RD-R173 114 EXTREME POINT METHODS IN THE STUDY OF CLASSES OF 1/1BIYARIATE DISTRIBUTIONS A .(U) PITTSBURGH UNIY PACENTER FOR MULTIYARIRTE ANALYSIS K SUBRAMANYAN ET AL.

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EXTREME POINT METHODS IN THE STUDY OF

0CLASSES OF BIVARIATE DISTRIBUTIONS AND SOME

wAPPLICATIONS TO CONTINGENCY TABLES

by

IK. SubramanyamDepartment of Mathematics and Statistics

*University of Pittsburgh

M. Bhaskara Rao

Department of Probability and StatisticsUniversity of Sheffield, UK

DTIC* $ELECTE 0

tB

Center for Multivariate Analysis

University of Pittsburgh

°ITRIBUTON STATEMENT A

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EXTREME POINT METHODS IN THE STUDY OF GLASSES OF Technical - April 1986BIVARIATE DISTRIBUTIONS AND SOME APPLICATIONS TO ~~R~MN R.R~OTMME

CONTINGENCY TABLES .ERIAftDGRLRT0uV

I. AUTHOR(a) 6. CONTRACT OR GRANT NUMUER(s)

K. Subramanyam and M4. Bhaskara Rao F49620-85-C-0008

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W. SUPPLEMENTARY NOTES

'i KEY WORDS (Cmnino.an rieverse aide if necesary~ aid Identify by block numbst)

Extreme point, convex set, compact set, bivariate distributions, positivequadrant dependence, negative quadrant dependence, power function andcontingency tables.

20 ASTRACT (CodaIgfl'a0an reverse aide If necessary and Identity by block rnusibw)

The set of all bivariate positive quadrant distributions is neither compact noronvex. But the set of all bivariate positive quadrant distributions with fixed

arginals is a convex set. These convex sets are compact in the case of discrete

ivariate distributions if the marginals have finite support. A simple method

o0 enumerate the extreme points of these convex sets is given.In the context of contingency tables for testing the null hypothesis independencgainst the alternative positive quadrant dependence one can use the method ofxtreme point analysis to comp are the-pro e npn 1nj

DO 1JA73 1473 Unclassified

SECURITY CLASSIFICATION Of THIS PAGE (WhSen DlS 1010*e60)

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63

REFERENCES

[1]. AGRESTI, A. (1984), Analysis of Ordinal Categorical Data, John Wiley,New York.

[2]. EATON, M.L. (1982), A Review of Selected Topics in Multivariate ProbabilityInequalities, Ann. Stat., 10, 11-43.

[3]. KEMP Jr., L.F. (1968), Construction of -Joint Probability Distributions,Ann. Math. Stat., 39, 1354-1357.

[4]. LEHMANN, E.L. (1966), Some Concepts of Dependence, Ann. Math. Stat., 37,1137-1153.

[5]. PHELPS, R.R. (1966), Lectures of Choquet's Theorem, Van Nostrand, New York.

[6], BHASKARA RAO, M., KRISHNAIAH, P.R. and SUBRAMANYAM (1985), A StructureTheorem on Bivariate Positive Quadrant Dependent Distributions andTests for Independence in Two-way Contingency Tables, Technical ReportNo. 85-48, Center for Multivariate Analysis, University of Pittsburgh,Pittsburgh.

.

a~ 5'

* . S ~.?: 4' b

EXTREME POINT METHODS IN THE STUDY OFCLASSES OF BIVARIATE DISTRIBUTIONS AND SOME

APPLICATIONS TO CONTINGENCY TABLES -

by --. 1

K. SubramanyamDepartment of Mathematics and Statistics 4:

University of Pittsburgh

M. Bhaskara RaoDepartment of Probability and Statistics

University of Sheffield, UK

DTICA LECTE

April 1986 JUL 2 3 W6

B%Technical Report 86-12

Center for Multivariate AnalysisFifth Floor, Thackeray HallUniversity of PittsburghPittsburgh, PA 15260

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L$.ixtrbudmo Unlimited MAl$C"4F.Ch~,..ief. Iflf arseetgDlyisio

*Research sponsored by the Air Force Office of Scientific Research (AFSC) under

Contract F49620-85-C-0008. The United States Government is authorized to repro-duce and distribute reprints for governmental purposes notwithstanding anycopyright notation hereon.

IQ .*

EXTREME POINT METHODS IN THE STUDY OFCLASSES OF BIVARIATE DISTRIBUTIONS AND SOME

APPLICATIONS TO CONTINGENCY TABLES

by

K. SubramanyamDepartment of Mathematics and Statistics

University of Pittsburgh

M. Bhaskara RaoDepartment of Probability and Statistics

University of Sheffield, UK

Abstract

The set of all bivariate positive quadrant distributions is neither compact

nor convex. But the set of all bivariate positive quadrant distributions with

fixed marginals is a convex set. These convex sets are compact in the case of

discrete bivariate distributions if the marginals have finite support. A simple

method to enumerate the extreme points of these convex sets is given.

In the context of contingency tables for testing the null hypothesis indepen-

dence against the alternative positive quadrant dependence one can use the method

of extreme point analysis to compare the performance of various tests.

AMS 1980 subject classification: Primary: 62H05 Secondary 62H17

Key words and phrases: Extreme point, convex set; compact set, bivariate distri-butions; positive quadrant dependence, negative quadrant dependence, power functionand contingency tables.

1. Introduction. Let M2 be the collection of all probability measures defined

on the Borel a-field of the two-dimensional Euclidean space, R 2 . For p in M2,

let v and 12 denote the marginal probability measures on the Borel a-field1 2

of R , i.e., U 1(A) =- p(A x R) and P 2 (A) = p(R x A) for every Borel

subset of R . A u in M2 is said to be positively (negatively) quadrant dependent2I(PQD) (NQD) if P{[c, Wa) X [d, -")} > (< ) p 1 fc, -)I P2 {Ld, -) } for all c, d

in R . Let M(PQD) denote the collection of all probability measures in M2 which

are PQD and M(NQD) the collection of those in M2 which are NQD . The main

objective of this paper is to study the structure of M(PQD) and M(NQD) . Using

the structure theorem given in Section 3, we give some applications in the area of

testing hypotheses in Contingency Tables in Section 4. For further information on

properties of measures which are PQD or NQD, see Lehman [4] or Eaton

t' '" ....

4LA 1I

* w

2

2. Preliminary results. The sets M(PQD) and M(NQD) are not convex sets.

Examples are easy to provide. We give a simple sufficient condition under which

convex combinations of measures in M(PQJ) are again in M(PQD). Let M1 denote

the collection of all probability measures on the Borel a -field of R. For v , n

in Ml, v is said to be stochastically smaller than n if '{C, -)) < f{c, a))

for all c in R. If v is stochastically smaller than r , we use the notation

v <st n . The relation " < st is a partial order on MI. For , i in 2

we say that V is stochastically smaller than X if w < st Xl and

< st X In this case, we again use the same notation < st X . The

relation " < st " on M2 is only a pre-order. See Definition 2.1 of Eaton

p.lfl.

Proposition 1. If p,X e M(PQD) and either V < st X or X < st

then ap + (1 - a)X e M(PQD) for any O< a < 1.

Proof. We need to show that +ap + (1- a)3{L, -) x d, c)) >

[a 1 + (1 - a) Xlf3[9,-)' Va12 + (1 - a) X21{[4,)) for all c, d in R. We

note that EV + (1 - a)X3{ , -) x Ld, -)) > al{c, C') 2 { E ,) +

(1 - a) X1 { C, X 2{ C )) and ap1{ {CC, P2{, {E)d +

( - a) X ,1 {c cc X2 C, - 2-i{F c, {[) U2{

a(l - a) v1{Lc, X 2 a(l - a, 2 Ld, ) 1 EU, -

(l - a) 2 X 1{CC, X) 2 c)Ed, a (1 - a)[: p{C' c)) p{d, cc) +

{C, -c)) { ,)) -X {)I p {. c,)) - I ,{-) X) oPE., Xf-

- a(l - a) Epl{Ec cc)- A1 fc)jJLi 2 {1d, cc) f [2 d, c > 0, if

U < st X or A < st V . This completes the proof.

An analogous result holds for measures which are NQD as declared in the

the following proposition.

3

Proposition 2. Let p , M(NQD). If ' 1 < st X 1 and X 2 <st p 2 or

1 <st p 1 and P 2 <st X 2' then a + (1 - a) X E M(NQD) for any

0 <a< 1.

Corollary 3. Let p ,X e M2 have the same marginals, i.e., U 1 and

p2 = 2" If p , X M(PQD), then ap + (1 -a )k X M(PQD) for any

0 < a < 1. If p , £ M(NQD), then ap + (1 -a )X E M(NQD).

In order to determine the structure of M(PQD) and M(NQD), Corollary 3

provides a pointer. Let and be two given probability measures on the

Borel a -field of the real line R. Let

and

MNQD( I''2 ) = X e M(NQD) ; i = 1i and X 2 = 1 2 )

MPQD( Vp 2 ) is the collection of all those probability measures X which are

PQD and having the same marginals p 1 and p 2 This set is more tractable than

M(PQD) as the following corollary points out.

Corollary 4. MpQD(P 12 12) and MNQD(' p 2) are convex sets.

Our next objective is to determine the extreme points of the above convex

sets. If MpQD( V 2 ) is a compact set in some relevant nice topology, then one

can write every X in MpQD ( , 2) as a mixtur- of its extreme points. See

Phelps [ ). Then a knowledge of all extreme points of the set MpQD( 1' p 2 ) provides

us an insight into the structure of MQ(U ,p 2). In the sequel, we concernMQD 1

ourselves exclusively with the convex set MPQD(P' il, 2 The convex set MNQD( 1 ' 2'

4

can be studied in a similar vein. The determination of all extreme points of the

set MpD( 1i U2 ) seems to be difficult in the general case. However, in Section 3,

we give a complete description of all extreme points of MpQD( i p 2 ) when PI

and V.2 are some special discrete measures with finite support. The method given in.

Section 3 can be adapted for any discrete measures 1 and U with finite*1 2

support. Before sailing into the next section, we identify one particular measure

as an extreme point of MpQD( ' 12) in the general case. For probability measures

d I and I2 on the Borel a-field of the real line, let 1 x 2 denote the12 1 2

product probability measure in M2.

Proposition 5. The product probability measure 1 x V 2 is an extreme point

of V 1' 2) and also of MNQD( ' )

Proof. One can prove this result by using the elementary fact that if c > a,

d a, 0 < a < 1 and ac + (1 - a)d = a, then c = a and d = a, and the

fact that any measure in M2 is determined by its values on sets of the form

[c, -) x [d, ®) c, d real.

Remark. The above result is in marked contrast to the following result of Kemp [3].

