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This is to certify that the

dissertation entitled

First Order-Allocation

presented by

William Noble

has been accepted towards fulfillment
of the requirements for

 

PhoD. degree in Stat'iSt'iCS

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Major professor /

Date July 18. 1991

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emmut

 

 

 

-——____—_

FIRST ORDER ALLOCATION

By

William Noble

A DISSERTATION

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

DOCTOR OF PHILOSOPHY

Department of Statistics and Probability

1991

'41
J

J‘:

L/

‘ "7}

ABSTRACT
FIRST ORDER ALLOCATION
By

William Noble

The problem under consideration is that of allocating a fixed number of
observations among several populations with the goal of reducing the mean square
error of an estimator. Such allocations have been recommended in several special
cases, such as when one is estimating linear combinations or products of

population means or when one is estimating quotients of Bernoulli proportions.

In this dissertation, a general allocation procedure, called first order
allocation, is investigated. This allocation has the above allocations as special
cases but can also be applied in many other situations. It is shown that under
general conditions, first order allocations asymptotically minimize the mean
square error. Since the case of the quotients of population means (not necessarily
Bernoulli proportions) is both unexplored and of practical interest, it is studied in
detail. Also, since first order allocations typically depend on unknown parameters,
a two-stage allocation procedure is developed and recommendations are made for

the size of the pilot sample.

ACKNOWLEDGMENTS

Many people have contributed to the successful completion of this degree.
My thanks go out to the following individuals: to my thesis adviser, Professor
Connie Page, for the guidance and unerringly good advice which she has provided
throughout the preparation of this thesis; to Professor Habib Salehi for recruiting
me into the graduate program at MSU and for serving on my guidance committee;
to Professors Joseph Gardiner and Dennis Gilliland for also serving on my
guidance committee; to Professor James Hannan for making many helpful
suggestions which have led to a number of improvements in my dissertation; and
to Professor James Stapleton, who recommended me for the position of consultant
in the Statistical Consulting Service and from whom I learned much about the art

and science of consulting.

I would also like to thank the Department of Statistics and Probability for
their financial support during my first two years, and the College of Natural
Science for the fellowship which enabled me to concentrate on thesis research

during my last year.

Most of all, I want to thank my wife, Joyce, and children, Jennifer and

Daniel, for their support, patience, and understanding.

iii

TABLE OF CONTENTS

Chapter
1. FIRST ORDER ALLOCATION

1.1. Introduction

1.2. The Coal and the Method

1.3. An Approximation to the Mean Square Error

1.4. The Minimizer of the Approximation to the Mean Square Error
1.5. Asymptotic Optimality of First Order Allocations
1.6. Application

1.7. An Example Where 0 Does Not Hold

. EXPANSIONS OF E#[g(Uk) — g(p)]r

2.1: Introduction

2.2: General Moment Expansions

2.3: Conditions Sufficient for Hypothesis H4 to Hold
2.4: Examples

. QUOTIENTS OF POPULATION MEANS

3.1: Introduction

3.2: Asymptotic Optimality of CV-Allocations

3.3: Examples

iv

WRIQOSOVN

10
11
16
20

25
26
28

4. A TWO-STAGE ALLOCATION PROCEDURE FOR PRODUCTS
4.1: Introduction
4.2: Preliminary Result
4.3: MSE Expansion
4.4: Asymptotic Optimality
4.5: Square Root Rule for Pilot Sample Size

4.6: Examples

BIBLIOGRAPHY

30
33
35
38
39
40

44

CHAPTER 1
FIRST ORDER ALLOCATION

1.1. Introduction

The problem under consideration in this dissertation is that of allocating a
fixed number of observations among several populations with the goal of reducing
the mean square error of an estimator. This problem has been considered from
both classical and Bayesian perspectives, but in this dissertation only the classical
situation will be considered. In the classical setting, such allocations have been
developed in several special cases, such as when one is estimating linear
combinations or products of population means (see Cochran (1977) and Page
(1990)). An allocation has also been recommended when one is estimating a
quotient of Bernoulli proportions (see Brittain and Schlesselman (1982)).

Typically, all of these allocations depend on unknown parameters.

In Chapter 1, a general allocation procedure is developed which has the
above allocations as special cases, but which can also be applied in many other
situations. This allocation is obtained by minimizing the first order terms in an
expansion of the mean square error, and hence is called first order allocation. It is
shown that under quite general conditions, first order allocations asymptotically

minimize the mean square error.

In Chapter 2, technical results are developed which will be used in
Chapters 3 and 4. The reader may proceed directly from Chapter 1 to Chapter 3

or 4 and refer back to Chapter 2 as necessary.

In Chapter 3, the results of Chapters 1 and 2 are applied to the estimation

of quotients of population means (not restricted to quotients of Bernoulli

2
proportions). It is shown that the allocations developed in Page (1990) for
products of population means also asymptotically minimize the mean square error

when estimating quotients of population means.

The first order allocations developed in Chapters 1 and 3 usually depend on
unknown parameters. In Chapter 4, a two—stage allocation procedure is developed
for the special case in which one is estimating the product of two population
means. This procedure does not depend on unknown parameters and is shown to
do asymptotically as well as optimal one-stage procedures for which the
parameters are known. Recommendations are also given for the size of the first

stage in the two-stage procedure.

1.2. The Coal and the Method

Let N be the set of positive integers and for k e N, let Nk be the set of k-
tuples of positive integers. Let 16 N be such that I Z 2. Let szI —+ R be Borel
measurable and let 2 = (01, ,oI) 5R1. Let 1 = {1,2, ,1}. For iel, let {Zik:
k 6 N} be a sequence of estimators of 0i. Then f(_Q) can be estimated by f(ZQ) =
f(Z1n1, ’ZInI) with mean square error

MSEQa) = EQ[f(Zn) — fun?
where n 2 (n1, ,nI). The goal is to asymptotically minimize MSE0(n), subject
to the constraint Eini = N. (From this point on, unless otherwise indicated, the
range on indices for sums and products will be from 1 to 1. Thus, 2i and Hi will
be abbreviations for Eilzl and nilzl’ respectively.) Note that Q, {Zik}’ and f are

given and only the selection of n = (n1, ,n1) is to be determined.

The method for achieving the goal of asymptotically minimizing the mean

square can be outlined as follows.

1. Obtain an approximation to MSEQQ).

3

2. Minimize the approximation to MSEQ(n), subject to the constraint
Eini = N.

3. Show that as N —» co, the minimizer of the approximation to MSE-6(a)
does asymptotically as well as the minimizer of MSEQ(n).

Note that the vector n above implicitly depends on N. This dependence is
made explicit in the following definitions. Also, unless otherwise indicated, all
limits, limit infimums, and limit supremums are to be taken either as N —’ 00 or

as k —» 00. It should always be clear from context which of the two is being used.

Definitions: 1. An allocation is a sequence {Q(N)} = {(n1(N), ,nI(N))} in N1
such that for all N, zinim) = N.
2. An allocation {am} is 99m if for all N,
MSEQ(n(N)) = min {MSEQ(m) : 1116 NI, Zimi = N}.
3. An allocation {3(N)} is W mm; if
MSEQ(n(N))/MSEQ(111.(N) -+ 1,

where {m(N)} is an optimal allocation.

Remarks: 1. Allocations are not random. Also, ni(N) 2 1 for all i and N, so that
each population always gets at least one observation.

2. Optimal and asymptotically optimal allocations are parameter dependent in
that if Q is changed, then the optimal and asymptotically optimal allocations may
change also.

3. Optimal allocations always exist. Indeed, if N is fixed, then there are only
finitely many ways to choose [:1 = (m1, ,ml) 6 NI so that Eimi = N. For one
of these, the mean square error MSE9(_m_) = E9[f(Zm) —f(Q)]2 will be a minimum.

4. In the definition of asymptotically optimal allocation, the ratio of mean square

4
errors is always larger than or equal to one, because {m(N)} is assumed to be
optimal. Thus, in order to prove that {n(N)} is asymptotically optimal, it suffices
to show that
lim sup MSEQ(n(N))/MSEQ(m(N) g 1.

Sometimes, as in the following examples, the dependence of an allocation
on N will be suppressed; in this case the allocation may be written as {n} instead

of {n(N)} and the components may be written as {ni} instead of {ni(N)}.

It will also be assumed in the following examples and throughout this
dissertation that there is independence between populations. In terms of random
variables, this means that if {Uij} is a doubly indexed sequence of random

variables, then Uij is independent of Ui’j’ Whenever i ;é i’ .

M: For iel, let {Xikz k e N} be i.i.d. with mean ”i’ variance 0? > O, and

coefficient of variation Ci = oi/Ipil (if ”i 36 0). For i 61, let 9i = ”i and let
Zlk = Xik = (XII. "l' ... + Xik)/k.
1. Linear Combinations: Estimate f(Q) = ziaifli by f(Z_Q) = Eiaix- where ai

,
In;

 

¢ 0 for all i. Then Neyman allocation, defined by
Iii/N = lailai/ Zjlajlaj ,
is optimal. (See Section 5.5 of Cochran (1977).)
2. Products: Estimate f(Q) = nil‘i ;£ 0 by f(Zn) =I'Iixin3 Then CV-allocation,
defined by
ni/N = Ci/chj ,
is asymptotically optimal. (See Page (1990).)

