ON THE DISTRIBUTION OF A LINEAR
COMBINATION OF tDISTRIBUTED 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 BehrensFisher
Problem . .. . . . 1
1.2 A Sequential Solution to the
BehrensPisher Problem . . . 4
1.3 Statement of the Problem . . 6
1.4 Notation . . . . . . 8
2 THE PREt 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 tVARIABLES . . . . . .. 22
3.1 Simplification . . . . .. 22
3.2 The Distribution of T Using the
Q Matrix . . . . . ... .23
3.3 The Q1 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 TSTATISTIC . . .
4.1 Introduction . . .
4.2 General Results . . . . .
4.3 Applications: Some Examples and
Counterexamples . . . . .
4.4 The Power of the Pret 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 + (la)X .
5 COMPARISONS WITH OTHER METHODS . .
5.1 Ghosh's Results . . . .
5.2 The Normal Approximation . .
6 APPLICATIONS . . . . . . .
6.1 The BehrensFisher 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 LOWERTRIANGULAR ELEMENTS OF THE FIRST
11 ROWS AND COLUMNS OF . . . . ... 27
3.2 THE LOWERTRIANGULAR ELEMENTS OF THE FIRST
11 ROWS AND COLUMNS OF Q1 .. . . .. 28
3.3 THE ELEMENTS OF n FOR THE AVERAGE OF TWO
iid Pret R.V.'s WITH 2m+l D.F. . . . 36
3.4 ELEMENTS OF n WHEN T = aX + (la)X . 38
3.5 VALUES OF A SUCH THAT PR{aX + (la)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 + (la)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 tDISTRIBUTED 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 Studentt random variable is expressible
in closed form. The.characteristic function of an
arbitrary linear combination of independent tvariables
is then derived and the distribution function is ob
tained, itself expressible as a weighted sum of
Studentt 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 trandom variable.
One application is in a twostage sampling procedure
to solve the BehrensFisher problem. The test statistic
has the distribution of a weighted sum of trandom
variables. It is shown how to choose the weights for
maximum power.
CHAPTER I
INTRODUCTION
1.1 Motivation: The BehrensFisher 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 = x1x2)
1 1
n1 n2
2 (nll)s + 8(n2l)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 Studentt
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 BehrensFisher problem.
Under BehrensFisher conditions no criterion with
distribution as simple as the t above is available.
Fisher proposed a fiducial test based on the socalled
BehrensFisher 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], WelchAspin [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 Studentt
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
chisquare 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 tvariables,
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 nontabulated 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 twostage sampling
scheme was devised by Stein [15], the power of the
resulting ttest being independent of the nuisance
parameters. This approach may be modified to give a
solution to the BehrensFisher problem (see Sections 1.2
and 6.1), and in so doing, the test criterion has the
distribution of a weighted difference of Studentt 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
tvariables with different weights has not been investi
gated in the literature. Furthermore, if the degrees of
freedom associated with the two tvariables 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
tvariables with arbitrary weights and degrees of
freedom were known.
1.2 A Sequential Solution to the BehrensFisher 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 : lP2=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 twostage 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= 1l2 it is shown in section 6.1 that T
may be represented as
a+ a1 a2
T 1 + 2
1 Irv 2
where vi=nil 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 tvariables. Letting X (al,a2,nl,n2)
be the 100(1) percentage point of the distribution of
T, we reject H0 in favor of the onesided 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 Studentt 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 Studentt 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
BehrensFisher 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{XY}
*
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
Studentt r.v. on v d.f.
pret r.v. (tv/i)
chisquare r.v. on v d.f.
standard normal r.v.
density of X2j+1
cdf of'X2j+
CHAPTER 2
THE PREt DISTRIBUTION
2.1 Density and Moments
We denote by t a random variable (r.v.) having
the Studentt 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 "pret 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 v2k
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 chisquare r.v. with v d.f., we may
Z 2 2
represent XV by . If v2 > v and X and X
denote chisquare 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,
sinexh(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) = ez (n+,k)(2z)k
k=0
(n+k)
where (n+,k) = k!(nk)! k=0,1,2,...,n. Substituting,
k (n+k)! ()k
S Kn+(z) ezk nk
2z 2z k=O
or,
K n(z) z k(n k) (2z)k n=0,1,2,...
n + 22 kCkTnkY)
(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
1s(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(26 )m lel (m+k)! (2 )k
(2m)! e. k!(mk)! 1
Slel m! m (m+k)! mk
= (2mTk kTnmk!(21 or finally,
(e6;2m+l) = e11 m! (2mk)!(2l)k (2.2.5)
(2m)!ko k!rnk)!
mi (2mk) i k
Denoting qm(z) = m!T !(2mk)!(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) = ele
(e;3) = e1 e ( + l el )
0(8;5) = e 1( + 1 + 1e 2) (2.2.7)
(e;7) = el l (1 + le 8 + ]2 15 + el3)
W(e69) = e (1 + le + 310 2 + l 3+l 1]4)
2.3 The Cumulative Distribution Function
After an integration by parts,
cos2m dO = cos2mlsinO + 2
I 2m si 2m
and,
o2m2 d
cos 0 de
2 1
cos 6 dO = 2(sin6cos6 + e).