Let M( Pip P ) denote the collection of all probability measures X in M2

satisfying Ul = 1l and X= 2. Then M( p1 , l 2 ) is a convex set, and

P1 x 42 is an extreme point of M( pi 1 2) if and only if either iis

0-1 valued or p2 is 0-1 valued..d2

- - - -- . - - . - P. 7. .. T. - -

5

3. Extreme points. In this section, we assume that the probability measures

P and p 2 have finite support. For simplicity, assume that the support of P1 2

is {l,2,3,...,m} and that of 11 is {1,2,3,...,n) . Any probability2

measure in M2 with support contained in {(ij) ; i = 1 to m and j = 1 to n}

and with marginals p i and pi can be identified as a matrix P = {p ij}

of order m x n satisfying pi > 0 for all i and j, Z m = (nf a i=l ij 2

for all j = 1 to n, and Z n Pi=pl~j=l 1 ij7" 1(t)foal i 1tom Le

j ({i}), i 1 to m and q. = 1 ({j}), j = 1 to n. The objects of interest are1 J 2

the convex sets MpQD(1 ,) and (NQD1 9) We denote p )QD(I by

MPQD(PlP 2 ,... pm;qlgq2,...,qn ) and MNQD(PlP2, ...,pm;qlpq21...,qn). These

convex sets are, obviosly, compact. Moreover, they have each a finite number

of extreme points. Each of these sets is determined by a finite number of hyper-

planes in R m + n . Now, we describe a method of generating extreme points of

MPQD(PlP 2 ,. ..,pm;ql,q2,...,qn) in some special cases of m and n. The method

so described generalizes to for any m and n.

The case of 2 x 2 tables.

Let p, q be two numbers in the open interval (0, 1). Let p = p

P2 = 1 - p, q1 = q and q2 = 1 - q. A matrix P = {pij} of order 2 x 2 with

the properties pij > 0 for all i and j, p1 1 + p1 2 = p, P 21 + p2 2 = 1 -p,

pl + P2 1 = q and P1 2 + P2 2 = 1 - q is a member of MQD(PlP2 ;qlq 2 if

and only if P2q2 < P2 2 < p2Aq 2, a Ab denotes the minimum of the two numbers

a and b. The extreme points of this convex set MpQD(PlP2;ql,q2) are given by

V- - F- pw u jF1'J w V- WW V -w-4 --v- .- .-- ' . ~ * * -- ~

6

plql plq 2 qq2 - P22 and 0 P 2 Aq 2 when P2A q2 =P2;

Ap2ql p 2 q 2 P2 0

and when p2A q2 q 2 '[P~q1 P2 q2 P2 - p2 /\q 2 P 2 Aq 2

1

If p = q = 1/2, then the extreme points of MPQD(1/2,1/2;1/2,1/2) are given by

1iI-/4 1/4 2 012

and,' 1L/4 1/4 1 1/2

On the other hand, the extreme points of the set !'*i/2,1/2;1/2,1/2), the set of

all matrices P = {p ij} of order 2 x 2 with the properties pij > 0 for all

i and j, Pll + P1 2 = P2 1 + P2 2 = pll + P2 1 = P1 2 + P 2 2 = 1/2, are given by

and

The case of 2 x 3 tables

Let pl,P2;ql,q 2,q3 be five positive numbers satisfying p1 + p2

q1 + q2 + q 13 = 1. We now discuss the problem of identifying the extreme points

of 'QD(PlP2;qljq2,q3 ) This convex set is precisely the collection of all matric

P = iP 1]

P21 P22 p2 3

7

satisfying Pl + P1 2 + P1 3 = PI' p2 1 + P 22 + P2 3 = P 2 ' pll + P2 1 = ql

P12+ P2 2 = q 2 ' P1 3 + P 23 = q 3 9 P2 3 P 2q3, P2 2 + P 23 P 2q 2 + p2 q3 and

each pij is non-negative.The problem of identifying the extreme points of

this convex set can be solved both graphically and algebraically. The following

results are helpful towards this goal.

Theorem 6. Let P = (pij) be a matrix of order 2 x 3. If P is a member of

M pQD(PlP2 ;qlq 2,q3) then

(1) P2q3 < P2 3 < P2Aq3

(2) (P 2 q 2 + P 2 q 3 ) V P 2 3 P P 2 2 + P 2 3 < P2 A (q2 + P 2 3 )

(3) P1 2 = q2 - P 2 2 ' P1 3 =3 - P 2 3 ' P2 1 = P2 - P 2 2 - P 2 3 and

Pll = Pl - P12 - P1 3 = q, - P2 1

Proof. Since P is positively quadrant dependent, p23 2 P2q3 and P2 2 + p2 3 >

p2q2 + p2q3. Since P has the prescribed marginals and non-negative entries,

P2 q3' P22 + P 23 > P2 3 ' P2 2 + P2 3 < P2 and P2 2 + P 23 < q2 + P2 3 . This

proves (1) and (2). (3) is obvious. (Note. For any two real numbers a and

b, aV b means the maximum of a and b.)

In the above theorem, the inequalities (1) and (2) are very

important. They help to reduce the problem of identifying the extreme points of

MpQD (PlP2 ;qlq2q3) to a two-dimensional graphical problem. The following

result amplifies this remark.

Theorem 7. Let P2 2 and P2 3 be two real numbers satisfying

(1) P 2q 3 < P2 3 < P2 Aq 3 and

(2) P 2 3 V(P 2 q2 + p 2 q 3 ) < P 2 2 + p 2 3 < p 2 A(q 2 + P23

I I I I I.I.I.I.'. , .

*o u . -. . . .. -. * + + • . + • • t + + " + " ... .. . . .. .i

8

Let p12 = 2- P22 ' P13 = q3 - P23 and P21 = P2 -P 22 - P23. Then

PI - P12 - P13 = - P21 = plI' say, and the matrix P = (pij) is a member

of MQD(PlP2;qlq2,q3).

Proof. First, we show that P1 - P1 2 - P1 3 = 1 - P2 1. Note that p1 2 + pI3 1

q2 +q 3 - (P22 + P2 3) and q + P12

+ P1 3 = - (P22 + P2 3 )

= p 1 + P2 + P21 - P2

= p1 + P21' Consequently, p1 - (P12 + P q3) - p2 1. Next, we show that

each Pij is non-negative. From (1), P23 > 0 and q3- P23 = P1 3 >

- 0. From (2)

P2 2 > 0 ' P2 -P 2 2 - P2 3

= P2 1 > 0 and q02 - P22 = P12 > 0. It remains to be

shown that pl1 is non-negative. It is clear that pll + P1 2 + P1 3 + P2 1 + P2 2 +

P23 = 1. It suffices to show that P<2 + P 3 + P2 1 + P22 + P2 3 1 . Note that

P12 + P1 3 + P 21 + P2 2 + P2 3 = q 2 + q3 - (P2 2 + P 23 ) + P2 - (P2 2 + P 23 ) + (P2 2 + P2 3)

= q2 + q3 + P 2 - (P2 2 + P 2 3 ) _ q 2 + q3 + P 2 - P2q2 - P2 q3 (by (2)) = q2 + q3

+ P2 (1 - q2 - q3 ) = 1 - q, + P2q1 1 - q(1 - p2) = 1 - plq1 < 1. This shows

that p1, is always strictly positive.

It is obvious that the matrix P = (ptj) has the prescribed marginals

pl,p 2 and ql,q 2,q3 . From (1) and (2), it follows that P is positively

quadrant dependent.

The purport of the above theorem is that it is enough to determine two

numbers p2 2 and P23 satisfying (1) and (2) from which we can build a

matrix P = (pij) belonging to MQD(PlP2;qlq2,q3) . We call (1) and (2)

to be the determining inequalities. This theorem also provides modus operandi

to construct extreme points of MQD(PlP2;ql,q2,q3 ) . Intuitively, one can

proceed by looking at those pairs P2 2 and P2 3 of numbers for which equality

holds in (1) and (2) either on the right or on the left to yield extreme

points. This method would not exhaust all the extreme points of (PlP2;qq2q.9 2 2 --

9

The reason is that the bounds in (2) are variable and depend on the value of

P23" If we can partition the interval [p 2 q3 , p 2 /q 3] into sub-intervals so that

the bounds in (2) become stable over each sub-interval, then we can isolate

the extreme points. We explain this method graphically and algebraically as

follows.

Graphical solution

Determining inequalities

(1) P 2 q 3 <-- p 2 3 L- P 2 Aq 3

(2) P2 3 V(P 2q 2 + P2 q3 ) <_P2 2 + P2 3 <- p 2 A(q 2 + P23 ) .

Case 1 P2q2 + P2q3 >-P2 A q3 and q2 - P2. Then the determining inequalities

become

(1) P 2q 3 < P2 3 < P 2 A q3 and

(2) P2 q2 + p2 q3 <-P 2 2 + P2 3 <p 2.

These inequalities yield the following graph., P22 '

P2 q2 + P 2 q3 3

P 2 q3 P2 A q3 P2 3

, './ ', I" . " "-,*I '-- **,",~ I*'g . "., - ..- ./ ' .. '. .,. . , ./ . - ; .o ., . - _.' ' .£ - ' - .; . . -/ -. .-.

10

Comments. The extreme points, in this case, are obtained by setting each

expression in the centre of (1) and (2) equal to the quantities either on

the left or on the right of (1) and (2). This will give us 4 extreme points.

Case 2. p2q2 + p2q3 > p2/%q3 and q2 < P2 "

Then either (A) P2 - q2 < P2 q3 or (B) P2q3 P2 - q2 < P2 Aq 3 or

(C) P2 - q2 P2 Aq 3 holds. The determining inequalities and the corresponding

graphs then respectively work out as follows.

Case 2(A). Inequalities

(1) p2q3 < P23 -1 P2 Aq3 and

(2) P2q 2 + P2 q 3 -- P2 2 + P2 3 < P2

Graph. Same as in Case 1.

Case 2(B). Inequalities

(1) p2q3 <p 2 3 < P2 - q2 and

(2) P 2 q 2 + P2q3 < P2 2 + P2 3 < q2 + P23

or,

(1)' P2 - q2 < P2 3 < P2 A q3 and

(2)' P2 q 2 + P2q3 < P2 2 + P2 3 < q 2 P 2 3 .

11

P22Grph

P2 q2 + P2 q

P2 q3 p2 -q2 P2 Aq 3 P2P 2 3

Comments. q2 can be above p2 q2 + p2 q3 . The number of extreme points

is either four or five.

Case 2(C). Inequalities

(1) P2q3 < P2 3 <- P2 A q 3 and

(2) P2q 2 + P2 q3 -- P2 2 + P2 3 q2 +P 2 3

,'-I i " '". " .".-" ,,:, "- ,." , F .--. '."..-, ' "., , ' .. '.-.-., "-,.'.....- -". -".," :' . . .' "--.'

. t

12

Graph

P2 2

p 2 q2 + p2 q

qi.2

qI.

p 3 P2 A q% p23

Comments. q2 can be above p2q2 + p2q3. The number of extreme points

is four.

Case 3 P2q2 + P2q3 <-p2A q3 and P2 I q2 "

The determining inequalities become

(1) P2q3 <-- P2 3 < P2q2 + P2q3 and

(2) P2q 2 + P2q3 f< P2 2 + P2 3 < P2

or,

(1) P2q2 + P2q3 < P2 3 < P2Aq3 and

(2) P23-- P22 + P23 -1 P2

13

GraphA

P22

p q2 + p q3

P2q3 P2q2+p2q 3 2 P2 P23

Comment. The number of extreme points is five.

