge so as to make the
standard deviation 1 through out. Koziol and Green (1976) considered the
above—mentioned testing problem for the censored data and discussed the much
simpler situation when the hazard functions of the survival and the censoring
distributions are proportional (herein referred to as prOportional hazards model)
and the integrating measure is the finite measure dFo. Even in this case it is
not clear how they prove the tightness.

In this dissertation, the result Of Anderson and Darling (1952) is
extended to random censorship model, where there is an additional unknown
function G. Even if the integrating measure depends on F, the limiting
distribution of corresponding goodness-of—fit test statistic may not be
distribution—free in the case of randomly censored data. However, if one
chooses the measure suitably, perhaps depending on the data and varying with
n, one can get asymptotically distribution—free (A.D.F.) tests. Two
Anderson—Darling type statistics are considered and it is proved that these
statistics are A.D.F. and hence can be used for tests regarding F. The
limiting null distribution in both cases is same as that of the
Anderson-Darling statistic Of the uncensored case.

Now we will state the problem more precisely. Let )(,,...,Xn be i.i.d.
random variables (r.v.’s) with distribution function (d.f.) F on [0, an), and
Y1,...,Yn be i.i.d. r.v.’s independent of Xi’s with unknown and possibly
defective d.f. G (that is, G may assign positive mass to m) on [0, m].
We observe the pairs {(Xi A Y1), I(Xi g Yi)}, 1 _<_ i g n, where a A b
denotes min(a,b) and I(A) denotes the indicator function Of the set A.
If we Observe Y1: that is, if I(Xi 5 Y1) = 0, we say the observation is
censored. Otherwise it is referred to as an uncensored observation. The

general problem is to make inferences about F.

4

Let TH denote the supremum Of the support of X1 A Y1 and Wn(t)

denote the scaled error J6 (Fn(t) — F(t)), t 5 TH. Gill (1980) proved that
W“ = W in D[0, T] for every 1' < 1' , where W is a Gaussian process
depending on F and G, and D[0, 1] denotes the set of all bounded
CADLAG functions on [0, r]. Yang (1988) extended this result to the entire
set [0, TH].

Let p be a a—finite measure on [0, TH]. It is shown in this thesis
that, under suitable conditions on F, G and p, Wn belongs to
L2([0, TH], p) and that wn converges weakly to w in L2([O, TH], u).
For infinite p, Wu may not belong to L2([0, TB], [1) which is the reason
for considering the modified version W“. This is discussed in more detail in
Remark 1.1. For the L2— weak convergence of W“, the key step is to show
that Rn, the mean of an estimate of the survival ratio (1-F)/(l—G) is
uniformly bounded on [0, TH].

The material is organized as follows. In Chapter 1, further notation and
assumptions are stated and some preliminary results are proved. In Chapter 2
it is proved that Wn converges weakly to W on L2[0, r] for every
7 < TH. Chapter 3 is devoted to proving that if the underlying model is
prOportional hazards model with hazard ratio less than 1, then Rn is
uniformly bounded on [0, 7H]. In Chapter 4, the weak convergence of Wu
to W in L2([0, TH], u) under the prOportional hazards model is proved.
In Chapter 5, it is proved that the above weak convergence holds under more
general assumptions on F and G. Some examples, where prOportional

hazards model does not hold, are given. Two A.D.F. tests are constructed in

Chapter 6 and the variance Of the limiting distribution is computed.

CHAPTER 1
NOTATION, ASSUMPTIONS AND PRELIMINARIES

First, we need to introduce some general notation.
(i) For any real—valued function f, f denotes l-f, ||f||b denotes

supb |f(t)|, f _:(t)= lim f(s) whenever the limit exists and
t

f(sr 3) .= f _( H.) Unless8 otherwise mentioned, {— 1(t) = 1/f(t) for any
real—valued function f. For any d.f. K, let Ts denote the supremum Of
the support of K. For any set A, I(A) denotes the indicator function of
A. For any interval 1, D(I) denotes the set of all bounded CADLAG
functions on I equipped with the supremum norm.

Next, notation and assumptions for some frequently used functions and
r.v.’s are introduced.
(ii) Let F be a continuous d. f. on [0, m), G be a possibly defective
censoring d.f. on [0, m] such that TF 5 7G; H = F G; )(,,...,)(n be i.i.d.
F and Y,,...,Yn be i.i.d. G such that Xi’s and Yi’s are independent;
Zi := Xi A Y1 and 61 := I(Xi 5 Yi). Let Z(i,’s denote the order
statistics of Zi’s, 6m’s denote the order statistics induced on 6i’s by

Z -,’s and TH denote Z(n,. Let

(1
cm == j; FTC: K(t) == cam/(Hum.

Next the PLE’s of F and G, two versions Of the standardized

 

Kaplan—Meier Processes (KMP’s) and some other related stochastic processes
are defined.
(iii) Define the PLE E, or F by

6

n
7 n-' a.
as) == H (at, m,
1"!
ZIjlst

and Fu by

Fn(t) := I(t < Tn) 1",,(t), t 3 TH.
The PLE Gn Of G is defined by

Gnu) := H (fig—f)" 5m, t 5 TE.
5'1

ZmSt
The standardized KMP’S are
Wilt) == t/i (inc) - F(t)), Wn(t) == t5 (Fn(t) - F(t)),

for t g r . We shall also need the following estimators Of C

H
t , t an
0.0) == 1 715-, (2.0) == J —.
o F CL 0 FnFn_Gn_

and the corresponding estimators Of K
we ;= C.(t)/(1+c.(t)) and Kim == Che/(Hoke) for t s TH.
Some rescaled KMP’s are defined by
as) == (mo/norms). as) == (Kan/Fauna),
:30) == (K;(t)/F.(t))-w.(t). ts TH-
Another sequence Of stochastic processes and their mean functions are
defined next.

(iv) Define an estimate of the survival ratio and its mean by

 

Q () FL“) I T R E

t := t < , nt := nt , t g .

11 Unit) ( - n) () (Q ( )) 73
Observe that Qn(t) is bounded for each n and each t 3 TH, so the

expectation is guaranteed to be finite.

Next, two key Gaussian processes are defined.

7

(v) Let B be a Brownian Motion on [0, w), B0 be a Brownian Bridge
on [0, l],

W(t) := F(t) B(C(t)), t < TH.

Finally, we state the condition on the integrating measure with respect

to which the L2 space is defined.
(vi) For any set S, any measure 7 on S and any subset A of S, let
L2(A, 7) be the set of all real-valued functions on A which are square
integrable with respect to 7. Let p be a a—finite measure on [0, TH]

such that

T
JHF2Cdfl<CIL
0

£31153; 1.1. Recall that, if the last observation is censored, that is, if

6“,, = 0, then for t z Tn, Fn(t) = Fem.) < 1 so that 13‘n is a
defective distribution. Consequently, Wu may not be in L2([0, TH], p),

for it will be a non-zero constant function of t on [Tm TH] and hence
non-integrable if p is an infinite measure. To overcome this difficulty, as is
customary in applications, we assign all the left-over mass to Zn“, the last
observation, whether or not it is censored, making F1n a regular distribution.
This makes Fn a 1 on [Tm TH] and enables us to prove that, under (vi),
Wn belongs to L2([0, TH], p) . Note that Fn(t) and Fn(t) differ only
for t 2 Tu. As we have the condition that 7F 5 TG’ Fn('rH) = F(rH) = l
and the change in the definition of PLE will not affect the uniform
convergence of Fn to F on [0, TH]. This becomes clearer after

Lemma 1.1. On the other hand Gn should be defined without the indicator

function Fn has, to assure us of the uniform convergence on [0, TH] when

7F < r0. Gn(rH) cannot converge to G(rH) if Gn('rH) is always 1

and G(TH) < 1.
We shall state and prove a few lemmas that are frequently used in this

dissertation.

E ILA .L

(i) ll Fn — F ":3 —+ o a.s..

(ii) II Gn - G ":3 _.P o.

PBQQF.
(i) From the Proposition of Wang (1987), since F(rH-) = l, we get,
(1.1) sup Irina) - F(t)| _. o a.s..
t<‘rH

Since F(Tn) —. 1 a.s., (3) of Wang (1987) gives Fn(Tn) -+ 1 a.s..
$0

(1.2) sup an(t) - Fn(t)| = II - Fn(Tn)| —o 0 a.s..
(TH
Now, (1.1) and (1.2) together with the fact that Fn(rH) = F(rn) = 1 imply

(i)-

(ii) Apply (2) of Wang (1987) to Cu; F and G are
interchangeable for the results in Wang (1987), so

sup | Gn(t) - G(t) | —o" 0.
t<‘rH

By (3) of Wang (1987) applied to Cu, Gn(Tn) —+p G(rH) and hence

Gn(‘rH) -»p G(TH). D
Biting 1.2. All the results in Wang (1987) go through even if G is
defective.

