SOME INCOMPLETE AND BOUNDEDLY COMPLETE FAMILIES OF DISTRIBUTIONS*
by
Wassily Hoeffding
Department of Statistias
Univepsity of Nopth Carolina at Chapel, Hill,
Institute of Statistics Mimeo Series No. 984
March~
1975
* Research supported in part by the Mathematics Division of the Air Force
Office of Scientific Research under Grant No. AFOSR-72-2386.
AMS 1970 Subject Classifications:, Primary 62G05. 62G10; Secondary 62G30 •
Key Words and Phrases: Complete (incomplete) families of distributions.
Boundedly complete families of distributions. Completeness relative to
the permutation group. Invariance under permutations. Symmetric
unbiased estimator. Similar tests.
..
SUMMARY
SOME INCOMPLETE AND BOUNDEDLY COMPLETE FAMLIES OF DISTRIBUTIONS
Let
Jui
dP
P be a family of distributions on a measurable space such that
= ci
'
= 1,
i
••• ,k , for all
P
€
(t)
P , and which is sufficiently
P consists of all distributions dominated by a a-finite
rich; for example,
measure and satisfying
(t).
It is known that when conditions
(t)
are not
present, no nontrivial symmetric unbiased estimator of zero (s.u.e.z.) based
on a random sample of any size n
g(x l , •.• ,xn )
Here it is shown that
(I)
if
is a s.u.e.z. then there exist symmetric functions
h.1 (Xl' ••• ,xn- 1)
= 1,
i
k
g(xl' ... ,xn )
exists.
.•• ,k , such that
n
= L L
i=l j=l
{u. (x.) - c.} h. (xl' ... ,x; ... !, x;+l' ..• ,xn ) ;
1
J
1
1
J
J
and (II) if every nontrivial linear combination of u I ' ••• ,~ is unbounded
then no bounded nontrivial s.u.e.z. exists. Applications to
unbiased estimation and similar tests are discussed.
SOME INCOMPLETE AND BOUNDEDLY
COMPLETE FAMILIES OF DISTRIBUTIONS l
by Wassily Hoeffding
..
Univexosity of Noxoth Car'oZina at ChepeZ HiU
[Abbreviated title: Incomplete and boundedly complete families.]
1.
Introduction and Statement of Results.
a set X, and let
on
(X,A)
(1.1)
Let A be a a-field of subsets of
P be a family of distributions (probability measures)
P
which satisfy the conditions
I u.
1
dP
= c.1
where k is a positive integer,
i
= 1,
•.. ,k
u l ' ••• ,uk are given A-measurable
functions~
and c l '
,c k are given real numbers. Let A(n) be the a-field of subsets
of Xn generated by the (cylinder) sets in An, and let pen) = {pn: P € P}
denote the family of the n-fold product measures pn on (Xn , A(n)) •
A family Q of distributions on
(Xn , A(n))
will be said to be complete
relative to the permutation group if the condition that the A(n)-measurable
sYmmetric real-valued function
g satisfies
f
g dQ
=0
for all
Q€ Q
IResearch supported in part:by the Mathematics Division of the Air Force Office
of Scientific Research, Grant No. AFOSR-72-2386.
AMS 1970 Subject Classifications: Primary 62GOS, 62GIO; Secondary 62G30 •
Key Words and Phrases: Complete (incomplete) families of distributions.
Boundedly complete families of distributions. Completeness relative to
the permutation group. Invariance under permutations. Symmetric
unbiased estimator. Similar tests.
-2-
implies
g(X l , ••• ,Xn ) = 0 a.e. (Q)
Here g is called symmetric if it
is invariant under all permutations of its arguments. The family Q will be
said to be boundedly complete relative to the permutation group if the same
conclusion holds under the additional condition that
Informally,
g is bounded.
Q is [boundedly] complete relative to the permutation group if
there is no nontrivial [bounded] symmetric unbiased estimator of zero.
(This
definition relates to the well-known notion of a [boundedly] complete family
[5] as follows.
and let
Let T be a maximal invariant under the permutation group
QT = {QT: Q £ Q}
be the family of distributions of T induced by
the distributions in Q.
Then Q is [boundedly] complete relative to the
permutation group iff the family QT is [bounded1y] complete.)
It is well known (Halmos 1946, Fraser 1953) that if the conditions (1.1)
are absent and
P is sufficiently rich then
the permutation group.
pen)
is complete relative to
This is not true in the presence of conditions (1.1)
(unless the ui and c i are such that the conditions impose no restriction).
Indeed, if hI' ••. ,h are any A(n-l)-measurable symmetric functions such
k
that
Ihildpn-l < ~, i = 1, ••• ,k , for all P £ P , then the function g
J
defined by
k
(1.2)
n
g(x l , •.• ,xn) = 2 2 {u.1 (x.)
- c.}
h.1 (Xl' .•• ,x.J- l' x.J+ l' ••• ,Xn )
J
1
i=l j=l
is a symmetric unbiased estimator of zero.
In this paper two theorems (each in two versions) are proved.
The first
theorem shows that if P is sufficiently rich then a symmetric unbiased
estimator of zero is necessarily of the form (1.2).
The second theorem shows
-3-
that although
pen)
is not complete relative to the permutation group it is
boundedly complete if all nontrivial linear combinations of uI ' ••. ,Uk
unbounded.
To state the theorems, we introduce the following notation.
tains the one-point sets, let
a a-finite measure on
If A con-
Po be the family of all distributions
centrated on finite subsets of X which satisfy conditions (1.1).