1.3. An Approximation to the Mean Square Error
Let ‘3’ and A: denote the following conditions.
‘3’: f is a polynomial of degree di in the ith variable.

A: (Moment Expansions) For i = 1, , I, there is a 6i 2 0 such that

_ (XI/k), J = 1,
E9.(Zik — ail] ={ (Siz/k + 0(1/1‘), 5 = 2,
I 0(1/1‘): 3 S l S 2di'

Condition .Ala is satisfied, for example, if Zik = Xik (as long as the
individual Xik’s have sufficiently many finite moments). The greater generality

will be used later when estimating quotients of population means.

Theorem: (MSE MM) If EP and A: hold, then
where fi = fi(Q) is the partial derivative of f with respect to the ith variable,
evaluated at Q.
Proof: Since f is a polynomial, then
[f(Ze) — fun"- : amaze... - 9.)?
+ 2,5,5]: fi(_0_)fj(.Q)(Zini " 9;)(Zjnj — 93')
+ higher order terms.

Now take expectations as follows.
1. For the squared terms, use the j = 2 case of A.
2. For the second order cross terms, use the j = 1 case of A together with

the assumption of independence between populations.

6

3. For the higher order terms, use .Ala together with the assumption of
independence between populations to see that the expectation of the

higher order terms is included in 2io(1/ni) + 2: 0(1/ninj).

iifij
Thus, an approximation to MSEQ(n) is given by
MSEQ(n) 2: first order terms = Eifgciiz/ni.

1.4. The Minimizer of the Approximation to the Mean Square Error

Let $25 denote the following condition.
%:fi 7‘: Oanddi > Oforalli,
where 6i is the constant appearing in condition A. If 25 holds, then the first order
terms EifizdiZ/ni, subject to the constraint Eini = N, have a minimum of

(Zilfi|6i)2/N when ni/N = |fi|6i/ zj|fj|6j. An allocation {Q(N)} is said to be a

first order allocation if

 

ni(N)/N _. |fi|6i/2j|fj|6j .

Examples: Refer to the examples in Section 1.2.
1. Lima; mm: Neyman Allocation is a First Order Allocation.
2. Products: CV-Allocation is a First Order Allocation.

1.5. Asymptotic Optimality of First Order Allocations

The following theorem is the primary result in this chapter. It says that
when 5P, %, and .Me hold, a first order allocation will do asymptotically as well as

any allocation whose components tend to infinity.

W: Assume that ‘EP, 23, and A. hold. If {n(N)} is a first order allocation
and if {mm} is any allocation such that mi(N) _. 00 for all i, then
lim sup [MSEQ(n(N))/MSEQ(rn(N))l s 1.
lam: By the MSE Expansion,
um N.MSEQ(n(N)) = (Eilf,(fl)l6i)2-
But also by the MSE Expansion and the fact that the components of {m(N)} tend
to infinity,
MSEQenm» = 229429631 + awn/mam,
where ei(N) —> 0 as N —+ co. Thus for any 6 > 0,
1imian-MSE0(m(N)) = lim inf N(1 _ e):ifi(fl)26i2/mi(N)
_ 2 um ian(1 - e)(2,lfi(fl)l6i)2/N
= (1 - e)(2ilfi(fl)l6i)2-

Since 6 > 0 was arbitrary, then the theorem has been proved.

Let 0 denote the following condition.

0: If {gl(N)} is optimal, then mi(N) —» 00 for all i.
The following corollary gives a method for determining if a first order allocation is

asymptotically optimal.

Corollary: If 5?, A, 523, and 0 hold, then first order allocations are asymptotically

optimal.

1.6. Application

The theory developed in the previous sections will now be used to obtain a

theorem from Page (1990) on the asymptotic optimality of CV-allocations when

8
estimating a product of population means by a product of sample means. As in
Example 2 in Section 1.2, let Zik = Xi]: and f(Q) = Hioi = Hipi 96 0.
m: Condition 0 holds, i.e., optimal allocations have components tending to
infinity.
P;o_gf: The contrapositive of 0 will be proved. First observe that MSEQ(n) —v 0 if
each ni —. 00, so that the mean square error must tend to zero for optimal
allocations. Suppose that {n} is an allocation such that for some j E 1, nj + 00.
Write
MSEQm) = Egmii'm, -— [1,1192 = niEflfi) - nip?

= mm? + viz/mi) - my?

: (Hip?)[1'li(1 + c?/n,) - 11

= (nip?)[2:ici2/ni + h(n)],
where h(n) > 0 for all neNI. Since h(n) is positive and since no 4» 00, then

.l
MSE0(n) + 0. Thus, {n} cannot be optimal.

Tm (Page, 1990): Let Zik = Yik and f(Q) = I'IiOi = nipi’ where 9i = ”i ¢
0 for all i. Then CV-allocations are asymptotically optimal.

Pm: Since 0 has already been proved, then it remains to be proved that ‘3’, %,
and .111: hold. Since f is a polynomial, then ‘EP holds. As seen in Section 1.2, ail:
holds for sample means with 0i = ”i and 6i = oi. Since 0i 7t 0 for all i, then fi(_0_)
¢ 0 for all i. Since also 6i > 0 for all i, then 1'25 holds.

This result is slightly more general than the result in Page (1990) in that
here it is only required that the parameters be nonzero, whereas the original result

required that the parameters be positive.

9
1.7. An Example Where 0 Does Not Hold

Do optimal allocations always have components which tend to infinity?
The answer to this question is no, as the following example indicates.
m3 Let f(91,92) = {(#pflgl = (#1 " 1)#1(#1 + 1) + #2: Where #1 75
:tl/fi. Let Zik = Xik and let {Xlkz k e N} be i.i.d. uniform over {p1 — 1, [11, #1
+ 1}. It can be shown that when Q = (p1,;12) = (0,0),
MSE (a) = ('X’ - D2763 (x + l)2 + o 2/n
Q lnl 1n1 1n1 2 2'

Since the first term is zero if n1 = 1 and positive otherwise, then MSE0(n) is

minimized when n1 = 1.

CHAPTER 2
EXPANSIONS OF E,,[g(Uk) — g(#)]r

2.1. Introduction

In Chapters 3 and 4, it will be necessary to have expansions of expressions
of the form Ep[g(Uk) - g(lt)]r, where gle _. R is a Borel measurable function,
{Uk} is a sequence of random variables, and p 6 IR. More specifically, in order to
verify that condition .111: from Section 1.3 holds, it will be necessary in Section 3.3
to have expansions of the form

Enlewk) - em] = 0(1/k). Enigmk) - e001? = gram/k + 00/19”)
for the cases g(x) = x and g(x) = 1/x; in Section 4.6, in order to obtain
expansions of the mean square error for a two-stage allocation procedure, it will be
necessary to have expansions of the form

Enlgwk) - em] = (1/2)g"(u)a2/k + 0(1/k)
for the cases g(x) = x. g(x) = 1/x. g(x) = xl/Z, g(x) = 1/x1/ 2, g(x) =
1 2 1 2
[(l—x)/x] / ,and e<x>=lx/(1-x>l / .

In this chapter sufficient conditions are given for such expansions to hold.
The results obtained are perhaps of greater generality than is needed for the
examples in Chapters 3 and 4, but it is hoped that the results proved here will be

useful in examples other than those considered in this dissertation.

The results in this chapter are technical in nature. It is possible at this
point to immediately begin reading Chapter 3 and to refer back to Chapter 2 as

necessary.

10

11

2.2. General Moment Expansions

In order to obtain expansions, the following specializations of results from

Section 4.4 of Fabian and Hannan (1985) are useful.

M123 (Minn m in Fabian and Hm): Let r and t be positive integers
such that r 5 t. Let g(k) denote the kth derivative of g. A(r,t) will denote the

following set of conditions:
t 2
1. E,.IUk — alt = 0(1/k/ ),
(t — r +1) . . . _
2. g ex15ts and IS bounded on an interval G — (p — 61, p + 62),

3. Ennetuk) — g(p)lr-1{Uk ¢ Gil = (XI/kt”).

Ihgprgrn (fIfheorern 43,2 in Map and Ham): Suppose that A(r,t) holds. Let
P be the (t — r)th degree Taylor polynomial for g at p. Let Q = P — g(p) and
let Lrt be the terms of degree r through t — 1 in Qt. Then

Enlgwk) - to)?r = EnL,.(Uk) + 00/13”).