LEMMA 1.2. For every 1' < TH,
W11 = W on D[0,T].

1113991, By Theorem 1.1 of Gill (1983),
Wu = W on D[0,r].
llWa - Wu"; = M (F11 - fin)"; = Iv‘i (1 - f‘a(Ta))| I(Tn s t s 'r)
-o 0 as.
because Tn —o TH as. and r < TH. :1

Emmy; 1.3. It may be interesting to know the conditions under which
(1.3) Wn = W on D[0,'rH].
In Theorem 2.2 of Yang (1988), it is proved that Wn = W on D[O,1'H]

under the condition

(1.4) (TB EF- < m

o G_
. TH
SO it suffices to examine ll Wn — Wn . This quantity is equal to

 

 

,5 INT.)- But by Gill (1933; Remark 2.2), under (1.4), F(t) B(C(t)) _. o

as. as t -o TH. So the aforementioned weak convergence result of Yang

also tells us that fii [Pn(Tn) - F(Tn)] —+p 0, since Tn —o TH as. as

n -o m. Therefore we need to find when J17 F(Tn) —op 0. in Lemma 1.3,
the condition under which this happens is proven to be

(1.5) F(t)/G(t)) —9 0 as t —o TH.

So (1.4) and (1.5) together imply (1.3). Note that, though both (1.4) and
(1.5) say that F(t) should approach 1 as t —o r somewhat faster than

H
G(t) does, neither of these conditions implies the other.

10

LEMMA 1.3. A necessary and sufficient for Jii' F(Tn) _.p 0 is (1.5).

PM; fiF(Tn)—op0 t: P{JiF(Tn)> £}—' 0 V e>0
4: Hr, < r'l(1—{e/,/n))} —» o v e > 0
ea P{Tn g E‘1(1—[e/,/n])} _. o v e > o
where F.1(t) = inf{x: F(x) 2 t}. Fix 6 > o.
P{Tr t F‘lu—{e/mn = ants-Imam
= [1 - (e/m G(F'1(1-[e/fil))]“-
Now recall that for a sequence {0n} such that 0 5 on g l, [1 - 0n]11 -+ 0
if and only if n on -o a). Thus v‘i F(Tn) _.p 0 if and only if
,5 G(F'lu-{e/Jn‘n) _. a v e > o. If M is the smallest integer greater
than or equal to (1/e)2, then e/Jr'i 2 INME and hence
«5 G(F’ltHe/«n» 2 «5 G(F'IU-{l/t/MED)
= (WM) «Ma C(F"1(1—[1/t/MFD)-
Therefore, since F(F_l(x)) = x and TF = TH,
«a atria—WE) -» . v e > o e «5 G(F‘lu—n/m» -+ .
=0 C(P'1(x)/(1-x)) —. a as x .4 1
:9 C(t)/(1—F(t)) —0 m as t —+ TH.
This proves the desired result. a

LEMMA 1.4. Let A e (0, m] and let g: [0, A] o—-v [0, m] be such that
g(t)<an forall t<A and g(t)—00€[0,m]ast—+A. Letpbe
an infinite measure on [0, A] such that p[0, t] < or for all t < A. Then

I.
([40, t])-1 Jo g (1;: —o 0 as t —0 A.

PEEP. Case 1. 0 < or. Given 6 > 0, 3 M > 0 such that for all

11
t>MJfifi—H<e.%
t d M d t d
has -Joss+JMsn
implies
—1 M —1 t
(no. t1) )0 sh + (0-e){u[M. ti/uto. t1} 5 (no. t1) [0 8 dn

5 (no, t1)“ {:3 as + a + e

As t —o A, p[0, t] —o m and {p[M, t]/u[0, t]} —o 1 so by taking liminf
in the first inequality and limsup in the second inequality we get the result.

Case 2. 0 = m. Proof is similar. a

CHAPTER 2

CONVERGENCE OF W. IN L’ [0, r], r < TH

In this Chapter 7 will denote an arbitrary but fixed number less than
TH. All the conditions of Chapter 1 are assumed to hold, though all the
functions and measures that are originally defined on [0, TH] are now
restricted to [0, 7]. Theorem 2.1 states that Wn = W on L2([0, r], p)
where )1 is a a—finite measure satisfying 1(vi). TO prove this, we are going
to show that a bias—adjusted version of Wu satisfies all the conditions of
Lemma 9 of Koul (1984), which in turn is based on Theorem 2.2 of
Parthasarathy (1967). For that, we will state and prove a few lemmas.

For p = 1, 2, Lp(p) is short for Lp([0, r], )1). As L104) n L2(p)
is dense in L201), there exists a sequence {en: 11 E II} g L101) n L2(p)

such that {en: 11 6 It} is a complete orthonormal set for L201). Define

7 r
U,(t) := I(t 2 Tn) “:2; )0), x,,(t) .= Wn(t) — ,5 Un(t), t 3 TH.

 

LEMMA 2.1.

E(X.(t)) = 0. v<x.(t)) = PM)

where Rn is as in 1(iv).

thF

n

01"r3

 

,t<r

PRQQP. From the identity Fn(t) = Fun) + Ecru) I(t 2 Tu) we get
as) - F(t) - Una) = as) + 1(t 2 T.) its.) - F(t) - v.0)

= as) .. F(t) + 1(t a Tn) Rm) [1 -(F(t)/F<'r.))1

12

 

 

T.(T.)IT0) - F(T.))]

_ I(t 2 Tu) [ F(T)

+10 2 T.) TAT.) [1 — (T0)/T(T.))1
(from 3.2.15 of Gill (1980))

 

(2.1) = F(t) I

where 11,, = F C

n n, 1,, =1(E,,_ > 0), and
Mn(t) = i I(zj g t, 5,. = 1) — I; (11 amp) dF.
j 1

(Note that in Gill (1980), n Hn_ is denoted by Ya.) By (2.1) and an
argument similar to 3.2.20 of Gill (1980),

dF

 

(l-fin)3 Jn
E(Xn(t)) = 0: V(xn(‘)) = F20) I; E[ H J

 

F-—3', t S TH.
Moreover,
(l—T._0))“’ J.0) = (l-TL0))210 s T.)

= m.) 10 s T.)

= fires) 0.0)-
by 1(iv). Thus

2 t RndF
V(x,,(t)) = F (t) [0 F3 . o

LEMMA 2.2. As 11 -+ at,
V(x,,(t)) _. F2(t) C(t) v t 3 TH

and

14

JTV(Xn(t)) dp(t) _. (Tran) C(t) dn(t).
0 0

PgQQF. By Lemma 1.1, Fn(t) —o F(t) a.s. and Gn(t) -+p G(t) for each

t E [0, 1']. Therefore Qn(t) —op 1G1?) for each fixed t e [0, T]. We shall
I:

show that Qn(t) is uniformly integrable by showing E(Q:(t)) 5 C r for a
constant C 7 depending only on 7. As 11 Hn_(t) ~ Bin(n, H_(t)),

J. 0) 2

n Hn_(t)

= n2 ‘23“ C? [H.0)]’ [H_(t)]”"

is =1

 

E0230» 5 n2 s[

] n—i—l

_ r2 :0“) 2 C... [IL 0)]i+1 [H_ 0)
= n3 H _(t) E[1 + Bin(n—1,H_(t))]_3

Sn H_(t)[H_0)] (300-1)

3 4s Hj2(r), t e [0, 1']
because E[1 + Bin(n, p)]_r 5 r! (np)—' by Moment Lemma, Section 7 of
Koul, Susarla and Van Ryzin (1981).

 

Consequently,
= —+ m
Rn“) - E(9.0)) 3-0),
and
11.0) , _,
F3m [F (t)U__(t)] for all t e [0, 1].
Now,

11.0) = E(Q.0)) s [E(Q§0)1* s 7 [H_(rn'l to. an t e [0. 71.
Therefore by the Bounded Convergence Theorem and by 1(ii),

15

t RndF
o F3
Hence by Lemma 2.1,

V(x,,(t)) _. F2(t)C(t) for all te [0, 1'].

t
—. Jom’(r)c_(r)]'1 dF(s) a C(t).

Now,
v<x.0)) s 7 [H_(r)1’1F20) If)!” «m = 7 me)“ T0);
by 1(vi) and the fact that

T 1 T 1 TF2
Jo T0) «110) s ijo F(t)1’(t)du(t)Sflfljo 0) <20) <00)

F(t) is integrable on [0, r] with respect to )1. SO by the Dominated

 

 

Convergence Theorem,

J;V(xn(t)) dp(t) —. £1520) C(t) dp.(t). n

LEMMA 2.3. For every m E II,

Joxne,dn, Joxne,dh,...,Joxne.dp =t [Joli/6M» J0We2dp,...,J0We.dp].