Case 4 P2q2 + p2q3 < p2 Aq 3 and P2 > q2 "

Then either (A) p - q2 > q or Bp2q3 < P2 - q2 < P2A q3 or (C)

P2 - q2 <p--

2 q3 holds. We now discuss each of these cases.

Case 4(A). Inequalities

(1) P2q3 <-P 2 3 < P2q2 + P2q3 and

(2) p2q2 + P2q3 <-p2 2 + P2 3 < q2 + P2 3

or,

( 2) P2q2 + P2q3 < P2 3 <- p2A q3 and

(2) P23 <p 22 + p23 <q, + p23.. ,,',,V'-'_',P,3 P2 P23 ,.23-,,- - -.-. -.. .. . . .. . .. *.. . . .

14

Graph

p2 q2 + p2 q3

P2 q3 P2 q 2'P2q3 p2 Aq 3 2

Comments. q 2 can be below P2 q2 + P2 q 3. There are five extreme points.

Case 4(B). Assume, without loss of generality, that p2 q q2 2 pq 2 + P2 q 3

Inequalities

(1) pq 3 2 ~ 2 q2 and

(2) P2q2 Pq31 22+P3 q2+ 3

or,

(1) p2 - q 2 pq 2 pq and

(2) p 2 + p2q3 <Sp2 2 + p23 <p 2

or,

(1) p2 q2 + p2 q3 ' p23 L P2 A q3 and

(2) p2 p 2 + p2 <p2

15

Graph

q

p2 q2+P2 q3

P2 q3 p2 -q 2 p2 q2+p2 q3 p2 Aq 3 p2 p23

Comment. The number of extreme points is six.

Case 4(C). Inequalities

(1) P2 q3 L P23 1-2 pq 2 + p2 q3 and

(2) 'P2q 2 + p 2q 3 < p 22 + p 23 L p2

or,

(1) P2 q2 + p2 q3 < p23 :L P2A q 3 and

(2) P23 <-p22 + P23 <p 2 .*

Graph. It is similar to the one given under Case 3.

16

Algebraic solution

The expression (P2q2 + p2q3 )V P23 induces a partition of the

interval [p2q3, p2A q3 ] into two sub-intervals. These intervals are given by

AI = 'P23 C[P2q3 p2 Aq 3] ;(P2q2 + P2 q3 )V p 2 3 P2q2 + P2 q3 }

and

A2 = {P2 3 C [P2 q3, P 2 Aq 3] ; (P 2 q2 + p2 q3 )VP 2 3 =P231

A and A2 may have a common end point. In a similar way, the expression

P2 A(q 2 + P23

) induces a partition B and B2 of the interval [p2q3, p2A q3]

into two sub-intervals. The partitions {Al, A2) and {B1 , B2 } of [p2q3, P2Aq 3

induce a finer partition {CI , C2 9 C3 } Of [P2q3' p2A q3 ]. Let

Cl = [p 2 q3 , a1] ; C2 = [al, a 2] ; C3 = [a 2 , P 2 Aq 3]•

We now consider three sets of inequalities.

(A) (1) P 2 q 3 < P2 3 < a1 and

(2) p 2 3 V (p2 q2 + P2 q3 ) < P22 + P23 < p2 A (q 2 + P23 ) "

(B) (1) a1 < P23 < a2 and

(2) P23 V(P 2 q2 + P2 q3 ) < P2 2 + P2 3 <p 2 A(q 2 + P23 )

(C) (1) a2 <- p 2 3 < P2 A q3 and

(2) p2 3 V (p2 q2 22 P2 + P23 < p2 A (q 2 + P23 ) "

In each set, in the inequality (2), the max and min symbols V , A would

disappear after due simplification. From each set, the central expression

in each inequality is set equal to the quantity either on the right or on the

left. These equations are solved for unknown values of P22 and P23. The set

of matrices so obtained after using Theorem 7 contains the set of all extreme

points of MQD(PlP2;ql~q2,q3) .

S.. . . . . . .- ~ . . .-

17

A numerical illustration

Let p1 = 0.4, p2 = 0.6; q = 0.3, q2 = 0.2 and q = 0.5. The

inequalities (1) and (2) of Theorem 7 translate into

(1) 0.3 < P23 < 0 .5 and

(2) P2 3V0.42 <<P2 2 + P23 <0.6 A(0.2 +

This set of inequalities come under Case 4(B) and is equivalent to

(1) 0.3 < P23 < 0.4 and

(2) 0.42 <p 2 2 + P2 3 < (0.2 +

or,

(1) 0.4 < p23 < 0.42 and

(2) 0.42 <P 2 2 + P2 3 <0.6,

or,

(1) 0.42 < p2 3 <0.5 and

(2) P2 3 <p 2 2 + P23 <0.6

Graph p .0.g2

0.42

(vi v)

(vli

0.. . 2 0.6 p

(ii)

V.42 1

18

Extreme points as marked on the graph

(i) F0"12 0.208 ° (v) 023 0. 0].T

1 8 0 .0.( 0 .2 0

(iii) 2 0.2 0 8l (vi) 3.2 0 0 ]18~ 0 0.51]42. 0.2 0.]

10.1 0 0. 50.1 0.2 0 .

The algebraic method explained above gives a1 = 0.4 and a2 =0.42.

Solving equations, we obtain the following matrices in addition to those listed

above.

(vii) j 12 0.18 O (vi3) 0.02 0.0 8]

1.18 0.02 O0 0.18 0.4j

These matrices are not extreme points. Non-extreme points can easily be weeded

out from the list of points provided by the algebraic method.

The case of 2 x n tables

Let P1 9 P2 ; q1 9 q2, "'" qn be positive numbers such that pl + P2 -

q + q2 + ... + q = 1 If a matrix P = (Pij of order 2 x n with non-negative

entries and marginals pil + Pi2 + ... + P in = Pi, i = 1,2 and pl + p2j = qj,

j = 1,2,...,n, is positively quadrant dependent, then the following inequalities

hold.

6- 37 77 Rk .

19

(1) P~n <- P2n < P 2 Aqn

(2) P2nV (P2 n-1 + P2n ) <-P2n-l + P2n <- P2 n- + P2n

(3) (P2n-i + P2n)V (P 2 q n-2 + p 2 qn- 1 + P 2 qn) <P 2 n_ 2 + P2n-l + P2n <

P2 A(qn- 2 + P2n-l + P2n)

(n-i) (p 2 3 + P 2 4 + "'" + P2n ) V (p 2 q 2 + p 2 q 3 + ... + P 2 qn) <

P22 + P23 + "'" + P2n - p 2 A(q 2 + P2 3 + P 2 4 + "' + P2n )

These inequalities provide a basis for generating members of

QD (PP2 ;qlq2,...,q n ) . The following theorem provides a method for finding

members of PQD(PP2 ;ql~q2,...,qn ) .

Theorem 8 Let P22' P2 3 ' .' 9 P2n be (n-1) numbers satisfying the

inequalities (1), (2), ... , (n-i) above. Let Plj = qj - P2j' j = 2,3,...,n

and P21 = P 2 - (P2 2 + P 2 3 + +'. P2n " Then

q - P21 = PI - (P1 2 + P1 3 + "'. P l n ) - pl, say,

and the matrix P = (p1V) of order 2 x n belongs to MQD(PlP2;qlpq2,...,qn) .

Proof. It is similar to the one presented for Theorem 7.

Now, we set about obtaining extreme points of MQD(PlP2;qlq2, . . .,qn

algebraically.

20

We develop some sets of inequalities equivalent to the inequalities

(1), (2),...,(n-1) described above with the following properties.

(M) Each set contains (n-i) equalities

(ii) The max and min symbols V , A disappear from each

side of each inequality in each set.

Step 1. The interval [P2qn, P2 Aqn] of variation for P2n is split into

3 sub-intervals using the inequality (2). The expression

P2n/(p2qn-1) + P2qn ) gives a partition of the interval [p2q, P2Aqn]

into two sub-intervals as follows. Let

A1 = {P2n [P2qn , p2 Jq n ];

P2n V(P2qn-i + P2qn) = P2n }

and

A2 = fP2n ' [p2qn, P2 Aqn];

P2n/(P2q n-I + P2qn) = P2qn-i + P2qn}

Note that A and A2 are closed intervals with possibly a common

end point and

AIUA 2 = [p2qn, P2Aqn "

(One of the A.'s could be a null set.)1

Similarly, the expression pA(q + P) gives a partition {Bl,B2}P2 n-l 2n

of the interval qp2qn, p2 qn] into two closed sub-intervals. The

partitions {AI,A 2 } and {B1 ,B2 } induce a finer partition {CI,C 2,C3 } of

[p2 qn p2A q n] such that each Ci is either a closed interval or a

null

set and

CIUC2UC3 = 2[p2qn, P2 Aqni.

For example, if A1 = [al,a 21, A2[a2 a3], B1 = [bl,b 21, B2 = [b2,b3] with

a1 = b = P2 qno a3 = b3 P2 h qn and a 2 < b2, then we can take C1 = [al,a 21,

3 3 2 . 2 2' . . . . .

21

C2 = [a2,b2] and C3 = [b2,b3 ]. To maintain a conformable notation,write a1 = a[i], a 2 = aial, ,2 = a [21' b2 2 a[3] b ba-a' 3 ] . Thus

we have

[p2 qn) P2 Aq I = [a,,a] [13 a~1 L.Ja 1

(1)r a2~ p- q2 all =2 andada n

(2) 2- n'1 [ ''21- 2 3n- + 2n

are equivalent to

[] a < P2 a and

(2)p V(p q + pq )[12 P2n 2 (Pl2- + n) P2n- + P2n

< P 2 A (qn-_ + P~n)

or

[2]a < <' anda a[21 - P2n -_ a[2

[21](2) P2n"(P2qnl + V2qn ) _ P2n-i + P2n

2 P2 n (qn-1 + P2n)

or

[31a 1 1 <p 2 ' and[2) a[2] <- P2n S_ a[3]

[3](2) P2nV(P2qn-1 + P2qn) < P2n-l + P2n

< P2 A (qn-1 + P2n ) "

In each of the inequalities [1](2), [2](2) and [3](2), the symbols

V and A would disappear when P2n is confined to the limits imposed by

[11, [2] and [31 respectively. To mark this freedom from the symbols

V and A , let us rewrite the above inequalities as follows.

- 22

[1] aIll < P2n <_ a ll and[11~ (2)a]a

[11(2) a[l](2) 5_ P2n-i + P2n <[1(2)'

or,

[2] a[2] < P2n < a 2] and

[21(2) a[ 2 ]( 2 )< - P2n-i + P2n _ a[2](2) ,

or,

[3] a[3] < p2 <a and

[3]1(3) a[3 ](2) _ + <

Step 2. Using the inequality (3), we sub-divide each of the intervals

given by the inequalities [1](2), [2](2) and [3](2) onto three sub-

intervals or inequalities. Let us concentrate on [1](2). The procedure

outlined below is similar to the one given in Step 1. The expression

(P2n-1 + P2n ) V ( P 2 q n - 2 + P2 qn-i + P2qn) bifurcates the

inequality [1](2) into two inequalities. The sets

D1 = 'P2n-l + P2n E[a[,]( 2 ) a[] (2 )1;

(P2n-i + P2n ) V ( P2 q n -

2 + P2qn-i + P 2 qn )

' P2n-i + P2n) and

D = {P2n-i + P2n C [a[ 1 ](2), a[1 ](2 )];

(P2n-l + P2n ) /(P2 qn-2 + P2qn-i + P2qn )

p 2qn- 2 + p 2qn-l + P2 qn}

do give the necessary bifurcation of the inequality [1](2) with P2n

satisfying the inequality [1]. Similarly, the expression p2 A (qn-2 + P2n-i + P2n )

bifurcates the inequality [1](2) into two inequalities. These two bi-

furcations would give three inequalities equivalent to [1](2). Let these

inequalities be denoted by [11], [12], [13]. Let us write these inequalities

* ;"-

- -2 3

including [1] as follows.