Example 1: (a) Calculate L12. In this case,
P(x) = (t - r)th degree Taylor polynomial for g at p
= 1St degree Taylor polynomial for g at p
= 80!) + g’(ll)(X - It),
Q(X) = P(X) - 30‘) = g’(u)(x - It).
and
L12(x) = terms of degree I through t—l in Qr
= Q(X) = S'U‘XX - fl)-
Thus, if A(1,2) holds, then
Eplg(Uk) - 301)] = 8’(F)Ep(Uk - II) + 0(1/k)-

12
(b) Calculate L23. In this case,
P(x) = (t - r)th degree Taylor polynomial for g at p
= lst degree Taylor polynomial for g at p
= 80‘) + s’(#)(x - fl),
Q(X) = P(X) - g(#) = g’(lu)(x - l1),
and
L23(x) = terms of degree r through t — 1 in Qr
= terms of degree 2 in Q2
= s’(u)2(x - I02-
Thus, if A(2,3) holds, then
I 2
Enlgwk) - 30012 = g(lu)2Ep(Uk - M2 + 0(1/k3/ )-
(c) Calculate L13. In this case,
P(x) = (t - r)th degree Taylor polynomial for g at p
= 2nd degree Taylor polynomial for g at p
= 301) + g’(lu)(x — p) +(1/2)s”(u)(x - ”)2,
00c) ——- P(x) - g(a) = g’(lu)(x - a) +(1/2)s”(#)(x — m2,
and
L13(x) = terms of degree r through t—l in Qr
= 000 = s’(#)(x - a) +(1/2)s”(#)(x - it)?
Thus, if A(1,3) holds, then
I n t 2
E#[g(Uk) - 800] = g(fllEp(Uk — II) + (1/2)g (I‘lEp(Uk "' ”)2 + 0(1/1‘ / ).

Lemma 1 (Lemma 4.4.2 i_n_ abian and Hannan): A(r,t) implies A(r-1,t—1) if r

>1.

m 2: Suppose that A(2,3) holds. Then by Lemma 1, A(1,2) also holds.
Thus, from Examples 1(a) and 1(b), both of the expansions

13
Eplgwk) - 501)] = 8’(#)Ep(Uk - I‘) + O(1/k)
and

Enlgtvk) - 30012 = s’(#)2Ep(Uk - 102 + 00/19”)
hold.

Throughout this chapter, {Vk} will be a sequence of i.i.d. random variables

2

with mean it and variance 0 > 0. If p ;6 0, then the coefficient of variation is c

= a/Ipl. If g(p) is estimated by g(Vk), where Vk = (V1 + + Vk)/k, then it is
sometimes the case that g(Vk) is undefined with positive probability, such as
when g(x) = l/x and P #(Vk = 0) > 0. This difficulty can sometimes be avoided
by replacing Vk by Vk = aka + bk’ where ak and bk are constants. The
notation A(r,t) will be used to denote A(r,t) with Uk = Vk. Lemma 3 below

gives sufficient conditions for moment expansions to hold when Uk = Vk‘

Notation: The notation H(r,t) is used to denote the following set of hypotheses.
H1. The Vk’s have finite tth moments.
H2. ak = 1 + O(l/k), bk = O(l/k).
(t — r + 1) . . . _
H3. g ex1sts and is bounded on an interval G — (ll—61, n+62),

where 61, 62 > 0.

H4. Bantam) - emf-1W1, t G}] = (la/kt”).
Hypothesis H4 is further subdivided as follows.

H4... EpIIng) - emf-1W1. s a - 61}] = 00/13”).

~ ~ t 2
me Enllgtvk) — g(#)|r-1{Vk a + 62}] = O(I/k/ ).
The notation H’ (r,t) is used to denote H(r,t) with hypothesis H2 replaced by the

IV

following stronger hypothesis.

14
H2’. ak = 1 + o(1/k), bk = 0(1/k).

Lem; 2: If H1 and H2 hold, then for any 6 2 0,
~ ~ t 2

e‘Pnuvk - pl 2 e) s Enlvk - ult = 0(1/k/ ).
Proof: The inequality is the Markov inequality. For the equality, use Corollary
10.3.2 of Chow and Teicher (1988), which says that

EuI'V'k - ult = 0(1/kt/2)-
Let || || denote the Lt norm. Then by the triangle inequality,
Eank - ult = :le - ,“t = lam — u) + bk + ulak -1)nt
sllakl Wk - all + Ibk + H(ak — 1)llt

= [0(1/k1/2) + 00/101t = 00/13”).

Lemma 3: If H(r,t) holds, then A(r,t) holds and hence,

~ ~ t 2
Enlglvk) -- etallr = Banalvk) + O(I/k/ )-
Proof: This follows immediately from the definition of A(r,t) and from Lemma 2.

 

Lemma 4: (a) If H2 holds, then
Em - a) = O(I/k). Em - u)? = 0% + 00/13).
(b) If H2’ holds, then
Enlvk - a) = 0(1/k). Ean - a)? = 0% + 00/13).
M: If H2 holds, then
EMVk - fl) = Eplaka - 1‘) + bk + #(ak - 1)] = 0(1/k)
and
Entvk - ”)2 = Elem - p) + bk + #(ak - 012
= afiEpfif-k — p)2 + [bk + g(ak — 1)]2
= [1 + O(1/k)]a2/k + O(1/k2) = 02/1: + O(1/k2).

15
The proof of (b) is similar.

The next two examples will be used in Chapters 3 and 4. More
specifically, the results involving H(2,3) will be used when applying the theorem
in Section 3.2 and the results involving H’ (1,3) will be used when applying the

theorem in Section 4.6.

Example 3: (a) If H(2,3) holds, then by Lemma 3, Example 2, and Lemma 4(a),
both of the expansions
Eping) - 3(a)] = 0(1/k)
and
En[g(\7k) - g(lu)]2 = g’(#)202/k + 0(1/k3/2)
hold.
(b) If H’(1,3) holds, then by Lemma 3, Example 1(c), and Lemma 4(b),
Elem) - em] = g”(#)02/2k + owls).
Thus,
Enlelvkl/elpll = 1 + s”(#)02/2s(#)k + O(I/k)-

In particular, if g(X) = X”. where p i6 0, then s”(lu)/g(#) = p(p- 1W

Ep[g(Vk)/g(/1)] = EnWfi/u”) =1 + p(p—1)c2/2k + 0(1/k)-

2and

Example 4: (a) Suppose that H(2,3) holds with g(x) = l/x. Then g’(x) = —1/x2
so that by Example 3(b),
E,,(1/\7k) = l/p + O(1/k)
and
E iv -12— 2 4k 0113/2
p(/ k M) —0/# + (/ )-
(b) Suppose that H’(1,3) holds with g(x) = xl/Z. Then by Example 3(b) with p

16
= 1/2,
E,,(v11‘/2/p1/2) = 1 - c2/81t + 0(1/k).
(c) Suppose that H’ (1,3) holds with g(x) = 1/x1/2. Then by Example 3(b) with p
= —1/2.
E,,(p1/2/\7i/2) =1 + 3c2/8k + 0(1/k).
(d) Suppose that H’(1,3) holds with g(x) = [(1 - x)/x]1/2 and suppose that o <
ilk < 1 as. Then by Example 3(b),
Ep[g(Vk)/g(lu) =1 + s”(#)02/23(lt)k + 0(1/k)
= 1 + (3-4p)02/8u2(1—p)2k + 0(1/k).
(e) Suppose that H’(1,3) holds with g(x) = [x/(l -x)]1/2 and suppose that 0 <
Vk < 1 as. Then by Example 3(b),
Epistvkl/sml = 1 + g”(#)02/Zs(#)k + 0(1/k)
= 1 + (4p-1)a’2/8p2(1—p)2k + 0(1/k).

2.3. Conditions Sufficient for Hypothesis H4 to Hold

In practice, hypothesis H4 is usually the most difficult to verify. In this
section, conditions are derived which are sufficient for H4 to hold. Applications of

the results of this section to specific examples can be found in Section 2.4.

Lemma 1: Suppose that H1 and H2 hold.
(a) If

IA

Enllslvk)I’-1lvk a — 61}] = ell/kt”).
then H4a holds.

(b) If

Epllslvkll‘llvk a + 62}] = 00/13”),

then H4b holds.

IV

17
PM: By the inequality
Ix + ylZ 5 224(leZ + Iylz), xsy. zeRzal.
and Lemma 2 in Section 2.2,

Eplls(Vk)-g(fl)lr-1{Vk s a — 61H
2"1Ep[(ls(Vk)I’+Ig(lt)lr)-1{\7k .<. u — 61H
2"1(Ep[|s(Vk)lr-l{‘7kSit-61H
+ lslmermsu-all)
= 0(l/ht/2) + omit/2) -_— O(1/kt/2).

IA

IA

This proves part (a). The proof of part (b) is similar.

Low 2: Suppose that H1 and H2 hold.

(a) If g is bounded on (—oo, p— 61], then H4a holds.

(b) If g is bounded on [p +62, oo), then H4b holds.

Proof: If g is increasing on (- 00, p — 61], then by Lemma 2 in Section 2.2,
Epllswkll‘ltvk s a - 61H s (a-sll‘rlvk s a - 61) = ell/kt”).