 

 

Pmr. By Lemma 1.2, we know that Wn =) W on D[0, 7]. Next,
observe that on [Tn 5 t], F(t) 5 F(Tn) and hence Un(t) 5 I(Tn 5 t). So,
sup NE Un(t)| g (E sup |I(Tn g t)| 5 J6 I(Tn g r).

t51’ t51’

Since Tn —-t TH a.s. and r < TH, for ac. w, I(Tn(w) 5 t) = 0 for
n sufficiently large. Therefore,

sup Iv‘i Un(t)| —o 0 a.s..

t5r
Thus X1: = Wn — y‘i Un = W on D[0, 1].

Define F: D[0, r] r—o km by

I‘(x) = [Jox e, an , Jox e2 an, ..... , J01: em as].

16

Then I‘ is a continuous function since ei’s are in L1(p). Therefore,

r(x,,) =4 F(W) on Rm. 0

LEMMA 2.4. For all i E III,
2

r 2 r
aneidfl aEIWeidp.
0 0

m Fix iell. Let dtb = eidp. Then (i isafinitesigned

E

 

measure and we need to show that

r 2 r 2
J Xu as: _. E I w as .

0 0
By Theorem 3.4 of Gardiner, Susarla and Van Ryzin (1985) and Theorem 1 of
L0 and Singh (1986),

E

 

(2.2) m0) = n-* 3} tr0) + t/firr(t). t s r.
where {i’s are i.i.d. uniformly bounded r.v.’s with mean zero, (2.3)

sup E|rn(t)|p = 0(n'p)
t5r

and

(2.4) 22p |rn(t)| = 0(n'3/4 (log n)3/4) a.s..

From (2.2), (2.4) and Lemma 1.2,

Wn(t) t: n-* i1: 4,0) = w on D[0,1']

and hence
T~ T
(2.5) I w, as = I w an.
o 0
Now,
7'... *n 7' n
J wn d4) = n- )3 J g,(t) d¢(t) = n-* s B, (say)
0 l o 1

where Bi’s are i.i.d. bounded mean zero r.v’s. SO by the Central Limit

17

Theorem,

T
(2.6) (w, as =4 Normal (0, V(B,))

0
where V(X) indicates variance of X. By (2.5) and (2.6),
T T 2
Iwas] = E“ Wdtb] .
0 0

As W = Wn + (5 rn, by Jensen’s inequality and (2.3),

1", 2
(2.7) E JoWndt/J = v

 

 

 

 

 

JTWnd¢ = V(B,) = v
0

2 2
(2.8)E[J;Wndw - KWndul = n E[I;rnd¢] g k, n J;E(r§)d|¢| 5 k n"1

for some k1 and k. From (2.7) and (2.8) we get

(2.9) E[ Kw, d¢]2 —» El Kw (Mir.

Now we shall show that
2

-00

 

r

(2.10) E[ J (Wu - Xa) dd:
0

which will prove the lemma in view of (2.9). Note that W,I -- Xn = ,5 Un.

Now,

T 2 T T
r «a j Undtb s k n j 2411?.)de s k n J [H_(rnndlto)
0 0 0
= k n urn [H_(rn‘ - o

for some k by Jensen’s inequality since Un(t) 5 I(Tn 5 r) for t 5 r and
P(Tll 5 r) = [H_(r)]". Hence (2.10). a

 

MEN 2.1. Assume 1(i)—1(vi) hold. Then,
wn =4 w on L201).

M All the conditions of Lemma 9 of Koul (1984) are verified by
Lemmas 2.1 through 2.4; so it follows that Xn =9 W on L2(p). Now all

18
T 2
that needs verification is that J n Un du _.p 0.
0

We know that T,I —4 TH a.s. so for almost all u, I(Tn < r) is
equal to zero if n is sufficiently large. Therefore for almost all a), 3 N at
such that 4/5 Un(w) = 0 for all n 2 N (12' Therefore for almost all as, 3

7.
Nu such that JnUgw) dp = 0 forall nZNw. Thus
0

T 2
J'nUn du —o 0 a.s.. a
0

CHAPTER 3

R
UNIFORM BOUNDEDNESS OF —L
1"It?

In view of Theorem 2.1, to show the weak convergence of Wu in
L2[0, TH], we need to show that W1. is ‘tight at 711’ in the sense of
Theorem 4.2 of Billingsley (1968). A sufficient condition for this to hold is

that the sequence Rn 5 k FIG for some constant k. In this chapter it is
R
proved that I’D—G is uniformly bounded on [0, TH] under the assumption

that
(3.1) G = F“ for some a 6 [0, 1).
We begin with several Lemmas.

EMMA .1. Let X ~ F and Y ~ G be two independent r.v.’s. Let u

be (F+G)/2. Let f=gg and g=ca1%. Let Z denote XAY and
f(It) G.(X)

f(x)c.(x)+r(x)F(x)

Then ¢(Z) is a version of P(X 5 YIZ).

¢(X) =

 

£3991; As ¢(Z) is a<Z> measurable, all we need to check is

JAM) dP = (Arm 5 Y) dP
for all sets A of the form {Z 5 z}, where P is the probability measure
with respect to which the distributions of X and Y are F and G
respectively.
JAI(X5Y)dP = P(X g Y, z 3 z)

= I JdF(s) dG(t)
s5t,s5z

19

20
= (“3(a) dF(s).
(Am) dP = I J ¢(s)dF(s)dG(t) + I (4)0) dF(s) dG(t)

s<t,s5z s>t,t5z

= l 90) 6.0) 4T0) + J 90) T0) «160)
852 t5z

= Imus) [H_(srts) + T0)r0)1 Ms)
J" G_(s)f(s) (19(4)
852

J G_(s) dF(s).

852
This proves the result. a

LEMMA 3.2. Let {(Zi, 3i), 1 5 i 5 n} be i.i.d. two—dimensional random
vectors with the d.f. of 2, continuous. Let 2(1): 1 5 i 5 n be the order
statistics of Zi’s, 3(1) be the corresponding induced order statistics of Ei’s
and G, denote the conditional d.f. of 3 given 2 = 2. Given

{2,, 15i5n}, 3(1): 1 5 i 5 n, are conditionally independent with d.f.’s

Gztif
PMF. See Lemma 1 Of Bhattacharya (1974). n
LEMMA 3.3. Let Xi, Yr: 2,, 6i’s and 6(1) ’s be as in 1(ii). Assume (3.1)

holds. Then 5(1) ’s are i.i.d Bernoulli (p) and are independent of Z(i,’s,

where p = Tia.

P395211. From Lemmas 3.1 and 3.2, all we need to show is that the function

It of Lemma 3.1 is identically equal to p when G = F0.

First note that g := a F“-1 f is a possible version of 3%— because

21

far“ r or = Jta r‘” dF = 1 — r°(t) = G(t).
0 0

Thus

 

 

= = _ = =p 0
¢ fG-l-gF rr“+ar“‘tr 1+“

In what follows, D = {0, 1} and Dj = {0, 1}5, 1 g j g n. For
g 6 Dn, (1k is the kth entry of d and 51(0) and (1(1) respectively denote
the number of zeros and ones in d. By Bni(q), we mean P(X = i)

when X ~ Binomial (n, q) and p will always denote fil-"E'

LEMMA 3.4. Let Qn and RI. be as in 1(iv). Let ano = 1 and

i e e
a... = H [Mfi-h) + (rhea-’31)]
i '1

for 1 5 i 5 n—l. Then under (3.1),

Rn(t) = 2 Bni(H(t))ani°

i=0

P3552132 Note that
i . 2d.—1
- J
9.0) = H (.7311)
i=1
on {Zm < t 5 Zn”), 6(k) = dk’ 1 5 k 5 n — 1}. Hence,
Rn“) = E(Qn(t))
n-l i . 2d-1
_ n- l
- 2 2 H‘r-RT)
i=0 536an jsi

.P{Z(i) < t S Z(I*1)’ 6(j) = dj’ 1 S j S 11-1}

22

= Z A... (see)
i=0

By Emma 3.3 and by elementary prOperties of order statistics,
P{z(1)<t$ Ztim: 5m=dkt 19511-1} = P{Z(i) < t 5 211m}

.P{6(k) = dk’ 1S k S n-l}

( )
= B..(H0)) pe<1>(1_p)e 0

Therefore,

d(0) i n_. 211—1
A... = B..(H0)) 295'“ 1)0—9)" [1 (air)
is!

QEDn-t
Write Dn_1 as D, 1: Du-” so that d is written correspondingly as
(£11: 92) Then, d1(0) + d2(0) = (1(0) and 191(1) + 92(1) = $30)- Hence

=2 pd(1)( l-p)g(o ) H (n—EJ‘Ll-fdj—
j'

~.g€1)n_1
91(1) 91(0) ,_4 my—l 92(1) 94(0)
2 9 (HT) 110,311) 2 P (l-P)
g IEDI j a (1261);] -1-i
«1(1_1(1) )gom
= 2” UNIT-1+ +I)2dj-
giébi

because the last sum is 1. Now, by breaking up Di as Di,1 1: D and

proceeding as in the earlier step,

ani = 2 Pd(l)(1"P)~(0) I: (11—411)”).

dlEDi -
rl((1)l d(0) ' _. 2d-—1 d(1) d(0) _. 2d—1
=29‘0-91Hgfii—f)’ Zr 0-9) (river)
(11591-1 in den

 

d(1)( 4.0) ,_ 1...
=2 9 0-9)“ H(,—_,:l7)m_ 1,s,—rlr()+0 p)( +1)1

dfiDi-l

23

n—i+1)].

