Title: On the distribution of a linear combination of t-distributed variables
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Title: On the distribution of a linear combination of t-distributed variables
Alternate Title: Linear combination of t-distributed variables
Physical Description: vii, 106 leaves : ; 28 cm.
Language: English
Creator: Walker, Glenn Alan, 1950-
Copyright Date: 1977
 Subjects
Subject: Variables (Mathematics)   ( lcsh )
Statistics thesis Ph. D
Dissertations, Academic -- Statistics -- UF
Genre: bibliography   ( marcgt )
non-fiction   ( marcgt )
 Notes
Statement of Responsibility: by Glenn Alan Walker.
Thesis: Thesis--University of Florida.
Bibliography: Bibliography: leaves 104-105.
General Note: Typescript.
General Note: Vita.
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Bibliographic ID: UF00098924
Volume ID: VID00001
Source Institution: University of Florida
Holding Location: University of Florida
Rights Management: All rights reserved by the source institution and holding location.
Resource Identifier: alephbibnum - 000063604
oclc - 04214071
notis - AAG8803

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ON THE DISTRIBUTION OF A LINEAR
COMBINATION OF t-DISTRIBUTED VARIABLES







By

Glenn Alan Walker


A DISSERTATION PRESENTED TO THE GRADUATE COUNCIL OF
THE UNIVERSITY OF FLORIDA
IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE
DEGREE OF DOCTOR OF PHILOSOPHY







UNIVERSITY OF FLORIDA

1977

















ACKNOWLEDGMENTS


I wish to express my deepest gratitude to

John G. Saw for originally suggesting the topic and

providing a continued source of motivation in my

graduate years and especially in this endeavor. His

perseverance and research expertise will always be

remembered.

Appreciation is expressed also to Jim Boyett

for his helpful comments while proofing the rough

draft. I thank also the other members of my

supervisory committee for their suggestions and

interest. The faculty and graduate students who

have provided a stimulating atmosphere for learning

deserve a special thanks.

Finally, my appreciation goes to Betty Rovira

for her patient help in typing the final draft.
















TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS . . . . . . . .. ii

LIST OF TABLES . . . ... . . . . v

ABSTRACT ... . . . .. . . . vi

CHAPTER . . . . . . . . .. .

1 INTRODUCTION . . . . . . . 1

1.1 Motivation: The Behrens-Fisher
Problem . .. . . . 1
1.2 A Sequential Solution to the
Behrens-Pisher Problem . . . 4
1.3 Statement of the Problem . . 6
1.4 Notation . . . . . . 8

2 THE PRE-t DISTRIBUTION . . . .. 10

2.1 Density and Moments . . . .. 10
2.2 The Characteristic Function of X 11
2.3 The Cumulative Distribution
Function . . . . . .. 15
2.4 Additional Properties . ... . 16

3 THE DISTRIBUTION OF A LINEAR COMBINATION
OF t-VARIABLES . . . . . .. 22

3.1 Simplification . . . . .. 22
3.2 The Distribution of T Using the
Q Matrix . . . . . ... .23
3.3 The Q-1 Matrix . . . . . 26
3.4 Examples . . . . . . 34
3.5 The Case of Even Degrees of Freedom 37










TABLE OF CONTENTS Continued


CHAPTER

4 THE POWER OF THE T-STATISTIC . . .

4.1 Introduction . . .
4.2 General Results . . . . .
4.3 Applications: Some Examples and
Counterexamples . . . . .
4.4 The Power of the Pre-t R.V . .
4.5 General Results Based on Intuition
4.6 Properties of the Relation "(4
4.7 Some Counterintuitive Examples
4.8 The Power of aX + (l-a)X .

5 COMPARISONS WITH OTHER METHODS . .

5.1 Ghosh's Results . . . .
5.2 The Normal Approximation . .

6 APPLICATIONS . . . . . . .

6.1 The Behrens-Fisher Problem . .
6.2 A Nonstationary First Order
Autoregressive Process . . .

BIBLIOGRAPHY . . . . . . . . .

BIOGRAPHICAL SKETCH . . . . . . . .


Page


44

44
46

49
55
67
73
76
84

86

86
88

91

91

100

104

106














LIST OF TABLES


TABLE Page

3.1 THE LOWER-TRIANGULAR ELEMENTS OF THE FIRST
11 ROWS AND COLUMNS OF . . . . ... 27

3.2 THE LOWER-TRIANGULAR ELEMENTS OF THE FIRST
11 ROWS AND COLUMNS OF Q-1 .. . . .. 28

3.3 THE ELEMENTS OF n FOR THE AVERAGE OF TWO
iid Pre-t R.V.'s WITH 2m+l D.F. . . . 36

3.4 ELEMENTS OF n WHEN T = aX + (l-a)X . 38

3.5 VALUES OF A SUCH THAT PR{aX + (l-a)X >A}=4 40

4.1 VALUES OF Sk(a) FOLLOWED BY hk( k(a)) . 58

4.2 VALUES OF HkHk+ (x) . . . . . . 61

4.3 POWER OF THE TEST aX + (l-a)X .2 ...... 69

5.1 COMPARISON OF THE CDF OF X' +X" WITH A
NORMAL CDF . . . . . . 90















Abstract of Dissertation Presented to the
Graduate Council of the University of Florida in Partial
Fulfillment of the Requirements for the
Degree of Doctor of Philosophy



ON THE DISTRIBUTION OF A LINEAR
COMBINATION OF t-DISTRIBUTED VARIABLES

By

Glenn Alan Walker

June 1977

Chairman: Dr. John G. Saw
Major Department: Statistics

For odd degrees of freedom the characteristic

function of a Student-t random variable is expressible

in closed form. The.characteristic function of an

arbitrary linear combination of independent t-variables

is then derived and the distribution function is ob-

tained, itself expressible as a weighted sum of

Student-t distribution functions. An easy method of

obtaining the weights is demonstrated.

If U1,U2,...,Un are independent random variables

and Xi =d +Ui, 1=1,2,...,n are observable random

variables, we investigate the choice of al,a2,...,an to

maximize the power of tests of the form alX1 X+a2X2 +...

+ anX for testing HG: d =0 against H1: d > 0. Some

general results and examples are given. Of particular










interest is the case when X. is a t-random variable.

One application is in a two-stage sampling procedure

to solve the Behrens-Fisher problem. The test statistic

has the distribution of a weighted sum of t-random

variables. It is shown how to choose the weights for

maximum power.
















CHAPTER I

INTRODUCTION


1.1 Motivation: The Behrens-Fisher Problem

Given that the distribution of the population II.

is normal with mean p. and variance o. for i=1,2, let
1
it be required to test the equality of p1 and P2 based

on data independently drawn from H1 and H2. If we know
2 2
the value of 9 = 1/o2, the test criterion takes the

form

t = x1-x2)
1 1
n1 n2
2 (nl-l)s + 8(n2-l)s2
where s = nl+ n2- 2 and ni, xi and si are

the sample size, mean and variance, respectively, of

the sample taken from Hi. If it is true that ipl = 2

then t will be distributed according to the Student-t

probability law based on n+ n2- 2 degrees of freedom.

Under standard analysis of variance assumptions, 6 is

taken to be unity, which may or may not be a valid

assumption. If not, the resulting test will be biased

to a degree commensurable with the deviation from that

assumption. When the value of 6 is unknown, the test

criterion has an unknown distribution. This situation









is commonly referred to as the Behrens-Fisher problem.

Under Behrens-Fisher conditions no criterion with

distribution as simple as the t above is available.

Fisher proposed a fiducial test based on the so-called

Behrens-Fisher statistic,

(x 2)
D = 2 2 (1.1.1)
S 2 2 1
1 + 2
nl n2

and Sukhatme [16] calculated approximate percentage

points. Other solutions have been proposed, notably by

Welch [17], Scheff6 [13], Welch-Aspin [18], McCullough,

et al. [9], and Cochran [3], to name a few. Many of

the proposed solutions are based on an approximation or

asymptotic expansion of the distribution of the statistic

D. Fisher rewrote this quantity as

D = t cosO -t2sin(1.1.2)

-lI 1s2/v2\
where e = tan- S and ti ( =1,2) are Student-t
s'l/n;1

random variables. Ray and Pitman [11] developed an

explicit form for the density function of D. Using a

series expansion for the density of a weighted sum of

chi-square random variables, they produced a very

complex formula representing the density of D and

showed how one might compute percentage points. Using
2 2
the fact that, conditional on s~ and s2, D is the








weighted difference of two independent t-variables,

Rahman and Saleh [10] expressed the density in terms

of intergrals of Appell functions and used numerical

quadrature techniques to table critical values. In

each of these attempts an approximation was used some-

where in the development before any values could be

tabulated. In addition, it is a very complex procedure

to obtain non-tabulated critical values using any of

these methods.

Although it is still not known whether there exists

a fixed sample size test whose significance level is

independent of the unknown parameters, Dantzig [4]

proved there exists no fixed sample size test whose
2
power is independent of o.. A two-stage sampling

scheme was devised by Stein [15], the power of the

resulting t-test being independent of the nuisance

parameters. This approach may be modified to give a

solution to the Behrens-Fisher problem (see Sections 1.2

and 6.1), and in so doing, the test criterion has the

distribution of a weighted difference of Student-t vari-

ables. Chapman [2] tabulated some values of the cumula-

tive distribution function for this weighted difference

but only for the special case of equal weights and equal

degrees of freedom. Ghosh [7] also investigated this

distribution for the case of equal weights. He gave

explicit formulas for finding the distribution function










when the degrees of freedom are equal and less than 5,

and tabulated a few values of the probability integral

when the degrees of freedom are equal and greater than

5. In addition he used a series expansion to obtain an

approximation to the cumulative distribution function

when the degrees of freedom differ. Ghosh, incidentally,

pointed out that many of Chapman's values were incorrect.

Thus far, the distribution of a difference of two

t-variables with different weights has not been investi-

gated in the literature. Furthermore, if the degrees of

freedom associated with the two t-variables differ, the

exact distribution has not been derived. The Behrens-

Fisher problem would be solvable (section 1.2) if the

distribution of a weighted difference of independent

t-variables with arbitrary weights and degrees of

freedom were known.


1.2 A Sequential Solution to the Behrens-Fisher Problem

Given two normal populations, ~1 and 12, where the

mean and variance of i. are i. and a respectively (i=1,2),

we desire to test the hypothesis

H0 : l-P2=d0

To this end one might take ni observations from Hi

and use the statistic D in (1.1.1) as test criterion.

If the sample sizes are small however, the usual normal

approximation may not be accurate.

Consider the following two-stage sampling scheme:









sample variates {X i, j=l,2,...,ni, from I., i=1,2,

and compute
n.
li
Xi = iJ
j=l ni

n.
1
SS. = (X i- )2
j=1


Ni = max n + 1,ai J + 1


where al, a2 and T are preassigned constants and the

square brackets indicate "the integer part of."

The second stage consists of sampling the additional

observations {Xij}, j=n.+1 ,n+ 2 ,...,N. from II (i=1,2).

The test criterion for testing H0 will be


T = a aX2 (1.2.1)
J= i j=l
where ({a j, j=1,2,...,N., are constants satisfying


N.
1
ii). aij = 1 (1.2.2)
j=l
N.
Ni i SS.
iii). I a2 = where f. 2 2
j=1 i T a.T
i

Such constants may always be found since the minimal
Ni 1 1 1
value of a2 is 1 and N >fi so that .< T.
j=l ij 1 1I

Letting d= 1--l2 it is shown in section 6.1 that T

may be represented as










a+ a1 a2
T 1 + 2
1 Irv 2

where vi=ni-l and ti (i=1,2) are independent t random

variables on vi degrees of freedom. Thus, when H0: d=0

is true, the test criterion has the distribution of a

weighted sum of t-variables. Letting X (al,a2,nl,n2)

be the 100(1-) percentage point of the distribution of

T, we reject H0 in favor of the one-sided alternative,

HI: d>0, whenever T> (al,a2,nl,n2). The experimenter

is free to select the constants {a ij, j=l,2,...,N., a.

and ni for i=1,2, as well as T. The initial sample

sizes, ni, influence the expected total sample size,

E(N +N2) while T and the a 's influence the power

function of the test T. A discussion of a discreet

choice of T, nl, n2, and the aij's is deferred until

section 6.1. Presently we are concerned with the selec-

tion of al and a2. Two problems of immediate concern,

then, are

i. that of obtaining the percentage points, ,,

of the distribution of T, and

ii. the method of selecting aland a2 to maximize

the power of T.


1.3. Statement of the Problem

Let w1,w2,...,wn be an arbitrary set of constants

and, for i=1,2,...,n, let t. be Student-t random variables

independently distributed on v. degrees of freedom. For










d>0, denote by Td the linear combination

Td = w (tl+d) + w2(t2+d) +...+ w(t +d) (1.3.1)

which may be written as

n
Td = U + d wi (1.3.2)
i=1
n
where U = i w.t.. The first problem above may be
1=1 1
i=l

generalized to that of finding the distribution of Td.