(X,A) , let Po(p)
absolutely continuous with respect to
are
P con-
If p is
be the family of all distributions
p whose densities
dP/dp
are simple
functions (finite linear combinations of indicator functions of sets in A)
and which satisfy conditions (1.1).
Theorem lA: Let A contain the one-point sets and let P be a convex family
of distributions on
(X,A)
which satisfy conditions (1.1), such that Po c P •
If g is a symmetric A(n)-measurable function such that
g dpn = 0 for all
P
€
f
P then there exist k symmetric A(n-l)-measurable functions
which are pn-I-integrable for all
(xl' ... ,xn )
€
P
€
hI'
,hk
P , such that (1.2) is satisfied for all
Xn .
Theorem 2A. If the conditions of Theorem lA are satisfied and if g is
bounded while every nontrivial linear combination of u l ' .•• ,Uk
then g(x l , ••• ,xn ) = 0 for all (Xl' ••• ,xn ) € Xn •
is unbounded
The following analogs of the two theorems hold for dominated families of
distributions.
-4-
We shall say that a A-measurable function u is P-unbounded if for every
real number c there is a P in
p(lu(x)1 > c) ~ 0 •
P such that
Theorem lB. Let P be a convex family of distributions absolutely continuous
~
with respect to a a-finite measure
(1.1), such that
such that
f
PO(~)
c
P.
g dpn = 0 for all
on
(X,A)
which satisfy conditions
If g is a symmetric A(n)-measurable function
P
€
P then there exist
hI' ••• ,hk which are
such that (l. 2) holds a. e~ (p(n)) .
measurable functions
Theorem 2B. If the conditions of Theorem
k symmetric A(n-l)_
pn-I-integrable for all
P
18 are satisfied and if
g is
,~
is P-
bounded while every nontrivial linear combination of u l ' .••
unbounded then g(x , ••. ,x ) = 0 a.e. (p(n)).
n
l
€
P ,
Remark 1. The assumption that the family P is convex is used only to prove
that there are versions of the functions
the families
PO'
PO(~)
tinuous with respect to
Remark 2. Theorems
hi
that are integrable.
, and the family of all
Note that
P which are absolutely con-
p and satisfy conditions (1.1), are convex.
18 and 2B remain true if
Po(~)
is defined as the family
of all distributions absolutely continuous with respect to
~
whose densities
are finite linear combinations of indicator functions of sets in a ring which
generates the a-field A; compare Fraser (1953).
Remark 3. If the assumptions of Theorems IA or
(1.1) are absent then the family pen)
e-
18 are satisfied but conditions
is complete relative to the permutation
-5-
group.
Here the assumption that
P is convex is not needed.
This is essen-
tially known (as noted above) and is easily seen from the proofs.
The theorems are proved in Sections 3 - 6.
Section 2 contains lemmas
that are used in the proofs.
This section is concluded with three examples of applications of Theorems
land 2.
Example 1. Let Xl' ••• ,Xn be independent real-valued random variables with
common probability density p
= ci
fXip(X)dX
wCP)
'
i
= 1,
= dP/dx
and suppose that the first
••• ,k , are known.
k moments,
Nothing else is assumed.
Let
= pm{(X l ,
••• ,Xm) € A} be the probability of some event involving
Xl' ••• ,Xm ' where m S n. It is reasonable to require that an estimator W
of w(P) satisfy 0 S ~ S 1. It is easy to see that a symmetric unbiased
A
estimator,
unique.
wO' with this property exists.
Theorem 2B implies that it is
(Note that the moment conditions may restrict the range of w(P)
proper subset S of
values of
[0,1] , via Chebyshev-type inequalities.
$0 are not confined to S.
to a
Typically, the
In such a case the use of an unbiased
estimator can not be recommended.)
Example 2. Let Xl' ... ,Xn be independent real-valued
common distribution
sis
f x dP.O
P whose variance is known.
against the alternatives
every a
€
(0,1) , and every
size a
against the alternatives
E
> 0
I
r~dom
variables with
Consider testing the hypothe-
x dP > O.
For every n
~
1 ,
there exists a strictly unbiased test of
f x dP ~
E.
(In Hoeffding (1956), p. 112,
-6-
a test is exhibited which (after a suitable change in notation) is strictly
unbiased against
against
f
f
x dP
~
I
x dP =
€.)
This test can be shown to be strictly unbiased
€.
Theorem 2 implies that against the alternatives
x dP > 0 no nontrivial unbiased test exists.
unbiased test is similar; see [5].
(One first shows that every
We may assume that the test is symmetric.
By Theorem 2 the only symmetric similar test of size a
is trivial.)
Example 3. Let the assumptions of Theorem 1 (A or B) be satisfied. If W(P)
admits an unbiased estimator, then the difference of any two symmetric unbiased
estimators is given by (1.2).
Let
wCP) =
I
w dP.
We discuss only the simplest case,
Then any unbiased estimator t(x)
n =1 •
is given by
k
t(x) = w(x) +
l
h.{u.(x) - c 1,}
k=l
1
where hI' ••• ,hk are arbitrary constants.
have finite second moments. Then
k
varp(t) = varp(w) + 2
l
i=l
Suppose that w, u l ' .•• ,uk
k
h.C.(P) +
1
,
1
1
k
l l
i=l j=l
h.h.O .. (P)
1
J
1)
where Ci(P) = covp(w,ui ) and 0ij(P) = covp(ui,u j )
It is straightforward
to minimize varp(t) with respect to hI' .•. ,hk • Let Q be a distribution
in
P such that the matrix
its inverse.