 

= ani- l[p(fi:l'—'f) + (l—PX
Since “110 = 1, iteration of the above relation gives

i . .
a... = Hug.) + (Inc—21%)] = .,,.
i *1

n-l
Thus A... = B..(110)) a... ... 11.0) = 211.4110» 1.. n

i=0

LEMMA 3.5. Let
- - 1
1(x)= log lag—I) + 0-p)(:_:+ )1. . . [0.14).
Then for M 5 n,

 

J:f(x) dx = (n+1?) log[(n-I-fl)2 + 7] — (n-M-I-fi) log[(n-M+fl)2 + 7]

+ (n—M) log(n-M) + (n—M+l) log(n—M+1) - n log n

m We follow the convention that x log x = 0 if x = 0.
p(n x--)2 + (1—p)(n-x+1):
f(x)= log[( n —x) 2(n—x-I-l)
=log [(n—x)2 + 2fi(n—x) + 13] — log(n—x) - log(n—x+l).

SO

I n 2 n
Jf(x)dx=Jlog[y +2fly+flldy-J
0 n-M

n-M
= (1) - (11) - (111) (8111')-

(I) = JnlgsKHfllz + 111-11)] 1y

n
losydy-Jlosydy
n-M+l

n+fl 2
= I log 2 +7) dz
n—M+

= W(n-l-fl) - W(n—M+fl)

24

where

\Ilz=zl 22 2 t-lz—-—z.

() 08(+'7)+{1/'7an(fi) }
This will give us
(I) = (n+5) log [(1441)2 + 71 -(1r1-M+fi)1<>s[(n-M+16)2 + 71
2 1'1 EL -t"1n-M+ -2M.
+ m an (J7 ) an (Fr—2)}
(II) + (III) is easily seen to be
11 log n — (n—M) log(n-M) + (n+1) log(n+1) — (n—M+1) log(n-M+l)

- 2M.
From these relations, the lemma follows. 0
LEMMA §.§. There is a constant K such that ani 5 K (n--i)2"'l n1-2p for

0 g i g n - l, where ani’s are as in Lemma 3.4.

PMF. Let f be as in Lemma 3.5. One can easily see that
sgn(f’(x)) = sgn(x - n + 1 )

/p7(I-p) -1

so that f, (x) = 0 for at most one x E [0, 11). Hence f has at most one

 

local minimum in [0, n). Thus it follows that

32 if’ ( i+1f dx
(.1 E(1)-jo (x) .

Note that LHS of (3.2) = log ani; therefore,
(3.3) an, s exp(jg+11(x) dx).
Now, tan-1(a) -tan—1(b) 5 J; for any non—negative a and b; so
- 1 —1 n-i— 1r/2.
(3.4) exp [ ZJEII—pfltan 13:1} - tan {——P—} ]5 e
VEI I‘D) Vi“ I‘D)

Note that

 

1 2 1- n 1-2 “
[@+-fign+~1‘)p( p1 =[1+TB++nn+U]

25

 

 

 

 

 

s [ 1 + 2532]
(3-5) S 620-9);
(3,) [on-pg? p(l-p)11"’ S 4,141»;
and since
M-p< M2 M—p
[(M-p)2+p(l-p)] (M-p)2]
= [1-(p/M)]2("’M).
we have
M2 M) —4
3?? (M—p)”+p(1—p)] “Hp/2))
Therefore,
Ln—i-l)““'1(n-i)“‘f =[ (n—i )2 ]”“‘”(n_—1;1)n.1-l(n_i)m
[(n-i-p)2+p(1-p)]“""’ [(n—i-p)2+p(1—p)] ““
(3.7) g 16 (n—i)2p-1.

From (3.4), (3.5), (3.6), (3.7) and Lemma 3.5 we get
i+l
exp(J f(x) dx) S 64e2'Hr/2 nl-zp (n—i)2p_l.
0

Therefore by (3.3),

a s K n1-2p(n—i)2p—l. u

ni

THEOREM 3.1. Assume 1(i) - 1(vi). Then under (3.1),

(3.8) -:—/&(t)<K t 5 TH, 11 e

where K as in Lemma 3.6.

111.923.: From the convexity of the map x .—. xllr for x e [0, m),

r E (0, 1], and Jensen’s inequality, it follows that for any non-negative r.v.

26

Y. we)?” 3 Earl/r). Applying this to r = 2,... and y = xzp—l

where X is Binomial (11, H0»: one gets

B(K’H) s [n Ha)

Now, by Lemmas 3.4 and 3.6,
n-l
11.0) = 2 3.411(0) a...
“3-.
s K 2 B..(H(t» (xx-02"" n”?
i=0

n
= K Baum» 12‘” n1...
is!

2p—1
] n1—2p

s K [n Hm

= K [MVP—1

Now, since p = TL, H = F G and (3.1) holds,

+0
(3.3).

] 2p—l

H2p_1 = F/C. Hence

CHAPTER 4
CONVERGENCE OF wn ON L2[0, TH]

This chapter proves the weak convergence of Wn to W in
L2([0, TH], 11) under (3.1).

LEMMA 4.1. Under the assumptions of Chapter 1,
2
W E L ([0, TH], 11) a.s..
Moreover, for every 11,

Wu 6 L2([0, TH], 11) a.s..

PgQQF. That W E L2([0, TH], 11) a.s. is obvious since by 1(vi) its second

moment is integrable with respect to p.

r
H W33w)dp < m for
0

almost all u. Let {rnz n E N} be an enumeration of rationals in [0, TH).

Let

Let no 6 N be fixed. We shall show that I

1'

Rm: {111an W12,O(w) (111 < m }.
0
By Theorem 3.1, Wu 6 L2([0, r], )1) almost surely for all 1' < TH, so
I

P(flm) = l for all n and hence 12,, := n (In, has probability 1. Since
11:

F is continuous,

fln2 := {M T110”) < TH}

has probability 1 for each n and hence so does 002 := q 9112- Let
n:

n, := 11,, n no, so that P(flo) = 1. Let (1) e no. Tn(w) < TH so 3
o

n(w) E N such that rum, 6 (Tngw), TH) and rum, > 5F, where IcF is

27

28

the infimum of the support of F. Since 11) e Slum, 1, the following hold

when evaluated at 1.2.

First of all,
r
J n W: (11‘ < m.
o 0
Also

1' r
JHW2 dp =nJHF2du

n In

11100.11" 1:“ 1'2 as

<0.

1.
Thus I H W§(w)dp < m for all m E 11,. a
o o

LEMMA 4.2. If (b, defined in Lemma 3.1, is bounded away from zero near

TH, then
'r
n J H F2dp -op 0.

Tn

T
Pgogr. Let Vn = n L“ F2 dp. Since Tn —. TH a.s., P(Tn < a) J o
n

for all a < TH. So it is enough to show that 3 a e (0, TH) such that

r
H 2
:= nJ F du —op 0.
max(Tn1 3)

I

V

Define

f(x) = sup n x", x E [0, 1].
hell

Simple calculus techniques show that
f(x) 3 —(log x)"1 g (l—x)-l.

29

Hence
(4.1) new)“ 5 (Hon-1, [0. r31
Note that
C(t) = I; t dH'l
and use Lemma 1.4 to get H C is bounded away from zero near 7-H. So
by the continuity of F 3 a e (0, TH) and k e (0, m) such that
11‘1 5 1r C

on [a, TB]. This along with (4.1) gives us

1112(1) (110))" s 1: 1%) G(t)
v n and v t e [a, r3]. Since for every t e [a, r31, 11 F20) (11(0)n
converges to 0 as n -o a) and is dominated by the integrable function

k F2(t) C(t), Dominated Convergence Theorem gives

11 [TH F2(t)(H(t))n dp(t) —. 0.

This quantity is precisely the expectation of VIZ. 1:1

LEMMA 4.3. If ¢ is bounded away from zero near TH, then

r
JHnUgdu -vp 0.

 

 

0
rH Uzd _ rH Ffi(rn)1«*2(t)d
2.9.931 J0 n n u — JTnn F2(Tn) p(t)
Fin.) TB 2
= F2(T..) Tn n F(t) du(t).

2
(
By Theorem 3.2.1 of Gill (1980), —‘2'—3—
F (r

is Op(l) and by Lemma 4.2,

30

1.
J H n F2(t) du(t) —¢" 0. Hence the product goes to zero in probability. a
T

T_H_EQ§§M 4.1. Assume 1(i) - 1(vi) and (3.1) hold. Then
wn = w on L2([0, rH], u).