The second problem becomes that of selecting weights,

{wi}, i=1,2,...,n, to maximize the power of the test Td

for testing

H: d=0

vs. H : d>0 .

The purpose of this current work is to investigate

the solutions to these problems. We show in particular

that the probability density of Td may be written as a

linear combination of Student-t density functions whose

weights depend on {w },i=l,2,...,n, {vi},i=1,2,...,n,

and n. The power of Td depends also on these quantities.

If 1=V 2=.. vn it is shown that no linear combination

of the tie's has uniformly greater power than for the

case when w=w2= ...=wn. If vi5vj for at least one pair,

ifj, no distinct set of weights, {wi}, exists which will

maximize the power of Td for all size tests.

Chapter 2 is an investigation of the properties of

the distribution of t. including derivation of the

characteristic function and distribution function in
characteristic function and distribution function in









closed form for odd v.. Chapter 3 includes the deriva-

tion of the distribution of Td for odd vi; the case of

even degrees of freedom is considered separately. Fpr

the case n=2, tables of the percentage points of Td are

given. Some general theorems for comparing the power of

two tests are given in Chapter 4 along with some applica-

tions, including several examples and counterexamples.

In particular it is shown that t. has greater power


than t. whenever v.>v.. Chapter 5 gives a comparison

of results with the normal approximation and with Ghosh's

results. Finally, along with the solution to the

Behrens-Fisher problem, another application is indicated

in Chapter 6.


1.4 Notation

The following notation will be used whenever con-

venient:

a. Abbreviations


cdf

d.f.

iid

iff

wp p

r.v.

b. Symbols

(A)ij


cumulative distribution function

degrees of freedom

independent, identically distributed

if and only if

with probability p

random variable



the element of row i, column j of
matrix A









(A).

(A)

E(X)



Pr{A}

r{x}

B(x,y)

J{X-Y}

*
F

C (P)

[k]

[k:l]







V







c. Generic

t

X
V
2
Xv

Z

h.

Hj


th
the i row of matrix A

the jth column of matrix A

expected value of X

variance of X

the probability of event A

the gamma function

the beta function, Ffx+yl

the jacobian of the transformation
from X to Y
-1
F the inverse function of F

the coefficient of xr in P

reference k in bibliography

page or line 1 in reference k

implies

is equivalent to

is distributed as

for all

such that

therefore

section

random variables and their distributions

Student-t r.v. on v d.f.

pre-t r.v. (tv/i)

chi-square r.v. on v d.f.

standard normal r.v.

density of X2j+1

cdf of'X2j+















CHAPTER 2

THE PRE-t DISTRIBUTION


2.1 Density and Moments

We denote by t a random variable (r.v.) having

the Student-t distribution on v degrees of freedom

(d.f.). The density of t is

v+l
t2 2
f(t) = (1+
B1 v v


Making the transformation

t
X -

we obtain

J{tV ]x =

so the density of X is

v+l
1 2 2
h(x) = 1 (1+ x )


The r.v. X shall be referred to as the "pre-t r.v.

on v d.f." We use this in place of the t in the

ensuing theoretical development to simplify the math-

ematical formulations. It is clear that h(x) is

symmetric about x=0 and decreasing as (x( increases.

Hence, all odd moments of X are zero. For k=1,2,3,...,







k t Pr{ }r2k+j v-2k
2k Ev 2k 1 2k 2 2
E(X ) = E(-) E(t2 ) =
vv k v 1 r{
V 2

v>2k (2.1.1)

In particular,

E(X ) = 0

2v>2.
and V(X V) v>2.

If Z represents a standard normal r.v. and X ,

an independent chi-square r.v. with v d.f., we may
Z 2 2
represent XV by -. If v2 > v and X and X

denote chi-square r.v.'s independently distributed on

v1 and v2- 1 d.f., respectively, then

Z2 2 2 2 -1
> 2 (X2 + with probability 1.
X2 > v1 2 V
V1

Thus for any t>0,

Pr{IX Vl>t} > Pr{|X v2>t} whenever v1
That is, IX is stochastically less than X I.

The remainder of this chapter will be devoted to

deriving the characteristic function and cdf of XV and

exploring other useful properties of its distribution.


2.2 The Characteristic Function of X

We may use a property of the Bessel functions to

give a closed form representation of the characteristic

function of X in the case when v is odd. The modified

sperical Bessel function of the third kind is given

from [1: 9.6.25] as









F{f+ }(2z) f
Kf(xz) = f
7T x
1 T
for f> and arg zl < .-
and z=1, x=lel, we obtain

v/2
K- e} 2
122
'2~ x 6 9


Scos(xt) dt
0 (t + z2)f+

Letting f= (v>-1)



S cos(e t) dt

o (l+t2)(v+1)
(2.2.1)


so that


2 cos(t ) (l+t2)-(v+1)
o (,2)2


2r 1 K (lel ).
7 /2

The characteristic function of X is

O(e;v) = E(eieXv)
= E(coseX +i sinOX )
= E(coseX ),


since E(sinexv) =

symmetric about zero
have, therefore,


sinex-h(x)dx = 0, h(-) being

and sin(ex) an odd function. We


(e;v) -= cosex (+x2-(+)
B(2'2)


(2.2.2)










= 2 cos(xje|) 2 (v+ )
=.2 (1+x ) dx ,
O B(2' )

or, from (2.2.2),


(;,v) r( 2 K (le) (2.2.3)
2'2

By [1: 10.2.15, 10.2.11],



K (z) =- e-z (n+,k)(2z)-k
k=0

(n+k)
where (n+,k) = k!(n-k)! k=0,1,2,...,n. Substituting,



k (n+k)! ()-k
S Kn+(z) e-zk nk
2z 2z k=O
or,
K n(z) z k(n k) (2z)-k n=0,1,2,...
n + 22 k-CkTnkY)
(2.2.4)
Letting v=2m+l for m=0,1,2,..., and using the identity


r{2m+1 1(2m+l) (2m)!/ 2
r 2 2 "
2m! 2m

(2.2.3) becomes


1-s(2m+l)
(6e2m+l) -2m+ Km+m (l! )
2 \








-i( i(2m+l) 2 r
S2 m! m 6|
[. I l z 2 e- l


k 0 k!(-k !(21e)


S(2|6 )m -lel (m+k)! (2| )-k
(2m)! e. k!(m-k)! 1


S-lel m! m (m+k)! m-k
= (2mTk kTnmk!(21 or finally,


(e6;2m+l) = e-11 m! (2m-k)!(2l)k (2.2.5)
(2m)!ko k!rn-k)!

mi (2m-k) i k
Denoting qm(z) = m!T !(2m-k)!(2z )k we may
k=0
simply write

0(( ;2m+l) = e- q m(l el ) m=0,1,2,...
(2.2.6)
In particular, we have

$(e;l) = e-le

(e;3) = e-1 e ( + l el )

0(8;5) = e- 1( + |1 + 1e 2) (2.2.7)

(e;7) = e-l l (1 + le 8 + ]2 15 + el3)

W(e69) = e- (1 + le| + 3|10 2 + l 3+l 1]4)









2.3 The Cumulative Distribution Function

After an integration by parts,


cos2m dO = cos2m-lsinO + 2-
I 2m si 2m


and,


o2m-2 d
cos 0 de


2 1
cos 6 dO = 2(sin6cos6 + e).


Therefore,


2 1 2m-1 2m-1 2m-3.
cos2me de = 1 cos2m- sine + 2m- cos 2 sin6
2m 2m(2m-2)


+ (2m-1)(2m-3) 2m-5
2m(2m-2)(2m-4) c os OsinO +

(2m-1)(2m-3)-. (5)(3) 2
2m(2m-2)(2m-4)-. (6)(4) cose d

sin os2m-1 + 2m-1 2m-3 (2m-1)(2m-3) C2m-50
2m + m-2 cos (2m-2)(2m-4)

(2m-1)(2m-3) .. (5)(3)
+ (2m-2)(2m-4). ..( )(4)(2)cos


+ (2m-1)(2m-3)...(5)(3)
2m(2m-2)... (4)(2)


(2.3.1)


By Gauss duplication,


Fr{m+} = 2 rF{2m)
2mFm}


so that


1 F{m+1) mi
(1, 2m+l li 2m+l
s(2- r-I2r{---} 2 r m+}


(2.3.2)








2m {(m-1)!2m-1 2
wT (2m-1)!
2m (2m-2)(2m-4)---(6)(4)(2)
T (2m-1)(2m-3) I (5)(3)(1)

Denote the density of X2m+1 by hm and its cdf by H ,

m=0,1,2,... For t > 0 and e = tan-lt,

1 2 (1 2 -(m+1)
Hm(t) = + 21 (l+x ) dx
0 2 2

1 1 2m
S+ 1 m+ cos udu
2 T 2

1 +1 2m(2m-2)(2m-n4)) i-(6)(4)(2)]
2 IT (2m-1)(2m-3)--- (5)(3)(1)

where I represents the right-hand side of equation (2.3.1).

The closed form expression for the cdf of X2m+l is

therefore


Hm(t) = + e + sin e {cos + ^cos3 + (2 os5

+ (2)(4)***(2m-4)(2m-2) 2m-1
(3)(5)... (2m-3)(2m-1)cos } ,


where, 0 = tan-1t, t>0. (2.3.4)


2.4 Additional Properties

The following properties about the distribution of

X will be useful in later sections:

Property 1. h(t) t(v+l) 2 -(v+3)
Property 1. (l+t )
at iv(^)
(2'2

Property 2. Hm(t)-H (t) = hm(t), m=0,1,2,..
m+l m 2mt>
t>o









Property 3.



Property 4.



Property 5.


m-1 h.(t)
H(t)-Hk(t) = t Z j
j=k 2j+l


, m >.k.


30(8;2m+l) -_ 9(e;2m-l)
96 2m-1 m=1,2,3,...


62 2(e;21-l)
4(6;2m+3) = (6e;2m+l) + 62(2 ;2m-1l)
(2m-1)(2m+1)


m=1,2,3,...

Property 1 follows by differentiating both sides of



h(t) (l+t2)
1+v


To prove Property 2 we may use the results of

section 2.3:


Hm+l(t)-H(t) =


1sin6 (2)(4)(6)' (22m) os2m+l
r (1)(3)(5) ---(2m+ )


(6=tan- t)


1 (m!2m)2 -(m+l)
- 2m t(l+t2)
71 (2m+l) l


t r{m+l}
2m+l {N1


22m {m} 2 -(m+1)
r{2m}2 (+t
Y{2mn}2,/


Using (2.3.2), we obtain


Hm+1(t)-Hm(t) =


t F{m+l}
T2mtI) 2{}r +l
7 2


(l+t2)-(m+)


t
2m+1 hr(t) m=0,1,2,...








If m> k,

m-1
Hm(t)-Hk(t)= H m+k.(t)-Hk.j(t)}
j -
j=k

m-1
m -i h (t)}
j=k 2m+2k-2j-l m+k-j-1

m-1
= t 1 h (t)
j=k 2j+l ]

which is Property 3. To demonstrate Property 4, we

may begin by writing, from (2.2.6),

e e;2 (-1)! m 1 (2m-k-2)! (1 k
e a e(0;2m-l) = 2 2 k k! (m-k-1) (21 1) ,
k=O

a polynomial of degree m-1 in ]e|. Also,


(;2m+) = e-(2m-k)! )k
()'1 km(Y-k) (2161)

so that


el6| 8 0(;2m+l) m! m (2m-k)! 2(k elk-l k
S36(2m)! k 0 k!(m-k)! l -I I),

a polynomial of degree m in 16 Using Cr to denote
x
"the coefficient of xr in", we have


r 3 0 (6;e;2m+l) = e-1 m! f (2m-r-l)! 2r+l(r+l)
6 O9 T6 (2mF! (r+l)!(m-r-l)!

(2m-r)!2r
r!(m-r)!










e- 2 m! (2mr l)!2r {2(r+1)(m-r)-(2m-r)(r+l)}
(2m)l (r+P!m-r)l


= I m (2m-r-1)!2r
(2m)!(r-1)!(m-r)! r=1,2,3,...,m. (2.4.1)

Also,


r 0 n 1 -Ie 2|( m
i{- t 4(e;2m-l)} = (2m-1) (2m-2)!

x (2m-r-l) 2r-i
(r-l) (m-r)!


= _e-1I m!r(2m-r-1)! 2r
(2m) (r-1) !(m-r)! r= m

which is identical to (2.4.1). When r=0, we have


CO 3(e6;2m+l) -|6| m! (2m-1)!2 (2m)!
S(2m)! (m-1)! (m)!