[11 ] +<P~ < a'[ 1:- -~ [ill -

a[11] <P2n-l + P2n i- 11]'

or

[121 a[1 2] -- P2n-1 + P2n <- a[12]'

or

[13] a[1 31 <-- P2n-l + P2n < a'l3],

where a' = a [12] and a' = a By bifurcating the inequality [1][1 [1 121 [13P*

further, necessary, ensure that a'l, < a'l,] as P2n- P2n-i P2n

be guaranteed. This procedure is applied to each of the inequalities

[2](2) and [3](2).

We can now claim that the inequalities (1), (2) and (3) are

equivalent to the following 9 sets of inequalities. (We describe these

inequalities using a suggestive notation.)

[1] a[1] < P2n < a'a -l 2n- [1]'

[11] a[l-11 < P2n-l + P2n < a 1ll] and

[11](3) (P2n-1 + P2n ) %/(p2q n-

2 + p2qn-l + p2qn)

< P2n-2 + P2n-1 + P2n

_ P2 A (qn- 2 + P2n-l + P2n '

or

[1] a[1] <- P2n _ aill ,

[12] a [12] 5_ P2n-i + P2n < a[1 2 ] and

[12](3) (P2n-i + P2n ) / ( p2 q n -2 + P2qn-i + P2 qn)

<- P2n-2 + P2n-l + P2n

- P2A (q + P2n-l + P2n '

• "-" "'". -" .. . . . . . .... -.. . . . . . . . . . . . .." " " " -° " ." "" " < .

'" nl I , , I -t -

24

or,

*[1] a[Ill . p2 n_ a ill,

[13] a [13 1 1 p2n-1 + P2n L a j131 and

[13](3) (P2n-1 + P 2 n)/(P 2 q n 2 + P 2 q + P2 qn)

<p2n-2 + P2n-1 + P2n

P2 A (q 2 + p 2n-1 + n)

or,

[21] a[2 ai and[21] [21] - P2n-1 + P2n < 2 1 1 an

[21](3) (p2n-1 + P~)VP - + P2 q n- + P2 qn)

P2n-2 + P2n-1 + P2n

P 2 A (q n 2 + P2n-1 + ~)

or,

[22] a [22] f- P2n- L a' [2 22 a

[22](3) (p2n-1 + P2n)v(P2 q n 2 + P2qnl+ P2 qn)

.f ~'2n-2 + Pnl+ n

< P'2 A(q n 2 + P2n-1 + Pn

or,

[2] a[(2] -f <~ _a' [1

[23] a f23 a pnd[23 <2n-1 + 2n 123

[23](3 ( p2n-1 + P)V (p2q 2 + p 2 q + P2 qn)

P-~2n-2 + P2n-l + P2n

P ~2 J\(q n 2 + P2n-1 + n)

25

or,[3] a[3]-< P'2n -<a'3

[31 [3)[3n [31]3

(31] a[3 11 < P2n- + P2n< a' [31 and

[31(3) (p2n-1 + P2n) \/(p 2qn- 2 + P2 qn-i + P2qn)

_P2n-2 +P2n-i + P2n

n- 2 A (q 2 + P2n-i + P2n

or,

[3] a[3) < P2n < a'- - [3]

[32) a <a' anda321 < P2n-i + P2n - 132]

[32](3) (P2n-1 + P2n ) V(P2qn-2 + P2qn-i + P2qn)

-P2n-2 + P2n-i + P2n

P2 A (qn-2 + P2n-I 1 P2n )

or,

[3] a[3] ' p2 <a'[31 f - [31

[33] a [33 P2 1- + P2n _ a33] and

[33] (3) (P2n-i + P2n ) V pqn-2 + p2qn-l + p2qn)

< P2n-2 + P2n-1 + P2n

P2 (qn- 2 + P2qn-1 + p2n) "

The following facts should be borne in mind before proceeding to

Step 3.

(M) ail,, = a[12]; a 121 a [13]; a 211 = [221;

a ~ a a =a1221 [23) aj311 a[321; a1321 = a[33]L

(i) ; < a' and a' <a

- - [21] [3] "[311.

I4

.I

e- " gI

26

(iii) In each of the nine inequalities [11](3), [12)(3),

[131(3), [21](3), [22](3), [23](3), [31)(3), [32](3)

and [33](3), the symbols V,A would disappear as long

asp 2n and p2n-l + P2n are confined to the limits of

the corresponding inequalities within each set.

(iv) The inequalities [1], [2) and [3] obtained in Step 1

can further be bifurcated to accomodate (iii). This is

helpful in conjunction with the inequalities [11], (12],

[13], [21], [221, [23], [311, [32] and [33] in each

appropriate category to get rid of the symbols V and

in the inequality (3).

Step 3. Using the inequality (4), we split the last inequality in each

set above into three inequalities. The procedure is similar to the one

explained in Step 2.

Step (n-2). Using the inequality (n- ), we split the last inequality

* in each set into three inequalities. At the beginning of this step,

each set has (n- 2) inequalities.

At the conclusion of Step (n-2), the inequalities can be arranged

pyramidically as in the following diagram, assuming that at every

stage we split each inequality into three inequalities.

"10J

27

- n

~ L Een

C4

4

1N aq Nq

.4

4

4

4.

4+q.

4 .LLC6

28

Each inequality at the final Step (n-2) is coupled with the in-

equality (n -1). Thus we have 3n - 2 sets of inequalities with each set

containing (n-1) inequalities including the inequality (n-1). In the

last inequality (n-i), the symbols V and A would disappear. The

following are examples of two sets of inequalities when n =6, where at

every stage each inequality gave rise to exactly three inequalities.

Example 1.

[1] a[1 ] < P26 < a111,

[121 a 1 1 a'+1 a12 11 25 + P26 [12]

[121] [121] :-- P24 + P25 + P26 <- [121]'

[1213] a[1 2 1 31 <- P2 3 + P24 + P2 5 + P26 < a[1 2 1 31

and

(5) (P2 3 + P2 4 + P25 + P2 6 ) V(P 2 q 2 + P 2 q3 + P2 q4 + P 2 q 5 + P2 q6 )

:--P22 +P23 + P24+ P2 5 + P 2 6

< P2A(q2 + P2 3 + P2 4 + P2 5 + P26 ) "

Example 2.

[31 a[3 1 < P26 < a 3]'

[32] a[ 3 2 ]-- P25 + P2 6 <[321

[321] a[3 2 1 1 ] < P 2 4 + P2 5 + P 2 6 < a1 3 2 1 1 1 '

[32121 a[3 2 1 21 < P23 P24 + P25 P2 6 -a[ 3 2 1 21

and

(5) (P2 3 + P2 4 + P25 + P2 6 ) V(p 2 q 2 + P2 q3 + P 2 q4 + P 2 q5 + P 2 q 6 )

<- P22 +23 + P24 + P 2 5 + p2 6

< P2 A(q 2 + P 2 3 + P2 4 + P 2 5 + P26 ) "

7I- . . 2. .. . .

-- *.'' -'''" ' ' " i " "r 'P r - ' "- . . " :'" .- "**.%'-" ".-." *." .s. *"..% ... . . . "

[I 29

Enumeration of extreme points of MPQD(PlP 2; qlq2,...,qn)

Let J denote the collection of all those matrices P = (p ij) in

MPQD(PlP 2 ; ql,q 2 ,...,q.) for which equality holds either on the left

or on the right in all the inequality expressions in some set of the

3n- 2 sets described above. Let Ji. i = 2,3,...,n, denote the collection

of all matrices P = (pij) in MPQD(PlP 2 ; q1,q2,q3) for which strict

,inequality holds in (i - 1) inequality expressions and equality holds

either on the left or on the right in the remaining (n-l) - (i-l) = n-i

inequality expressions in some set of the 3n - 2 sets described above.

Obviously, n

MPQD(Pl'P 2 ; qlqm = U Ji.

i=l

Theorem 9. The set of all extreme points of MPQD(PlP 2; ql,q2,...qn )

is a subset of Jl'

Proof. It suffices to show that every matrix in MPQD(PlP 2; q 2,...qn )

can be expressed as a convex combination of members of J1 ' Every member

of Jl, trivially, has the above property. Let P = (Pij) be any member

of J2 " We express P as a convex combination of two members of Jl'

Since P c J2' strict inequality holds in exactly one inequality expression

of some set of inequalities. For simplicity, let the strict inequality

occur in [23], i.e.,

[23] < P2n-l + P2n < a1 2 31

and equality holds in the remaining (n- 2) inequality expressions either

on the left or on the right. Let Q = (qij) be the matrix in

MPQD(PlP2; ql,q 2,.., qn) for which equality holds on the left in [23]

and equality holds in the remaining (n-2) inequality expressions in

exactly the same place as that of P in the same set of inequalities.

30

Let S = (sij) be the matrix in MPQD(PlP2; ql,q 2 ,...,qn) for which

equality holds on the right in [23] and equality holds in the remaining

(n-2) inequality expressions in exactly the same place as that of P in

the same set of inequalities. The existence of Q and S is guaranteed

by Theorem 8. We also observe that q2n-1 < P2n-i < S2n-l" There exists

0 < a < 1 such that P2n-1 = aq2n-l + l-c)s 2n-l Indeed, after laborious

calculations, one can check that P =ctQ + (l-t)S.

Let P be a member of J We can express P as a convex combination

of two members of J2 " This then would imply that P is a convex com-

bination of members of J1. Since P e J3' strict inequality holds in 2

inequality expressions and equality holds either on the left or on the

right in ea_.h of the remaining (n-3) inequality expressions of some

set of inequalities. For simplicity, let the strict inequality hold

in [123] and [23112], i.e.,

a' and[123] < P2n-2 + P2n-l + P2n < a123

a[23112] < P2n-4 + P2n-3 + P2n-2 + P2n-l + P2n < a[23112]"

Let Q = (qij) be the matrix in MPQD(PlP 2; ql~q2,...,qn ) such that q2n = p 2n ,

q 2n--l P2n- ' q2n-2 P2n-2' equality holds on left of [23112] and equality

holds in the remaining inequality expressions in exactly the same place

as that of P in the same set of inequalities. Let S = (sij) be the matrix

in MPQD(PlP 2 ; ql,q2,...,qn) such that s2n = P2n' S2n-i = P2n-l,

S2n-2 ' P2n-2' equality holds on the right of [23112] and equality holds

in all the remaining inequality expressions in exactly the same place as

that of P in the same set of inequalities. One can check that P = cQ + (1-ct)S

for some 0 < a < 1.