(Dij(Q))
is nonsingular, and let
be
Then the unbiased estimator which has minimum variance when the
distribution is Q is t(x)
with hi
=tj
P = Q is varQ(w) - EEOij(Q)Ci(Q)Cj(Q) •
e-
(oij(Q))
Oij(Q)Cj(Q) , and its variance at
-7-
2.
lemmas. The following lemmas will be used in the proofs of the theorems.
A homogeneous polynomial on Rn which is zero on a nondegenerate n-dimensional interval, is identically zero on Rn •
~emma~:
(Halmos).
The proof is by induction on n, just as Halmos' proof in [2] for the
case where the n-dimensional interval is the positive orthant.
lemma 2:
If a 1.)
.. (i
= 1,
an index h (I s h s m)
n
l
(2.1)
j=l
(~.
)
= 1,
.•. ,m; j
... ,n)
are real numbers, there exists
such that the inequalities
- a1...)p.
))
;;?;
i
0,
= 1,
... ,m ,
are satisfied in a nondegenerate n-dimensional sub-interval of
Proof: The inequality (2.l) with i
PI' ••• 'Pn'
=h
If two rows in the matrix
is trivially true for all
A = (a ij ) are equal, the correspon-
ding two inequalities in (2.1) are equivalent.
We may therefore assume that
A has no equal rows. Under this assumption it will be shown that there exist
an index h and strictly positive numbers PI' .•• 'Pn
inequalities (2.1) with
n
(2.2)
r
j=l
(~.
)
- a ..
i
~
h are strict:
)p. > 0 ,
1.))
such that the
i
= 1,
... ,h - 1, h + 1, ... ,m •
This implies, by continuity, that the same is true in a neighborhood of the
point
(PI' ••• ,Pn ) , and the lemma will be proved.
-8-
Let
A
have no equal rows.
integers m(l), •.• ,mer)
m = m(O)
~
Then there exist an integer r ,
1 S r S m
such that
m(l) > m(2) > ••• > mer) = 1
and columns C1 , ••• ,Cr of the matrix A such that column C1 has exactly
mel) maximal elements (the specified elements in column C1 ) and, for
g = 1, ..• ,r - 1 , the meg)
elements in column Cg+ 1 which are in the same
specified elements in column Cg contain exactly meg + 1)
rows as the meg)
maximal elements (the specified elements in column Cg+ 1 ).
We
may assume that Cg is the g-th column of the matrix A and that the
.
elements in column Cg are the first
column, for g = 1, ••• ,r. Then
specified meg)
a 1 ,g
for
g
=
= 1,
= am(g),g > a.1,g
..• ,r.
Hence if i
i = meg)
= meg)
+
meg)
elements of that
1, ... ,meg - 1) ,
+ 1, ...
,meg - 1)
,r ,
g = 1, ...
then
n
(2.3)
I
n
j=l
(a 1 · - a . . )p. = I (a l · - a 1• J·)pJ· ,
,J
1,J J
j=g
J
where a 1 - a. > 0 • Thus if Pr + 1 , ••. 'Pn are any fixed positive numbers
g
19
and, having defined Pj for j ~ g + 1 , we take Pg > 0 so large that the
right side of (2.3) is strictly positive, with g
the inequalities
(2~2)
= r,
r - 1, ••. ,1 , then
are satisfied with h = 1 .
We denote by u the column vector with k components u1 ' ••.
e-
,~
.
-9-
Lemma 3A: If, for (xl'
,X ) E
n
k
(2.4)
Xn ,
n
r
g(x l , ••• ,xn) = I
u. (x.)h. (xl' .••
i=l j"l 1 J 1
where each h.1
is symmetric in its n - 1
k points in X such that the k
x
k
,X.
J-
l' x. l' .•. ,x ) ,
n
J+
and if
arguments~
zl' ..• ,zk are
matrix
(2.5)
is nonsingular, then, for
n
(xl' ••• ,xn) EX,
n-l
(_l)n-m-l T
(xl' . .. ,x ) ,
n
m=O
n,m
r
(2.6)
where
k
r
(2.7)
..... it
·n-m=1
••• I
(2.8)
and
vex)
rm,n-m
integers
= U-1
Z.
g(xj , . .. , x. , z.1 , ••.
Jm
1
1
)v. (X.
1 n _m
11
Jm+l
)
u(x),
•
, J. n
denotes summation over those permutations iI'
1, ••• ,n for which jl < ••• < jm and
jm+l < •••
Note that representation (2.6) of g(x l , ••• ,xn )
,hk which appear in (2.4).
the functions hI'
Remark:
<
of the
in •
does not involve
Proof: From (2.4) and (2.8) we have
k
(2.9)
=
n
r r v. (x.)f.
(xl'
i=l j=l
J
1
1
••• ,x '- l ' x '+ 1 ' ••• ,xn ) ,
J
J
-10-
where
(2.10)
(The superscript T stands for transpose.)
(2.11)
Hence, for
(2.12)
=0
v. (z.)