=0


= iel'- T T(8;2m-1)

which proves Property 4.
Property 5 may be proved by induction. From
(2.2.7) we have

(06;5) = e-l (1+ |e + 11e2)

W(6;3) = e-l (1+ le|)


4(O;1) = e-161









so that.

2
9(e;5) = Q(e;3) + Ti(y (e;1)

which shows that Property 5 holds for m=l. The
induction hypothesis is

2
(69;2m+l) = (69;2m-l) + (2m-3)2m-l) (6;2m-3)

(2.4.2)

Writing

(2m+l) 6 1) 2 -
(6;2m-) = --- (;2m-1) (;2m-1) ,

(2.4.2) becomes


0(O;.2m+l) 6 2 0(6;2m-3)
(2m+l) (2m-lT0-(;2m-1) + (2m-l)(2m+l) (2m-3)


(2m-2m+ 2m)
(2m-l)(2m+lj)(e;2m-l)


(2.4.3)


From Property 4,


0(e;2m+3) =" (6;2m+l) +


(2
(2m-l)(2m+l)(0;2m-i)


if and only if for all 6,


6e(6;2m+l) St(6;2m-l) + 3 (6;2m-3)
2m+l 2m-1 (2m-l)(2m+l) (2m-3)


26 -00 2 1
(2m-1)(2m+l)(@;2m-1)









which is equivalent to (2.4.3). Thus Property 5 is

established.

Property 4 is useful in demonstrating the infinite

divisibility of the distribution of X A function T

is the characteristic function of an infinitely

divisible distribution if and only if Y(6)=e-t(6),

where a has completely monotone derivative and w(0)=0

(see [6: p. 425]). Grosswald [8] has shown that

e6(e;2m-l)
-(6;2m+l) m=0,l,2,..

is a completely monotone function of 6. Using Property

4, we must have

3a (0;2m+l)
S(e;2m+) (-log (6 ;2m+l) ) is completely


monotone in e. That is, -log (6;2m+l) has completely

monotone derivative. Thus,


T(6) = elog (6;2m+l)= (86;2m+l)

is the characteristic function of an infinitely divisible

distribution. That is, the pre-t r.v. with odd degrees

of freedom has an infinitely divisible distribution.

















CHAPTER 3

THE DISTRIBUTION OF
A LINEAR COMBINATION OF t-VARIABLES



3.1 Simplification

Letting t denote a t r.v. on v d.f., the Cauchy

variable may be represented by tl and the standard

normal by tm. For an arbitrary set of constants,

c1,c2,...cn, the linear combination


n
T = citli
1 i=1l

has the Cauchy distribution, and


n
T2 c.t .
i=l1
has the normal distribution, where ti are lid as tv,

i=1,2,...,n. If 1 n
exists for the linear combination I c.t .
i=l
Here we consider the generalized linear combination


n
T' = i c.t
1=1 i

where the t's are independently, although not necessar-

ily identically distributed. The cumulative distribu-

tion function of T' is









n n
Pr{ i cit y'} = Pr{ c ViX V y '}
=1 i i=1 i

where X denote independent pre-t r.v.'s on vi d.f.

n c. ..
Letting = c.. a. =- and y =
i- 1 1 w a
n
(3.1.1) becomes Pr{ i a .X i=1 i

and vi = 2mi+l, the characteristic function of a.X

is, from (2.2.6),


(ai ;Vi ) = e q (jam( l) ,

so that aiX and [a.ilXv have identical distributions.
1 1

We therefore only need to consider the distribution of

n
T = a.X (3.1.2)
j=l J Vj
n
for the case aj 0, j=l,2,...,n, and l a. =1.
J=1


3.2 The Distribution of T Using the Q Matrix

The characteristic function of T is


YT(8) = EeiT = E exp{i Y a X }
j=l e j

n i~ajX,
= n Ee X J
j=1
n
II= (Oa ;v ) .
j=1

With V. =2m.+1 (mj =0,1,2,...), (2.2.6) gives us
J J 3









n -1ea.j -1 e n
T(6) =I e qj ( ea.) = e H q ( o6a ) .
j= J m
(3.2.1)

Denote by 4 and 0(x) the arrays

t(e;l) 1
S= (e;3) O(x) = jexJ
p(6;5) lex12
oex13



and 0 = 0(1). Let Q be a matrix whose element in row 1,

-joJ
column j is the coefficient of 'lele in (96;2i+l)

(or simply the coefficient of 6elJ in qi(|61) ), for

j=0,1,2,...; i=0,1,2,.... We may now rewrite (2.2.6) as

-lel
P = e 0Q (3.2.2)
Denoting the ith row of Q by (Q). we have

qr(16 ) = (Q) 0r and qm (Bea. ) = ( E) 6(a.),
j J
so that (3.2.1) is

-le1 n
j=1 j
oel
Clearly, e T(() is a polynomial in j9j, so we may
find a vector k such that

n
'0e = H (Q) 0(ae.) (3.2.4)
j=l j"
The matrix Q is lower triangular and is non-singular.

Now, (3.2.2) becomes -16
-1I = e 0









so that,

-lel
T(e) = e i' = P 'Q-l
or finally,

T(6) = n'i (3.2.5)
where r' = 'Q- Thus, we may write the characteristic

function of a linear combination of independent t's as

a linear combination of Student-t characteristic functions.

Since ()m has zeroes after the first m.+l elements
j J
(the first element, corresponding to column 0, is 1),

(Q)m 0(a ) is a polynomial of degree at most mj in 1e|;
n
H (Q) 0(a.), therefore, is a polynomial in e16 of
j=l j'-
n
degree at most s = m. This tells us that Z has at
j=l J
most s non-zero elements after the first element (which

is 1). Therefore, q is a vector with zeroes after its

first s+l elements.

The cdf of T may now be written as the linear

combination of pre-t cdf's:


H(y) = Pr{T< y}

= nOPr{X1
or simply, using the notation of section 2.3,



H(y) = n H2i+ (y) (3.2.6)
i=0

Setting 0 = 0 in (3.2.5) or y = 0 in (3.2.6), we









S
see that ni = 1. In addition, some preliminary
i=0
results and a computer investigation indicate that

ni is always non-negative, i=0,1,2,...,s.

The first eleven rows and columns of Q and

Q-I are given in Tables 3.1 and 3.2 for convenience

in calculating the ni's. An easy way to obtain the

elements of Q-1 is presented in the next section.


3.3 The Q-1 Matrix
-l
In order to obtain the elements of QQ one may

of course, proceed by solving QxQ-= I, the identity

matrix; Q being lower-triangular, this is easily

accomplished row by row in a sequential manner. A

more general method is given here.

For m and j in {0,1,2,... } let Q be a matrix

whose element in row m and column j is given by

(-l)m-J(2j)'(r+1)! j
U j S2m(m-j)!j!(2j-m+l)!
(Q m,j (3
0 otherwise
-i
It will be shown that Q = Q-

The (m-k)th element of Q is the coefficient of

IeIk in q (e ), for m,k in {0,1,2,... That is,


m!(2m-k)! 2k
m > k
(2m)!k!(m-k)!
(Q)m,k
0 otherwise
1 otherwise


y


3.1)


(3.3.2)






























































- Cr


- 00






- CO
C14





o o
a -
N
m


0O (N
-i N C
1- -





N (N C-
-7

(N


r
c- i
<0



i-i \ n n m in '-
^-i <* Cr' <
J > r- -





o^ m


- Lr) V) u')
.- ^ C -
a1


Lrf




CCO




,a 0 0





oI oo






c Ln
m c
In u cc
o co C




m cN




1 0 A0



-



N (NC'JN N
C- r coo







" CO







SCI cj CC 4 r-

o -

lo


Cm- (Ncr
- (NC'
* CC
cr -


Cr) CC4
CrcN NC

-- (C
-7 3-


Co C CO C C


0
cO w

I w


r"


H I
a)
ci




to








0

E-0
a a


o









"u o
ja




1 1










il
T-l


O C


























C, r-.



















NN (N OC io
L










I rf


,-i







0 0I

0







,- o o
IN













(1 0o v 0




m m1 0 0 0


- 0 0 0 0 0


lo o o r c

l r
I co m





(1 t0 O m lO
o 0 i Ur ir o
S0 N oo o

CN C CO
0




0 N1 N 0







e 0 0 0
0 0 C







-- 0 0 0








0 0 0 0


o .- CN M .I u \o0


0


i 0 Ca 0
,--1









The j trow of Q is given by


(Q) i!


(2J)!20 (2j-1) !2 (2j-2)!22
O!j! '"!(j-1)7 '2! j-2)! '...'


(3.3.3)
j!!
2!O 0, 0, ''"


Theth column of Q
The k column of Q is


* (2k)!
).k k!


0
0



0

(-l) (k+l)!
2kO!(k+l)!

(-1)1(k+2)!
2k+11!k!

(-1)2(k+3)!
2k+22!(k-1)!





(-) k+l(2k+2)!
22k+2(k+l)!O

0
0
0


k zeroes


I


(3.3.4)









Clearly,
( () if j < k
1 if j = k
o k(Q x(Q)k!, 1=j if j=k(3.3.5)

For j > k denote (Q) .x(*).k by ". If k j. .k.

(2k)!j !j-k i (2j-k-i)!(k+i+l)'
kT2j) I(-l) I!(j-k-i)!(k-i+l)'

and if j > 2k+l,


k+l
S (2k)"j (-1)i (2j-k-i) !(k++il)
k!(2j)! l i!(j-k-i)!(k-i+l)!
i=O

or more concisely,


(2k)!j! I )1) (2j-k-i)!(k+i+l)
Sk!(2j) i! (j-k-i) (k-i+) 3
i=O

j-k if k < j 2k+l
where 6 =
k+l if j > 2k+l

Q-1
Surely, Q = Q iff = 0.

The binomial and negative binomial expansions are,

respectively,


n
S(l i'(n-ii (-y)l and
i=O


il(n-1) i
(l-y)-n i (n= -l)!
1=0 i'n-l 7 y


Thus, for any integer 4,











(1-xz)@ = i (-l)i(xz) and
i=O


(1-x)-2o =


S(24+t-1)! t
t=0 t!(2 -l)!


After multiplying these quantities, we obtain


S(-1)i (2+t-l)!! zi t+i
i=0 t=0 i! (
(3. 3. 7)


Letting S(z) =


m+l __-xz)_
(1-x)2


and denoting by C ,"the
x


coefficient of xk in," we obtain


0(o-1)1


m+l
I (-1)i
i=0


i! (2q+m-i)! m+l+i
i!(4-i)!(m+1-i)!(2i-l)! z
(s m+l)

!(2(0+m-i)! m+l+i
i!((-i)!(m+1-1)!(2)-1)!
(4 m+l)

(3.3.8)


and,


S-m+l (z)l = ( (2+m-i)!(m+i+l)
-z x 2 i=O i!()-i)!(m+1-1)!


if < m+i
where 6 =j if sm+l
m+1 if m+1


(3.3.9)


(l-xz)
(1-x) 2


I


Cm+l z) =
x


But,










ky 3(Z) f Li m+1 (l-xz) }]
z=l (1-x)2 z=l

-(1+1)
= (1-x) {(m+l)-x(m+1+0)}


= {(m+l)-x(m+l+n)} [ i x
i=0

so that,



m+x 13 S(z) =1 (m+1)(O+m+1)! (m+l+4)(!+m)!
x z (m+1)!(!- m! !
z= (3.3.10)
= 0
Equating (3.3.9) and (3.3.10), we have

6 (20+m-i)!(m+i+l)
-1) i(-i)(m+l-i)! = 0 6 =min,m+
i=0 (3.3.11)

By letting m=k, =J-k in (3.3.11), we get & =0

in (3.3.6) demonstrating that Q given by (3.3.1) is
indeed the inverse of Q. The matrix Q-1 is now

completely specified; any element may be obtained

through a simple arithmetic calculation. The following

recursive method of generating the elements of Q-1

may also be useful:

( -l)m,k = (2m-l)(q-1 m-l,k- -m(m-2)(q-l)m-2,k-1

(3.3.12)
This is verified by obtaining from (3.3.1),









(- )m -(-1)m-k(2k-2)!m!
S, -i j-i r-i and
m- ,k-I 2m-(m-k)!(k-l)!(2k-m)!

(-1 2(-)m-k-(2k-2)!,k- (-l)
m-2,k- 2m-2(m-k-l)!(k-l)!(2k-m+l)!