Therefore,
2 1 2m1 2m1 2m3.
cos2me de = 1 cos2m sine + 2m cos 2 sin6
2m 2m(2m2)
+ (2m1)(2m3) 2m5
2m(2m2)(2m4) c os OsinO +
(2m1)(2m3). (5)(3) 2
2m(2m2)(2m4). (6)(4) cose d
sin os2m1 + 2m1 2m3 (2m1)(2m3) C2m50
2m + m2 cos (2m2)(2m4)
(2m1)(2m3) .. (5)(3)
+ (2m2)(2m4). ..( )(4)(2)cos
+ (2m1)(2m3)...(5)(3)
2m(2m2)... (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 rI2r{} 2 r m+}
(2.3.2)
2m {(m1)!2m1 2
wT (2m1)!
2m (2m2)(2m4)(6)(4)(2)
T (2m1)(2m3) 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 = tanlt,
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(2m2)(2mn4)) i(6)(4)(2)]
2 IT (2m1)(2m3) (5)(3)(1)
where I represents the righthand 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)***(2m4)(2m2) 2m1
(3)(5)... (2m3)(2m1)cos } ,
where, 0 = tan1t, 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.
m1 h.(t)
H(t)Hk(t) = t Z j
j=k 2j+l
, m >.k.
30(8;2m+l) _ 9(e;2ml)
96 2m1 m=1,2,3,...
62 2(e;21l)
4(6;2m+3) = (6e;2m+l) + 62(2 ;2m1l)
(2m1)(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,
m1
Hm(t)Hk(t)= H m+k.(t)Hk.j(t)}
j 
j=k
m1
m i h (t)}
j=k 2m+2k2jl m+kj1
m1
= 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 (2mk2)! (1 k
e a e(0;2ml) = 2 2 k k! (mk1) (21 1) ,
k=O
a polynomial of degree m1 in ]e. Also,
(;2m+) = e(2mk)! )k
()'1 km(Yk) (2161)
so that
el6 8 0(;2m+l) m! m (2mk)! 2(k elkl k
S36(2m)! k 0 k!(mk)! 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) = e1 m! f (2mrl)! 2r+l(r+l)
6 O9 T6 (2mF! (r+l)!(mrl)!
(2mr)!2r
r!(mr)!
e 2 m! (2mr l)!2r {2(r+1)(mr)(2mr)(r+l)}
(2m)l (r+P!mr)l
= I m (2mr1)!2r
(2m)!(r1)!(mr)! r=1,2,3,...,m. (2.4.1)
Also,
r 0 n 1 Ie 2( m
i{ t 4(e;2ml)} = (2m1) (2m2)!
x (2mrl) 2ri
(rl) (mr)!
= _e1I m!r(2mr1)! 2r
(2m) (r1) !(mr)! r= m
which is identical to (2.4.1). When r=0, we have
CO 3(e6;2m+l) 6 m! (2m1)!2 (2m)!
S(2m)! (m1)! (m)!
=0
= iel' T T(8;2m1)
which proves Property 4.
Property 5 may be proved by induction. From
(2.2.7) we have
(06;5) = el (1+ e + 11e2)
W(6;3) = el (1+ le)
4(O;1) = e161
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;2ml) + (2m3)2ml) (6;2m3)
(2.4.2)
Writing
(2m+l) 6 1) 2 
(6;2m) =  (;2m1) (;2m1) ,
(2.4.2) becomes
0(O;.2m+l) 6 2 0(6;2m3)
(2m+l) (2mlT0(;2m1) + (2ml)(2m+l) (2m3)
(2m2m+ 2m)
(2ml)(2m+lj)(e;2ml)
(2.4.3)
From Property 4,
0(e;2m+3) =" (6;2m+l) +
(2
(2ml)(2m+l)(0;2mi)
if and only if for all 6,
6e(6;2m+l) St(6;2ml) + 3 (6;2m3)
2m+l 2m1 (2ml)(2m+l) (2m3)
26 00 2 1
(2m1)(2m+l)(@;2m1)
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)=et(6),
where a has completely monotone derivative and w(0)=0
(see [6: p. 425]). Grosswald [8] has shown that
e6(e;2ml)
(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 pret r.v. with odd degrees
of freedom has an infinitely divisible distribution.