31

Proceeding this way, one can show that every member of J. is a1

convex combination of two members of Ji-1 for i = 2,3,...,n.

This completes the proof.

The above theorem helps to identify the extreme points of

MPQD(PlP 2; ql,q 2,..q n) as follows.

Step 1. Create 3n- 2 sets of inequalities as described above.

Step 2. For each set of inequalities, set the central expression in

each inequality equal to the quantity either on the left or on the right.

Solve the resultant equations for P2n' P2n-l'"'P23 and P22.

Step 3. Using Theorem 8, build a matrix P = (pij) in QD(PlP 2 ; qllq2,...,qn).

By Theorem 9, every extreme point of MPQD(PlP 2 ; ql,q 2,...,q n) arises

this way. There might be a duplication of matrices on a moderate scale.

Also, many non-extreme points might arise this way. We can weed them

from Jl' If it is a hard task, we can keep them. Retaining them will not

hamper us in discussing objectively merits of various tests for contingency

tables in the next section. Once the sets of inequalities are created,

solving equations can be programmed easily.

Intuitively, MPQD(PlP 2 ; ql,q 2,...,qn) can be viewed as a multi-fac-

eted diamond and the solutions obtained above give the points where

edges meet.

A numerical example

Let n=4, p, = 0.6, P2 = 0. 4, ql = 0 "3, q2 = 0.2, q3 =0.2 and q4 =0.3.

Determining inequalities

(1) P2 q4 < P2 4 < p2 A q4 "

(2) P2 4 V(P 2q 3 + P 2 q4 ) < P23 + P2 4 < P 2 ((q 3 + P24 ) "

(3) (P 2 3 + P2 4 ) V(p 2q 2 + P 2 q 3 + P2 q 4 ) P 2 2 + P 2 3 + P2 4 < P2 (q 2 +p 2 3 +p 2 4 ) "

32

Numerically, these inequalities become

(1) 0.12 < P2 4 < 0.3,

(2) P24 V(0.20) < p23 + P2 4 < 0.4 A (0.2 +

(3) (23 + P2 4) V (0.28) < P2 2 + P2 3 + P2 4 < 0.4 A (0.2 + P2 3 + P24

) "

The inequality (2) induces only two inequalit ies [1] and [2]

equivalent to (1).

[i] 0.12 < p24 < 0.2 and

[2] 0.20 < P2 4 < 0.3.

The inequality (2) in conjunction with the inequalities [1] and [2]

becomes

[1] 0.12 < P24 < 0.2 and

[1)(2) 0.20 < P2 3 + P2 4 < 0.2 + P24'

or,

[2] 0.2 < p24 < 0.3 and

[2](2) P24 -< P2 3 + P2 4 < 0.4.

The inequality (3) induces two inequalities equivalent to [1](2) and also

two inequalities equivalent to [2](2). We have altogether 4 sets of

inequalities listed below.

I Set

[1] 0.12 < p24 < 0.2,

[11] 0.20 < p23 + p2 4 < 0.28 and

(3) 0.28 < p22 + P23 + P24 < 0.4.

II Set

(11 0.12 < p 24 < 0.2,

[12] 0.28< p2 3 + p24 < 0.2 + P24 and

(3) p2 3 + p24 - P2 2 + P23 + P2 4 < 0.4.

-I. ~ ~'~***'* ..-....- -.. .- .. .. . ~ **~****%***i

33

III Set

[2] 0.2 < P24 < 0.3,

[21] p2 4 <- P2 3 + P 24 < 0.28 and

(3) (P23 + P24 ) V (0.28) < P2 2 + P2 3 + P24 < 0.4.

IV Set

[2] 0.2 < p2 4 < 0.3,

[21] 0.28 < P 23 + P2 4 < 0.4 and

(3) (P2 3 + P24 )V(0.28) < p22 + P2 3 + P2 4 < 0.4.

In the III set, the number on the left hand side of [2] is larger than

the number on the left hand side of [21]. We, further, bifurcate [2] to

set right this anomaly. After this bifurcation, we relabel the inequalities

and obtain the following sets.

Set I

[1] 0.12 < P24 < 0.2,

[11] 0.20 < P2 3 + P2 4 < 0.28 and

(3) 0.28 < P2 2 + P2 3 + P2 4 < 0.4.

Set II

[1] 0.12 < P24 < 0.2,

[12] 0.28 < P23 + P2 4 < 0.2 + P2 4 and

(3) P2 3 + P24

- - P2 2 + P23 + P24 0.4.

Set III

[21 0.2 _ P2 4 < 0.28,

[21] P2 4 ! P 23 + P2 4 < 0.28 and

(3) 0.28 < P2 2 + P2 3 + P2 4 < 0.4.

-~~~ ~ . . .. . + P2 + , . . ..4

34

Set IV

[2] 0.20 < P24 < 0.28,

[22] 0.28 < p2 3 + P 24 < 0.4 and

5(3) <0-4., (3) P23 + P24 < P22 + P23 + P24 !04

S Set V

[31 0.28 < P2 4 < 0.3

[31] P 24 < P2 3 + P 24 < 0.4 and

(3) < 0.4.3) P23 + P24 < P22 + P23 + P24 <- 04

Set the central expression in each inequality in each set to the

quantity either on the right or on the left and solve the resultant

equations for p22 ' P23 ' p2 4. The distinct matrices in J, of Theorem 9

are listed below.

(1) 0.18 0.12 0.12 0.18 (2) 0.3 0.0 0.12 0.18

0.12 0.08 0.08 0.12 0.0 0.2 0.08 0.12

(3) 0.18 0.2 0.04 0.18 (4) 0.3 0.08 0.04 0.18

0.12 0.0 0.16 0.12 0.0 0.12 0.16 0.12

(5) j 0.18 0.12 0.2 0.1 (6) 0.3 0.0 0.2 0.1

0.12 0.08 0.0 0.2 0.0 0.2 0.0 0.2

(7) 0.18 0.2 0.12 0.1 (8) 0.3 0.08 0.12 0.1

0.12 0.0 0.08 0.2 0.0 0.12 0.08 0.2

(9) 0.22 0.2 0.0 0.18 (10) 030.12 0.0 0.18

(0.08 ) . 01 . 0.08 0.2 0.12

(.1) 0.3 0.2 0.0 0.1 (12) 0.18 0.12 0.2 0.1

0.0 0.0 0.2 0.2 0.12 0.08 0.0 0.2

0?.-. . . .. 2.. . 1) 0.2 01 . .

35

(13) 0.18 0.2 0.2 0.02 (14) j0.3 0.08 0.2 0.02

0.12 0.0 0.0 0.28 0.0 0.12 0.0 0.28

(15) 0.3 0.2 0.08 0.02 (16) 0.2 0.2 0.2 0.0

0.0 0.0 0.12 0.28 0.1 0.0 0.0 0.3

(17) 0.30.1 0. 0.3 (18) 0.30.2 0.1 0.0

0.0 0.1 0.0 0.3 0.0 0.0 0.1 0.3

From the above collection of matrices belonging to Jl, we can weed

out those matrices which are not extreme points of MQD(0.6, 0.4; 0.3,

0.2, 0.2, 0.3). (4) is not an extreme point. For, we can write

1 2(4) 3(i2) + -(18). (15) is not an extreme point. For, we can write

(15) =-1(11) + A(18). (7) is not an extreme point. For, we can write

1 11 1(7)= (3) + 2(13). (8) is not an extreme point. For, (8) = -1(4) + 1(14).

The number of extreme points, therefore, of "PQD(0. 6 , 0.4; 0.3, 0.2, 0.2, 0.3)

is 14.

36

Now, it is time to discuss the general m x n case. But the

arguments involved and the notation used are so complex, it is not

really worth the effort. Instead, we offer a thorough discussion of

3 x 3 case from which one can make a successful analysis of any m x n

case. The analysis of 3x 3 case is not exactly in tune with the 2x n

case. A study of 2x n and 3x 3 cases in tandem should help to map out

a successful strategy to analyze mxn case in general.

The case of 3x 3 tables

Let p1. P22 P3 ql' q2, q3, be six positive numbers such that

+ P 2 + P3 = q1 + q2 + q3= 1. If a matrix P = (pij) of order 3x 3

belongs to MPQD(PlP 2 ,P3 ; q1,q2,q3 ), then one can check that

(1) p3q3 < P 3 3 <-- P3 A q3,

(2) P3 3 V (p 2 q 3 + p3 q3 ) < p2 3 + P3 3

< q3 A (P2 + P33 ) '

(3) p3 3 V (P3q2 + p3 q3 ) < p3 2 + P3 3

S< P3 t\ (q 2 + P33 ) '

(4) (P2 3 + P3 2 + P3 3 ) (P 2 q 2 + P2 q3 + P 3 q 2 + P3 q3 )

<- P 22 + P23 + P32 + P3 3

< (P2 + P3 2 + P33) A (q2 + P2 3 + P 3 3 )

hold. P is PQD yields (1) p3 3 > P3 q3 , (2) P2 3 + P3 3 > P2 q3 + P3 q3 ,

(3) P 3 2 + P 33 > P 3 q 2 + P 3 q 3 and (4) P2 2 + P2 3 + P32 + P 3 3 > P 2 q 2 + P 2 q 3 + P 3 q 2 + P 3 q 3 .

The non-negativity of the entries of P, and the marginality conditions

on the entries of P coupled with P being PQD yield the above inequalities.

These inequalities are important in the sense that if there are four

numbers P2 2 ' P23' P3 2 ' P3 3 satisfying (1), (2), (3), (4) above, then we

can build a matrix P belonging to MPQD(PlP2,P3; ql,q 2,q3). We make this

precise in the following theorem.

• "

37

Theorem 10 Let p2 2 1 P23 ' P3 2 ' P33 be four numbers satisfying the

inequalities (1), (2), (3), (4) above. Define

P21 =P2 - P22 - P23'

P31 - P3 - P32 - P33'

P12 - q2 - P 2 2 - P3 2 ' and

P13 = q3 - P23 - P33"

Then

Pl - P12 - P13 = ql - P21 - P31 = p1l1 say

and the matrix P = (pij)E MPQD(PlP 2 ,P3 ; ql,q 2,q3).

Proof. The proof is analogous to the one given for Theorem 7.

Determination of the extreme points of M- (plp 2,P3 ; q1,q2 ,q3)

As explained in the analysis of 2 xn tables, inequality (2) gives

rise to at most 3 inequalities equivalent to (1). Equivalent inequalities

are given below for each of the cases.

Case A P3 A q 3 < P2 q3 + P3q 3 andq3 -P 2 <0

(1) P 3 q 3 < P3 3 < P 3 A q3 and

(2) P2q3 + P3q3 < P23 + P3 3 < q3 "

Case B P3 A q 3 < P 2 q3

+ p3q3 a nd 0 < q 3 - P2 < P3q3

(1) P3q 3 < P3 3 < P3 A q3 and

(2) P2q3 + p3q3 < P2 3 + P3 3 < q3

Case C P3 A q3 - - P2 q3 + p 3q3 and p3q3 < q3 - P2 < P3 A q3

() P3q3 < P3 3 <q 3 - P2 and

(2) P2q3 + p3q3 < P2 3 + P3 3 < P3 3 + P20

or

() 3 - P2 < P3 3 < P3 A q3 and

(2) P 2 q 3 + P3 q 3 < P 2 3 + P3 3 < q 3 "

P O~

38

Case D P3 A q3 <p 2 q3 + P3q3 and P3 A q3 -- q 3 - P2

(1) P3q3 < P33 P3 q3 and

(2) P2q3 + P3q3 < P2 3 + P3 3 < P2 + P3 3.