Now, the right-hand side of (3.3.12) is

(-1) -k(2k-2)!(m-l)! m(2m-1) m(m-2)1
2m-2(m-k-i)!(k-1)!(2k-m)! 2 m-k) 2k-m+l

(-l)m-k(2k)!(m+l)! 4(m-k)k(2k-m+l) (m+l)(2k-1)
2m(m-k)!k!(2k-m+l)! 2k(2k-1)(m+l) 2(m-k)(2k-m+l)

(-l)m-k(2k)!(m+l)! -1
2m(m-k)!k!(2k-m+l)! m,k

again using (3.3.1). In addition, a recursive formula

for obtaining the elements of Q is


(2m-k)
m,k k(2m-) m-l,k-1 (3.3.13)


Using (3.3.2) we have

k-1
(2m-k) (2m-k)(m-1)!(2m-k-l)!2k
k(2m-l) (m-l,k-1 k(2m-1)(2m-2)!(k-1)!(m-k)!

m!(2m-k)! 2k 2m 1
(2m)!k!(m-k)! m 2

m! (2m-k)! 2k
(2m) k!(m-k)!


= (Q)m,k









verifying (3.3.13).

3.4 Examples
The following examples will illustrate the devel-
opments of the two preceding sections.
i. Let it be required to find the distribution of
1 1
T = X3 + X3

The characteristic function of T is

| | 2
T(e) = {e (1+ +1el)} (using (2.2.7)
-lel
=e (1 + 6l + e2)

i -el -lel1 2
= e (1 + 1e) + e (l +je +1|e2)

1 )
1 3 (e) + 4 ,5(e)

Thus, Pr{Tstl = -Pr{X t} + 3PrX5 st}
3 5-

and the density of T is given by
-2 -3
fT(t) = ( +t ) + 3( +t2)
T 4(1, ) 46(1, )


ii. It is desired to find the distribution of
S5 1 1
T = 1X + X + X5 .
12 1 3 3 IFX

The characteristic function of T is

-(( ) = 5(( 1e; )-.4 (-e;3). (( e;5)

= e ( i+ |9el)(l+|+ 1 | + le|2)
-16V












-l e
= e r'0


using (3.2.4).


Thus, V' = ( 1 0 0 0
12 y yITT 0 0 )

Using Table 3.2, we get

-1 60 42 27 15
S= Q = ( rTW i- T 4T

so that

Pr{T < t} = 0Pr{X t + Pr{X < t} + 27Pr{X5 t
ITTPrfX st} + T4 3 1 r4T4 5

+ 15Pr{X t} .


iii. Consider the case when T represents the average

of n lid pre-t r.v.'s, each with 2m+l d.f. It may be

verified that for n= 2, the (i +)st non-zero element

of ri is given by


m (2m+2i)!(2m-2i)!
i (m+i)!(m-i)!
qi+m = m
l (i, (2m+2j)!(2m-2j)!
j 0 i (m+j)!(m-j)!


, l=0,1,2,...,m

(3.4.1)


For m=0,1,2,...,7, Table 3.3 gives the values of n.


iv. It will be convenient for future reference to

calculate the n-vectors for some common cases. Letting

n=2 in (3.1.2), we have


= e (1+ 7 5 2 + 1 1e3)















Table 3.3

THE ELEMENTS OF n FOR THE AVERAGE OF TWO iid
PRE-t R.V.'s WITH 2m +1 D.F.


m 0


1 2 3 4 5 6 7


1 0

.2500

.7500


0

0

.0625

.2083

.7292


0

0

0

.0156

.0656

.1969

.7219


0

0

0

0

.0039

.0201

.0663

.1915

.7182


0

0

0

0

0

.0010

.0060

.0222

.0665

.1884

.7160


0

0

0

0

0

0

.0002

.0017

.0072

.0234

.0666

.1864

.7145


0

0

0

0

0

0

0

.0001

.0005

.0023

.0080

.0241

.0666

.1850

.7134









T = aX + (1-a)X 2 (3.4,1)


9 4 3 2 1 1 1 1 1
For oa = 5' 3' 2 y , and T, and for


v1 =1(2)9, v2 =1(2)9, p was computed and tabulated

in Table 3.4. The upper percentage-points of the

distribution of T were then calculated. For 0 =.10,

.05, .025 and .01, Table 3.5 gives values of 1 (a,vlv2)

such that

Pr{T> 0(a,V1 V2) =

Thus, for example,

2 1
Pr{X3 + X > 1.277} = .025


3.5 The Case of Even Degrees of Freedom

We have thus far restricted attention to t or

pre-t r.v.'s having odd d.f. This has been necessitated

by the form of Kn+ in (2.2.5), which gives no indica-

tion of the form of 0(e;v) when v is even. The character-

istic function of a t r.v. with even d.f. is unknown.

If T in (3.1.2) contains a X for which v. is

even, we may put bounds on its distribution using our

knowledge of the distribution of T when v. is odd for
J
all j. To do this we need some properties of stochastic

ordering.

A r.v. X is said to be stochastically larger than

the r.v. Y (X >Y) iff Pr{X > a} Pr{Y > a} for all a.

We may define an "absolute stochastic ordering" in a










Table 3.4

ELEMENTS OF n WHEN T = OX + (1-a)X2


(v1, 2) 9/10 ./5 3/4 2/3 1/2 1/3 1/4 1/s 1/10

(1,3) .9000 .8000 .7500 .6667 .5000 .3333 .2500 .2000 .1000
.1000 .2000 .2500 .3333 .5000 .6667 .7500 .8000 .9000

(1,5) .9000 .8000 .7500 .6667 .5000 .3333 .2500 .2000 .1000
.0900 .1600 .1875 .2222 .2500 .2222 .1875 .1600 .0900
.0100 .0400 .0625 .1111 .2500 .4444 .5625 .6400 .8100

(1,7) .9000 .8000 .7500 .6667 .5000 .3333 .2500 .2000 .1000
.0882 .1563 .1781 .2074 .2250 .1926 .1594 .1344 .0738
.0108 .0384 .0563 .0889 .1500 .1778 .1688 .1536 .0972
.0010 .0080 .0156 .0370 .1250 .2963 .4219 .5120 .7290

(1,9) .9000 .8000 .7500 .6667 .5000 .3333 .2500 .2000 .1000
.0874 .1509 .1741 .2010 .2143 .1799 .1473 .1234 .0669
.0112 .0384 .0552 .0847 .1339 .1481 .1356 .1207 .0729
.0013 .0091 .0167 .0353 .0893 .1411 .1507 .1463 .1039
.0001 .0016 .0039 .0123 .0625 .1975" .3164 .4096 .6563

(3,3) 0 0 0 0 0 0 0 0 0
.7300 .5200 .4375 ,3333 .2500 .3333 .4375 .5200 .7300
.2700 .4800 .5625 .6667 .7500 .6667 .5625 .4800 .2700

(3,5) 0 0 0 0 0 0 0 0 0
.7290 .3120 .4219 .2963 .1250 .0370 .0156 .0080 .0010
.2260 .3280 .3438 .3333 .2500 .2222 .2813 .3520 .5940
.0450 .1600 .2344 .3704 .6250 .7407 .7031 .6400 .4050

(3,7) 0 0 0 0 0 0 0 0 0
.7290 .5120 .4219 .2963 .1250 .0370 .0156 .0080 .0010
.2186 .3072 .3164 .2063 .1875 .0741 .0352 .0192 .0027
.0461 .1360 .1797 .2364 .2500 .1975 .2109 .2560 .4860
.0063 .0448 .0820 .1728 .4385 .6914 .7383 .7168 .5103

(3,9) 0 0 0 0 0 0 0 0 0
.7290 .5120 .4219 .2963 .1250 .0370 .0156 .0080 .0010
.2157 .2984 .3051 .2822 .1741 .0670 .0314 .0170 .0026
.0462 .1317 .1695 .2116 .2009 .1058 .0565 .0329 .0021
.0091 .0464 .0771 .1358 .2188 .1975 .1846 .2048 .4084
.0000 .0115 .0264 .0741 .2812 .5926 .7119 .7373 .5859

(5.5) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
.5905 .3280 .2383 .1358 .0625 .1358 .2383 .3280 .5905
.3150 .3733 .3516 .2881 .2083 .2881 .3516 .3733 .3150
.0945 .2987 .4102 .5761 .7202 .5761 .4102 .2987 .0945










Table 3.4 Continued



(v1, 2) 9/10 4/5 3/4 2/3 1/2 1/3 1/4 1/5 1/10

(5,7) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
.5905 .3277 .2373 .1317 .0313 .0041 .0010 .0003 .0000
.3012 .3408 .3115 .2346 .0938 .0741 .1377 .2112 .4784
.0913 .2240 .2665 .2881 .2187 .2305 .3076 .3584 .3685
.0170 .1075 .1846 .3457 .6562 .6913 .5537 .4301 .1531

(5,9) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
.5905 .3277 .2373 .1317 .0313 .0041 .0010 .0003 .0000
.2952 .3277 .2066 .2195 .0781 .0137 .0036 .0013 .0000
.0916 .2102 .2419 .2455 .1406 .0604 .0857 .1382 .3876
.0200 .1006 .1516 .2222 .2344 .1975 .2571 .3195 .3958
.0027 .0338 .0725 .1811 .5156 .7243 .6526 .5407 .2166

(7,7) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0. 0 0 0
0 0 0 0 0 0 0 0 0
.4783 .2097 .1335 .0590 .0156 .0590 .1335 .2097 .4783
.3482 .3172 .2574 .1575 .0656 .1575 .2574 .3172 .3482
.1398 .2839 .3045 .2765 .1969 .2765 .3045 .2839 .1309
.0337 .1892 .3045 .5070 .7219 .5070 .3045 .1892 .0337

(7,9) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
.4783 .2097 .1335 .0585 .0078 .0005 .0000 .0000 .0000
.3396 .3020 .2419 .1431 .0313 .0283 .0755 .1343 .3874
.1379 .2585 .2656 .2189 .0844 .1021 .1957 .2710 .3698
.0380 .1595 .2175 .2656 .2062 .2414 .3046 .3136 .1865
.0062 .0703 .1414 .3139 .6703 .6277 .4242 .2811 .0563

(9,9) 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
.3874 .1342 .0751 .0261 .0038 .0261 .0751 .1342 .3874
.3585 .2554 .1811 .0859 .0201 .0859 .1811 .2554 .3585
.1816 .2770 .2540 .1707 .0663 .1707 .2540 .2770 .1816
.0605 .2129 .2626 .2690 .1916 .2690 .2626 .2129 .0605
.0120 .1205 .2272 .4484 .7182 .4484 .2272 .1205 .0120










Table 3.5
VALUES OF A SUCH THAT PR{aX + (1-a)X > } =

((&i,v2) given in parentheses)

.10 .05 .025 .01 .10 .05 .025 .01
a (1,3) (3,3)
.900 2.774 5.684 11.437 28.640 0.861 1.234 1.664 2.369
.800 2.478 5.060 10.170 25.459 0.798 1.135 1.519 2.145
.750 2.335 4.751 9.538 22.869 0.775 1.095 1.460 2.050
.667 2.106 4.242 8.488 21.222 0.746 1.047 1.385 1.924
.500 1.695 3.267 6.416 15.936 0.723 1.009 1.324 1.821
.333 1.366 2.413 4.444 10.701 0.746 1.047 1.385 1.924
.250 1.234 2.062 3.561 8.154 0.775 1.095 1.460 2.050
.200 1.165 1.881 3.093 6.689 0.798 1.135 1.519 2.145
.100 1.044 1.583 2.342 4.157 0.861 1.234 1.664 2.369

(1,5) (3,5)
.900 2.771 5.683 11.436 28.640 0.855 1.227 1.657 2.362
.800 2.467 5.054 10.166 25.458 0.773 1.103 1.484 2.109
.750 2.317 4.740 9.532 23.867 0.736 1.046 1.401 1.979
.667 2.069 4.218 8.476 21.217 0.683 0.961 1.277 1.790
.500 1.593 3.185 6.367 15.916 0.606 0.835 1.081 1.463
.333 1.178 2.195 4.291 10.728 0.573 0.778 0.989 1.295
.250 1.006 1.743 3.251 7.980 0.573 0.777 0.983 1.278
.200 0.918 1.508 2.665 6.402 0.579 0.785 0.996 1.293
.100 0.772 1.142 1.689 3.339 0.607 0.827 1.052 1.373

(1,7) (3,7)
.900 2.771 5.683 11.436 28.640 0.853 1.225 1.'656 2.361
.800 2.465 5.053 10.166 25.458 0.766 1.097 1.478 2.104
.750 2.313 4.738 9.531 23.867 0.726 1.035 1.392 1.978
.667 2.062 4.215 8.474 21.217 0.663 0.939 1.254 1.771
.500 1.570 3.173 6.361 15.914 0.562 0.777 1.012 1.388
.333 1.114 2.149 4.256 10.615 0.501 0.676 0.853 1.111
.250 0.919 1.659 3.201 7.891 0.487 0.653 0.816 1.042
.200 0.819 1.394 2.597 6.385 0.485 0.648 0.808 1.024
.100 0.655 0.971 1.485 3.241 0.497 0.664 0.827 1.046