CHAPTER 3
THE DISTRIBUTION OF
A LINEAR COMBINATION OF tVARIABLES
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 pret 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 nonsingular.
Now, (3.2.2) becomes 16
1I = e 0
so that,
lel
T(e) = e i' = P 'Ql
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 Studentt 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 nonzero 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 pret 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 nonnegative, i=0,1,2,...,s.
The first eleven rows and columns of Q and
QI are given in Tables 3.1 and 3.2 for convenience
in calculating the ni's. An easy way to obtain the
elements of Q1 is presented in the next section.
3.3 The Q1 Matrix
l
In order to obtain the elements of QQ one may
of course, proceed by solving QxQ= I, the identity
matrix; Q being lowertriangular, 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)mJ(2j)'(r+1)! j
U j
S2m(mj)!j!(2jm+l)!
(Q m,j (3
0 otherwise
i
It will be shown that Q = Q
The (mk)th element of Q is the coefficient of
IeIk in q (e ), for m,k in {0,1,2,... That is,
m!(2mk)! 2k
m > k
(2m)!k!(mk)!
(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
ii \ 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
E0
a a
o
"u o
ja
1 1
il
Tl
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 (2j1) !2 (2j2)!22
O!j! '"!(j1)7 '2! j2)! '...'
(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!(k1)!
() 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 !jk i (2jki)!(k+i+l)'
kT2j) I(l) I!(jki)!(ki+l)'
and if j > 2k+l,
k+l
S (2k)"j (1)i (2jki) !(k++il)
k!(2j)! l i!(jki)!(ki+l)!
i=O
or more concisely,
(2k)!j! I )1) (2jki)!(k+i+l)
Sk!(2j) i! (jki) (ki+) 3
i=O
jk if k < j 2k+l
where 6 =
k+l if j > 2k+l
Q1
Surely, Q = Q iff = 0.
The binomial and negative binomial expansions are,
respectively,
n
S(l i'(nii (y)l and
i=O
il(n1) i
(ly)n i (n= l)!
1=0 i'nl 7 y
Thus, for any integer 4,
(1xz)@ = i (l)i(xz) and
i=O
(1x)2o =
S(24+t1)! t
t=0 t!(2 l)!
After multiplying these quantities, we obtain
S(1)i (2+tl)!! zi t+i
i=0 t=0 i! (
(3. 3. 7)
Letting S(z) =
m+l __xz)_
(1x)2
and denoting by C ,"the
x
coefficient of xk in," we obtain
0(o1)1
m+l
I (1)i
i=0
i! (2q+mi)! m+l+i
i!(4i)!(m+1i)!(2il)! z
(s m+l)
!(2(0+mi)! m+l+i
i!((i)!(m+11)!(2)1)!
(4 m+l)
(3.3.8)
and,
Sm+l (z)l = ( (2+mi)!(m+i+l)
z x 2 i=O i!()i)!(m+11)!
if < m+i
where 6 =j if sm+l
m+1 if m+1
(3.3.9)
(lxz)
(1x) 2
I
Cm+l z) =
x
But,
ky 3(Z) f Li m+1 (lxz) }]
z=l (1x)2 z=l
(1+1)
= (1x) {(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+mi)!(m+i+l)
1) i(i)(m+li)! = 0 6 =min,m+
i=0 (3.3.11)
By letting m=k, =Jk 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 Q1 is now
completely specified; any element may be obtained
through a simple arithmetic calculation. The following
recursive method of generating the elements of Q1
may also be useful:
( l)m,k = (2ml)(q1 ml,k m(m2)(ql)m2,k1
(3.3.12)
This is verified by obtaining from (3.3.1),
( )m (1)mk(2k2)!m!
S, i ji ri and
m ,kI 2m(mk)!(kl)!(2km)!
(1 2()mk(2k2)!,k (l)
m2,k 2m2(mkl)!(kl)!(2km+l)!
Now, the righthand side of (3.3.12) is
(1) k(2k2)!(ml)! m(2m1) m(m2)1
2m2(mki)!(k1)!(2km)! 2 mk) 2km+l
(l)mk(2k)!(m+l)! 4(mk)k(2km+l) (m+l)(2k1)
2m(mk)!k!(2km+l)! 2k(2k1)(m+l) 2(mk)(2km+l)
(l)mk(2k)!(m+l)! 1
2m(mk)!k!(2km+l)! m,k
again using (3.3.1). In addition, a recursive formula
for obtaining the elements of Q is
(2mk)
m,k k(2m) ml,k1 (3.3.13)
Using (3.3.2) we have
k1
(2mk) (2mk)(m1)!(2mkl)!2k
k(2ml) (ml,k1 k(2m1)(2m2)!(k1)!(mk)!
m!(2mk)! 2k 2m 1
(2m)!k!(mk)! m 2
m! (2mk)! 2k
(2m) k!(mk)!