PMP. Since (3.1) holds, by Lemma 3.3, ¢ 5 p so that the condition in
Lemmas 4.2 and 4.3 is satisfied. Since Theorem 2.1 has already shown the

weak convergence on L2([0, r], p) for every 1 < TH, as mentioned at the

beginning of Chapter 3, it is enough to show ‘tightness at 73’ in the sense

of Theorem 4.2 of Billingsley (1968).

Define
X,n(t) = Wn(t) if t 5 r A Tn
= 0 otherwise.
and
X,(t) = W(t) ift 5 r
= 0 otherwise.

Wn = W on L2([0, r], p) by Theorem 3.1. Now if p1 denotes the

norm on L2([0, 1'], p),
pftw... x...) = If we - W..(MT..)]2 am)
0
= 111‘. < r) J;1w.(t)- w..(T.)12 aw)

—9 0 a.s.

since Tn -t TH a.s. and 1' < TH. Therefore,

x =1 w on L2([0, 1'], ,1) asn —+ m.

111

This is same as saying V 1' < TH,

X,“ = X, on L2([0, TH],p) asn —-o m.

Now we shall show that as 'r —o TH, X1 =9 W on L2([0, TH], 11)

31
by showing that p(X,,W) —op 0.

Elp2(X..W)] = E[ I? (X;- W)2 111‘]

= J7“ B(W2(t)) ass)

1.

= IT“ F2(t)C(t) one)

T

-o 0 as 1' -+ TH.
Therefore p(X1,W) -o 0 in probability.
It remains to show that

(4.2) lim limsup P{p(Xm,Wn) > s} = 0.

T—‘T n-—Om
H

But observe that

p2(X...W.) = f“ (X..- W.)2 as + [TH (X..- “1.)” s...
o Tn

TH 2
The second term = I Wn(t) du(t)
T

11 IT“ F2(t) du(t)
Tn

(4.3) _.P o

by Lemma 4.2. As far as the first term is concerned,
'1‘ '1‘
J0“ (K...- we” as =10“. > r1] " Wis) «mm
r
'11, 2
= I(Tn > r) J Xn(t) one)
1'

(4.4) s 1.3 Kim (Mt)

because Xn = W for t 5 Tn. But by Lemma 2.1,

thF

E(x,2,(t)) = V(Xn(t)) = F2“) JO 1';

and by Theorem 3.1, Rn(t) 5 K F(t) C"l(t). Hence

 

32

(4.5) 1101.0» s K 1%) G(t).
Now (4.2) follows from (4.3), (4.4), (4.5) and 1(vi).

CHAPTER 5
MORE GENERAL CASES

In this Chapter we shall assume F and G are any two continuous
d.f’s on [0,m), not necessarily satisfying (3.1). Theorem 5.1 proves the
weak convergence of Wu in L2([0, TH], 11) under more general conditions.

1(i) - 1(vi) are assumed to hold.

LEMMA .1. Let Z1, Z2, .......... ,Zn be i.i.d. H and Z”), Z”), ......... ,Zm,
be the corresponding order statistics. Let u be as in Lemma 2.1 and h be
a density of H w.r.t. V. For 0 5 i 5 n — 1, define

Ara) == 1w: z...(w) < t s 2.....(w11.
Then for every i, the conditional density of (Z111: Z”), ......... :Zm): given
Ai(t), is given by

h§(xl,x2, ....... ,xi) =

 

1!
Iii—(5

where all the densities are w.r.t. the corresponding product measure.

1'1

P353217. We need to show that

(5.1) ("1"... J01 h§(x,,x,,; ..... ,x,) su(x,) du(x2) ........ dV(xi)

0 0
= P[Z(1) S 21, Z”, S 22, .......... Z(i) S zilAi(t)]'
Let hi denote the joint density of (Z111: Z”), .......... ,Zm). Using

the ideas of Section 2.2 of David (1970), we get that hi(x,,x,, ....... 1x1) is

equal to

i
(£7! Hn'4(xi)[l_[h(xj)] I(xl 5 x2 5 ....5 xi)-

.i'1

33

34

For i = l to n - 1, define

it
it
hi,,(xl,...., xi,t) := Jo hi,1(x1, ..... xi,s) du(s).

Note that

h:,1(x1,.....xi,t) = J (:[n—1-. [111108)] h(s) Riki—1(8)

-I(x1 5 x2 5 ..... 5 xi 5 8)] dV(s)

_ n—(‘TIY- (_.Lmfl'lhor )] I(xl 5 x,< _ ..... 5x1)

.111“ 1(s) dH(s)

5%,[Hhu )] I(xl 5 x,< _ 5 xi)

{H“‘(x.)- W(t)}

i
= hi(xv ..... xi) — (ll—3%! [Hh(xj)] I(x1 5 x2 5 ..... 5 xi).H"-i(t).

5“ . .
On the other hand, because P(Ai(t)) = “(gm H‘(t) Hn-'(t), one can rewrite

h:(xl,x2, ....... ,xi) as
I(xift) n i .
P(KiltH [(11—3ij [1_Ih(xj)JI(xl 5 x2 5 ..... 5 xi) H'H(t)].
i ' 1

Thus
I(x 5t) ,,
hti(x,,x2,.... .,xi)= P(X—if?” [hi(xv ..... xi) - hi+1(x1: ..... xi,t)].

I(xi5 t) 1:
=m 1h.(x.. .,)- J0 h...(x.. ..... es) du(s)l-

Therefore, LHS of (5.1)

= [p(Ai(t)]-1[J;1J;2J;i"‘[hi(xpx2,....-,xi) - J; hi,,(xl,....,xi,s)du(s)]llIdV(xj)]
.i ' 1

35

—1
= [P(Ai(t)] {Hi(Zl’zz’ ..... Zi-l’ZiAt) - Hi‘1(z1,z2’ ..... Zi-I’ZiAt,t)}
—1
= [P(Ar(t)] Plzm 5 ”1» Zn» 5 221 --------- z(i) 5 ”it Z(i) 5 t < z(i+1)]

= Plztt) 5 21’ Zn) 5 22: ---------- z(i) 5 zilAi(t)]
where Hi is the joint d.f. corresponding to hi. 0

LEMMA 5.2. Let Qn and Rn be as in 1(iv). Then

Rn“) = 2 Bni(H(t))ani(t)

where i t 0
.,,(t) = 13—23le Emlfih) to.) + eat—11sxsmdnxs.

P399132 Note that exactly as in Lemma 3.4,

R...) = "2" 2 fl eats“

isogEDn-1j-1

=12! 2 fl (fii‘fluj-l P{5(k) = dis» 1 5 k 5 n-llAi(t)} P(Ai(t))

i=0 (1601.4 jsi

213.1110» 2 Htfilr12dj'P{6m = d... 1 skin-1}

denn- 15.1

= 23..(H(t)1a..(t) (say).

i=0
We need to prove that ani(t) = ani(t). By writing Du.l = Di x Du-.-”

i o
a..(t) = 2 2 [lg—331.12%

531591 glzenn-H i=1
'P{6(k) = dk 1 S k S n—llAi(t)}

36

= 2 fl (n—EiiT)2dj—l 2 p(sm = dk, 1 5 k _<_ n—1|Ai(t)}

91531 in 925911-14
i
. 2d._1
- J ,

From Lemma 3.2, given Z(j,’s, 6 ,’s have conditional distribution

(1'
given by

l-d
P{6(j1 = dj: 1 5 j 5ilzii12Z (2)» ------ z=(i)} HW! “Z111” “Z111”

j- 21
Thus, by Lemma 5.1,

P{5(j1 = d)» 1 S j silAr(t)}

i
= E[ midis...11714112..-,111A.(t)}

Hiétyl'; Jail: i-1 “Han Nd .i(x 949-de 1)] HdH(xj) .

Therefore, to show 0511:3111: it is enough to show that

[1'1 97334112dj ] LII 11 11x 1114111: )1]

dIEDi jsl

=J1'[1(,—33111¢(x1 + ($1?) 605-)-
We shall show this by induction on i. For i = 1,

LHS = 2 [%]2d"11“(x.)¢?“‘(x.1

 

g: o .1
= [9-31] 11x.) + [n—Ef] to.)
= RHS.

Assume the result for i—1. Write Di = Di,1 x D and d as (9,, d).
Then LHS is equal to

37

i-l . 2d,_1 i-i _ . 2d—1 -
2 [H (11—351) ’ )[Hltdi(x.)¢”i(x.)1] {(5,317) 1“(x.1¢"“(x.)
5'1

i-1 , , _ . . -
= H [($311) ¢(x,-) + (ll—:41?) ¢(xj)] {($17) ¢(xi) 1, (11:11) 110:0}
in

 

= 1] [1.733111 10:.) + ($1115.11
1' =1

= RHS. c1

THEQREM 5.1. Assume 1(i) - 1(vi) hold. Then
wn =1 w in L2([0, TH], 11)
if any of the following conditions is satisfied.
(5.2) For some a E [0, l), U/l’n is non-decreasing and bounded
above.
(5.3) it is non—increasing and F‘H Cat is bounded above.