(1,9) (3,9)
.900 2.771 5.683 11.436 28.640 0.853 1.224 1.655 2.361
.800 2.464 5.052 10.166 25.458 0.764 1.094 1.476 2.102
.750 2.314 4.747 9.569 24.108 0.721 1.031 1.388 1.974
.667 2.059 4.213 8.473 21.217 0.654 0.929 1.245 1.764
.500 1.561 3.168 6.359 15.913 0.540 0.749 0.983 1.362
.333 1.088 2.139 4.267 10.719 0.461 0.622 0.788 1.035
.250 0.877 1.633 3.202 7.966 0.439 0.585 0.729 0.928
.200 0.766 1.347 2.579 6.379 0.431 0.572 0.708 0.891
.100 0.588 0.880 1.399 3.220 0.432 0.572 0.705 0.878












Table 3.5 Continued


a
.900
.800
.750
.667
.500
.333
.250
.200
.100


.900
.800
.750
.667
.500
.333
.250
.200
.100


.900
.800
.750
.667
.500
.333
.250
.200
.100


.10 .05 .025 .01

(7,7)
0.485 0.649 0.809 1.024
0.444 0.592 0.735 0.927
0.428 0.569 0.705 0.887
0.406 0.538 0.663 0.827
0.389 0.513 0.629 0.777
0.406 0.538 0.663 0.827
0.428 0.569 0.705 0.887
0.444 0.592 0.735 0.927
0.485 0.649 0.809 1.024

(7,9)
0.484 0.648 0.807 1.023
0.440 0.587 0.729 0.921
0.421 0.560 0.695 0.876
0.393 0.521 0.644 0.805
0.362 0.475 0.581 0.716
0.362 0.476 0.582 0.715
0.375 0.494 0.606 0.749
0.387 0.511 0.628 0.778
0.419 0.555 0.684 0.852


.10 .05 .025 .01

(5,5)
0.599 0.817 1.040 1.360
0.550 0.746 0.946 1.229
0.531 0.717 0.905 1.169
0.506 0.680 0.854 1.095
0.487 0.651 0.811 1.029
0.506 0.680 0.854 1.095
0.531 0.717 0.905 1.169
0.550 0.746 0.946 1.229
0.599 0.817 1.040 1.360

(5,7)
0.597 0.815 1.038 1.358
0.542 0.736 0.935 1.218
0.518 0.701 0.888 1.153
0.483 0.650 0.817 1.051
0.439 0.584 0.724 0.912
0.432 0.572 0.706 0.882
0.443 0.587 0.727 0.912
0.454 0.604 0.749 0.943
0.487 0.652 0.812 1.028

(5,9)
0.596 0.814 1.037 1.375
0.538 0.732 0.931 1.214
0.511 0.694 0.881 1.149
0.471 0.636 0.802 1.038
0.414 0.550 0.682 0.862
0.389 0.513 0.630 0.779
0.392 0.516 0.633 0.782
0.398 0.524 0.644 0.798
0.422 0.558 0.688 0.856








similar manner:

Definition: The r.v. X is said to be absolutely

stochastically larger than the r.v. Y (denoted XI Y)
iff Pr{IXI >a} > Pr{IYI > a) for all a (and strict
inequality for at least one a).
Notice that X1 Y iff (XIj IY
Lemma

Let X and Y be independent random variables with

XIY. If Z is independent of X and Y then X+Z Y +Z.
proof:
Xq Y ==> |X Y|
= Pr{ Xj >u) Pr{IY >u}, for all u
=t Pr{IXj
SPr{-u for all u.
Let Z have cdf F Then,


Pr{-w < X + Z < w) = Pr{-w- X sw-}dFz()


and,

Pr{-w
From (3.5.1) we have
Pr{-w-S < X < w-C} < Pr{-w-C < Y < w-C), for all w,C
=> Pr(-ws X +Z w} s Pr{-w < Y + Z w} for all w

IX+ZI >S Y+ZI --> X+ZY+Z E








THEOREM 3.1

Let Xi Yi, = 1,2, where Xl, X Y1 and Y2 are

independent r.v.'s. Then X1 + X2IYI +Y 2

proof:

X1 +x2 Y1 + 2 by Lemma

Y1 + X2 Y+ Y2 by Lemma

X +X IX2 Y +Y2 by transitivity U

From 2.1 we see that X X whenever v1
Therefore, from the theorem, the distribution of
n
T= aj.X wherein some of the v.' s are even, may be
j=1 j n
bounded by the distribution of T a.X and
n *=1J j
T2 = a.X IT where
j= 1 j
I
v. = v. = v. if v. is odd, and
J J J J
v. = v.-l, v. = v.+ if vj is even.

For example should one desire the distribution of

T = aX4 + (1 a)X

he may find the distributions of T1= aX3 + (1 a)X5

and T2 = aX5 + (1 -a)X5 and, noting that X3 1X4 lX5,

Theorem 3.1 implies that T IT1TYIT2, or alternatively,

Pr{ IT1 > x} Pr{ ICaX + (1 a)X5 > x) > Pr{IT2( > x},

for all x.

For one-sided approximations, we may remove the

absolute value signs since T., T2 and T are each

symmetrically distributed about 0.















CHAPTER 4

THE POWER OF THE T-STATISTIC



4.1 Introduction

Let u (nxl) be a vector of independently distri-

buted random variables with known distribution and

suppose x and u are related by


x =u +dl,

where 1 represents the (nxl) vector of unit elements.

Given x, suppose it is required to distinguish between

H0: d=0 and

HF: d>0

subject to Pr{incorrectly rejecting H0} =Q. We consider
n
tests of the type a'x= 1 a.x.. Given d, let Td(a)
i=1 1 d
denote a.r.v. with the same distribution as a'x. That

is,
n n n
Td(a) ~ aixi = d ai + J a.u.
i=l 1=1 1=1
Defining A (a) to be the critical value for rejection

in a size ( test, we have

Pr{To(a) > X (a)} = .

The power function of the test is

Pr{Td(a) > X (a)} (4.1.1)









The problem now confronting us is that of choosing

a to maximize (4.1.1). If the u.'s are symmetrically

distributed about 0, the distribution of T (a) is

invariant over sign changes of the elements of a. We

may assume therefore that a. 20, i=1,2,...,n. In

addition, for any positive constant K, we have


Pr(T0(Ka)> A(K)} = Pr{T (a)> >(a)} = (.


But

TO(Ka) = KT (a)

and
Td(Ka) = KTd(a)

so that

A (Ka) = KA (a)


and

Pr{T (Ka) > X (Ka)} = Pr{T (a) > 1 (Ka))
d d K

= Pr{Td(a) > A (a)}.

n n
We may therefore let ai =1, for if not, let K = / ai.
1=1 1
Our attention is restricted to the case

n
a. > 0, i=1,2,...,n and a a = 1.
a 1 .
i=l


If n=2 and u. =X, the statistic Td(a) becomes

the test criterion of 1.2. As a special case of the

above, therefore, is the choice of weights assigned to









the t-variable needed to maximize the power of the

test used to solve the Behrens-Fisher problem. This

chapter gives some general theorems on comparing the

power of two tests. Some examples are given, along

with some counterexamples to intuitive notions. It

is shown that the power of the pre-t r.v. increases

with d.f. and that there exists no distinct set

{a }, i=1,2,...,n, which yields maximum power of Td(a)

for all size tests when the d.f. differ.


4.2 General Results

The theorems which follow give a general method

of comparing the power functions of two tests about a

hypothesis of a location parameter and will be useful

in the ensuing investigation.

THEOREM 4.1

Let X and Y be absolutely continuous random variables

with distribution functions F and G respectively, and let

a and B satisfy F(a) =G((). If the inverse function of F,

say F exists, then the following conditions are

equivalent:

(i) F(a-d) > G(B-d) V d > 0

(ii) F G(x+d) >F G(x) +d V x, d>0
dF*G(x
(iii) d x) 1 V x where F G is differentiable
(iv)
(iv) 0









proof:

(i) iff (ii):


F( -d) > G( -d),

a-d > F G(B-d),

a = F G(B) > F G(B-d) +

F G(x +d) > F G(x) +d,


(ii) iff (iii):


F G(x +d) F G(x) > d,

F G(x +d) F*G(x) > ,


dF G(x)
dx -


Conversely, (iii)

Jx+d x+d
dF G(y) > dy
i x x

F G(x+d)-F G(x) > d


(iii) iff (iv):


Vd > 0
Vd>O

Vd>O

Vd, Vd

Vx d


Vx,d>0.


a = F G(.)


a dF G(B)
aB d6


( = x +d)


+









THEOREM 4.2

Let X and Y be absolutely continuous r.v.'s with

cdf's F and G respectively.. Suppose a and B satisfy

F(a) =G(B). Then, Y+d is a more powerful test of size

4 than X + d for testing

HO: d=0

vs. H : d> 0

iff 1. We shall indicate that this property holds

by writing F G (read "F is less in power than G").
(p)

proof:

1> =- F(a-d) > G(B-d), V d >

= -Pr{X a-d)} Pr{Y Y -d), V d> 0

=> PrX +d>a) > Pr{Y+d> 6, V d> 0

= Y + d is more powerful than X +d in a
test of H0 vs. H1.

If, in addition to the hypotheses of Theorem 4.1,

there exists a c such that F(c) =G(c), |c[
any of (i) thru (iv) in Theorem 4.1 holds, we have

a> (if >c). This is seen by writing

F G(x +d) >F G(x) +d => F G(c+d) >c+d, V d > 0

so that,

a = F G(5) >g

The next result gives a relationship between the

power of two tests and their moments, giving a necessary

condition for F G.
(p)








THEOREM 4.3

Let X and Y be absolutely continuous r.v.'s distrib-

uted symmetrically about 0, with cdf's F and G respec-

tively. Suppose that F is strictly increasing so the

inverse function, F exists and is uniquely defined.

If F 5 then E(Y2k) (p)
expectations exist.

proof:

F G = > 1 Theorem 4.2
(P)
*
SF G(x +d) >F G(x) +d, Vd > 0, x Theorem 4.1

x= 0 F G(d) F G(O)+d = d

or G(x)(x) F(x) Vx >0

= 1 G(x) < 1 -F(x) Vx > 0


= k fJx2k[l G(x)]dx< 1k fx2kl[l -F(x)]dx
0 0

=> 2 x2kdG< 2 x2kdF (integrate by parts)
0 0

2k iJ 2k
> 2kdG 5 x 2dF


or, E(Y2K) < E(X2K)
D

4.3 Applications: Some Examples and Counterexamples

Let W be a r.v. with cdf H(-) and Z a constant 3

0 1(p) 2
whenever Z < Z2. This is clear upon letting HZ (a) =
Z2(a), so that aZ= or Now invo1
Hz (B), so that aZS = 6Z2 or U ==-Z > 1. Now invoke








Theorem 4.2. This situation occurs whenever two

random variables have distribution differing only in

scale parameters. In particular, HZ < H.
(p)
Suppose now that Z is a r.v., 0< Z 1 with prob-

ability 1, and suppose it is still true that


H, z H. (4.3.1)
(P)

Of primary importance in the investigation into the

power of a linear combination of pre-t r.v.'s is whether

the power of a pre-t r.v. increases with d.f. Let


Y
2 2
(xl + x)

have the cdf H and


XI
Z X
(X2 + X22
2 2


2
where Y is a standard normal r.v. and Xi, =1,2 are

independent chi-square r.v.'s on v. d.f.

W then has the pre-t distribution on v1 +2 d.f.

and is independent of Z.
Y
W/Z = has the pre-t distribution on v1 d.f.
x 1
We therefore would reach the conclusion that the power

of the pre-t r.v. will increase with d.f. if (4.3.1)

held.









It is not true in general, however, that (4.3.1)

holds. The following is a counterexample.

Let

.4 ,0 lxl < wp
h(-x) = .1 1 < x 2 and Z =
0 elsewhere wp

Let a, be 9 H(a) =H(B) = .95

Then, a =2.0, = 1.5 since

HZ(a) = EZPr{W<2Z}

= H(1) +H(2)

= .95 = H(1.5) = H(B)

Taking d =.2 (say),

HZ(a -d) = H(l -) +H(2 d)

= LH(.9) +H(1.8) = .92

H( -d) = H(1.3) = .93

so that H(a -d)
H, N H.
(P)
It is of interest to investigate the conditions for

which it is true that H < H. Letting a, B be 3
(P)
HZ(a) =H(B), we have

EZH(aZ) = H(B)

EZZh(aZ)da = h(6)d8 or

a h(5)
Ez =E Zh(aZ)









By Theorem 4.2, H < H iff h(B) >EzZh(aZ). It should
(p)
be noted that H(B) = EZH(aZ) H(a) so that B a. The

following are examples for which HZ < H.
(p)
(i) Let Z Bin(l,p) and h(-) nonincreasing.