1
J
1 S ir S k
By (2.8),
i
,
;t
j
r = 1, ... ,n - m ; n - m = 1, 2, •.. ,n ,
, xm' z.1 ,
g(xl'
1
k
m
I I
= i=l
j=l
, z.
v.(x.)f.(x1 , •••
1
J
1
1
, zi
+ f.1 (xl' ••• , xm' z.1 ,
2
1
)
n-m
) + •••
n-m
, z.1
• .. ,xm'
) .
n-m-l
In particular, with m = n - 1 ,
f . (x 1"
1
1
•• , X
n-l) = g(x l , •••.. xn _1 ' Z1'l)
n-l
r
I v.(x,)f,(xl'
i=1 j=l
J
k
1
Inserted in (2.9) this gives
ewhere
1
.. , ,x '- l ' x '+l' ... ,xn _1 ' Zil) •
J
J
-11-
,x.
, z. )} .
I n -2
11
In general, by induction on m,
(_l)m-l T
+ (_l)m R
n,n-m
m
(2.13)
m = 0, 1, ... ,n - 1 ,
...
,x ) , and R differs from T
n
m
n,n-m-l only in that
) +
, z.1 ,
,z.
,z.1 ) is replaced by f. (
1
1
+
11
m+1
2
m1
1
+ f.
(
, z. ,
,z.1 )
In particular, by (2.12) with m = 0 , we
1
1 m+1
m
1
follows from (2.13) •
have R
and
(2.6)
n,
n-1 = TO'
where Tn,r
g( .•• ,z. ,
= Tn,r(Xl ,
...
...
...
...
...
...
.
Lemma 38: Let v be a finite measure on the measurable space (X,A), let g
be a An-measurable function such that
A-measurable functions such that
J
Igldv
J luildv
<
n
<
~
~
, and let
i
= 1,
k x k
(2.14)
matrix
Uv
=
(fB
U dv, ••.
l
is nonsingular, then, for all
n-l
(2.15)
~
L.
m=O
e-
..• ,k.
If there
hI' ... ,hk such that
can be represented in the form (2.4) for all
Xn , and if Bl , .•• ,Bk are k sets in A such that the
exist symmetric An-I-measurable functions
g(x , ••• ,xn )
l
(xl' .•• ,xn ) €
u1 ' ... ,uk be
,J
Bt
(xl' .•• ,xn )
U dV)
€
n
X ,
(_l)n-m-l T(v) (x
n,m l'
-12where T(v) (x
•.. ,x )
n,m l'
n
with g(x. ,
Jl
(2.16)
fB1.
is defined like Tn,m (xl' .•. ,xn)
• •• ,Z.
)
replaced by
1
n-m
1
1
,X. , Y1' ••• ,y
dv (Yn-m) g(x • ,
fB.
in (2.7) but
n-m)
Jm
Ji
n-m
and vex) = (VI (x),
v(V)(X) =
(2.17)
u-v l
u(x) •
(xl' .•. ,Xn)
in the two places where it occurs.
The same is true with the phrase "for all
by
Ii
a.e. v(n)
II
Xn " replaced
€
The proof of Lemma 3B closely parallels that of Lemma 3A.
ference is that any substitution, in a function
f( ••• ,xi' •.• ) , of Zj
x.
in the proof of Lemma 3A is replaced by integration over
to
dv(x.)
•
1
1
The only dif-
B.
J
for
with respect
Incidentally, Lemma 3B.contains Lemma 3A.
The functions
hI' ••. ,hk in representation (2.4) of g are not, in
general, uniquely determined by the functions g, u l ' .•• ,uk' For example,
if hl(x,y) ,
h 2 (x,y)
HI(X,y)
HZ (x,y)
where w(x)
satisfy (2.4) with
= hI (x,y)
= hZ(x,y)
is arbitrary.
+
k
=2
,
n
=3
, so do
w(y)u 2 (x) + w(x)u 2 (y)
- w(y)ul(X) - w(x)ul(y)
The following lemma records, for future reference,
a certain version of the functions
hI' ••• ,hk •
-13-
lemma 4: Suppose there exist symmetric functions hI' .•• ,hk such that g
has the representation (2.4).
Under the conditions of Lemma 3A,
can be so chosen that each h.(x
1
l , .•• ,xn- 1)
of terms of the form
(2.18) g(x. , ... ,x. , z. ,
JI
Jm 1 1
where the subscripts
,z.
1
jl'
, J•n-l
of Lemma 38, each hi(x l ,
nation of terms of the form
fB•
(2.19)
1
.xn _1 )
)u
n-m
r
is a finite linear combination
(x.
1
J m+l
)
u
r
are all different.
n-m-l
Proof:
jl'
(x.),
I n- l
Under the conditions
can be chosen as a finite linear combi-
d" (~ m) g (X ' , ... ,x. , Yl" ..•
J1
Jm
-
n-m
u
where
hI' ••. ,hk
r
n-m-l
(x.).
I n- l
are all different.
, J. n-l
Under the conditions of Lemma 3A,
g(x l , ... ,xn)
has the representa-
tion (2.6), where Tn,m(X l , •.. ,xn ) is defined in (2.7) and each Vi (x) is
a linear combination of ul(X) , .•. ,uk(x) • Hence g(x l , ••• ,Xn) can be
written as a linear combination of terms, each of which, for some i and some
j , is of the form u.(x.)
1
J
times a product of the form (2.18) which does not
involve xj • This fact and the symmetry of g(x l , •.• ,xn) imply the assertion of the lemma. Under the conditions of Lemma 3B the proof is analogous.