PROOF. First, we will prove that the condition U/F‘l is non-decreasing is

equivalent to 152%; a.s. u. Since F and G are continuous,

‘221-4175 a.s. V :1 [1 + (gli‘/l'(})]-1 2 1% a.s. V

«:1 gF/fG 5 a a.s. v
=1 11/6 5 06/13 1... u

b b
=1] (g/C)du gJ a(f/F)du for all a < h
:1 log C(a) - log C(b) 5 a[log F(a) - log F(b)]
for all a < b
e. G(a)/C(b) 5 mag/mm)“ for 111 a < b

1:1 (C/r")(1) 5 (C/r'°)(h) for all a < b
=1 (U/Fn) is non-decreasing.

38

¢ 2 T1? implies
($31711 + (11—) 17 < (fifiKfi—w (“3'“)(1301

 

¢ 2 LIFE a.s. V implies
i

ans) 5 fl{(,_3.1,11(11,)+ (”2+1111101}

i-1
from Lemma 5.2 and from the fact that

Hiét) I; El: -1 “I: JIIIdHOKj) = 1.

Now from Lemma 3.6 and proof of Theorem 3.1, it follows that
1—I_a

11.0) < K 111+— “.(t)

 

 

 

 

 

 

Therefore,
Itn E2 5;“ __ _
< K F 0‘ G +0 F 1G
F/C
~2a 2
= K rm cm
1 l 01
Now C 5 KO Fa implies Um 5 KE— F11?r
—a 1 1
implies Fl+a CI+a 5 KW
—2a 2 2
implies K F1” CH“ 5 K K?

R
implies Flu-_t:— 5 K, for some constant K,.

From the proof of Theorem 4.1, it is clear that this is all we need to show
that (5.2) is sufficient for the weak convergence of Wn to W in
12110, 7.11.11)-

To show that (5.3) is sufficient for the L2— weak convergence of Wu, as

earlier, we just need to show Ru 5 K1 FIG for some constant K,.

39

If t) is non—increasing, (TEE-T) ¢ + (gal-:1) ¢ is non-decreasing; so
for all j, l 5 j 5 n,
(flip) 1o.) + (“—32:11 11x.) 5 ($317110) + (“—;Ll}t1-117(t)

i

and therefore, as earlier, ani(t) 5 Ilka-r) ¢(t) + (BEE—1) 6(0}

i=1
Consequently Rn(t) 5 K H2¢(t)-1(t) as in Lemma 2.5 and Theorem 2.1.
Note that the bound in Lemma 2.5 is uniformly in p so the constant K is free

of t.
32¢(t)-1 = pz¢(t)-1(t) C”(t)_l(t)

._. 41.11am.) emf
= K K3
= K1
where K0 bounds 15245—1 db. :1

implies 3%“) < K r2¢<‘)-2(t) C’¢(‘)(t)

REMARK 5.1. Note that condition (ii) of Theorem 5.1 is the same as AF/AG
being non-increasing and AF - AH-(AF/AG) being bounded above, where
A’s and A’s are hazard functions and integrated hazard functions

respectively.

EXAMPLE 5.1. Let G be any distribution on [0, m), 6 E (1, m) and \P
be any bounded non-decreasing function on [0, m) with \II(0) = 1. Define
F by

F(t) = [G(t)/Km".
It is easy to verify that these F and G do not satisfy the prOportional
hazards model, but do satisfy (5.2).

CHAPTER 6
ASYMPTOTICALLY DISTRIBUTION FREE TESTS
In this chapter some Anderson—Darling type A.D.F. statistics that are
used to test Ho: F = Fo vs. H,: F # F0 are discussed. It is shown that

T 1
Jo“ 1K./K.1 [w./1=12 «K. = {onto/mm» .1.

and

r 1
H * * 2 ‘ 2
Jo 1K./K.1 1w./F.1 dK. =1 [0 Bp(t)/[t(l-t)] at.
where Kn and K; are as in 1(iii). Also discussed are the L2 - weak

convergence of Zn, £11 and 5; (as defined in 1(iii)) to W. A few

preliminary results are proved first.

Kn Kn
LEMMA 5.1 95 For each n, 0 5 F— 5 1 and F— is a monotonic

decreasing function on [0, TH].

PgQQP. From the definition both Cu and Kn are increasing and

non-negative functions. Moreover, for t 5 TH,

C (t) > Y Q = Em- so TlU—(t) < F(t) and hence I53(1) < 1.
” ‘ o F2 r ’ + n ‘ F "
For t 5 TH, we shall show that

F- t
in) = Jo Mn dF

where

 

Mn=-Cn+ 1
FE

which will prove that is increasing.

 

4O

41

Using integration by parts, as F is continuous and Cn(0) = 0,

 

Engn- = ENC, = F(t) Cn(t) + J; CndF

and hence

* F
J Mn dF = F(t) Cn(t) = ——(t).
o Kn

 

But
t
dF F 1
C(t)=J s—-(t)s——(t)
" o r2 C,_ r C,_ r C,_
and hence Mn(t) 2 0. n
11:
K2 K.
LEMMA 6.1 h. For each n, C 5 — 5 l and — is a monotonic
F!) n
decreasing function on [0, Tn].
111
I{ 0

Pmr. Note that on (Tn, TH], F“- is of the form —0-— so there, it is

n
111

K
defined to be far), which is well-defined. Now, 0; is same as the

function C defined in Gill (1983). The remarks that follow Theorem 1.2 of

Gill (1983) give the result. a
LEMMA Q2 (1.
K TH
(6.1) _E - K _.P o
F F o

 

 

 

Pmr. The proof is split into two cases. First is the case when

42

T
C(TH) > 0. In this case, "Cl;1 — G‘lllon _.p 0. The proof below is given
under the assumption that the convergence holds almost surely. If it holds
only in probability, a subsequence argument will yield the result.

All the statements in this paragraph hold on a probability 1 set.
1 .1 TH
Given 6 > 0,3 N such that Vnz N, "G; —G H0 56. If n z N,
t
1111) (0.11) — C(11)) s . F(t) I] 1'3 dF| s .
o
T
B. So it follows that ||F(Cn- C)||OH —O 0. By an application

of Lemma 1.4 with dp = dF’1 and g = C51, F(t) C(t) is seen to
Since F (1+Cn) —c F (1+C) uniformly

forall t5r

converge to C’IUH) as t —o 1H.
on [0,111] and F (1+C) is bounded away from zero near TB, (6.1) follows.

Next is the case when C(13) = 0. In this case F(t)C(t) -+ m as

t—or and hence g-(t)-+0. Now for each r<r,

 

 

 

 

 

 

 

 

H H
K T
__n - K —o 0 a.s.
F F 0
because
1 —o l uniformly on [0, r] a.s..
Cm _
(Note that C and F are bounded away from zero on [0, 7].) So 3 11,
Kn K T
such that P(flo) = 1 and for all u e 11,, ——(w) — — -+ 0 for all
F F o
r < TH. Then the following statements hold on 11,.
Kn K Kn K Kn Kn
—(0)=—(0)=1;—(T)=—(T)=0; 05—51;—is monotonic
F F F H F H F r

is continuous, by

decreasing; and E(t) ._. I-{-(t) pointwise. Since K
F F F

Polya’s Theorem the convergence is uniform on [0, TH]. 13

43

LEMMA 9.2 2.
K. TH

(6.2) __n — K —Op 0
Fn F o

 

 

 

 

m5 Again the proof is split into the same two cases as in Lemma 6.2 a.

Because of Lemma 6.1 b, the proof in the second case is exactly the same as

in the proof of Lemma 6.2 a so we will prove the result only in case 1.

Here, the proof will go along the same lines as in Lemma 6.2 a, with the

comment about a possible subsequence argument applicable. So all we need to

prove is that ||Fn(1+C;) — F(l+C) II —o 0. Note that we need to define

Fn(1+C;) to be a suitable left limit whenever it is of the form 0 times a.
By Lemma 6.1 b, (In 5 Kin/F11 5 l and Kn/Fn is increasing. Hence

(6.3) F.(1+c.) s 6.11

and for each n,

(6.4) Fn(1+Cn) in a non-decreasing function.

We know that G51 -+ C‘l on [0, TH] as C(13) > 0. Also we know

that F(t)(l+C(t)) ._. Clan) as t —. r and that, for any r < r ,

H H
Fn(l+Cn) -+ F(l+C) asn —-1 a) on [0, 7]. Let c> 0. 3 1'< TH

such that

(6.5) |F(t)(1+C(t)) —C-1(rH)| 5 e
for all t 2 1'. Also, 3 N such that V n 2 N,
(6.6) IIF.(1+C..) - F(l+C)"; s t
and

(6.7) 1111'.1 - 6'11) s ..