H1(a) = H(B) => a = h(a)
a h( ) h(6) h(B)
SEzZh(aZ) ph(a) h(a

(ii) Let h(S) = Xe-, S,A > 0.

H(B) = EZH(aZ)

I-e = 1 -E-AZ or, e- = EZe-


h(B) = AXe = Eze-A a E A EZe- A =EZZh(aZ)

aS h(B) 1
SB E Zh(aZ) -

(iii) Let h(-) be non-increasing and v(C) =Sh(C)

concave

i.e., Ve e (0,1) and any 1, 2

e v(51) + (1 ) v(52) v(851 + (1 e)2)

By Jensen's inequality, Ev(aZ)
= v(ap) where p =E(Z).


EzaZh(aZ) < aph(ap)

= EZZh(aZ) 5 ph(ap)









==> EzH(aZ) < H(ap).


But H() =E H(aZ) H(ap) = < ap

so that

h(B) > h(ap).

Now,

a h(B) h(B) h(B)
Ea EZZh(aZ) ph(ap) -h(ap 1.

An example of such a distribution is

h(S) 2(K- ) 0 < < K. The function h() is
K 2
K d 2h() -4
concave since d 2 2 < 0.
d K1

(iv) Let h() = p-1 ,0 < F 5 1. The cdf is



H(%) = p ,0 S 1
1 5 > 1


E Zh(aZ) = Zp(aZ)p-ldP
0
SI;min(l,l/a)
P (aZ)PdP
0

where P is the cdf of Z.

But

H(B) = EzH(aZ) =>

p min(1,1/a)
= (aZ)PdP +Pr{Z > l/a}









min(11/)(aZ)PdP
0

so,


E Zh(aZ) =


< p s P = pBp-1 = h(s)
a B


aa
->1
HR


(v) Let h() = p(l- ) 1 0 1 p 1.


0 ,
H( ) = (1 ) ,
1


The cdf is


5 0


H(6) = EzH(aZ)

min(1,1/a)
1 -1 )= [1 ( aZ) ]dP + Pr{Z > 1/a)}
0


= (1 aZ)PdP


Now,


E Zh(aZ) =
1


.min(1,1/a) Z( -Z)P-1dP

0


but,

min(l, 1/a)
I~ z


(1 aZ)P-ldP


rImin(l,i/a)
a (aZ) dP








1 p p-1
rmin(l,/a) p in(1,1/a) p-1 p-1 p
Z< P [(1 aZ) ] .dP



by H1lder's Inequality.

Thus,

EzZh(aZ)
1 p-l

min(1,1/ac) min(1,/a) p
1 j[(1-aZ)] dP



_p _0Z) d
Lc P-1
p ([ ,z)Pd h


P-1
= pI- B)- p = p(l- B)p-1 = h(B)



as -


4.4 The Power of the Pre-t R.V.

Let t have the Student-t distribution on v d.f.
v
and define X = t //-. For v =2j +1, j =0,1,2,...,

denote the cdf of X by H. ()=Pr{X2j+I 5 }), and the
d
pdf by h (S) =-H (j). Let .j be 3 H.(B ) = We wish

to demonstrate that H < H whenever k k(P) m
sufficient to show that

aBk
k 1 vo
m .


and then use Theorem 4.2.









Let k
m-1 h (S)
Hm()= Hk(2 + (4.4.1)
j=k


so that,


m-1
h() = hk(E) +
j=k


Sdhj (S)
2j + d *


h () = 2 1 (1+2(+I) +1) Property 1 of 2.4


dh. ( )
d ( =) -2 (j +1) h
do +ha 2
so that)

so that,


dh.(5)


252(j + 1)h.( )
= hj () 1 -2
(1+2)


(4.4.3)


Notice that


dSh. ()
dC


s 0 iff


(1 + E2) > 2 2(j + 1)


f 2 1<
1ff E (2j + 1)


Now,


Hm(m) = Hk(k)


38k_ h (Bm)
m8m hk( k


m-1
k m =k
hk jm k 2j 1


dhd .(E)
d $=- m ",(4.4.5)
k)


hk


With


(4.4 .2)


(4.4.4)


----~-L-









which results by substituting (4.4.2). It follows

from (2.1.2) that Sm< Bk, so that hk(m) 2hk(k).

From (4.4.5),


k
> 1


dSh. ()
h j > 0 .
d I S=s


j =k,k+l,...,m-l.


By (4.4.4) this will occur iff B < 1
m 2j+l
If .5 < s .75, then


H (B ) = a .75 = H0(1) = Pr{t1 s 1}


< Prt2+ < 1} =
2m+1


1 1
m /2- < -- j
m" 2m+ j +1


H 1
{-}


= k,k+l,.. .,m-l ,


and we have

3k
k- > 1.
m

This appears to be true also when a>> .75. For o=

.75(.05).95(.01).99 and k= 0,1,2,...,21 the values of

Sk have been computed and tabulated in Table 4.1.
Beneath each value of Sk in the table appears the value

of hk(Bk). It is seen that as k increases, hk( k)

increases also for each o. Thus,


Therefore,
















Table 4.1


VALUES OF Sk(C) FOLLOWED BY hk ( ())


k .75 .80 .85 .90 .95
k 75 .80 .85 .90 .95


0 1.000
0.159

1 0.441
0.446

2 0.325
0.628

3 0.269
0.770

4 0.234
0.891

5 0.210
0.998

6 0.193
1.093

7 0.179
1.182

8 0.167
1.264

9 0.158
1.343

10 0.150
1.415


1.376
0.110

0.565
0.366

0.411
0.531

0.339
0.660

0.294
0.768

0.264
0.864

0.241
0.950

0.224
1.029

0.210
1.101

0.197
1.171

0.187
1.236


1.962
0.066

0.722
0.275

0.517
0.417

0.423
0.527

0.367
0.620

0.328
0.700

0.300
0.773

0.277
0.841

0.259
0.902

0.244
0.961

0.232
1.015


2.800
0.036

0.946
0.177

0.660
0.287

0.535
0.373

0.461
0.444

0.411
0.507

0.374
0.563

0.346
0.614

0.323
0.663

0.305
0.706

0.289
0.749


2.800
0.036

1.359
0.079

0.901
0.143

0.716
0.195

0.611
0.238

0.542
0.276

0.491
0.311

0.453
0.342

0.422
0.371

0.397
0.397

0.376
0.423


.96 .97 .98 .99


2.800
0.036

1.504
0.060

0.980
0.113

0.773
0.156

0.657
0.193

0.581
0.225

0.527
0.254

0.485
0.281

0.452
0.305

0.424
0.328

0.402
0.349


2.800
0.036

1.704
0.042

1.083
0.083

0.847
0.117

0.717
0.146

0.632
0.172

0.571
0.195

0.525
0.217

0.489
0.236

0.459
0.254

0.434
0.271


2.800
0.036

2.010
0.025

1.233
0.053

0.951
0.077

0.799
0.098

0.702
0.117

0.633
0.134

0.581
0.149

0.540
0.163

0.506
0.176

0.478
0.188


2.800
0.036

2.621
0.010

0.505
0.024

1.133
0.037

0.940
0.049

0.819
0.059

0.735
0.069

0.672
0.077

0.622
0.085

0.583
0.092

0.549
0.099
















Table 4.1 Continued


k .75 .80 .85 .90 .95


11 0.143
1.486

12 0.137
1.552

13 0.132
1.615

14 0.127
1.678

15 0.123
1.735

16 0.119
1.794

17 0.115
1.847

18 0.112
1.901

19 0.109
1.955

20 0.106
2.006

21 0.104
2.055


0.179
1.296

0.171
1.356

0.164
1.414

0.158
1.469

0.153
1.520

0.148
1.570

0.144
1.620

0.140
1.667

0.136
1.713

0.133
1.761

0.130
1.804


0.221
1.068

0.212
1.118

0.203
1.164

0.196
1.209

0.189
1.253

0.183
1.297

0.178
1.339

0.173
1.378

0.168
1.418

0.164
1.453

0.160
1.492


0.275
0.789

0.263
0.826

0.253
0.863

0.244
0.897

0.235
0.931

0.228
0.965

0.221
0.995

0.214
1.027

0.209
1.056

0.203
1.084

0.199
1.110


0.357
0.447

0.342
0.470

0.328
0.493

0.316
0.512

0.305
0.532

0.295
0.552

0.285
0.572

0.277
0.589

0.270
0.608

0.263
0.623

0.257
0.640


.96 .97 .98 .99


0.382
0.369

0.365
0.388

0.350
0.407

0.337
0.425

0.325
0.442

0.314
0.458

0.305
0.474

0.296
0.490

0.288
0.504

0.280
0.519

0.273
0.532


0.413
0.287

0.394
0.303

0.378
0.317

0.363
0.331

0.351
0.345

0.339
0.358

0.328
0.371

0.319
0.383

0.310
0.395

0.302
0.406

0.295
0.416


0.454
0.200

0.433
0.211

0.415
0.221

0.399
0.232

0.385
0.241

0.372
0.250

0.361
0.259

0.350
0.269

0.340
0.277

0.331
0.285

0.323
0.293


0.521
0.106

0.497
0.112

0.476
0.118

0.457
0.124

0.441
0.129

0.426
0.134

0.412
0.139

0.400
0.144

0.388
0.149

0.378
0.154

0.368
0.158









a8k hh (f)
Bm hk (k 1
TV h-kk


for m >k.


(4 .4.6)


Since this holds for all a, we may write


H k Hm
(P)


for m > k.


(.4.7)


It is interesting to note that a stronger condition

is actually true, namely the function Hk H (x) is a

convex function of x. That this implies (4.4.6) follows

from Theorem 4.1 since, if Hk H is convex then
k m


(providing

Hk Hk+l(x)
The values


dHk H (x)
d n 1 Vx,
dx

dHk H (x)
dH- -- H x 1). For k=0,1,2,...,5,
has been computed for several values of x.
has been computed for several values of x.


Hk Hk+l(x +h) -Hk Hk+l(x)
h

are seen to increase with x. These values have been

tabulated in Table 4.2. In addition, for small values

of x the tables indicate Hk Hk+l(x) x or,


Hk H k+(h) -Hk H k+(0)
h

supporting the belief that

dHk Hk+l(x)
dx x=O










Table 4.2

VALUES OF Hk Hk+l(x)

k=0


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


Hk Hk+1(x)
0.0004
0.1001
0.2014
0.3046
0.4107
0.5210
0.6362
0.7578
0.8865
1.0235

1.1702
1.3272
1.4962
1.6780
1.8737
2.0845
2.3120
2.5566
2.8205
3.1039

3.4083
3.7349
4.0852
4.4595
4.8598
5.2869
5.7419
6.2263
6.7408
7.2865

7.8653
8.4773
9.1243
9.8072
10.5270
11.2856
12.0833
12.9214
13.8014
14.7235


FIRST DIFFERENCE

0.0004
0.0997
0.1013
0.1031
0.1061
0.1104
0.1152
0.1216
0.1287
0.1371

0.1466
0.1570
0.1690
0.1818
0.1957
0.2109
0.2274
0.2447
0.2639
0.2834

0.3045
0.3265
0.3503
0.3743
0.4004
0.4271
0.4551
0.4844
0.5145
0.5457

0.5788
0.6121
0.6470
0.6829
0.7198
0.7585
0.7977
0.8381
0.8800
0.9221










Table 4.2 Continued

k= 1


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


H H k+(x)
Hk Hk+l(X)
0.00024
0.0668
0.1337
0.2006
0.2688
0.3373
0.4067
0.4770
0.5487
0.6213

0.6962
0.7721
0.8497
0.9295
1.0107
1.0942
1.1797
1.2669
1.3565
1.4484

1.5425
1.6386
1.7374
1.8384
1.9414
2.0472
2.1549
2.2650
2.3775
2.4926

2.6099
2.7299
2.8520
2.9762
3.1031
3.2324
3.3635
3.4976
3.6336
3.7720


FIRST DIFFERENCE

0.0004
0.0664
0.0669
0.0669
0.0682
0.0685
0.0694
0.0704
0.0717
0.0728

0.0746
0.0760
0.0776
0.0797
0.0812
0.0835
0.0855
0.0872
0.0896
0.0919

0.0941
0.0960
0.0989
0.1010
0.1030
0.1058
0.1077
0.1101
0.1125
0.1152

0.1173
0.1200
0.1221
0.1242
0.1269
0.1293
0.1312
0.1341
0.1359
0.1384










Table 4.2 Continued

k = 2


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


Hk Hk+l(x)