= (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 +1e2)
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  + le2)
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 pret r.v.'s, each with 2m+l d.f. It may be
verified that for n= 2, the (i +)st nonzero element
of ri is given by
m (2m+2i)!(2m2i)!
i (m+i)!(mi)!
qi+m = m
l (i, (2m+2j)!(2m2j)!
j 0 i (m+j)!(mj)!
, 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 nvectors 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
PREt 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 + (1a)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 percentagepoints 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
pret 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 + (1a)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 + (1a)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{wS < X < wC} < Pr{wC < Y < wC), 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 onesided 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 TSTATISTIC
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 tvariable needed to maximize the power of the
test used to solve the BehrensFisher 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 pret 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(ad) > G(Bd) 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),
ad > F G(Bd),
a = F G(B) > F G(Bd) +
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(ad) > G(Bd), V d >
= Pr{X ad)} 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 pret r.v.'s is whether
the power of a pret 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 chisquare r.v.'s on v. d.f.
W then has the pret distribution on v1 +2 d.f.
and is independent of Z.
Y
W/Z = has the pret distribution on v1 d.f.
x 1
We therefore would reach the conclusion that the power
of the pret 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)
Ie = 1 EAZ or, e = EZe
h(B) = AXe = EzeA a E A EZe A =EZZh(aZ)
aS h(B) 1
SB E Zh(aZ) 
(iii) Let h() be nonincreasing 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() = p1 ,0 < F 5 1. The cdf is
H(%) = p ,0 S 1
1 5 > 1
E Zh(aZ) = Zp(aZ)pldP
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 = pBp1 = 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)P1dP
0
but,
min(l, 1/a)
I~ z
(1 aZ)PldP
rImin(l,i/a)
a (aZ) dP
1 p p1
rmin(l,/a) p in(1,1/a) p1 p1 p
Z< P [(1 aZ) ] .dP
by H1lder's Inequality.
Thus,
EzZh(aZ)
1 pl
min(1,1/ac) min(1,/a) p
1 j[(1aZ)] dP
_p _0Z) d
Lc P1
p ([ ,z)Pd h
P1
= pI B) p = p(l B)p1 = h(B)
as 
4.4 The Power of the Pret R.V.
Let t have the Studentt 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
m1 h (S)
Hm()= Hk(2 + (4.4.1)
j=k
so that,
m1
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
m1
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,...,ml.
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,.. .,ml ,
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 hkk
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 pret 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
nonincreasing 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 pret variables are
equal (a =l). When d.f. differ, the power will be
maximized at some point for which the pret 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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a o c o ce o' o 
to oacs m
t)
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o n c; r c;C;cmmc;o
c C; C; 3 1; 3 o
oocoooooo
C3000000c0
mm~~~m~mm
oruc~me\o~o~n
ct oo\~~o~rr.o~oi
~ch~r~~c~no~or
Nm3U1IDh~r(Ol
InlCCo~OlO\olOl~
C~OIDC~O\~OIO~
0\010\01010~C10~01
WITigm~D10~0~~o
r~7mClhC\DJO\
mmojl3.OlOICnCIm
OIOI~OI~CIPlOl(n
NOIh~NOLnm
OC=r?33Lnm~
= N = 9 m m c\D ~
c~o~no~oia~ma~m
u~\~o~mr~ooo~
o.~v~ru\omu\ul~
crrmmmmmr
~~ln~mmm~mm
m~~~i~~~~
m ilroor~o\o~\o
~L~r~mmm~cin
ocoocooo
i1000000000
0001OC~000
0 C'Ln 07~1Ln0 i~
0aa0C0000
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(ad) 2 G(6d), 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(ad) 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) ki times and
to (4.6.6) nk2 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 + (dic )Fk+l
i=1 i=k+2
k n
) il di )Fk+l + I (d Ci)Fi
(P) i=1 ik+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(xd) =G( 2d)
G( d) =G(yd), 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) +(ad)} = {(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(8d)
+ 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(ad) < {'(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
/2f 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)} = 1p.
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
S2h2(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 pret 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 2n2i+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 fi
and
S. = S
i ni '
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 tvariables, 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 (tx)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} = + tan1T
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 416p3+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 + (la)X (5.2.1)
and the quantity
2 2
v (1a)
v + 22 (5.2.2)
V12 v22
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
(2v4) 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,{2v4})
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 BehrensFisher Problem
Using the sampling scheme for the BehrensFisher
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,
plP2, 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, UN(0,R1), then
fvU t This may be demonstrated as follows:
Pr{/U x} = Pr{U x I R=rldPr{R
0
x 2 1
2 U 0ur 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 pret 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