Now Vn2N and Vt2r,
16.110) - F.(t)(1+c.(t))1 = 1151(1)- F.(t)(1+c.(t))
(by (6-3))

44

< Gila) - F.(r)(1+c.(r))
(by (6-4))
(14(1) — F(r)(1+C(r)) + 2c
5 36.
So by (6.5) and (6.7), Vn 2 N and V t 2 'r,
IFp(t)(1+C.(t)) - F(t)(1+C(t))l 5 st

IA

 

 

 

The above, together with (6.6), gives the result. :1
K1, Tn
LEMMA 5.3 g l — is bounded in probability. More specifically, for all
0
M E (1, 111),
T
P{ E n >M+1}sfi.
K o

 

 

 

 

PRQQF. Note that it is enough to show that

P1183:

 

 

n>M}51%I forall M€(1,m)

because

Kn 1+C

C
— — 1—-C— ( l .
K + n - + C;
Apply Lemma 2.6 of Gill (1983) to Gn to get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C Tn
P{ n' > M} 5 I)! for all M e (1,...)
C_ 0
Now,
C, Tn
CI, 2 Cu 1 ' g M
[ G- 0 ]
. (1F 11 TI.
1 _
2 — I 5M
MJ0F211_ [IL 0 1
C Tn
C n-
= 1 5M
M i U. o }

 

 

 

 

45

on [t 5 Tn]. Hence

 

 

 

 

 

 

 

 

 

n G T11
P(fi- 2M}5P "' 2M5fi. 1:)
n G_ 0
K; TH
LEMMA .3 . — is bounded in probability.
0

 

 

 

PMP. The proof will follow as in Lemma 6.3 a but we still need to show

t an
(6.8)

 

 

 

 

is ()p(1).

By (7.7.21) of Shorack and Wellner (1986),
"K;— K"; —. o a.s. v r < r

 

 

 

 

H
and hence, as K(r) > 0 for r < TH,
K; T
— —o 1 a.s., r < TH.
K 0
Therefore, we only need to show (6.8) near TH. Now, if i is such that
Z”, 5 t < Zn“), by explicit computation we can show that
t dF i - 5 j
(6.9) __"=X{.‘lai‘_l}”_1
o I“n J

1"!
Also, logx5x—l Vx2l, so

(6.10) 2[{+{-n_j__-+1}615’ - 1J2 1,2 log {n—--i—j—_'*'l}ls (j) 2 - log F11-

by the definition of Fn. By Lemma 2.6 of Gill (1983) and the relation
log (Fa/F) / (- log F11) 2 —1, it follows that

(6-11) [108 (Fa/F) / (-108 F10] = 0p(1)

near TH. Now (6.8) follows from (6.9), (6.10) and (6.11) n

46

LEMMA 5.4. Assume ¢ is bounded away from zero near TH a.s. V. Then
V 6 > 1/2,
(6.12) ,5 K5(Tn) _.P 0.

335321; Following along the same line as in the proof of Lemma 1.3, we get
that (6.12) is equivalent to

[1+C(t)]26 H(t) —-v u) as t —o TH.
So it is enough to prove that C(t) H(t) is bounded away from 0 near

T

11' Note that C(t) = It ¢ dH'l. So by Lemma 1.4, C(t) H(t) —o ¢(TH)
0

ast—orH. n

LEMMA Q5 9, Assume that 3 a e [0, 1) such that

(6.13) 1 21-11;,
and
(6.14) (G/Fn) is non-decreasing.

Then 3 fl 6 (0, 1/2) such that for a special construction of Xi’s, Yi’s and

Pmr. Under (6.14), 3 fi 6 (0,1/2) such that

JTH dF < m
o F2311

Use the construction of Theorem 7.1.1 of Shorack and Wellner (1986) and

B.
1-
Kn ‘ Wn _ Kl'flw

F F

TH
_.p 0
0

 

 

 

 

where W := F B(C).

(6.15)

apply the theorem with q(t) := (l-t)‘9 to get

47
T
W - W n
K n p
(6.16) — —o 0.
F KB K"(t)
By (6.13), Lemma 6.4 is applicable, and hence Wn in (6.16) can be replaced
by Wu. Also,

 

 

 

 

 

 

 

K W - W K-l(§) p
F K3 0
. K"(1)
s1nce F is bounded away from zero and Wn-W" _.p 0. Thus we
0
have
Kl‘fim —w1 Tn
(6.17) n _.P o.
F o

 

 

 

From (6.15) and from Remark 2.2 of Gill (1983), it follows that
F1_fl B(C(t)) —t 0 a.s. as t-o 7. Therefore

 

 

 

 

 

 

 

 

 

 

 

 

H
T
W H
(6.18) < co a.s..
175 o
and since K 5 F,
(6.19) K14 B(C(t)) —. o a.s. as t —. '11
From (6.12), (6.17) and (6.19) one gets
Kl"fl[Wn—W] TH
(6.20) —s 0
F 0
Use Lemma 6.3 a to get
1— T
Kn filwn-W] n _.p 0
F o
and use the same idea as above to conclude
1-13 1'
K W —W] H
(6.21) n l n —op 0
F o

 

 

 

 

So it is enough to show that

48

 

 

 

 

 

 

 

 

[Ki‘fl - KHIW “'11
_.p 0
II F o
1K3."9 - K”) ’n
From Lemma 6.2 a, 151-5 o —+9 0. Now the result follows
from (6.18). a

LEMMA 65 5. Assume (6.13) and (6.14) hold. Then for the fl and the
special construction in Lemma 6.5 a,
t _

F F 0
P3525217. Follows exactly as in Lemma 6.5 a by using b type lemmas instead

TH
--9p 0.

 

 

 

 

of a type whenever applicable. :1

BQMAQ §.2. Note that if G is continuous, we need to assume only one of
(6.13) and (6.14), since they are equivalent in that case by the proof of

Theorem 5.1. The remark is applicable to Lemma 6.5 C also.

LEMMA a.s. Fix A > 0 and let {X(t): t e [0, A]} be a stochastic
process such that X(t) —o 0 a.s. as t -+ 0 and "X": < a) a.s.. Let
{Yn(t): t E [0, A]} be a sequence of stochastic processes such that

”11,113 5 1) for a r.v. 1) and "11,111L —. o a.s. for all s > 0. Then
"X Yn”: —. 0 a.s.. If the convergence of "Ya": holds in probability, so

does the convergence in the conclusion.

LEM First, note that the probability 1 set where "Ya": -+ 0 can be
chosen free of e by taking countable intersection of probability 1 sets. So
3 11,, P(flo) = 1, such that for all u E 11,, X(t,w) —o 0 as t -o 0 and

49

||Yn(w)||‘: —o 0 for every t > 0. Now the statements in the following
paragraph hold on 11,.
Let 651. 3 e > 0 such that for all t 5 e, |X(t)| 5 6.
"H.113 s 11x1r..11:,+11111.11;L
s a D + 11x11: "v.11:
Taking limsup as n —» p, limsup ||XYn||3 g 6 D. Since 6 is

11—91:)

arbitrary, it follows that IIXYnll’; —o 0.

Thus we have proved that on no, ||XYn||3 —o 0. Hence the first part.
For the second part, take any subsequence {nk} of 11. By a diagonal
argument, 3 a further subsequence {nkj} of {nk} through which

"Y'all: —. 0 a.s. simultaneously for all e > 0. Apply the first part of the

theorem to this subsequence and this will prove the second assertion. :1
Now we shall prove a variation of Helly-Bray Lemma.

LEMMA 5.1. Let a, b E [0, m] such that a 5 b and let fn be a
sequence of functions on [a, b] such that "fa-fl]: —o 0 and f is
continuous and bounded on [a, b]. Let Fn be a sequence of monotonic
functions on [a, b] such that 0 5 Fu 5 1, Fa converges pointwise to F

on [a, b] and F is continuous on [a, h]. Then

an _, I de.

I fn
(a,b] (a,b]

PROOF. By integration by parts formula,

Ja blf or, = f(b) F.(b) — f(a) Fn(a) — I(a.b]F" di

because f is continuous. Also

[($in dF = f(b) F(b) — f(a) F(a) - 1(1th df.

50

Since by Polya’s Theorem "Fa-F": ._. o,

Fn(b) _. F(b), Fn(a) -o F(a) and [(11 b111,, df _.

[(a,b]]? df.

I(ablf an —. [(a,b]r dF

Let r > 0. 3 N1 such that for all n 2 N,, urn-in: 5 r/2. a N,
such that for all n 2 N ,,

LetN=N,VN,. Nowforall n2N,

fn an - I(a blf dFl 5

 

 

 

fn an - [(8.be an

] f dFn —] f dFI
(a,b] (a,b]
|F11(b) - F11(3)| (6/2) + e/2

6. D

I(a,b] (a,b

+

 

IA

IA

Now we shall state and prove a theorem that will provide an A.D.F. test

for Ho: F = F0 vs. 11,: F 11 F0

THEOREM 6.1 9, Assume that i(i) - 1(vi), (6.11) and (6.12) hold. Then

I :11 [Kn/Kn] [Wu/F12 dKn =1 J;B%(t)/[t(14)] dt.