0.0004
0. 0599
0.1204
0.1803
0.2411
0.3019
0.3632
0.4253
0.4877
0.5512

0.6151
0.6799
0.7453
0.8118
0.8790
0.9476
1.0171
1.0875
1.1586
1.2310

1.3043
1.3789
1.4540
1.5308
1.6084
1.6870
1.7667
1.8475
1.9294
2.0122

2.0959
2.1808
2.2666
2.3533
2.4414
2.5302
2.6198
2.7110
2.8025
2.8952


FIRST DIFFERENCE

0.0004
0.0595
0.0605
0.0600
0.0608
0.0608
0.0613
0.0621
0.0724
0.0634

0.0639
0.0648
0.0654
0.0666
0.0672
0.0685
0.0695
0.0704
0.0712
0.0723

0.0733
0.0746
0.0751
0.0768
0.0776
0.0786
0.0797
0.0807
0.0819
0.0829

0.0837
0.0848
0.0858
0.0867
0.0882
0.0888
0.0896
0.0911
0.0916
0.0926









Table 4.2 Continued

k= 3


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.4o
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


Hk Hk+l(x)

0.0004
0.0569
0.1143
0.1716
0.2291
0.2873
0.3451
0.4037
0.4624
0.5215

0.5815
0.6418
0.7026
0.7639
0.8260
0.8886
0.9520
1.0158
1.0806
1.1456

1.2117
1.2781
1.3453
1.4130
1.4815
1.5511
1.6210
1.6913
1.7625
1.8342

1.9064
1.9798
2.0531
2.1274
2.2026
2.2778
2.3538
2.4308
2.5080
2.5862


FIRST DIFFERENCE

0.0004
0.0565
0.0573
0.0573
0.0575
0.0582
0.0578
0.0587
0.0587
0.0592

0.0600
0.0603
0.0608
0.0613
0.0621
0.0626
0.0634
0.0638
0.0648
0.0651

0.0661
0.0664
0.0672
0.0677
0.0685
0.0695
0.0699
0.0704
0.0712
0.0717

0.0722
0.0733
0.0733
0.0743
0.0752
0.0752
0.0760
0.0770
0.0772
0.0783










Table 4.2 Continued

k =4


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


Hk Hk+l(x)
0.0004
0.0556
0.1113
0.1670
0.2225
0.2787
0.3352
0.3917
0.4482
0.5056

0.5629
0.6210
0.6794
0.7380
0.7972
0.8566
0.9169
0.9774
1.0387
1.0998

1.1619
1.2245
1.2875
1.3509
1.41L8
1.4791
1.5442
1.6097
1.6758
1.7415

1.8083
1.8755
1.9434
2.0107
2.0786
2.1491
2.2164
2.2867
2.3594
2.4273


FIRST DIFFERENCE

0.0004
0.0552
0.0557
0.0557
0.0555
0.0562
0.0565
0.0565
0.0565
0.0573

0.0573
0.0582
0.0583
0.0587
0.0592
0.0595
0.0603
0.0605
0.0613
0.0611

0.0621
0.0626
0.0629
0.0634
0.0689
0.0643
0.0651
0.0656
0.0661
0.0657

0.0668
0.0672
0.0679
0.0673
0.0680
0.0705
0.0673
0.0703
0.0727
0.0680









Table 4.2 Continued

k= 5


x

0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45

0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95

1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45

1.50
1.55
1.60
1.65
1.70
1.75
1.80
1.85
1.90
1.95


Hk Hk+l(x)
0.0004
0.0548
0.1092
0.1638
0.2187
0.2734
0.3286
0.3840
0.4395
0.4952

0.5512
0.6077
0.6647
0.7215
0.7791
0.8367
0.8950
0.9537
1.0123
1.0718

1.1315
1.1914
1.2517
1.3125
1.3738
1.4351
1.4972
1.5596
1.6220
1.6850

1.7486
1.8142
1.8761
1.9378
2.0012
2.0738
2.1336
2.2013
2.2692
2.2956


FIRST DIFFERENCE

0.0004
0.0544
0.0544
0.0547
0.0549
0.0547
0.0552
0.0554
0.0555
0.0557

0.0560
0.0565
0.0570
0.0568
0.0575
0.0576
0.0583
0.0587
0.0587
0.0595

0.0597
0.0600
0.0602
0.0608
0.0613
0.0613
0.0621
0.0624
0.0624
0.0630

0.0636
0.0656
0.0619
0.0617
0.0634
0.0727
0.0597
0.0677
0.0680
0.0264








X *
Since Hk Hk+l is convex, k= 0,1,2,... then Hk Hm is

convex providing m > k. This follows by writing Hk H

as

Hk Hm(x) = Hk Hk+l Hk+Hk+2 HmiH m(x)

and noting that the composition of a convex increasing

function with another convex function is itself convex.


4.5 General Results Based on Intuition

An intuitive generalization of (4.3.1') is the

following: Let {F Y be a sequence of cdf's and

suppose that F < F whenever yl y (p) Y2 1 2 1
Y2 be r.v.'s taking values in Q. Then,


E F E YF (4.5.1)
1 (p) 2 2


providing Y2 Y That (4.3.1) is a special case of

this is clear upon letting Y2 =1 w.p.l, 0 5 Y1 w.p.1

and F (x) =Fl(xy).

Letting X .i be iid pre-t r.v.'s on v d.f.,

i =1,2,...,n (4.5.1) points out how to choose a set

of weights {a.} i= which will maximize the power of

T= ] a.X Using the notations of 4.4, we obtain
i=1 V'i

n n s s
c H.< d Hj providing I c.> d (4.5.2)
j=l j (p)j=l j=1 J j=1ij

for s =1,2,...,n and {c.},{d.} chosen to satisfy c. 0,
n n a
d. 0 and c. =1= = d.. This is seen by letting
Sj=1 j=1









Y1 j w.p. c. and Y2 = j w.p. da, j =1,2,...,n, in
s s
(4.5.1). Clearly, Pr{Y < s= c. d =Pr{Y2 : s,
j=1 j=1 J
so that Y2 Y1. As seen in 54.4, Hi < Hk whenever
(p)
i k. (4.5.1) now becomes (4.5.2).

It is observed from Table 3.4 that if v1 =2 =v,
S
then for s =0,1,2,...,v the quantities n.l are
j=1
non-increasing as a approaches . This would indicate

that the maximum power is attained when equal weights

are assigned to each Xvi in T. That is, we would
1
choose a. =-, j = 1,2,...,n. A preliminary investiga-

tion of the power function of T corroborates this

finding. In a .05 size test the power of T for n=2

is computed for alternative values ranging from .00

to 2.00 and tabulated in Table 4.3. It is seen that

when the degrees of freedom are equal, maximum power

occurs when the weights of the pre-t variables are

equal (a =l). When d.f. differ, the power will be

maximized at some point for which the pre-t variable

with the larger d.f. has the larger weight. It will

be shown subsequently that that point depends not only

on the d.f. but also on the size of the test.

Notice that if i < k and

1 if j = i 1 if j = k
c: = d. =
0 if j =i 0 if j k

in (4.5.2), we obtain the result Hi < Hk for i (p)
which was shown in 4.4.






































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Unfortunately, (4.5.1) has already been shown not

to hold in general by the counterexample to (4.3.1);

It is of interest, however, to investigate what addi-

tional conditions need be invoked to ensure (4.5.1)

(and hence, (4.3.1)). This is done by example in 4.7.

n < TI
4.6 Properties of the Relation

Let F and G be cdf's such that F G, as defined
(p)
in Theorem 4.2. Thus, for any oc (0,1), there exists

constants a and 3 satisfying

F(N) = G(B) = o

and

> i1
5B

We look at some properties of the relation < ":
(p)
For cdf's F, G and H, let F < G and G < H.
(p) (P)
Then F < H transitivityy of < ). This is obvious
(p) (P)
by letting a, S and y be such that F(a) =G() =H(y),

so that


ay _ B -

Since

F(a) = G(B) = F(a-d) 2 G(6-d), Vd>0 (4.6.1)


is equivalent to


F < G ,
(p)


(4.6.2)









we may write for any positive constant c,


cF < cG


(4.6.3)


to mean

cF(a)= cG(B) = cF(a-d) cG( -d) Vd>0


which is equivalent to (4.6.1). Thus (4.6.2) and (4.6.3)

have identical meanings. Assuming also that we may add

the same quantity to both sides of < we have
(p)


F G == F+H G +H
(p) (P)


and the following properties hold:


(i) F. < G i = 1,2 F1 +F2 < G1 + G.
(p) (P)

proof:

F <0 =G F +F 2 G +F by (4
F1 (P)1 1 2 +F2 by (
(p) (p)

F 2 G2 = G0 +F2 < G +G2 by (4
(P) 2 1 2 (P) 1 2
SF1 +F2 (< G0 +GG2 by transitivit
(P)


(4.6.4)


1.6.4)


.6.4)


:y.


(ii) For 0 d < c< 1,


F 5 G
(p)

proof:


= cF + (1 c)G s dF+ (l -d)G
(p)


(c -d) > 0 =


(c -d)F s (c -d)G = (1 -d)G- (1 -c)G
(p)









:. cF+ (1 -c)G < dF + (1 -d)G
(p)


after 2 applications
of (4.6.4)


(iii) Let c i},{di i =1,2,...,n satisfy


n n
c >0, di 0, c = di
i=l i=1


1, and for some k,


for i = 1,2,...,k


for i =k+l,...,n.


If F. < F. for i j then
(p) J


n

i=1


n
< diF .
(p) i 1


proof:

For i =1;2,...,k, (c. -d.) 0 so that
I1


(. -d.)F. i (ci -d)Fk+ .
(P)


(4.6.5)


For i =k+2,...,n, (di ci) 0 so that


(di -c )Fk+l ( di )F.
(p)


(4.6.6)


Applying property (i) to (4.6.5) k-i times and

to (4.6.6) n-k-2 times, we obtain


k k
I (c -d )Fi 5 ( -di)Fk+l
i=1 (p) i=1


(4.6.7)


and

n n
S(di c)Fk+1 5 (d -c )Fi. (4.6.8)
i=k+2 (p) i=k+2


c >2 di


while


C. < d.
I I









Applying property (i) again to (4.6.7) and (4.6.8),

k n
S(c -d.)Fi + (di-c )Fk+l
i=1 i=k+2

k n
) il di )Fk+l + I (d Ci)Fi
(P) i=1 i-k+2

which, by (4.6.4), becomes

k n n
ScF + cF i k+
i=1 i=k+2 1=1
iik+l
(4.6.9)
k n n
S diF + I diFi d F
(p) i=l i=k+2 i=1 k+l"
ik+l


Noting that

n n
ck+ = 1 I, dk+ = 1- di
i=1 i=1
i/k+l ifk+l


(4.6.9) gives the result.


4.7 Some Counterintuitive Examples

From Table 3.4 it is seen that if v =v2 =v, the

weights {ni(a )} corresponding to the linear combina-

tion T= aX + (- a)X are such that, for some k,
vV2


ni(a) > ni() i = 0,1,.. ,k

and


n (at) (-) i = k+l,k+2,...,v .









Thus by the results of 4.4 and property (iii) of 4.6,

we would have


Sni(a)Hi < ni()Hi (4.7.1)
i=1 (p) i=1


which is to say maximum power occurs when a = An
n
extension to T= [ aiXV would suggest a choice of
1 i=1 1
ai = i = 1,2,...,n whenever vi = v, Vi.

These results are very concordant with our intui-

tion. It seems reasonable to weight equally those

variables carrying equal amounts of information and to

weight more heavily those which carry a greater store

of information. The result (4.7.1) rests on the assump-

tion (4.6.4), which itself appears very much in agree-

ment with intuition. If sampling from the distribution

G results in a more powerful test than does sampling

from the distribution F, the natural inclination would

be to sample from a mixture pG + (1 p)H given a choice

between it and the mixture pF + (1 -p)H. Unfortunately,

this is not always the best thing to do; i.e., (4.6.4)

is not true in general. Although a computer study

indicates (4.7.1) is still valid, the extra conditions

which guarantee the validity of (4.6.4) are elusive.

The remainder of this section is an exploration of

some counterexamples to (4.6.4).








Counterexamples to (4.6.4):

Let F and G be cdf's and F 5 G. The assumption
(p)
of (4.6.4) is for any cdf H, we may declare F+H
(p)
G+H. The examples which follow take H as either

F or G.

(i) Let G be the cdf of a r.v. whose density is

.4c1 Ix < c
g(x) = lc- c < x < 2c
0 otherwise

and let F(x)=G( ). Choose x and y such that

F(x) =G(y)= a. For d>0, F(x-d) =G( 2d)

G( -d) =G(y-d), so that F G. Let a, B be
(p)
such that

{F(a) +G(a)} = {G(B) +G()} = .95.

We have

G(B)= .95 = = c

{F(a) +0(a)}= .95 = a=2c.