Thus, by Lemma 1, H4a holds. This proves part (a). The proof of (b) is similar.

Definitiop: The function g is said to be locally Lipschitz it p _o_rr it; set E if there
is a constant M 2 0 such that for all x e F, |g(x) - g(p)| _<_ Mlx - pl.

lamp 3: Suppose that hypotheses H1 and H2 hold.

(a) If g is locally Lipschitz at p on the interval (— oo, p], then H4a holds.

(b) If g is locally Lipschitz at p on the interval [p, 00), then H4b holds.

Prpgf: Let M 2 0 be such that for all x g p, |g(x) - g(p)| _<_ Mlx — p|. Then
Epllslvp) - slaw-107k .<. a - 61H s M’Epllvk- lat-107k .<. a — 61}]-

Now consider two cases. If r = t, then by Lemma 2 in Section 2.2,

~ ~ ~ t 2
Epllvk- plr-1{Vk S I‘ — 51H 5 Eplvk" I‘lt = O(l/k/ )

18
If r < t, then by Holder’s inequality with p = t/r and q = t/(t—r) and by
Lemma 2 in Section 2.2,
Eyllvk" tilt-1671‘ S It - 51“
= [Eplvk- #ltlr/t'PW’k S It - 51)(t—r)/t

= O(l/kt/2)r/t0(1/kt/2)(t—r)/t = O(l/kt/Z).

This proves part (a). The proof of part (b) is similar.

Conditions are now developed which are sufficient for H4a or H4b to hold
when the Vk’s are bounded below or above. The main tools used to obtain these
conditions are Lemmas 4 and 5 below. Lemma 4 is an inequality originally proved
by Bennett (1962) and was subsequently sharpened by Hoeffding (1963). Lemma

5 adapts Bennett’s inequality to the situation at hand.

In Lemmas 4 and 5, P is a fixed probability measure and Ep is the

corresponding expectation.

Lemrpa 4: Let {Uk} be a sequence of independent random variables having mean
zero and finite variance. Let Uk 5 b = constant as. [P] for all k. Then for any 0

< a < b, there is a constant ,6 = 3(a) > 0 such that for all k, P(Uk 2 a) _<_
_ flk
e .

Lemma 4 is now modified to fit the examples under consideration in this

dissertation.

Lemma 5: Suppose that hypothesis H2 holds. Suppose also that with respect to P,

{Wk} is a sequence of independent random variables having mean u and variance

19
02. Let Wk = aka + bk, where Wk = (W1 + - - - + Wk)/k. Let 6 and A be
constants such that 0 < 6 < A.

(a) If Wk 2 p — A as. [P] for all k, then there is a constant ,3 = [3(6) > 0 such

that
P(Wk 5 h ._ a) = O(e-flk).
(b If W 5 p + A as. P for all k, then there is a constant B = fl 6 > 0 such
k I‘
that

P(V'vk z p + 5) = 0(e'5k).
Prpgf: (a) Let Uk = p — Wk and Uk = (U1 + + Uk)/k. Then EpUk = 0 and
Uk 5 A. Let 0 < 6 < A. Since ak -» 1, then there is a kg 6 N such that ak > 0
for all k 2 k0. Since Uk = p — Wk, then for k 2 k0,
P(Wk g p — 6) =P(aka + bk 3 p — 6)
=Plak(fl - Up) + bk 5 u — 6]
= P[Uk 2 (akp — p + bk + 6)/ak].
LetO < 6’ < 6. Since (akp — p + bk + 6)/ak—»6 > Oask—+oo,thenfork
sufficiently large, (akp — p + bk + 6)/ak 2 6’. By Lemma 4, there is a
constant 6 = 3(6’) > 0 such that for all k, P(Uk 2 6’) 5 e-flk. But then for k
sufficiently large,
P(Uk S It - 5) = Plfik .>. (akfl - II + bk + 6)/ak]
5 P(Uk z 6’) 5 e'flk.
(b) Let Uk = Wk — p and proceed as in part (a).

The following result is not the most general that can be obtained, but it

suffices for all of the examples to be considered in this dissertation.

Lemma 6: Suppose that hypothesis H2 holds.

20
(a) If the Vk’s are bounded below by p — A < p and if there are numbers 6 > 0
and m > 0 such that
Ep[|g(Vk)l”‘°l{Vk s a - 61}] = 006“).
then there is a 61 e (0, A) such that H4a holds.
(b) If the Vk’s are bounded above by p + A > p and if there are numbers 5 > 0
and m > 0 such that
E#[|g(Vk)|r+E-1{\7k 2 II + 52“ = O(km)a
then there is a 62 e (0, A) such that H4b holds.
Proof: (a) By Hiilder’s inequality with p = 1 + e/r and q = 1 + r/e,
E#[Ig(\7k)lr-1{Vk S I‘ - all]
s [Enllslvk)l”‘-1{vksp — spy/“+0.34%. - 61)” (W)
___ O[kmr/(r+e)]o[e—Bek/(r+e)]

(b) Similar to (a).

Lemma 1: Suppose that hypothesis H2 holds. If the Vk’s are bounded and there
are numbers 3 < 7 and e > 0 such that
~ k
Epls(Vk)lr+‘ = O(e" 4‘),
then H4 holds. In particular, H4 holds if E#|g(\7k)lr + 6 = O(eZkE/Sr).
Proof: This follows immediately from Lemma 6.

2.4. Examples

In order to check growth conditions such as those in Lemmas 6 and 7 in

Section 2.3, the following result of Cressie, et a1 (1981), can be used. It is also

21

useful for calculating biases.

Merge Mement Bepregentatign: Let V be a random variable that is almost surely
positive and let Mv(t) = E(etv) be the moment generating function of V. Then

for any a > 0,

00
EV'O‘ = P(a)'1{) ta‘lMV(-t)dt.
In particular, if V = Vk = aka + bk 2 0 a.s., then

.. oo
Eyvl: a = Nod-1% ta _ 1exp( - bkt)MV1( - akt/k)kdt.

Example I: Let g(x) = x. Then by Lemma 3, if H1 and H2 hold, then H(r,t)
holds.

Example 2: Let g(x) = 1/x. In each of the following, it is shown that H(2,3) holds
and the corresponding expansions are found.
(a) Let {Vk} be i.i.d. Gamma(a, ,3). Let ak = 1 and bk = 0 for all k, so that VI:
= Vk’ By Lemma 2(b), H4b holds. The Inverse Moment Representation can be
used to show that if ak > 3, then
EhlngH” 1 = Earp—3 = k3/fl3(ak—3)(ak—2)(ak- 1)

= 0(1) = O(e2k/5).
Thus, by Lemma 6(a), H4a holds. Since H1, H2, and H3 also hold, then H(2,3)
holds. In this example, ,u = 0,6 and 02 = (162, so that by Example 4(a) in
Section 2.2,

E#(1/Vk):1/p + O(l/k)

and

E,,(l/Vk _ l/p)2 =1/o3/32k + O(1/k3/2).

In this example, the bias of the estimator l/Vk of l/p is O(I/k). The

22
Inverse Moment Representation can be used to show that if Vk = Vk is replaced
by Vk = aka/(ak— 1), then an unbiased estimator is obtained. Since a
Gamma(a,,B) distribution is sometimes used to model the time until the ath
Poisson event, then in some cases a, and hence the CV-allocation, will be known.
(b) Let {Vk} be i.i.d. Poisson(;l). Let ak = 1 and bk = l/k for all k, so that Vk
= Vk + l/k. By Lemma 2(b), H4b holds. Since Vk 2 l/k, then
~ ~ -3 2k 5
E,,|g(vk)|1f+1 = Epvk 5 k3 = 0(e / ).
Thus, by Lemma 6(a), H4a holds. Since H1, H2, and H3 also hold, then H(2,3)

2

holds. In this example, it = a , so that by Example 4(a) in Section 2.2,

E,,(l/\7k) = up + O(l/k)
and

E,,(l/\7k _ l/p)2 = 1/p3k + O(1/k3/2).

As in part (a), Example 4(a) in Section 2.2 only yields that the bias of the
estimator 1/(Vk + 1/k) of l/p is O(1/k). The Inverse Moment Representation
pk

can be used to show that, in fact, the bias is e -
(c) Let {Vk} be i.i.d. Bernoulli(p), where 0 < p < 1. Let ak = k/(k+1) and bk
= 1/(k+1) for all k, so that Vk = (ka+1)/(k+1). Since Vk 2 1/(k+1), then
~ ~ —3 2k 5
Eplg(Vk)|’+1 = Envk s (k+1)3 = 0(e / )-
Since the Vk’s are bounded, then by Lemma 7, H4 holds. Since H1, H2, and H3
also hold, then H(2,3) holds. In this example, 02 2 p(1 — It), so that by Example
4(a) in Section 2.2,
E,(1/vk)=1/p + O(l/k)
and

Eta/V, - up)? = (1-#)/#3k + 0(1/k3/2)-

As in part (a), Example 4(a) in Section 2.2 only yields that the bias of the
estimator (k + 1)“ka + 1) of Up is O(l/k). The Inverse Moment

23

Representation can be used to show that, in fact, the bias is (1 - p)k + l/p.