-14-
3. Proof of Theorem lA. We may and shall assume that conditions (1.1) are
satisfied with
tions
c l = •.. = ck =
Thus
Po
is the family of all distribu-
P concentrated on finite subsets of X which satisfy the
f
(3.1)
and
o.
u
i dP
= 0,
i
= 1,
conditions~
•.. ,k ,
P is a convex family of distributions P on
(X,A)
which satisfy (3.1),
such that P :::> Po • Let g be a symmetric An-measurable function such that
f g dpn = 0 for all P E: P • We must show that there exist symmetric An - l _
measurable functions
<
(3.2)
and, for all
(3.3)
,hk
hI'
(xl' ••• ,xn )
i = 1, ... ,k ,
00
,
E:
n
X ,
if P
E:
P
k
n
,X ) = i:
i: u. (x.)h. (xl' •.. ,X.J- l' x'+
' .•• ,X ) •
i=l j=l 1 J 1
J l
n
n
g(x ,
l
Since
such that
Po
are points in
c
P , we have that if N is a positive integer, Xl' ••• ,xN
X, and PI' .•• 'PN are nonnegative numbers such that'
(3.4)
PI + ••. + PN = 1 ,
then
N
(3.6)
r
i n =1
,X. )p.
1
1
n
1
=0
•
-15-
It will be convenient to identify points in Rk with the corresponding
column vectors,
and 0 for
(0,
U
= (u l ,
,0)
T
,uk)
T
• We write
u(x)
for
(u l (x), ... , uk (x)
r
.
Let
u = {u(x):
x
€
Xl .
Conditions (3.4) and (3.5) show that the origin 0 is in the convex hull of
u.
First assume that 0 is in the interior of the convex hull of U.
there exist k + 1 points Yl' ••. 'Yk+l
in X such that 0 is in the
interior of the polytope whose vertices are
are strictly positive numbers
Then
U(Yl)' ..• ,U(Yk+l)'
AI' ••• ,A k+l
Thus there
such that
(3.7)
The solution of this non-singular system of k + 1 equations for
AI' ••• ,Ak+1
is
(3.8)
Ah
= ~/(dl
+ ••• + dk+l ) ,
h
= 1,
•.• ,k + I ,
where dh is the determinant
(3.9)
(The unspecified elements of the determinant do not affect its value.)
Ah > 0 for all
h, the dh are all of the same sign.
Since
We may and shall
-16-
assume that
h = 1, .•. ,k + 1 •
dh > 0,
(3.10)
For
Ish
~
i s k + I
dh in (3.9) when
U(Yi)
define D.1, h(U)
is replaced by U
as the determinant obtained from
Thus if i < h then
(3.11)
We note that
(3.12)
(3.13)
D.
1,
h(U(Y,))
1
1lI
dh , D.1, h(U(Yh)) --d.,
D. h(U.(Y )) = 0 if g ~ i,h .
1
1,
g
Now let xl' ... ,xn be any n points in X and let zl' .•• ,zk be
any k of the k + 1 points YI , •.. 'Yk+l' Occasionally we shall use the
alternative notation
Suppose that there are n + k nonnegative numbers PI' ... ,Pn+k such that
P1 + ••• + Pn+k
=1
and
Then, by (3.4) - (3.6) with N = n + k ,
n+k
(3.15)
L
j -1
n
g(x. ,
JI
p.
In
= 0 •
-17-
"
=1
Note that the restriction PI + .•. + Pn+k
is irrelevant since
equations (3.14) and (3.15) are not changed if all Pj
the same positive constant.
are multiplied with
Only the assumption that the Pj
are nonnegative
is needed for (3.14) to imply (3.15).
For PI' ••• 'Pn held fixed, consider (3.14) as a system of k equations
for Pn+l , ••. ,Pn+k' Condition (3.10) implies that the matrix
,U(Zk»)
is non-singular, and the solution of system (3.14) is
n
Pn+i = -j!l vi(xj)Pj'
(3.16)
where vex)
= 1,
... ,k ,
= U-lU(x)
(h)
Write Pn+i for Pn+i when
where h
i
= 1,
(zl' .•• ,zk)
= (Yl'···'Yh-l'Yh+l'·"'Yk+l)
Then, in the notation of (3.9) and (3.11),
.•• ,k + 1 •
n
r
J=I
Dh ,1·
i = 1, ... ,h - 1 ,
.r
J=l
Dh i +1 (u (x ' ) ) P · ,
J
J
i = h, ••• , k
.
(h)
(3.17)
~ Pn+i =
Since
~
> 0
n
for all
'
h, the condition
p~~i ~ 0, i = 1, ••• ,k , is
equivalent to
n
(3.18)
r
j=l
Dh . (U(x.»)p. ~ 0 ,
,1
J
J
i
= 1,
..• ,h - 1, h + 1, •.• ,k + 1 •
(Xl' .•. ,xn) in Xn there is an
such that (3.18) is satisfied in a nondegenerate
We now show that for every point
integer h
(1
~
h
~
k + 1)
n-dimensional sub-interval of {(PI' ••• ,Pn): PI > 0, •.. ,Pn > O}. It is
kn
enough to show that for every point (U l , ••• 'Un) in R
there is an h
-18such that the conditions
n
~
(3.19)
. 1
J=
Dh .(U.)p. ~ 0 ,
,1
J J
i = 1, ••. ,h - 1, h + 1, .•• ,k + 1 ,
are satisfied in such an interval.
k
For every point U in R there are real numbers bl(U), ••• ,bk+l(U)
such that
u
=
k+l
r b.
j=l
(u) U (y . )
J
J
By (3.11) and (3.13)
k+l
Dh l'(U)
,
= j=l
r
bJ.(U)Dh . (u(y.))