But for d = .2, c > .2, we have


t{F(a d)-d) +(a-d)} = {(c .1) +G(2c .2)

= 03 .02
= .95 -- < .95 G=(1.5c- .2) =G( -d)
c c

so it is not true that


F < 0 ==> F + G < +G
(P) (P)






79

(ii) Let G be the cdf of the uniform r.v. with density

= -1 g(x) =
otherwise


and let 0 be the cdf of the standard normal r.v.

Choose x and y such that I(x) =G(y) = o. Letting
(x) (x)
dx

dx g(y) _rex >1
dy t(x) 2 2

so that


(p)

Let a and 5 be such that


({ (a) + (()} = {4 ( ) + (B)B)} = .95.

If Z represents a number such that t(Z )= 1 -p,

we have

a = Z.5 = 1.645

and,

S= Z = 1.281
.10

For several values of d, (a -d) and {4(8-d)

+ G(8 -d)} were calculated as follows:

d D(a -d) {j(- -d) +G( d)}
.05 .9446 .9454
.10 .9388 .9406
.15 .9326 .9352
.20 .9258 .9300
.281 .9137 .9206









It is seen that

D(a-d) < {'(B -d) +G( -d)}


for the cases considered, so it is not true that


P < G ==> ( + 5 G +
(P) (P)

Interestingly enough, however, it is true that


D G =-> D G+G5 G + G
(P) (P)

To see this, let a, P be such that


{+(a) +G(c)} = G(B)

Then,

ac 2g()
Sg P(a) +g(a)


(iii) Let G be any cdf satisfying:


G(-c) = G(c) = and F < G,
(P)

where F is the cdf of a normal r.v. with mean 0

and variance a2. Let a and S be such that


{F(a)+F(a)) = {F(B)+G(B)} = 1- ,


where p is chosen so that S > c.


F(a)= 1 a= oZ

{F() +G(B)}= 1 -- B =oZ2









Now, letting f(x) dF(x)
dx


a 2 f(B)
B 2f( ) 1


For several

as follows:


.20
.10 1
.05 1
.025 1
.01 2
.005 2


Thus for

not true


values c


Z

.842
.282
.6145
.960
.327
.575


the cases

that


iff f(B) >2f(a)

82 2
iff e > 2e


iff a2 62 22 In2

iff Z -Z2 l 2 In 2 = 1.3863.

2 2
,f p, Z Z2 was calculated
4> 2((


z
Z2
.253
.842
1.282
1.645
2.054
2.327


2 2
Z -Z24

.645
.935
1.063
1.136
1.196
1.216


indicated, <1, so it is


F < G =- F+F < F+G .
(P) (P)
(iv) Let P be the cdf of the standard normal r.v.

with density 4(x) =1 exp(-_x2) and G the cdf
2/ 2
/2-f 1 2
of a normal r.v. with density g(x) =- exp(-A),
/Fy 2Y
i.e.,

i(x) = G(yx).


For 0 (P) dx 1
,(x)=G(y), then y=yx or dX y1 >i.
dy y


--








Let a and B be such that


{Q(a) + (a)} = {(6) +0G(B)} = 1-p.


Then, a = Z and since n4(nx) 0 as n -4, y may
I B0
be chosen small enough so that 1 (_) = 0.00000
and G(= 1.00000, we have
and (-))=G(B) = 1.00000, we have
Y


S=2p


(to 5 decimal places).


-a (1B) +g(B)
a6 2 (a)


For several values

follows:

p $(Z )

.200 .2799
.100 .1754
.050 .1031
.025 .0584
.010 .0266
.005 .0145


+(B) 1 B
)-_ () (8)
2o ( a) 2- (t (a)


of p, was computed as
28pa


(Z2p)
.3864
.2799
.1754
.1031
.0484
.0266


For each of these cases we see < 1, so it is

not true that


D 5 G == D + + G
(P) (P)

Like example (ii), however, we note that it is

true that for appropriately chosen y,


< G =4> P +G G +G .
(P) (p)

To see this, let a and B be such that









{(a) +G(a)} = G(B) = 1 -p,

so that 6 = yZ and, if y is chosen so that

1 (a)= 0.0000 and P(") = 1.00000,


a= Zp '


In this case,

aB 2g(B) -
2a 4(a) +g(a)

(v) Let Hi be the cdf of

j < k, we have

H. < H
S(P) k

Let j =2, k= 48 and


{Hj(a) +H.(a)} =
JWe

We find


2 _(Zp)
->- 1, Vp.


X2i+l for i c j,k}. If



by 94.4 .


a and 8 such that


{Hj () +Hk(B)} = .90


a= .65994

B = .41142

and

a_ h2(B) + h4() 53092 + .00185 9
2.9284 <1
S--2h2(a) 2(.28694)


So it is not true in general that


H Hk =H.+H. ! H. +H
(P) k J ( ) J k









4.8 The Power of aX + (i -a)XV


Write

T = X + (1 e)X2 for 6 e (0,1)


Using 52.1, we get


E(Tn) = (4.8.1)








If V 1 v2 this quantity is minimized by a value of 9,

say en, which depends on n. For example,

v1 2
01 -I + v2 2 '


while 0 2,6 ... are more complicated functions of v1

and v2. Thus, there exists no unique value of 0 which

minimizes E(T2n) for all values of n (1,2,3,...). From

Theorem 4.3, then, it follows that there is no linear

combination of pre-t r.v.'s, TO, which has greater power

than all other linear combinations for all size tests.

If v1 =v2 this may no longer be the case. The

value of 0 which minimizes (4.8.1) is found by solving


DE(T2n)
=- 0.
20






85


Letting v be the common value of v l and v2, we have

n (2n f2i+1 v} -2l l 2n-2i+1 r}-2ni+212 21
1- 21 2 2 2 2 )1-) in)

= 0.


Calling the product of the first five factors in the

summand G. and letting Si= (i --), we notice that


GI = Gni ,
1 f-i
and

S. = -S
i n-i '

so that 0 = is a root. There can exist no other roots

by the symmetry of G.
2n
Thus for all n (=1,2,3,...), E(T0 ) is minimized

by 8 =. Thus, the test T, is admissible when v1 = v2

since, by Theorem 4.3, there can exist no test TO,

(8' $) more powerful for all sizes.















CHAPTER 5

COMPARISONS WITH OTHER METHODS


5.1 Ghosh's Results

Ghosh [71 expressed the cdf of the sum of two

lid t-variables, say t and t, as a finite sum of

hypergeometric functions which reduces to explicit

formulae for odd degrees of freedom. To do this he

writes the density of t' +t" as


f(t) = f (x)f (t-x)dx


where f (y) is the density of t After a formal

series expansion, a term by term integration, and

use of the duplication formula, this becomes

-2v /(2v+l)
v+1 2v+1 '-(2V+1
f(t) x
2 2r +t(5.1.1))
2 2

1 2v+1 v+2 t2
H( 2 2 2
S' v+t


r{c} r{a+i}r{b+i} i
where, H(a,b,c; x) = r{blijF iTf{c+i) x


is the hypergeometric function. Integrating (5.1.1)

the cdf is expressible as






87


v+l r2i+l r2v+21+l
Pr{t + t t} I + 2 2 2
Pr{V,' V- + --2 V+v2i+2
v v 2 2vr2{}i=o i!r{'+ 2
t2
4v (5.1.2)

0 (1+u) (2v+2i+1) du


for t > 0. If v is odd this reduces to an expression
1i1v3 1 1 1 t
involving H(13; x) or H(2-'-; x) where x
4v+t2
If v is even the hypergeometric functions reduce to

expressions involving the complete elliptic integrals
of the first and second kinds given by
7T/2
2 b
K(x) = (1- x sin ) dB and
0
fT/2 2
E(x) = (1- xsin B) dB


In particular,

Pr{t' + t t} = + tan-1T

1 1
Pr{t2 +2 2

Pr{t + t3 < t} + 2/3t(18+t2 2 tan-( t
3 3 2 T (1(12+t2 ) 2/3

Pr{t4 +t 64p3/2

-(p 4-16p3+6p2+4p+2)K(p)}
2 2
t t
for t> 0 and q=--2 =
8+t 16+t










Ghosh gave tables of the cdf of t + t" using
v v
these formulae. For v >4, he uses numerical integra-

tion to evaluate (5.1.2). His values agree to four

decimal places with those found using the method of

Chapter 3. To approximate the distribution of t' +t"
V1 v2
when v1
and integrates term by term. For the few cases he

tabulated there appears to be fairly good agreement

with the exact values.


5.2 The Normal Approximation

Because tV has normal limiting distribution, so

does the statistic T given by (3.1.2). Of interest is

how large the vj's must be to adequately approximate

the distribution of T using the normal distribution.

For n=2, (3.1.2) becomes

T = aX + (l-a)X (5.2.1)


and the quantity

2 2
v (1-a)
v + 22 (5.2.2)
V1-2 v2-2

represents the variance of T. Letting Y be a normal

r.v. with zero mean and variance v, the cdf's of T

and Y are compared for several values of a,v1 and v2.

It was found that the best approximation occurred when

Sv2 = v and a=2. In this case (5.2.2) becomes

(2v-4)- and the comparison is given in Table 5.1.






89


The maximum error was .066 for v=3, .017 for v=5

and .010 for v=7,9 or 11. When v=13 or 15, the

largest absolute deviation is seen to be .006. The

approximation becomes somewhat better in the tail areas

with an average error of less than .002 for v> 9. *The

normal approximation appears adequate to two decimal

places when v 15. This may not be the case, however,

if the weights or d.f. differ.









Table 5.1
COMPARISON OF THE CDF OF X' +, X WITH A NORMAL CDF

P1(x) = cdf of 2X, +1X
I v -1
P2(x) cdf of Y -N(0,{2v-4})

x v P1 2 P2

.1 3 .579 .556 11 .671 .664
.2 .654 .611 .810 .802
.3 .721 .664 .903 .898
.4 .778 .714 .956 .955
.5 .826 .760 .982 .983
.6 .864 .802 .993 .995
.7 .894 .839 .997 .999
.8 .917 .871 .999 1.000
.9 .935 .899 1.000


5 .606 .597
.704 .688
.786 .769
.850 .836
.898 .890
.931 .929
.954 .957
.969 .975
.979 .986
7 .629 .624
.743 .737
.833 .829
.897 .897
.939 .943
.965 .971
.980 .987
.988 .994
.993 .998

9 .654 .646
.783 .773
.877 .869
.936 .933
.969 .969
.985 .988
.993 .996
.997 .999
.999 1.000


13 .686 .681
.832 .826
.923 .920
.969 .970
.989 .991
.996 .998
.999 1.000
1.000


15 .700 .695
.851 .846
.938 .9'37
.978 .979
.993 .995
.998 .999
1.000 1.000















Chapter 6

APPLICATIONS


6.1 The Behrens-Fisher Problem

Using the sampling scheme for the Behrens-Fisher

problem described in section 1.2, the test statistic

for

HO: d=0

vs. H1: d> ,


where d is the difference in normal population means,

pl-P2, is

N1 N2
T = { a ai. X ij (6.1.1)
j= i xij j=


To find the distribution of T, we use the property

that if R~X2 and, conditional on R, U-N(0,R-1), then

f-vU -t This may be demonstrated as follows:


Pr{/U x} = Pr{U -x I R=rldPr{R
0
x 2 1
2 U 0u-r -i 2r
= re r e dudr
(2) rF{}2v
0 -









II -_(9+1) i
J 2 } r -1)e- r(1+u2) dr du




Jt- a 2_-l
0



= ( + du
P{I{P}r



which is the cdf of t //-. Thus / U t (or U -X ).

From 1.2, we have


SS.
i 2
Xn.- 6.1.2)
1i 1

independently for i= 1,2, and, conditional on SS.,


1 iX -i ~ N(o, ) (6.1.3)
N 2(6.1.3)


independently for i= 1,2, so that

N.
S a .X. i X (6.1.4)
Tai \j=j ii i

independently for i =1,2, where v = n. -1. Therefore,

we may think of T as having the representation


T = aX aX + (6.1.5)
lv1 2v2 a

where X 's have independent pre-t distributions on

v. d.f.
1









To test

H0: d=0

vs. H1: d>0


we use T as test criterion, noting that when H0 is

true, T has the same distribution as


aX +a2X ,
v1 2


so that the methods of Chapter 3 are applicable.

Notice that the power function of T is given by


Pr{T>A I = Pr{a X +a2X > d
1 2 V T

2
which is independent of the unknown variances, 01 and
2
02. The critical values, A, satisfying


PrIa 7 +a > A = ,
Pr{al v +aa2X 2



are given in Table 3.5 for = .10, .05, .025, and .01.

Yet to be discussed is the selection of initial

sample sizes n1 and n2, and the constants T and {a e},

j = 1,... N If

SS.
22 1
a.T
1


we take N. =n. +1 and if
1 1




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