Case E P2q3 + P3q 3 < P3 A q3 and q3 - P2<-- 0

(1) P3q3 < P33 <- P2q3 + P3q3 and

(2) P2q3 + P3q3 < P2 3 + P3 3 < q3'

or

(1) P2q3 + P3q3 < P3 3 < P3 \ q3 and

(2) P3 3 < P23 + P3 3

< q3 "

Case F P2q3 + P3q 3 < P3 A q3 and 0 < q3 - P2 <P 3q3

The inequalities here are the same as those of Case E.

Case G P2q3 + P3q3 < P3 A q3 and p3q3 < q3 - P2 < P3 A q3 "

Assume, without loss of generality, that q3 - P2 < P2q3 + P3q3 "

() P 3q3 < P3 3 <- q3 - P2 and

(2) P2q3 + P3q3 < p23 + P3 3 < P2 + P33'

or,

(1) q3 - P2 <-- P3 3 <

- P2q3 + P3q3 and

(2) P2q3 + P3q3 < P2 3 + P33 < q3'

or,

(1) P2q3 + P3q3 < P33- - P3 A q3 and

(2) P3 3 :<- P23 + P3 3 < q3.

* * . - . -

''' 'w ~ * i ' -4 L .r Y . ;T- -V 7T71 a" w-*. 0-.. e.. -i -- w.

39

Case H P 2 q 3 + P3 q3 < P3 q 3 and P3 A q3 < q3 - P2.

(1) P3 q 3 - P3 3 < P 2q 3 + P 3 q3 and

(2) P2 q 3 + P 3 q 3 < P2 3 + P3 3 < P 2 + P33'

or,

(1) P2 q 3 + P 3 q3 <- P3 3 < P3 A q3 and

(2) P 3 3 <- P2 3 + P3 3 <- P2 + P3 3 "

Under each of the Cases A, B and D, the inequality (1) virtually

remains the same, but in the Cases C, E, F and H, the inequality (1)

is split into 2 inequalities. The most complicated case is Case G in

which the inequality (3) is split into 3 inequalities. This splitting

of (1) influences the inequality (2) in that the symbols V and Adisappear from each side of the inequality (2).

We assume that we have Case G to contend with. Note that we have

under Case G, p 3 q 3 < q3 - P 2 < P 2 q3 + P3 q3 < P 3 A q 3 "

Similarly, the inequality (3) gives rise to at most 3 inequalities

equivalent to (1). We assume that here also we have the most complicated

case of 3 inequalities equivalent to (1). Assume, without loss of

generality, that

P3q 3 < P 3q2 + P3 q3 < P 3 - q2 < P3 A q 3 "

Assume also that p3q3 < p2q2 + p2q3 + p3q3 + p3q3 < P3 A q3 and

P 3 q 3 < q3 - P2 < P 3 q2 + P3q3 < P3 - q2 < P2q3 + P3q 3

< P2 q2 + P 2 q 3 + P 3 q2 + P 3 q3 < P3 A q3 "

If some other order prevails among these seven numbers, we can analyze

this case with equal facility. We are dealing with the maximum number

of inequalities that can arise equivalent to (1) induced by the

inequalities (2) and (3) and the number p2q2 + p2q3 + p3q2 + p3q3•

%I

40

To round up the discussion initiated by the inequalities (1), (2)

and (3), we conclude that the inequalities (1), (2) and (3) are

equivalent to the following sets of inequalities.

[1 P3q3 P33 < q3 - P21

[1](2) P2 q3 + P3 q3 < P2 3 + P3 3 < P2 + P3 3 ' and

[1](3) P3q2 + P3q3 < P3 2 + P3 3 < q2 + P33 '

or,

[2] q3 - P2 < P3 3 _ P3q2 + P3 V

[2](2) P2 q3 + P3 q3 <-L P2 3 + P3 3 < q3' and

[2](3) P3q 2 + P3q3 < P3 2 + P 3 3 _ q2 +

or,

[3] P3q2 + P3q3 < P3 3 < P3 - q2

[3](2) P2 q3 + P3 q3 < P2 3 + P3 3 < q3 0 and ".[31 (3) P3 3

< P32 + P33< q2 +

or,

[4] P3 - q2 < P33 P2q3 + P3q3'

[4](2) P2q3 + P3q3 < P2 3 + P3 3 < q3, and "

P[4](3) P33- P32 + P33-< PP

or,

[5] P2q3 + P3q3 < P3 3 <-- P2q2 + P2q3 + P3q2 + P3q3 -

[5P(2) P3 3 < P23 + P3 3 < q3 ' and

[5](3) P3 3 -< P32 + P3 3 < P3 5-

or,

(6] P2q2 + P2q3 + P3q2 + P3q3 < P33 p A q3,

[61(2) P3 3 < P23 + P3 3 < q3 ' and

[6](3) P3 3 < P3 2 + P3 3 < P3.

We observe that the symbols V and A have disappeared from the

inequalities (2) and (3).

=e ,''"'' ''" " "",2 ' " " . " '- *'. " ",' -". ", '. ' ..'5' . .. . ", " . . ..' .''' - -"""""' . ." . - '"%","",""""' " ""

41

Now, we have to f it the inequality (4) into the above scheme of

inequalities equivalent to (1), (2) and (3). The primary goal is to

get rid of the symbols A and V from (4) before appending it to each

one of the above six sets of inequalities. Neutralization of the

symbol A on the right hand side of the inequality (4) can be effected

in the same way as its done in the inequalities (2) and (3) earlier.

* This may involve further splitting up the inequalities (2) and (3) in

each and every set of inequalities above. Or, A can be retained as it

is in (4). To neutralize the symbol V on the seft hand side of the

inequality (4), we introduce a new inequality intermediate to (2), (3)

* on one side and (4) on the other side. For each set of inequalities

given above, one can determine the minimum and maximum value of

P2 + P32 + P33 . Let these values be a* and a* respectively. There

are several possibilities.

Case 1 q q + q < a*P2q2 +P 2q3 +p 3q2 +P 3q3 V

En this case, there is no need to introduce an intermediate inequality.

Case 2 a* < pq 2 +pq 3 pq pq < a*.

We introduce an intermediate inequality

a1 - P23 + +3 3 < P2 q2 + p2 q3 + P3 q2 + P3 q3 '

or,

+ + q3 < a*p2q 2 +p 2 q3 +p 3 q2 +p 3 P23 +P 32 +p 33 - *

Case 3 a* < pq 2 +pq 3 +pq 2 +pq

In this case, there is no need to introduce an intermediate inequality.

A typical set of inequalities has the following appearance.

A (1) a1 3<a2

(2) bI 2+ 3<b2

(3) c 1 p +p 3 c

(4) d1 p2 2 + p3 2 +p 2 3 + P3 3 f d 2;

42

or,

B (1) a 1 P33 < a2

(2) b 1 <-P 2 3 +IP33 b 2

(3) C 1 <P 3 2 + P3 3 C 2

(4) d 1 P23 P32 P33 d 2

(5) e 1 -P 2 2 + P2 3 + P3 2 + P3 3 e2

In the first lot above, bit b2, cI and c2 could be either constants or

linear functions of P3 3. The same remarks apply to bi, b2, c1 and c2

in the second lot. In the first lot, d 1 P2 3 + P3 2 + P3 3 or

p2q2 + p2q3 + p3q2 + p3q3 and d2 = P2 + P3 2 + P3 3 or q2 + P2 3 + P33 .

In the second lot, d = minimum value of P2 3 + P3 2 + P3 3 subject to

(1), (2) and (3) or p2q2 + p2 q3 + p3q2 + p3q3 and d2 = p2q2 + p2q3 + p3q2 + p3q3

or the maximum value of P23 + P3 2 + P3 3 subject to (1), (2) and (3).

Further, eI = P2 3 + P3 2 + P3 3 or p2q2 + p2 q3 + p3q2 + p3q3 and

e= P2 + P3 2 + P3 3 or q2 + P23 + P3 3.

In conclusion, we have some sets of inequalities where each set is

one of the two types A or B described above. These sets of inequalities

are jointly equivalent to (1), (2), (3) and (4).

43

In order to find the extreme points of MPQD(PlP 2 ,P3; ql,q 2,q3),

we look at the following problem. Let A be the collection of all

vectors (P22, P2 3 ' P3 2 ' P33) satisfying

(1) a 1 P33 2'

(2) bb(2) bl --P23 + P33 <- 2'

(3) c < P3 2 + P3 3 < c2 '

1(4) d P<P23 + P3 2 + P3 3 < d2 and

(5) e < P22 + P2 3 + P3 2 + P33 < e2'

where a1 and a2 are given fixed numbers; bl, b2, C 2 are functions

of P33 ; dl, d2 are fixed numbers; and el, e2 are functions of P2 3 ' P3 2

and P3 3. The set A is, obviously a compact convex set with a finite

number of extreme points. We now describe a method of generating

extreme points of the set A. Let F 6 636465 denote the collection61' 2' 3' 4' 5

of all vectors (P22' P2 3 ' P3 2, P33) in A for which

equality holds in (i) on the left of (i) if 6, = 1,

equality holds in (i) on the right of (i) if 6i = -1,

strictly inequality holds in (i) if 6, = 0,

for i = 1,2,3,4,5. Each 6 takes one of the values 1, -1, 0. There

53 = 243 pairwise disjoint sets F 6116 6 a 's whose union is A.35 24 pairise dsjoin s 1' ,2' 3' 4' 5

Some of these sets could be vacuous. Assume that for each 61, 6 3,64,6 = +1or-1F61 . hisasum1'o 2,3646

there exists• 64 such that F 6 6 6 6 0. This assumption is

satisfied in the original problem under investigation. We define the

dimension of each of these sets as follows. We associate the following

vectors with each of the five inequalities (1), (2), (3), (4) and (5).

44

Inequality Vector

(1) (0,0,0,1)

(2) (0,1,0,1)

(3) (0,0,1,1)

(4) (0,1,1,1)

(5) (1,1,1,1)

Let li,2,'63,'4 and 65 be fixed. The dimension of F,,2,63,64,65 is

defined to be the number of 4 -(the number of linearly independent

vectors among (i)'s above for which 6 1 - +1 or -1). For example, the

set F, 1 has dimension 0. The F1 ,_II110 has dimension 1.

Each of the sets F is compact and convex. These

sets have the following property.

Theorem 11 Every member of any set of dimension i is a convex combina-

tion of members of some sets of lower dimension for i = 1,2,3,4.