Emple 3: Let {Vk} be i.i.d. Poisson(p), where p > 0. Let ak = 1 and bk =
1/k2 for all k, so that Vk = Vk + l/k2 and both H1 and H2’ hold. In each of the
following, it is shown that H’ (1,3) holds and the corresponding expansion is found.

(a) Let g(x) = x1/2

. Ifx > Oandx 95 p,then
ls<x> - S(I‘)l=lx1/2 - #1/2|=lx - 14/le + Ill/2| 5 IX - ul/lltl/ZI-
Thus, by Lemma 3, H4 holds. Since H3 also holds, then H’(1,3) holds. In this
example, 11 = 02, so that by Example 4(b) in Section 2.2,
Ep(\711(/2/p1/2) =1 _ c2/8k + 0(1/k) =1 _ c2/81t + 0(1/k).
(b) Let g(x) = 1/x1/2. By Lemma 2(b), H4b holds. Since v, z 1/k2, then
E#|g(\7k)lr+1/ 2 = Epvk‘3 g 1:6 = O(e2k/5).
Thus, by Lemma 6(a), H4a holds. Since H1, H2’, and H3 also hold, then H’(1,3)
holds. In this example, 02 = p and c2 = 1/02 = 1/p, so that by Example 4(c) in
Section 2.2,

Efl(p1/2/\711(/2) :1 + 3c2/8k + 0(1/k) :1 + 3/spk + 0(1/k).

Example 4: Let {Vk} be i...id Bernoulli(p), where 0 < [1 < 1. Let ak = k2/(k2
+ 2) and bk— _ 1/(1:2 + 2) for all k, so that vk= (152vk + 1)/(k2 +2) and both
H1 and H2’ hold. In each of the following, it is shown that H' (1,3) holds and the
corresponding expansion is found.
(a) Let g(x) = [(1 _ x)/x]1/2. Since 1/(k2 + 2) 5 v, < 1, then
Ig(\7k)l = [(1 - \7],)/\7k]1/2<1/V1/2

SO that

Elg(Vk)lr+1/2 < E V}: < (k2 + 213: Gem).
Since the Vk’ s are bounded, then by Lemma 7, H4 holds. Since H3 also holds,
then H’ (1,3) holds. Thus, by Example 4(d) in Section 2.2,

24
Ep[g(Vk)/s(fl)] =1 + (3-4#)02/8#2(1-#)2k + ou/k)
=1 + (3—4p)/3o2k + 0(1/k).
(b) Let g(x) = [x/(l — x)]1/2. Let Wk = 1 — Vk‘ Then {Wk} is i.i.d.
Bernoulli(1—p) and 1 - Vk = Wk = aka + bk’ where bk = 1 - ak - bk =
1/(k2 + 2) = bk' Since 1/(k2 + 2) 5 Wk < 1, then
lg(Vk)| = lip/(1 -\7k)]1/2 s 1/(1 - Yip”? = Vivi/2,
so that
~ 1 2 ~ -3 2k
Ease/pl” / s Epwk s (k2 + 2)3 = 0(e ’5).
As in part (a), H’ (1,3) holds, so that by Example 4(c) in Section 2.2,
Ep[s(Vk)/g(#)l = 1 + (4e-1)a2/8a2(1-u)2k + 0(1/k)
= 1 + (4p—1)/802k + 0(1/k).

CHAPTER 3
QUOTIENTS OF POPULATION MEANS

3.1. Introduction

In Chapter 1, a general theory of asymptotically optimal allocations was
developed and was used to obtain a previously known result on products of
population means. In this chapter, the theory in Chapter 1 is used to obtain new
results on quotients of population means. Earlier work on quotients of population
means can be found in Brittain and Schlesselman (1982), where a quotient p1 /p2
of Bernoulli proportions is estimated by a quotient Ylnl/X-an of sample means
and the allocation ni = M /(\Jq—1m + W)}n, where qi = 1 — pi, is
recommended. Since a Bernoulli(pi) population has mean pi and variance piqi,

then this recommended allocation is a CV-allocation.

In this chapter, a more general situation is considered in which a quotient
[11/112 of positive population means is estimated by a quotient of the form
(alnlxln1 +b1n1)/(a‘2nzx2n2+b2n2) _=.. xlnl/x2n2’ where the aik’s and bik’s are
constants. The quantity Xlnl/X2n2 is used as an estimator of 111/ [12 for three
reasons. First, if P (X = 0) > 0, then the more obvious estimator

92 2n2

Xlnl/Xan is undefined with positive probability. Second, the constants {aik}
and {bik} can sometimes be chosen to reduce the bias E(X1nl/X2n2'pl/fl2)'
Third, with Xlnl/XZn, as the estimator, it is possible to use the Inverse Moment

Representation of Section 2.4 to exactly calculate moments.

It will be shown in Section 3.2 that under certain conditions, CV-
allocations are asymptotically optimal in this more general situation. In Section
3.3, the results of Section 3.2 are applied to quotients of Gamma means and

quotients of Poisson rates, as well as to quotients of Bernoulli proportions.

25

26
3.2. Asymptotic Optimality of CV-Allocations

The theorem in this section gives sufficient conditions for CV-allocations to
be asymptotically optimal when a quotient p1 / p2 of population means is
estimated by a quotient XIII/5:232 of perturbed sample means. First, notation

similar to that in Chapters 1 and 2 is reintroduced. For i = 1, 2, let

1. {Xikz k e N} be a sequence of i.i.d. random variables with mean ”i at 0,

2

variance oi > 0, and coefficient of variation ci = ai/Ipil,
2. {aikz k e N} and {bik: k e N} be sequences of numbers such that aik = 1
+ O(l/k) and bik = O(l/k), and

Estimate [ll/[12 by X1111 /X2n2. In the notation of Chapter 1, this means that

(ii) f(2) = flopoa) = 0102 and 01 = up 02 = 1/112,

The definition of H(2,3) in the following theorem can be found in Section 2.2.

Theorem: If H(2,3) holds with g(x) = 1/x and Vk = sz, then CV-allocations are
asymptotically optimal.
Proof: Recall that conditions ‘3’, Ala, ‘25, and 0 from Chapter 1 are defined as

follows.

h

(:P: f is a polynomial of degree di in the it variable.

A: (Moment Expansions) For i = 1, , I,
(XI/k), j = 1

27
%: fi = fi(Q) ¢ 0 and 6i > 0 for all i.

0: If {m(N)} is optimal, then mi(N) —+ 00 for all i.
It is first shown that conditions 2?, Jlla, EB, and 0 hold. It then follows from the
corollary in Section 1.5 that first order allocations are asymptotically optimal.
Finally, first order allocations are calculated and shown to be CV-allocations.

Condition ‘sz Since f(Q) = 0102, then condition 9’ holds with d1 = d2 = 1.

Condition Alp: Verifying condition A consists of verifying the expansions
E91011, — 111? = O(I/k). 1' = 1
= 6%/k + ell/k). 5= 2
and
Bozo/>121, — 1/112)J= 00/10, 1' = 1
= og/k + 0(1/k), j = 2.
By Lemma 4 of Section 2.2 with Vk == Xlk’ the i = 1 expansions hold with 61 =
01. Since H(2,3) holds, then by Example 4(a) of Section 2.3, the i = 2

expansions hold with 62 = 02/113-
Condition ‘56: Since f1(Q) = 02 ¢ 0 and f2(Q) = 01 51$ 0, then the first part

of condition ‘Z holds. Since 61 > 0 and 62 > 0, then the second part of condition
% also holds.

Condition 0: Since f(Q) = 9102, the mean square error has the expansion
MSE (p) = 922 (z _ a )2 + 02E (2 _ o )2 + O(1/n n )
Q 2 Q, lnl 1 1 02 2n, 2 1 2’

By condition .1119, E91(Zini — 0i)2 —» 0 for i = 1, 2. Thus, MSEQ(p(N)) —. 0 for any
allocation {p(N)} with components tending to infinity. It follows that
MSE9(Q(N)) must tend to zero whenever {p(N)} is an optimal allocation. It will
now Be proved that the contrapositive of 0 holds. Suppose that {p(N)} is an

allocation such that {n1(N)} does not tend to infinity. Then {n1(Nj)} is constant

28
for some subsequence {Nj}' Along this subsequence, n2(Nj) —. oo, so that
2 2
However, since PQ-(Zik = 0i) < 1 for i = 1, 2, then the expectation on the right is
positive. Thus, {n(N)} is not optimal. Similarly, if {n2(N)} tends to infinity,

then {p(N)} is again not optimal. This proves that condition 0 holds.