,1
J
= bh(U)d.1
- b.(U)dh •
1
Therefore conditions (3.19) are equivalent to
n
~
j=l
(For
i
{bh(u.)/dh - b.(u.)/d.}p. ~ 0,
J
1 J
1
J
=h
i
= 1,
... ,k + 1 .
this inequality is trivially satisfied.)
The assertion in the
preceeding paragraph now follows from Lemma 2 with m = k + 1 and
a ..
IJ
= b.1 (U.)/d
..
J
1
The foregoing shows that the set Xn can be covered with
k + 1 measurable subsets
El , •.• ,E k+l such that if (xl' •.• ,xn) € Eh
then inequalities (3.18) are satisfied in a nondegenerate sub-interval I of
{(PI' •.. ,Pn): PI > 0, •.• 'Pn > O} .
some h
n
,xn) be a p01'nt l'n X • Th en (xl""
(which will be held fixed in what follows). Let
(Zl'
,zk)
Let
( xl""
= (Yl ,
••• 'Yh-l' Yh+l' •.• 'Yk+l) .
Then, for
,xn)
€
Eh f or
(PI' ••• ,Pn)
the values pn+ l' •.. ,pn+ k defined in (3.16) are non-negative. We shall
insert these values in the left side of (3.15). The resulting homogeneous
€
I
-19-
polynomial in PI' ..• 'Pn
zero.
is zero on
I
and hence, by Lemma 1, is identically
The sum in (3.15) may be written as
n+k
L
(3.20)
n
n
k
L ... I
jl=l
where
In
Jl
n
nf
• .• p.
g(x. ,
j =1
II
g(x. , ••. ,x. ,
jm=l
J1
PJ'
m i=l
Jm
zm stands for the vector with m equal components
m.
],
Pn+i '
z • Identity
(3.20) can be proved by induction on k •
On inserting the expressions (3.16) for Pn + 1 , .•. ,Pn+k in (3.20) we
obtain
n+k
n
r
(3.21)
g(x. ,
Jl
j =1
=
I
j n =1
n
where
r
(3.22)
>0
>0
m~0,m1- , ... ,m - :
k
m+m1+···+mk =n
g(x. ,
J1
Let
a~
.
Jl'··.,J n
(_l)n-m
nl
mlm1 1••• mk l
m+ .•. +m.
k
m1
~
~
, x. , zl ' •.• , Zk) II
II
v. (x. ) .
Jm
i=lr=m+ ••• +m. ·1~+1 ], J r
1-
denote the symmetric average of the values
over the n! permutations .
~.(j 1'" . ;~n)
of the integers
j 1 ' ... ,jn'
The
-20-
condition that the homogeneous polynomial in PI' .•• 'Pn on the right of
(3.21) is identically zero is equivalent to the condition that the coefficients
are all zero.
We can write
n
(3.23)
a*1
= I..t ( -1) n-m Sn m(x I , ••.• xn ) •
, •••• n
m=O
•
where
(3.24)
Sn. m(x I , ••• ,xn) =
L
m ~O, ••• '1l1<~0
I
mI+···+~=n-m
m
l
k
~
m+ ••• +mi
v. (x. ) ,
g(x. , ... ,x. , zI ' •.•• zk) II
II
Jl
Jm
i=I r=m+ •.. +m.1- 1+ 1 1 J r
and
t m
I..
m
m
denotes the sum over those permutations
• 1····' k
the integers
J' l' ... 'n
J'
of
1, .•.• n which satisfy
J'l < ••• < J' m'• J' m+I < ••• < J' m+m ,
1
< ••• < jn •
The condition a*1, .. . ,n = 0 is equivalent to
n-l
(_l)n-m-l S
(x
(3.25)
.X ) = t
L
n,m I'
n
m=O
For m = 0, I, ••• ,n - I , each term in the sum in (3.24) is. for some
i
and some
j
• the product of
v,(x.)
1.
J
and a factor not involving x. .
J
Moreover.
Sn.m(xI , ... ,xn) is symmetric in Xl' ••• ,xn ' These facts
imply that the right side of (3.25) can be written in the form
k
n
L L
. I ' I
1=
J=
where the functions
v.(x,)f,(x l , ..•
1
J
1.
fl' •.• ,fk are symmetric.
Expressing the v.1. in terms
-21-
of u l ' ••• ,uk' we obtain a representation of g(x l , ••• ,xn)
(3.3) with symmetric functions hI' .•. ,hk .
in the form
hI' ••• ,hk can be so chosen that they
satisfy the integrability condition (3.2). Let Po be the distribution which
We now show that the functions
assigns probabilities
AI' ••. ,A +l to the respective points Yl' ••• 'Yk+l '
k
as defined in (3.7) to (3.9). Let B. denote the set which consists of the
J.
single point y. , for
the matrix
(f:
= 1,
i
, IUd
U d Po'
Since the
,k.
po]
Bk
1
of Lemma 3B with v = Po are satisfied.
A.
J
are strictly positive,
is non-singular.
The conditions
By Lemma 4, the functions
hI' ••• ,hk in (3.3) can be so chosen that each hi(xl , .•• ,xn _l )
linear combination of terms of the form
d poet
IB•
n-m
,x. , t , ... ,t
) g(x. ,
Jl
l
Jm
1.
n-m
is a
)
n-m
u
Let
P. The ui
P be a distribution in
Hence to show that the h.1.