Proof The proof in all its details is very lengthy. We merely sketch

the details. Take any set of dimension 1, F 1,1,_1 1-_0, say. Look at

the sets FI,,- i and F These two sets are singleton

sets. If (P22 ,P23 ,p32,p33) ' FII,_I,_I,0 , (q22 ,q23,q32 ,q33)e

FI,-I,-, and (s22,s23,s32"33) C FI ,, then P33-q33's33;

P23 " q23 " '23; P32 , q32 " '32 and q22 < P22 < s22 " We can write

(P22,p23,P32,p33 as a convex combination of (q22 ,q23,q3 2,q3 3) and

(s22,s23's32,s33) Similarly, every member of a set of dimension 2

can be written as a convex combination of some two members of a well-

chosen set of dimension 1. The same argument can be carried out for

sets of higher dimension.

As a consequence of the above result, we can identify the extreme

points of A.

N% * % f N!*r 17.'* ..

45

Corollary 12 The set of extreme points of A is contained in the union

of all sets of dimension 0.

Theorem 11 and Corollary 12 are useful in enumerating extreme

points of PQD(Plp 2,p 3 ;ql,q 2 ,q3). For each set of inequalities, using

Theorems 11 and 12, enumerate the extreme points of the corresponding

set A defined as above. The totality of all these extreme points pooled

together from each set of inequalities constitutes the set of all

extreme points of MQD(PlP2,P3;qlsq2,q3) .

We illustrate the above procedure with some examples.

Example 13

Let p1 , P2 = P3 , ql - q2 = q3 = 1/3

Determining inequalities are:

(1) P3q3 f P33 f P3 A q3'

(2) P3 3 V(P 2q3 + P3q3) P23 + P3 3 5_ q3 A (P2 + P33 ) '

(3) P3 3 V (P 3 q2 + P 3 q3 ) < P3 2 + P 3 3 <- P3 A (q2 + P33 ) '

(4) (P2 3 + P3 2 + P3 3) V (P2q2 + P2q3 + P3q2 + P3q3 )

<P22 + P2 3 + P3 2 + P3 3

<(P 2 + P3 2 + P3 3) A(q 2 + P23 + P3 3)"

After substitution, these inequalities become

1 1(1) 1 1

(2) P3 3 / (Z9) <P23 + P3 3 .3L 1 (3 + P33 )

(3) P3 3 \/ (-) < P32 < (- 3 P3 3 ),

(4) (P 23 + P3 2 + P3 3 ) V () < P2 2 + P2 3 + P3 2 + P3 3

(3 + P32 + P33 ) A (3 + P23 + P3 3)"

The inequalities (2) and (3) induce a partition of the inequality (1)

into

1 2 2SP33 < and < P33 <

9 ~.p 3 3 ~~-,!

. . . . . . .. . . . - 1* . .* .

46

The inequalities (1), (2) and (3) are equivalent to

1 2[(I) 7 _ P33 < 9

2 <1[19(2) _ P23 + P33 - 3

2 <1[1](3) -1< P3 2 + P3 3 1

and

[21(1) 21[2] I) 9<--P33 <_

1

[21(2) P33 < P2 3 + P3 3 3_

[2](3) P3 3 -<- P3 2 + P3 3 3_

The maximum value of P23 + P32 + P33 when P23 ' P32 ' P33 satisfy5

[1](1), [1](2) and [1](3) is 1. The maximum value of p2 3 + P32 + P3 3

4when P23' P3 2 ' P33 satisfy [2](1), [2](2) and [2](3) is .. Inequality (4)

can be appended to each one of the sets above giving rise to 3 sets of

inequalities equivalent to (1), (2), (3) and (4).

1 2[() 1 3 --

[1](2) 2 <1- P23 P3 3 3

2 <1[19(3) P23 + P3 3 3

2 4[1(4) 9 < P2 3 + P3 2 + P3 3

<--

4 1

[1](5) 9 - P22 + P23 + P32 + P33 -- + P 32 + P3 3) A (- + P23 + P3 3),

or,

[2](1) 1 < P < 2

9 - 33 - 9

[2](2) 2 p 3 +19 P23 + P33 32 1

[2P(3) - P32 + P33 -3

4 5[2P(4 -+ P32 + P33 -9

K-' + + P P + P23+ +

(-I + P23 + P 3 3) (-j + P32+ P3 3

) '

47

or,

[3] (1) 2 1_P3 _

1

[31(2) P33 < P23 + P33 <-

[3](3) P33 < P32 + P33 <.

[3](4) + 3 + P32 + P3 3 < (3 + P2 3 + P3 3) A ( P + P32 + P33

Now, in each of the sets [1] and [2] and in each of the inequalities

(1), (2), (3) and (5), we set the central expression either equal to

the quantity on the right or on the left and then solve the resultant

set of equations in the unknowns P2 2 ' P2 3 ' P32 and P3 3 making sure

that the inequality (4) is satisfied. Using these numbers, we build

matrix belonging to MPD 111111". a atrix belo ging o i-M QD - 3 ' ; 3 3 ' 39 . Avoiding repititions, we

obtain the following matrices.

1 1 1 1 1 1

1. 2 9

1 1 1 2 1

1 1 1 2 1

9 9 9 9 9

1 2 2 13. 9 0 4. 9 0 9

1 2 0 2 10 9

9 9 9 9 9

1 1 1 2 15. 6. 0

1 2 09 99 9

1 2 2 10 0 -

999 9

48

0 8

1 2 3S1 -

9 9i 1 1 1 2

9 9 9 9 91 1 1 1 21. - 0

2 1 i

0 1 2 1 2o 0

i12 02 11. - 2g 12. - -g

2 1 2 17 0 90 9 9

1 2 1 2

2 1 313. - 0 -1 14. - 0 0

9 7 9

1 2 2 17 0 0 7

0 1 2 0 1 20 9 9 9 9

1.2 1 0 16. 0 09 9

1 2 1 27 0 90 9 9

2 1 2 105. 0 96

o.-. o - . . . . . . .

49

In the third lot of inequalities, set the central expression in

each inequality either equal to the quantity on the right or on the

left and then solve the resultant equations in P2 2 ' P2 3 ' P32 and P3 3.

We obtain the following matrices after omitting repititions.

i 2 317. 1 2 0 18. 0 0

9 99

2 1 32 0 0 0

0 0 9 0 0 -9 9

Note that the matrix 14 is a corvex combination of the matrices

16 and 18. All these seventeen matrices with the omission of matrix 14

are the extreme points of MPQD(3 , ' 1111' 3 3'

It is interesting to compare the extreme points of the set

wtQD(I 1 11 1 h the set M(-, - -' - of all matrices"pQ 3' 3 ' 3' 3' 3' 3-

1 11of order 3 x 3 with nonnegative entries, row sums 3' 33 and column

sums 3' 3' - The set M( 3' 39 3' 39 1 has only 6 extreme points

given by

3 3

00 0 03 i3

o 0 1 0 03 J

1 1

0 0 0 0

3 3oo 0 0 1

1 0 0 03 o 3

1 10 0 3 0 0 -

o 1 0 0 03 3

1 10 0 0 3 0

50

Note that the set 1111 11 is a subset of

3 3' 3, ;, ) and only one extreme point of M(3, 3' 3 ' 3

is an extreme point of 11QD1 1 1 3 1 3' 3)"

Example 14

1 1 1 1 1 3Let P1 =W P2 =-i' P3 ' ql q2 8and q3

The determining inequalities are:

() P3 q3 < P3 3 < P3 Aq 3

(2) P 33 VI(P2 q3 + P 3 q 3 ) < P2 3 + P3 3 - q3 A (P 2 + P 3 3 )

(3) P3 3 "v (P3q2 + P3q3 ) < P3 2 + P3 3 < P3 A (q2 + P33 )

(4) (P2 3 + P3 2 + P3 3) V (p2q2 + p2q3 + p3q2 + p3q3)

-- P22 + P2 3 + P32 + P3 3 < (P2 + P32 + P3 3) A (q2 + P23 +P33 )

After substitution, these inequalities become

3 1(i) 6 -- P3 3

-- 4

P33 V ( 16 -- P23 + P33 - 4 ( 2 + P33 )

~1 1(3) P33 V ( 3) <- P32 + P33 < A (1 + P33 )

( 4 ) ( p 2 3 + P 2 + P 3 2 1

( 23 + P3 2 + P33 () < P2 2 + P2 3 + P32 + P3 31 1< + + P3 3) A (-- + P2 3 + P33 ) "

The inequalities (1), (2) and (3) are equivalent to the following two

sets of inequalities.

6 7[i] i) 32 5_P33 < 32

9 1[1)(2) 16 _ 2 3 + P33 -_2 + P33

7 1[11(3) 2 <P 3 2 + P 3 3 <_,

*-***-*********** .Al d.It~

L, u.L4 . ° W•1• U .. i -

51

or,

[27(1) 832(I -_P33 3-2

9 1[2] (2) < p2 3 +P 3 3 <-j+ p3 3

[2]1(3) P 3 3 < P3 2 + P 3 3 <-

The maximum and minimum values of P2 3 + P3 2 + P3 3 are tabulated

below for each of the above two sets.

Sets of inequalities Expression Maximum Minimum

[11 24 18P23 + P32 + P33 32 32

[21 + 24 18[2] 23 + P32 + P33 32 32

We append the inequality (4) to each one of the above sets after

18 21splitting the range of P2 3 + P3 2 + P3 3 into two parts [ ' -] and

21 24[L, -]. We obtain four sets of inequalities equivalent to

(1), (2), (3) and (4).

6 7[1] (1) 32 2- P33 2-32

18 16[1](2) 1[i 2) P23 + P33 3162 + P33

7 8[1](3) 32 < P32 + P33 32

18 21[i](4) _32 P23 + + P33 32[1(5) 21 16 4

32< P2 2 +P 2 3 +P 3 2 + P3 3 < (32 + P3 2 + P 3 3) ( + P 2 3 +P 3 3 );

(2](1) 6 7[2(i) 3-2 _P3 3_ 32

18 16[21(2) 3232 2) -a 5_23 + P33 1362 + P33

7 8[2](3) 32 -- P32 + P 3 3 32

21 24[2](4) 2 < P2 3 + p32 + P3 3 -5_32

[2](5) P2 3 + P3 2 + P33 _ P2 2 + P2 3 + P3 2 + P33

1_6 + P 3 2 + P33 ) ( + p2 3 + P.~~~.3 23- 33A(-j 33)

.~~~~S . ....

52

,.7 8[31 (1) - P <

32 33 3218 16

[3] (2) 3 2 - P23 + P3 3 -2-6- + P3 3

[31(3) P33- P32 + P33 32

18 21[3(4) 32 -- P23 +32 + P33 - 32A

[3(5 (16 + ) /A 3+ + P33[34(5) P P2 2 + P2 3 + P32 + P3 3 p2 3 2 +P 3 3 32 P2333

32- P33 <-32

18 16[4](2) T2 5_ P2 3 + P3 2 < T2- + P33

814](3) P33 < P32 + P33 32

21 <24

[4]3(4) 3- P23 + P32 + P3 3 32

[4](5) P23 + P3 2 + P3 3 < P2 2 + P2 3 + P3 2 + P33

< ("L + p3 2 + P3 3) A( + P2 3 + P33

In each set, for each of the inequalities (1), (2), (3) and (5),

we set the central expression equal to either the quantity on the left

or on the right and then solve the resultant equations in P22 ' P2 3 ' P3 2

and P3 3 after making sure that the inequality (4) is satisfied. Using

P2 2 ' P2 3 ' P3 2 'P3 3' one can build P in MPQD(plp 2 ,P3 ql,q 2,q3) in a

natural way. These matrices are given below.