Since conditions 9P, .Alt, %, and 0 hold, then first order allocations are
asymptotically optimal. But
|f1(Q)I61/(lf1(fl)|61 + |f2(fl)|52)= l02|01/(|02|01 + lanes/11%)
= (01/|#2|)/(01/I#2| + halve/1%)
= cl/(cl + c2),

so that in this case, first order allocations are CV-allocations.

3.3. Examples

In this section, the results from Section 3.2 are used to obtain
asymptotically optimal allocations for the estimation of quotients of Gamma

means, Poisson rates, and Bernoulli proportions.

In the following examples, the sequences {aik} and {bik} have been
selected so that H(2,3) holds and so that the bias EQ(Xlnl/X2n2 — pl/pz) is
reduced. Example 4(a) in Section 2.2 shows that in general the bias is O(I/N),
but in the examples below, the constants have been chosen so that the bias is at

worst O(cN), whereO < c < 1.

Example 1 (Quotiept pf Mme Means): For i = 1, 2, let {Xik: keN} be i.i.d.
Gamma(ai,fli). Then ”i = aifli, a? = aifliz, and Ci = l/ail/Z. Let f(Zn) =

(02n2—1)Yln1/02n2X2n2. (Here, 31k = 1, blk = 0, 3'21! = 02112/(02112—1), and
b2k = 0.) By Example 2(a) in Section 4.4, f(ZQ) is an unbiased estimator of

29
111/112. By the same example, H(2,3) holds with g(x) = l/x and Vk = sz.
Thus, by the theorem in Section 3.2, CV—allocations are asymptotically optimal.
In this case, CV-allocations are defined by

 

l/tla‘t
”i(N)/N *1/ch + l/Wz”

Example 2 (Qpetjent pf Peiemn Ba_te_s_): For i = 1, 2, let {Xikz keN} be i.i.d.
Poisson(pi). Then a? = ”i and Ci = l/pil/Z. Let f(Zn) = XlnI/(X2n2 + 1/n2).
(Here, 3'1]: = 1, blk = 0, a2k = 1, and b2k = 1/k.) By Example 2(b) in Section
4.4, the bias of the estimator f(Zn) is ple-”2k. By the same example, H(2,3)
holds with g(x) = 1/x and VI: = X2k' Thus, by the theorem in Section 3.2, CV-
allocations are asymptotically optimal. For Poisson distributions, the mean and

the variance are equal, so that in this case CV-allocations are defined by

N N l/ai
“i( V “’1/o, +1/02°

 

Example 3 (Quotiepf pf Bernerllli Proportions): For i = 1, 2, let {Xikz k 6N} be

i.i.d. Bernoulli(pi), where 0 < pi < 1, and let qi = 1 — p-.

_ 2 _
I Then [1i — pi, 0i —

piqi, and Ci = qi/pi. Let f(Zn) = (n2+l)Y1nl/(n2X—2n2+l). (Here, all: = 1,
blk = 0, a2k = k/(k+1), and b2k = 1/(k+1).) By Example 2(c) in Section 4.4,
the bias of the estimator f(Zn) is -p1q12‘+1p2' 1. By the same example, H(2,3)
holds with g(x) = l/x and Vk = X2k' Thus, by the theorem in Section 3.2, CV-
allocations are asymptotically optimal and are defined by

.. _, tire
“NW lat/p1 +lqs/pa'

 

CHAPTER 4
A TWO-STAGE ALLOCATION PROCEDURE
FOR ESTIMATING PRODUCTS

4.1. Introduction

First order allocations typically depend on unknown parameters. In this
chapter a two-stage procedure is developed which does not depend on the
knowledge of unknown parameters. The procedure to be developed allocates N

observations in two stages:

Eirst Stage (or Pilot Example): Use m of the N observations to estimate
-th

the unknown parameters, with mi observations allocated to the 1

population. (Thus, 2% lmi = m.) If the unknown parameters are

functions of population means (pi), then the estimates of the unknown

parameters are frequently functions of sample means (vim) or perturbed
1

sample means (?imi)'

Seeope Smge: Use the information from the first stage to estimate the
parameters involved in first order allocation and hence to obtain an
approximation to first order allocation. Use this approximation to first
order allocation to allocate the remaining 11 = N — m observations, with
Ti observations allocated to the ith population. (Thus, Eilei = n.)
The notation Ti’ instead of ni, is being used because the number of

th

observations allocated to the i population in the second stage is random,

30

31
being a flmction of the first stage observations. If f(pl, ,pI) is the

quantity to be estimated, then it is frequently the case that f(X—ITI,

’XITI) or f(XlTl, ’XITI) can be used as the estimator.

It would, of course, be desirable to have a general theory for two-stage
allocation procedures analogous to the general theory for one-stage allocation
procedures developed in Chapter 1. Thus far, however, such a general theory has
proved to be elusive and it has only been possible to obtain results in specific
cases. In this chapter, the case of the product of two population means will be
considered, so that f(Q) = 0102 = #1112. As in previous chapters, the setting for
this two-stage procedure will be classical; for a Bayesian two-stage allocation

procedure for estimating products of normal means, see Rekab (1991).

The description of the first and second stages for this special case of the

product is now formalized.

P3191; Smnpie (m observations): For i = 1, 2, let {Yikz k e N} be i.i.d. with
mean ”i and variance 0‘? > 0. Estimate ”i by Vim, =

(Yi1+ - - - +Yimi)/mi’ where m1 = [m/ 2] = greatest integer less than or
equal to m/2, and m2 = m - m1.

3% Smge (n = N — m observations): For i = 1, 2, let {Xikz k E N} be
i.i.d. with mean ”i and variance 0?, and let Xik = (xil + - - - +xik)/k'

For i = 1, 2, let Ti be positive integer valued and measurable with respect

to the sigma field 9(m1,m2) = 0{Yik : k 5 mi, i = 1, 2}. (The exact

32
choice of T1 and T2 will be made in Section 4.3.) Estimate 111112 by

x1'1‘,X2T,-

Observe that an approximately balanced allocation is being used in the
pilot sample, i.e., mi 2 m/2. This assumption is made in order to simplify
several results later on. Most of the following results (with the appropriate
modifications) can, in fact, be proved if all that is required is that mi ... oo for all
i.

The mean square error for this two-stage procedure is E(X1TIX2T2 —
p1p2)2, which will be denoted by MSE(N). (Two remarks about notation are in
order at this point. First, throughout this chapter, dependence of expectations on
parameters will be suppressed. Thus, for example, Be and MSEO will be
abbreviated to E and MSE, respectively. Second, if the mean square error of a
(one—stage) allocation {p(N)} is being considered, then the notation that will be
used is MSE(p(N)). Thus, it will always be clear whether mean square error is

with respect to a one-stage procedure or a two-stage procedure.) As in Chapter 1,

an expansion of MSE(N) will be obtained. This expansion will be used to
1. show that the two-stage procedure with unknown parameters does
asymptotically as well as optimal one-stage procedures with known
parameters, and

2. make a recommendation for the size of the pilot sample (relative to N).

Assumptiops: 1. It will always be assumed that random variables corresponding to
different stages are independent. Using the notation above, this means that Yij is

independent of Xi’j’ for all i, j, i', and j’.

33
2. Unless otherwise indicated, sums will be from one to infinity, so that, for

example, 2k will be an abbreviation for Eiil.

4.2. Preliminary Result

In this section, the problem of obtaining a mean square error expansion for

the two~stage procedure is reduced to the problem of finding expansions for

E(1/T1), E(1/T2), and E(1/T1T2).

Theorem: The mean square error can be written
_ _ — 2
= o§h§E(1/T1) + p§o§E(1/T2) + c§o§E(1/Tlr2)
= p§pglc§E(1/T,) + cgEu/rz) + c§c3E(1/r,r2)).
Proof: For i = 1, 2, let Wi = XiT- — pi. The first step is to write
(En-621‘, ' ”1112)2 = Wiwi + 21‘1""1Wi + Zl‘zwiwz
+ Piwi + Fiwi + 2I“1I‘2W1W2
and the second step is to take expectations. But before beginning the second step,
three preliminary results will be established. In the following claims, ‘1? is a sigma
field which corresponds to the sigma field g(ml,m2) above and the random
variables Vlj and V2k correspond to second stage estimates KIT, and XZT; In

all three of the claims it will be assumed that
1. Vlj and V2k are independent of ff for all j and k,
2. V1 j is independent of V2k for all j and k,
3. there are numbers BI’ 32 6 IR such that for all j and k, EIVIJ-I 5 BI
and E|V2k| 5 B2, and