PREP. It suffices to Show that

(1) imam-[MW dK. = JzBi(K)/1K(1-K)1 dK

and

(2) J:HIY<../K.1-1w./F12 «K. = [:3 BitK)/iK(1—K)1dK

51

for some 1' e (0, TH) because then the result follows from (1) and (2)

together with a simple change of variable.

Proof of (1): Apply (7.7.9) of Shorack and Wellner (1986) with
q(t) = (l—t)fi to get that for every 13 E [0, 1/2), 3 a special construction of
Xi’s, Yi’s and W such that
Inw: - W’l/K’fiu; 4’ 0
because 0 5 I{i(r) 5 $0) 5 1 for t e [0, r].
1+C '
is” = [#060 3*”
1' ‘ dF ‘ dF
5 1+Cn T
I ‘ )1 Jo Fe: ’ J Fia—
s [1+c.(r)1w’(r)c_(r)1’1
and Cn(1') —+ C(r) a.s. Therefore for all t e [0, 1'], g—(t) 5 Br for
n

 

some r.v. 13, free of n and t. Thus "[wfi — w21/Kfil’u; _.P 0. Now
we shall show that

(6.22) utw’IKi" - K251"; —»9 0.
As
2 219
W2 213_K2 1' = W K _1’
"I (K. ”mo W Wu

 

 

2
we can invoke Lemma 6.6 with X = gm and Yu = (Kw/K313) - 1. We

will now verify the conditions of Lemma 6.4. As

W = F B(C) = F B(K/(l-K)),
an application of the Law of Iterated Logarithm for Brownian motion (see
Theorem 12.29 of Breiman (1968)) will show that X satisfies the conditions
of Lemma 6.6. Since K? -o X”9 uniformly a.s. on [0, r] by (7.7.21)

52

of Shorack and Wellner (1986) and K” is bounded away from zero on

[6. T],

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

K25 T
— l —o 0 a.s..
1?? e
K K” E 2;?
A3 Kit)5B1forall tE[0,T], Riv—10$ B1 +1. Hence
(6.22).
Since F is bounded away from zero on [0, r], it follows that
2 2 1'
W
K n — K W _.p 0.
I "PK? FK29 o
Kn Wu 2 K 2
. _ _ W
Invoke Lemma 6.7 Wlth {n — n W , f -- w W- ,
Fn = K? and F = K”. This gives us
2 2
T Kn wn dKzfi _.p IT K W dKzfi.
o 779 l'i‘Kfin n 0 TB FK

 

 

 

 

That is,

Kim/Kn] [w.m’ dK. =4 J;B§(K)/[K(l-K)] ax.

Proof of (2): From Lemma 6.3 and the fact that K is bounded away from

zero, it follows that

 

 

 

 

 

 

[Ki-5 Wn]2 [Kl-fl wlz Tn p
_. —+ 0.
F2 Kn F2 K o
Kl-fi W 2 l—fl 2
Apply Lemma 6.7. again with fu = l n 2 n] and f = LIE—211,1 s
25 F Kn 2s F K

26

F1. = l - K2? and F = l - K . By (6.19), we are assured that f

satisfies the required conditions. Now the result follows. a

53
W Assume 1(i) — 1(vi), (6.11) and (6.12) hold. Then
T * "' t 1
Jon [Kn/Kn] [Wu/Fnlz dKIll = Jo B%(t)/[t(1-t)] dt.

Pmr. Follows exactly as in Theorem 6.1 a. 0

Now we shall prove the weak convergence of 5“, En and 5; in the

L2 — space. See 1(iii) for notation.

W Assume 1(i) - 1(vi) hold. If (5.2) or (5.3) holds, then on
Line. TH]. u).

6) 5n =’ B0(K),

(ii) 3n = B0(K).

(iii) :3 = Bose.

where B0 is a Brownian Bridge.

m First, note that (K/F) W has the same distribution as B0(K).
Now, by Lemma 6.1 a, {n E L2([0, TB], )4). By Theorem 5.1, there is a
construction of WI. and W such that Wn - W -o 0 a.s. in

L2([0, TH], p). So by Lemma 6.1 a,

Kn
(6.23) F— (Wu — W) —+ 0 a.s.
in L2([0, TH], p). Moreover, by Lemma 6.2 a,
Kn
(6.24) F— — g] w _.P o.

in L2([0, TH], s). Now (i) follows from (6.23) and (6.24).

Proof of (ii) and (iii) are similar. a

54

Denote the limiting r.v. in Theorems 6.1 a and b by X. For a representation
of X as a mixture of independent chi—square r.v.’s, see Anderson and
Darling (1952). Also given in the paper mentioned above is the characteristic
fimction of X. A more direct computation of the mean and the variance of
X is now discussed. The mean is easily seen to be 1 by taking the
expectation inside the integration. Calculation of the variance is done in the

following lemma.
LEMMA 6.8. X has mean 1 and variance (212/3) - 6.

Law; As all the quantities involved are non-negative, Fubini’s Theorem
justifies interchange of expectation and integration and thus the mean of X
is 1. Let us denote B§(t)/t(1-t) by B’(t) and E(X’) by Q so that
V(X) = Q -1.

1* 1*

Q = E[JB(s)ds JB(t)dt]
101 ,.. *0

= E[JOJ0B(8)B(t)dsdt]

1 t s at:
= 2] J E[B(s)B(t)]dsdt

0 0
If (X,Y) has bivariate normal distribution with mean 0, variance 1 and
correlation p, then E(XY) = 1 + 2p2. As (Bo(s), Bo(t)), for s 5 t, is
bivariate normal with dispersion matrix 2 given by

2 _ s l-s s l-t
’ s l-t tl-t ’

it follows that
l t
Q = 21 j {1 + [2s(1-t)/(l-s)t]} ds dt.
0 0

1
The above expression simplifies to 4 I (—log x) (x/(l-x)) dx - l.
o

55
Integrating by parts and changing variables we get
Ewes x) (x/(l—x» dx = 1 + J} ecu-y) 103(1-exp(-y)) dy
Using the Taylor expansion for log x, integrating term by term, and making

(I)
use of the fact that )i‘. n-2 = 12/6, we will get that the expression above
is equal to 1.2/s — 1. Thus Q = 4( 3/6 -1) - 1 = (2.2/3) — 5 and
hence V(X) = (212/3) — s. o

W5.-
Anderson, T. W. and Darling, D. A. (1952). Asymptotic theory of certain

goodness of fit criteria based on stochastic processes. Annals of Statistics
23 193-212

Bhattacharya, P. K. (1974). Sums of order statistics. Annals of Statistics 2
1034—1039

Billingsley, P. (1968). Convergence of probability measures. John Wiley and
Sons, New York

Breiman, L. (1968). Probability. Addison Wesley.

David, H. A. (1970). order Statistics. John Wiley and Sons, New York.

Gardiner, J. C., Susarla, V. and Van Ryzin, J. (1985). Estimation of the
median survival time under random censorship. Adaptive Statistical
Procedures and Related Topics. IMS Lecture Notes 8 350—364

Gill, R. D. (1980). Censoring and stochastic integrals. Mathematical Centre
Tracts 124, Am3terdam.

Gill, IL D. (1983). Large sample behavior of the PLE on the whole line.
Annals of Statistics 11 49-58

Kaplan, E. L. and Meier, P. (1958). Nonparametric estimation from incomplete
samples. Journal of American Statistical Association 53, 457-481.

Koul, H. L., Susarla, V., and Van Ryzin, J. (1981). Regression analysis with
randomly right-censored data. Annals of Statistics 9 1276—1288.

Koul, H. L. (1984). Tests of goodness-of—fit in linear regression. Colloquia
Mathematico societatis Janos Bolyai 45.

Koziol, J. A. and Green, S. B. (1976). A Cramér—von Mises statistic for
randomly censored data. Biometrika 63 465-474

56

57

Lo, S. — H. (1986). The Product Limit Estimator and the Bootstrap: Some
Asymptotic Representations. Probability Theory and Related Fields 71,
455-465.

Parthasarathy, K. R. (1967). Probability Measures on Metric Spaces. Academic
Press, New York.

Shorack, G. R. and Wellner, J. A. (1986). Empirical Processes with
Applications to Statistics. John Wiley and Sons, New York

Wang, J. G. (1987). A note on the uniform consistency of the Kaplan - Meier
estimator. Annals of Statistics 15 1313—1316.

Yang, S. (1988). Minimum Hellinger distance Estimation of parameters in

Random Censoring Models. Ph. D. Thesis: Michigan State University.

“lllllllllllllll