Xn
Igi
d(p~-mpm)
for m = 0, 1, ..• ,n - 1 and all
By (3.7) the distribution Po
in
) . .• u
J m+ l
d Qn
< m.
But
r n - m- l
(x.) •
In-l
are P-integrable by assumption.
P
€
< m
P •
is in
Po and hence in P.
P, so is Q = ~(Po + P) , due to the convexity of P.
I Igi
e-
(x.
are pn-I-integrable it is sufficient to show that
I
(3.26)
rl
I Igi
If P is
Hence
dQn can be written as a linear combination with
positive coefficients of the integrals in (3.26) •.
Thus (3.26) is true.
This
-22completes the proof under the assumption that the origin 0 is in the interior
of the convex hull of U.
Now suppose that the origin is a boundary point of the convex hull of U.
Then there are real numbers b l ,
,bk ' not all zero, such that
blul(x) + ••• + bk~(x) = 0 for all x € X. Therefore one of the conditions
(3.1) is implied by the others.
In this way the problem can be reduced to one
of these two: (I) a problem of the same structure, with k replaced by k' ,
I s k' < k , such that the origin of k'-space is in the interior of the convex
hull of the set corresponding to
no restrictions (3.1) present.
U; (II) the same kind of problem but with
In case (I), the conclusion of the theorem
follows from the first part of the proof. In case II, equality (3.6) with
n
N = n and arbitrary (Xl' •.• ,xn ) € X holds for all positive PI' •.• 'Pn '
so that, by Lemma 1, g(x l , .•. ,xn ) = O. (This is, essentially, Halmos'
Lemma 2 in [2].)
Theorem lA is proved.
-23-
4. Proof of Theorem 2A. Let the conditions of Theorem lA be satisfied, and
suppose that
ul '
(xl'
g is bounded while every nontrivial linear combination of
,uk is unbounded.
,x )
n
We must show that
g(x l , •.. ,xn)
=0
for all
n
X .
EO
We again assume that
cl
= ••• = ck = o.
Since every nontrivial linear
combination of u l ' .•. ,uk is unbounded, there exist k points zl'
in X such that the k x k matrix (U(zl)' ..• ,U(zk)) is nonsingular.
(xl'. .• ,xn)
Hence, by Theorem lA and Lermna 3A, we have for all
(4.1)
g ( Xl' ..• ,xn )
where (we now
€
,zk
n
X
n-l
= Lt (_l)n-m-l Tn,m (x l'
m=O
exhibit~the
dependence of Tn,m on
k
k
= Lm, n-m
(4.2)
. =1
L
1.
1
g)
i
L
n-m=1
Here each of VI' ... ,vk is a nontrivial linear combination of u l ' .•• ,uk
and hence is unbounded.
The theorem will be proved by induction on k and, for each k, by
induction on n .
For n
g(X)
=1
= hlul(x)
+
and k arbitrary we have, by Theorem lA,
+
hkuk(x)
side is bounded only if hI
this case.
J
where hI'
= ... = hk = 0
,hk are constants.
The right
, so that the theorem is true in
-24Now let
(4.3)
k
= I.
By (4.2),
•.
g) =
Tn,m (Xl"
Lm,n-m
g(x. ,
J1
,X. ,
Jm
z, ... ,z)
veX.
J m+ l
where
z
= zi
' and v
= vI
is unbounded.
)
veX. ) ,
In
There is a sequence {YN} in X
such that
Divide both sides of (4.3) by v (xn) , set sn = YN and let N ~ w. The
terms on the right of (4.3) with jm = n , divided by v(xn ) = v(yN) , converge
to zero, and we obtain
L
= m n-I-m g(xJ. , . .. , x.J , z, .•. ,z)
,
1
m
vex.
J m+ l
•
, Xn-I' g
)
veX.
)
I n-l
(1))
,
g(l) (Xl' ••. ,x _ ) = g(x , •.• ,x _ ' z) , for m = 0, ... ,n - 2 .
1
n l
n l
For m = n - 1 , the limit is g(1)(X , .•• ,xn _l ) • Thus if we set xn = YN
1
in (4.1), divide by v (YN) and let N ~ m , we obtain
where
g (I)(X I , ••• ,xn _l )
= n-2
t
m~O
(_1)n-m-2 Tn-l,m (Xl' ••• ,x _ ; g (1)) •
n l
It follo\1s!by'·,indy.ction onor.n that the theorem is true for k
Now let k
by k - 1 .
~
=1
2 , and suppose that the theorem is true with k
.
replaced
-25Since vk is unbounded, there is a sequence {YN} in X such that
Ivk(yN)I ~ 00 as N + 0 0 . There is a subsequence {YN,} of {Y } such that
N
VI (yN,)/vk(YN,) tends to a limit AI' _00 ~ Al ~ 0 0 . Repeating this argument,
Ivk(yN)I ~ 00 and
_00 ~ Ai ~ 00
for
we see that there is a sequence {YN} in X such that
vi(YN)/vk(YN) ~ Ai' i = 1, ... ,k , as N ~ 00 , where
= 1, •.. ,k
IA.I
= for
1
i
- 1.
i
00
Suppose that
= 1,
AI' ••• ,A k _I are not all finite, say
IA.1 I < 00 for i ~ r + I. Then
... ,r;
vk(YN)/Vr(YN) ~ 0 , hence Ivr(yN)I ~
or 1, for
i
~
~
A!1
~
induction that there is an index
j
,
v.1 (YN')/vr
such that
AI'
...