4. T1 and T2 are 9-measurable and take values in N.

34
Claim 1: For i = 1, 2, E(ViT.|‘EF) = 2kI{Ti=k}EVik.
Preef: To reduce notation, write T instead of Ti and Vk instead of Vik‘ Since
l2§=1vkllr=kll .<. EkIVkll{T=k}
and
E2k|Vk|1{T=k} = EkEIVk|1{T=k} = EkElelpiT=kl 5 Bl < 00,
then by the dominated convergence theorem for conditional expectations (see
Theorem 6.5.5 of Ash (1972)),
E(VTI°I) = ElzkvkllT=klm = EkE(Vk1{T=k}I°I)
= Ek1{T=k}E(Vk|‘J) = Ekl{T=k}E(Vk).
Cl_ajm 2: For i = 1, 2, E[Wi|€I(m1,m2)] = 0 and E[Wi2|‘:'f(ml,m2)] = ai2/Ti.
Proof: By Claim 1 with ViTi = Nil-Ill and 9? = c.l‘(m1m2),
EIW,|‘3F(m1.m2)] = EkliTi=k}E(Xik - #5) = 0-
Also,
E[Wi2|?I(m1,m2)] -_- zkl{ri=k}E(xik .. pi)2 = 022k(1/k)I{Ti=k} = oZ/ri.
Clmm 3: E(V1T1V2T2|‘:T) = E(V1T1|€I)E(V2T2|€F)
me: Since
IZJLlVljliT1=llijzlvgkliT2=kil s 23,-2lele|V2k|1{T1=i}1{T2=k}
and
E22,-:kIVljIIV2kI1{T1=j}1{T2=k} = Z; 21(13anllvgklliT1=jlliT2=ki
= Ej:kElvlj|E|V2k|P{Tl=j9T2=k}
S B1B2 < 00,
then by the dominated convergence theorem for conditional expectations and by
Claim 1,
= 2,- 2kE(V1jV2k1{T1=j}1{T2=k}|€}')
= E J-Ek1{T1=l}1{T2=k}E(V1j)E(V2k)

35
Qantinnfiiengfnmicfthmz By Claim 3, fori, k =1,2,
E(W11Wtf) = E{E[WJ1W12‘|€F(m1,m2)]} = E{E[W’1|€F(m1,m2)]E[W12‘|‘-F(mpm2)]}-

Thus, by Claim 2,

E(leW12‘) = 0,5 = 1 or k = 1,
and

E(w§w§) = e§o§E(1 /T1T2)
Also by Claim 2, for i = 1, 2,

Ewi2 = E{E[Wi2|‘:T(m1,m2)]} = ai2E(1/Ti).

Now use the expansion of (XITIXBT, - p1p2)2 from the first step of the proof.

4.3. MSE Expansion

From Section 1.6 it is known that CV—allocations are asymptotically
optimal for estimating products. Thus, it would be natural to choose T1 and T2 so

that they are approximately CV-allocations, i.e., Ti : Ain = [ii/(61+62)]n,

where 6i is a ”{Yij : 1 5 j 5 mi} measurable estimator of the coefficient of

variation ci. There are, however, some technical difficulties which arise from the

following restrictions on T1 and T2.
1. T1 and T2 must be integer valued.
2. T1 and T2 must sum to n, i.e., T1 + T2 = n.
3. At least one observation must be allocated to each population, i.e., it

must be the case that 1 5 T1,T2 5 n— 1.

Keeping in mind these restrictions on T1 and T2, set
T1=1{A1n < 1} + [Aln]1{A1n Z 1}, T2 = n — T1,

where [Ain] is the greatest integer less than or equal to Ain.

The next result says that l/Ti is approximately equal to l/Ain. This is

36

important because E(1/Ain) is easier to work with than E(1/Ti).

Clejm: Ifé, > 0 and 62 > 0, then fori = 1, 2, Il/Ti — l/Ainl 5 2/Ai2n2.
m: First consider the i = 1 case. Let I, = 1{A,n<1}, 12 =
1{15A1n 5n—1},I3 = 1{A1n >n—1}, and I23 =12 + 13 =1{A,n21}. Write
l/Tl = I, + {l/[Aln]}123 = {1/([A,n]+1)}I, + {1/[A1n]}I23
= l/Aln + {1/([A1n]+1)—1/A,n}11 + {1/[A1n]—1/A,n}I23.
Since
|1/([A,n]+1) _ 1/A,n)l, 5 {|A,n—[A,n]—l|/A,n([A,n]+1)} 5 1/A§n2
and
|1/[A,n] — 1/A1n|123 = {(Aln—[Aln])/A1n[A,n]}123
s (l/AinZHAln/[Alnblzg
s (I/Ain2){([A1n] +1)/[A1n]}123
= (I/Ainznuu/lAlnlnIp, s 2/A%n2,

then the result follows for i = 1.

Now consider the i = 2 case. Let J, = 1{A2n<1}, J2 =
1{15A2n5n-1}, and J3 = 1{A2n>n—1}. Observe that J, = 13, J2 :12, and
J3 = 1,. Write
T2 = n _ T, = n — (I, + [Aln]12 + (n—1)I3)
= J, + (n-[Aln])J2 + (n—1)J3
= ([A2n]+1)J, + (n—[Aln])J2 + [Azn]J3,
SO that
1/T2 = {1/([A2n]+1)}J, + {1/(n-[A1n])}J2 + {1/[A2n]}J3
= 1/A2n + {1/([A2n]+1) — 1/A2n}J,
+ {1/(n—[A1n]) — l/Azn}J2 + {1/[A2n] — 1/A2n}J3.

As in the i z: 1 case,

37
|1/([A2n]+1) _ 1/A2nll, _<_ 1/Agn2
and
|1/[A2n] — 1/A2nIJ3 5 2/A3n2.
But also
|1/(n—[A,n]) _ 1/A2nIJ2 = {IA2n+[A2n]—n|/(n—[A,n])A2n}J2
{1/(n—[A1n])A2n}J2 5 {1/[A2n]A2n}J2
(l/Agn2){A2n/[A2n]}J2
s (1/A§n2){([A2n] +1)/[A2n]}J2
= (1/lotgn2){1+(1/[A2n))}12 5 2/Agn2.
Thus, the result also holds for i = 2.

IA

IA

Theorem (MSE Expansion): Suppose that for i = 1, 2,
1. E(e§) < 00, Email) < co,

3. mi = m/2.
Then
MSE(N) = p§p§[(c,+c2)2 + 2c,c25/m + o(1/m)]/n + 0(1/n2),
where{= p, + p2 + ¢1+ 452.
Proof: Since
1/A‘f = 1 + 4&2/c, + sag/c? + slag/oi} + tag/8,1,
then by the independence of 61 and 62 and by assumption 2 above, E(1/A%) < 00.
Similarly, E(1/A§) < 00. Since E(1/A;1) < oo, then E(1/|Ai|3) < 00 and
E(1/Ai2) < 00. Since E(1/Ai2) < oo, then by the claim,
_ 2
E(1/Ti) — E(1/Ain) + O(l/n ).
Butalsobth 1' 2.. 22 33 44
y e calm, Il/Ti l/Ain | 5 4(1/Ain + l/Ain ), so that
E(1/T?‘) = O(1/n2). Thus, by Schwarz’s inequality,

38
E(1/T,T2) = O(1/n2).
It now follows from the theorem in Section 4.2 that
MSE(N) = p§p§[c§E(1/T,) + cgEu/TZ) + c§c§E(1/T,T2)]
= p%p%[c%E(1/A1) + c§E(1/A2))/n + O(1/n2).
Since Ai = (ii/(é, +62), then
By the independence of 6, and 62,
E(1/A,) = 1 +E(62)E(1/é,), E(1/A2) =1 +E(é,)E(1/&2).
Thus,
MSE(N) = 2 2[(c2 c2) c2E(& )E(1/<‘: ) + c2E(& )E(1/c )]/n + O(1/n2).
“W2 1 + 2 + 1 2 1 2 1 2
Now by assumptions 3 and 4,
c%E(c2)E(1/c,) = c1c2E(&2/c2)E(c,/&,)

= C1C2[1 + p2/m2 + o(1/m2)]-[l + ¢1/m1 + o(1/m1)]

= c1c2[1 + 2p2/m + o(1/m)]-[1 + 2¢,/m + o(1/m)]

= clc2[1 + 2(p2 + ¢1)/m] + o(l/m).
Similarly,

c§B(c2)E(1/c,) .—. c1c2[1 + 2(p1 + ¢2)/m] + o(l/m).
Thus,
c§E(c2)E(1/c,) + c§E(c,)E(1/c2) = 2c,c2 + 2c1c2§/m + o(1/m)

and the result follows.

4.4. Asymptotic Optimality

In the following theorem, let m(N) = m,(N) + m2(N) be the total number
of observations allocated to the first stage and let n(N) = N - m(N) be the total
number of observations allocated to the second stage. Recall that MSE(N) is the

mean square error for a two-stage procedure and MSE(M(N)) is the mean square

39

error for a one-stage procedure.