1. 1 6 2 6, 32 32 32 32 0 32

2 2 12 1 3 1232 32 32 32 32 32

,32 32 32 -2 T 32

53

116 2. 6

3 1 12 2 3 12

02 6 2 _ 6

1 1 6 2 632 352 2 32 32

2 3 11 1 4 1132 2 3j2 32 2 32

1 7 17

1 1 6 2 6 L

32 3 2 352 32 0 32

3 2 11 2 3 1132 3 2 '32 3 2 3 2 32

0 1 7 0 2 7

3 3 2 4 2 23 232 32 352 32

0 0 16 0 0 16

1 1 6 2 6-2 32 2 32 0

4C 3 1 3 4 132 32 32 32 32 32

0 0 16 0 0 16

0 1 7 1 0 7

54

1 1 6 4 432 32 32 32 32

3 3 10 0 0 16

32 3 " 32

8 80 32 0 0 3

2 632 32

2 4 1032 32 '32

832

These are the extreme points of MpQD( 1, 1 .1 1, 3M-, 4, 4,s,

-- - -W- t -

55

4. On the power of tests for testing independence against positive

quadrant dependence

In this section, we apply the results of the previous sections to

study power of tests in the context of contingency tables. The

problem, basically, is the following. Let X and Y be two random .

variables and each takes finitely many numerical values. The distribu-

tions of each of X and Y are known but their joint distribution is

unknown. For simplicity, assume that X takes values 1,2,...,m with

probabilities plp 2,...,pm respectively and Y takes values 1,2,....n

with probabilities ql,q 2 ,...,cn respectively. Let pij = P{x=i, y=j},

i = 1,2,...,m and j = 1,2,...,n. We say that X and Y are strictly

positive quadrant dependent if X and Y are positively quadrant dependent

but X and Y are not independent. We want to test the validity of the

null hypothesis

H0 : X and Y are independent

against

H1 : X and Y are strictly positive quadrant

dependent

based on a random sample of size N on (X,Y). Let nij = number in the

sample with x = i and y = j, i = 1,2,...,m and j = 1,2,...,n. The

data can be tabulated ,in the form of a contingency table.

x1 2 ... n

1 n1 1 n1 2 ... nin

2 n2 1 n2 2 . 2n

m nml nm2 ... nmn

m nHere, I n nlj N.

i=1 J=i

56

Let T be any test proposed for testing H0 against H1 for a given

level of significance, a. T is a function of the random variables

nij i = 1,2,...,m; j = 1,2,...n. The performance of the test T can be

judged based on its power function. More precisely, let

PQD(PlP2, ...,pm; q1 ,q2 ,..,%) be the collection of all matrices P of

order m x n with non-negative entries, row marginal sums plP2,...,pm,

column marginal sums ql,ql,...,qn and positive quadrant dependent.th

Let P0 be the matrix of order m x n whose (i,j) element is given by

Piqj , i = 1,2 ,...,m and j = 1,2,...,n. The parameter space of the

above problem is MQD(PlP2,...,pm; ql,q 2 ,...,qn). The hypotheses

H0 and H1 translate into as follows. Let P stand for the generic

symbol in MQD(PlP2 , . . . , p o ; ql'q 2 '. ' ' q n ) deemed as the joint

distribution of X and Y. The relevant hypotheses are:

H0 P P0

H1 : P #P 0

The power function 0(.) of T is defined by

ST(P) = Pr {T rejects H0 / the joint distribution of

X and Y is P),

P C MPQD(PlP2, ... ,pm; qlq 29,...,9qnd .

Obviously, 8T(P0) = a. The following results advocate the necessity

of enumerating the extreme points of MPQD(PlP 2 ,.. .,pm; ql'q2 '. ' ' q n ) "

Theorem 4.1 Let P e MPQD(PlP2,...Pm ; ql,q 2 ,...qn) be a convex

combination of some matrices PIP12,.'P k in MQD(PlP2 , . . . , p m ; ql'q 2' ' ' ' q n ) '

i.e. k

P Pi=l

kfor some al,c2 ,...,ak > 0 with I ai = 1. Then

i=l

kaT (P) = a iT (Pi)"

57

Proof. Obvious.

The above theorem establishes the fact that 8(.) is an affine

function on the compact convex set MPQD(PlP2 ,...,pm ; q1,q2'''' )

Corollary 4.2 Let E (plP 2,...,pm ; ql,q 2 ,...,qn) = E, say, be the set of

all extreme points of MPQD(PlP 2 ,...,pm; ql,q 2,. .qn). Let E

consist of some k matrices PI1P2,.',pk* Then for any P in

MQD(PlP2,...pm; ql,q2...,qn) , 8p(T) is a convex combination of

a T(PI1), a T(P 2,., T(Pk).

The above corollary provides us a criterion by which we can

compare the performance of any two given tests T1 and T2 proposed to

discriminate between H0 and H if PP 2 P ...,Pk are the extreme

points of MQD(PlP2,..,pm; ql,q2,...qn), then we compare the two

sets of numbers

T 1 (P1 sT1(P2 a T1(P k

and

a T2 (P 8T2 (P 2 ,.. T2(P k

for a showdown between T and T2 as to which one is more effective in

discriminating between H0 and H1 .

In a paper by Bhaskara Rao, Krishnaiah and Subramanyam [61, a detailed

comparison of performance of several tests is carried out for some cases

of contingency tables.

p'.

58

5. Comparison of tests

In this section, we actually embark on comparing the performance

of several tests for testing the null hypothesis H against the alternative

E 1 specified in the previous section in the case of 2 x 2 and 2 x 3

contingency tables. First, we deal with the 2 x 2 case.

We now discuss the power function of the test based on Goodman-

Kruskal's Gamma Ratio. Let

Pll P12 Pl

P21 p22 p2

ql" q2 1

be the joint distribution of X and Y with specified marginal distributions

pP 2 and qlIq 2 respectively. The Goodman-Kruskal's Gamma Ratio, y,

is defined by (see Agresti [lj)

PllP22 - P12P21

PllP22 + P12P21

The following properties of y are easy to check.

i) -i < y < 1.

(ii) X and Y are independent if and only if y 0.

(iii) X and Y are positive quadrant dependent if and only if Y 0.

(iv) X and Y are strictly positive quadrant dependent if and only

if Y > 0.

From the above properties, we can assert that y is a measure of positive

quadrant dependence between X and Y. (For the validity of (iii), we have

used the fact that the marginal distributions are specified.) We also

observe the following. The extreme points of the convex set MpQD(PlP2;qlfq2

59

are given by

P p1q pq 2 and P2 1 2 p2 .)

qp2q1 p2 2 L p2

assuming that p2 A q 2 P2. Under Pie i.e., the joint distribution

of X and Y is Pt y =0 and under P2 p y =l.

A natural estimator of y is given by

n1 ln2 2 nl 2 n 1 (5.2)

n1 1n2 2 nl2n 2 1

Test of H against H1 based on T.

Reject H in favour of H if and only if y >_ a,

where a is chosen satisfying Pr(y > a / H ) a, the given level

of significance. One can work out the power function of this test explicitly.

Let

8T(P) = Pr(the test T rejects H / P)0

= Pr(y > a / P)

for P in MpQD(p1,P2;qlfq2), where the above probability is computed

when X and Y have the joint distribution P. As has been pointed out in

Section 4, aT(P) is connected to 8T(P1 ) and ST(P2) as follows. Write

P = AP1 + (I-)P 2

for some O < X < 1. Then

(P) =A13Tp') --.. l=)9 (P )

T .T. .-lT-

Obviously, T(P 1 a. Under P2F n12n21 = 0 almost surely and

, (P ) - . . , ,' 12 21

-[V T V v 1-

60

consequently Y - 1 amost surely. This implies t-hat T(P 2 ) 2 1. Thus

we have proved the following theorem.

Theorem 5.1 The power function ST( ) of the test T of size a is

given by

B (P) . Xa + (iX)IT

for P in M PQD (p1,P2 ;ql,q 2 ), where P = P 1 + (1- )P2'

The above result yields the following result.

Theorem 5.2 Let T* be any test proposed for testing H against H

specified in the previous section. Then

ST ( P ) Z 6T* (P)

for every P in MPQD (plP2 ;qlfq2).

Thus we see that in the case of 2 x 2 contingency tables, the

test based on Goodman-Kruskal's Gamma Ratio is superior to any test one

can propose for testing the hypothesis of independence against the hypothesis

of ;strict positive quadrant dependence.

Let us examine how the test based on Goodman-Kruskal's Gamma

Ratio fares against another test proposed in the case of 2 x 3 contingency

tables.

Let the joint distribution of X and Y be given by

1i1 P1 2 P1 3 Pli

j21 P2 2 P2 3 P2

Iql q2 q3 1

.j..,-- .. . " . '- ", (-'. " :u 4. -. -. ". - - - - - --- ') ' '- '' ', ' '' ' - - r *. " " : ~ .'' '- ' " - - .- ' - ''

761

where p1 'P2 and ql ,q2 ,q3 are the marginal distributions of X and Y

respectively. The Goodman-Kruskal's Gamma Ratio, y , is defined by

Tc - Td

where 11c MP + P ) + P and =

1 1 (p22 p2 3 1223 "d P1 3 (p21 p2 2 ' pl2p21*

See Agresti El, p.160]. We propose another measure of association between

X and Y as follows.

2 2 2(P1 1P2 2 p1 2p2 1) (P1 1P2 3 - P1 3p 2 1) (P12 P2 3 - P 13p 2 2)K q q q

PlP 2qlq 2 P lP2qlq 3 P lP 2q2 q 3

ofThis measure of association is motivated by the definition Spearman's rho

for 2 x 2 contingency tables. The following properties of y and c

are easy to verify.

. -~< Y <.

2. If X and Y are independent, then y = 0.

3. If X and Y are positive quadrant dependent, then K > 0.

4. 0 <K <.

5. K - 0 -if and only if X and Y are independent.

One can build tests based on the Gamma Ratio and the kappa

criterion.

62

Test based on the Ga-ma Ratio T

Let C = the total number of concordant pairs in the data

n 1 (n22 + n23) n n12n23

and D = the total number of discordant pairs in the data

n n13 (n21 + n 22 ) + n 12 n21.

An estimate of y is given by

" C - D-Y C +D

Test based on y is given by

Reject the null hypothesis H in favour of H if Y > a,

where a is given by Pr(; 2 a / Bo) = 0 , the given level of significance.

Let 8T( ) be the power function of the test T.

Test based on the Kappa value T*

An estimate of K based on the given data is given by2 22

- (nl1n22 n 12n21) (n11 n23 n 12 n21 (n 12 n23 n 13 n2244 4N 4 Pp2q1 q2 N p1P2q1q3 N p1p2q2q3

Test based on K is given by

Reject H in favour of H1 if and only if K > a,

where a is given by Pr(K > a / H ) = 0, the given level of significance.Le o

Let 8T* ( ) be the power function of the test T*.

.~i- ... A ~ .2. - ~ - - - - - -. -, - - -M - - -