Iheerem: In addition to the assumptions of the previous section, assume that
1. n(N)/N —> 1, and

2. m(N) —+ co.
Then the two-stage allocation does asymptotically as well as an optimal one-stage
allocation, i.e.,
lim sup [MSE(N)/MSE(M(N))] S 1,
where {M(N)} is an optimal one-stage allocation.
m: Let {M(N)} be an optimal one-stage allocation and let {M’(N)} be a CV-
allocation. From the results of Chapter 3,
lim sup [MSE(N)/MSE(M(N))] = lim sup [MSE(N)/MSE(M’(N))].
Also from Chapter 3,
lim N-MSE(_M’(N))1=(22a1 + slag)? = 11%ch + c212.
But from the MSE Expansion in the previous section,
lim sup N-MSE(N)
= 11m {lN/n(N)l-a%a§~l(c1+c2)2 + 260162/m(N)
+ o(1/m(N))] + Noe/um?»

= 11%(01 + 02)2

4.5. Square Root Rule for Pilot Sample Size

In this section a recommendation is made for m - the size of the pilot
sample. From the previous section, it is known that if n/N —+ 1 and if certain
other conditions hold, then an asymptotically optimal two-stage allocation is
obtained. But if n/ N —» 1, then m = o(n). Hence I/n2 will be small compared to
l/mn. This motivates the dropping of the O(l/nz) term in the MSE Expansion

40
in the previous theorem and writing
MSE(N) e page, + c2)2 + 2c,c25/m + o(1/m)]/n.
The expression on the right, subject to the constraint m + n = N, is minimized
when
m=.]—ATN—+—A) _ them + A) - Bat—:7),
where
1 = 2c,c25/(c, + c2)2.
Assuming that N > A, then an approximate minimum occurs when
m = {TN .
This equation constitutes the M ro_ot_ ml_ - a recommendation for the size of

the pilot sample.

4.6. Examples

In all of the examples in this section, the estimator 6i of the coefficient of
variation Ci is a function of first stage sample means. The theorem below
specializes the MSE Expansion of Section 4.3 to this situation. The following

lemma uses the notation of Chapter 2.

Lemma: Let h:R —. R. Suppose that H’(1,3) holds with g = h and also that
H’(1,3) holds with g = 1/h. Then
E[h(Vk)/h(n) + h(u)/h(\7k)] = 2 + h’(lu)202/h(lu)2k + o(l/k)-
Proof: Since H’ (1,3) holds with g = h, then by Example 3(b) in Section 2.2,
E[h(Vk)/h(n)l = 1 + [h”(#)/h(u)102/2k + 00/10.

a s = 1/h.then s”(#)/g(u) = 2111102116)? — ham/ho). Thus, Hus)
holds with g = l/h, then again by Example 3(b) in Section 2.2,

E[h(lu)/h(\7k)] =1 + [2h’(#)2/h(u)2 - h”(#)/h(#)102/2k + 00/10.

41
Thus,

E[h(Vk)/h(lt) + h(u)/h(\7k)] = 2 + h’(2)202/h(2)2k + 00/2).

Ire-mam (MSE Expapsipn): Let M? .. IR and for i = 1, 2, let a, = h(Y,mi).
Suppose that for i = 1, 2,

1. E(c;1) < oo, E(1/é;1) < oo,

2. H'(1,3) holds with g = h, Vk = Yik and H’(1,3) holds with g = l/h,

Vk = Yik’ and

3. mi = m/2.
Then

MSE(N) = h§p§[(c,+c2)2 + 2c,c25/m + o(1/m)]/n + O(1/n2),

who. 5 = h’(#1)20f/h(#1)2 + h'(#2)20%/h(#2)2-
Preef: By the lemma, for i = 1, 2,

Elam/110,) + h(lti)/h(Y,k)] = 2 + edifice/110,12 + o(l/k).
Thus, using the notation of the theorem in section 4.3,

p, + 2, = h’(ui)20i2/h(lu,)2-

The result now follows from the theorem in Section 4.3.

Remark: A special case of the previous theorem is when h(x) = xp, p at 0. Then

h’(#i)/h(lu,) = p/lui.
so that

t: h’(#1)20f/h(#1)2 + h’(#2)20§/h(112)2

= Vivi/11% + 2203/23 = 1020:? + 63)-

Example 1 (Kpowe Yariances): Assume that the variances 0% and 0% are known

and the coefficients of variation are estimated by E, = 0i/?im-' Suppose that
l

42
H’(1,3) holds with g(x) = h(x) = x, Vk = Yik and that H’(1,3) holds with g(x) =
1/h(x) = l/x, Vk = Yik' Then by the remark after the theorem,
6: p2(c% + 0%) = c? + c%.
The square root rule 111 = {AN is now used to make a recommendation for the size
of the pilot sample. Since 6 = c? + cg, then
A = 2c,c25/(c, + 02)2 = 2c1c2[1 + (cl/c2)2]/[1 + (cl/c2)]2.
Since for all x 2 0,
1/2 s (1 + x2)/(1 + x)2 s 1,
then
c1c2 5 A 5 2c1c2.
The recommendation for the size of the pilot sample is, therefore,
W s m s W-

For example, for estimating a product of Gamma means when the variance

is known, the recommendation for the size of the pilot sample is
(1/a1a211/4W s m s (1/a1a211/4C2N .

This recommendation is useful, for example, if there are known upper and lower

bounds on a, and a2.

Having chosen m = size of the pilot sample, use half of these observations
to estimate 11, by Ylm,’ where m, = m/2, and the other half of the observations
to estimate 112 by 172,112, where m2 = m/2. Now estimate the coefficient of
variation Ci by (‘3, = ai/Yim, and allocate the remaining 11 = N — m second stage
observations as follows: to the first population, allocate T, = [El/(61+62)]n
observations, and to the second population, allocate T2 = [62/(61 +62)]n

observations. Finally, estimate 111112 by X',T1X’2T2.

Ex_amp_1e 2 (Poissop Means): For i = 1, 2, let {Yikz keN} be i.i.d. Poisson(/1i).

 

43
2 _ _ 1
Then oi — ”i and Ci - l/pi
H’(1,3) holds with g(x) = h(x) = xl/z, Vk = Yik and that H’(1,3) holds with g(x)
= 1/h(x) = 1 /x1/ 2, Vk = Yik' Then by the remark after the theorem,

6: p2((:% + c%) = (c? + Cg)/4.

/2. Estimate Ci by 6i = Viv/$143. Suppose that

As in the previous section, it can be shown that
C1C2/4 S A S C1C2/2.

Since Ci = 1/0,, then the recommendation for the size of the pilot sample is

,lN/40102 5 m 5 ,lN/20'102 .

Now proceed as in the previous example, replacing 6i = oi/Y- by 6- =

1m; 1
~1/2
1/Yim, '

Exempe 3 (Bemmm Propertions): For i = 1, 2, let {Yikz keN} be i.i.d.
Bernoulli(pi). Then oi? = pi(l—pi) and ci -_- [(1_p,)/p,]1/2. Estimate ci by c, =
[(l—Yimi)/Yimi]1/2. Suppose that H'(1,3) holds with g(x) = h(x) =
[(1-x)/x]1/2, vk = Yik and that H’(1,3) holds with g(x) = 1/h(x)
[x/(l-x)]1/2, vk = Yik. Since h'(,l,)2/h(h,)2 = 1/4p§3(1_p,)2 = 1/4o31, then by

the theorem,

t: h'lel)2a%/h(al)2 + h’(#2)2a§/h(22)2 =(1/4)(1/a% + meg),
so that

A = 2c1c26/(c, + c2)2 = - -- = (0% + 0%)/20’10’2(0’1 + 02)2.

Reasoning as in Examples 1 and 2,
1/40102 5 A 5 1/20102,

so that the recommendation for the pilot sample size is

,lN/40102 5 m 5 ,lN/20102.

Now proceed as in Example 1, replacing 6i = ai/Yim- by 6. =
l

[(1 — i.....)mmll/Z.

BIBLIOGRAPHY

Ash, RB. (1972), Real Analysis and Probability, Academic Press, New York.

Bennett, G. (1962), Probability inequalities for the sum of independent random
variables, Journal of the American Statistical Association, 57, 33-45.

Brittain, E and J.J. Schlesselman (1982), Optimal allocation for comparison of
proportions, Biometrics, 38, 1003-1009.

Chow, Y.S. and H. Teicher (1988), Probability Theory: Independence,
Interchangeability, Martingales, 2nd Ed., Springer-Verlag, New York.

Cochran, W.G. (1977), Sampling Techniques, 3rd Ed., John Wiley & Sons, New
York.

Cressie, N., A.S. Davis, J.L. Folks, and CE. Policello (1981), The moment-
generating function and negative integer moments, The American Statistician, 35,

148-150.

Fabian, V. and J. Hannan (1985), Introduction to Probability and Mathematical
Statistics, John Wiley and Sons, New York.

Hoeffding, W. (1963), Probability inequalities for sums of bounded random
variables, Journal of the American Statistical Association, 58, 13-30.

Page, C.F. (1990), Allocation proportional to coefficients of variation when
estimating the product of parameters, Journal of the American Statistical
Association, 85, 1134-1139.

Rekab, Kamel (1991), A nearly optimal two stage procedure, Preprint.

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