(YN')~
r.
and vi(yN)/vr(YN) ~ Ai with Ai = 0
Also, there is a subsequence {YN,} of {YN} such that
00
A!1 , with
IV j (YN) I ~
00
_00
00
for
,
1
~
and v. (YN)/v. (YN)
1
J
j
i
~
~
(4.4)
r - 1
,
and a sequence {YN} in X
i = 1,
'Ik , where
k
A.1
We may assume that
,Ak are all finite.
~
It now follows by
...
j
=k
IA.I
<
1
, so that
00,
i = 1, ... ,k - 1 •
Divide both sides of (4.2) by vk(xn) , set xn = YN ' and let N ~ 00
After this operation the right side of (4.2) is modified as follows. Each
term with
ji
=n
for some
i
~
m is replaced by zero.
In each of the
On rearranging
remaining terms the factor v. (x) is replaced by A1'
Ig n
g
terms and taking account of the sYmmetry of g, we obtain
where
k
(4.6)
g(l)(x , ••• ,xn _1)
I
= l
i=I
A· g(x I , ... ,x l' z.) .
1
n1
•
-26-
If in (4.1) we divide by v (Xn) , set xn = Y ' and let N + 00 , the
N
limit of the left side is 0, and since Tn- 1 , n~ - l(x , ..• ,xn- 1; gel)) =
l
,X _ ) , we obtain
n l
gel) (xl' ...
(4.7)
It follows
n-2
,X
n-
1)
= m=O
1:
(_I)n-m-2 T 1
(xl' ..•
n-.•m
in the 5ame way that if
we define
,X
n-
1; g(l)) .
gO). £(2), •.• ,g(n-l)
by
(4.6) and
(4.8)
g
(s+l) (
k
xl' ... ,xn - s - l ) = 1: >..g
. 1 1
1=
(s)
(xl'
,Xn- s - l'
z.)
1
s = 1, ••• ,n - 2 ,
then
(4.9)
g
for
(n-s) (
=
xl'
s-l
l:
m=O
s = n - 1, n - 2, ••• ,1.
g(n-l)(x)
= T1,0
(_l)s-l-m T
(xl' ... , Xs'• g(n-s))
s,m
In particular,
(x; g(n-l))
k
= . l: 1
g(n-l)(z.)v.(x) •
1=
1
1
Hence
g (n-l) (x)
(4.10)
= 0 ,
all
X ~ X •
(n-s+l)
_
We now show that g
(xl' ... ,x s _I ) - 0 for all
,xs _l ) .." Xs - l implies g (n-s) ( xl' .•. ,xs ) = 0 for all
s = 2, ..• ,n. Suppose that
(4.11)
g
(n-s+l) ( xl' ... ,x _ ) =
S 1
o
for
..27By (4.8)
g (n-s+l) (xl' ••• ,xs _l )
Hence, as
(4.12)
g
Ak
=1
(n-s) (
= ~l. ,
1\.
i=l
1.
g(n-s)(x l' ••. 's-l'
X
zi ) .
,
k-l
= - . ~l. 1
xl'
1.=
By (4.9),
g (n-s) ( xl' ... ,x )
s
A.1 g
(n-s)
(xl' . .. ,xs- l'
is a sum involving the terms
k
(n-s) )
, s' g
(4.13)
X
r=1
•
i
,z.1
) °
s-m
s-m
v. (x.
11
J m+ l
)
v.1 _ (X.)
J
s m
s
with m = 0, 1, .•• ,s - 1 .
Let
(4.14)
and let
T~,m(XI'
... ,xn ; g) be defined as Tn,m(x l , ... ,xn ; g)" ; but with
k, vl(o), •.• ,vk(o) replaced by k - 1, w (·), ••• ,wk_l(o). If we
l
eliminate zk from the right side of (4.13) by using (4.12), we obtain
(4.15)
for
m = 0, 1, •.•
g
,5 -
1.
(n-s) )
* (xl'. ..
= Ts,m
,X ;
5
g
(n-s) )
Note that any nontrivial linear combination of
WI' •.. ,wk _l is unbounded. It now follows from (4.9), (4.15) and the induc•
XS .
tion hypothesis that g(n-s) (xl'
for all ( Xl' ••• ,xs ) 1n
, X5 ) = 0
Thus g(x l , •.• ,xn) = 0 for all (Xl' .•• ,Xn) E: Xn •
-28-
5. Proof of Theorem lB. We again assume that c l =
= ck = O. Let ~
(X,A), let P be a convex
be a a-finite measure on the measurable space
family of distributions which are absolutely continuous with respect to
PO(~)
satisfy conditions (3.1), and let
P.
c
for all
P € P. We
k symmetric A(n-l)-measurable real-valued func-
must show that there exist
,hk
g be a symmetric A(n)_
J g dpn = 0
measurable real-valued function such that
tions hI'
Let
and
~
such that
(5.1)
i
= 1,
... ,k , if P
€
P
and
(5.2)
rk
g(x l , .•• ,xn) =
u. (x.)h. (xl' ... ,x,!' x. I' ... ,xn) a.e. (P
i=l 1 J 1
JJ+
Let
(n)
AO be the class of all sets A in A such that
~ (A)
(5.3)
+
I JA Iu. I d~
i=l
<
00
•
1
AI' .•.
,~
raJ'
II (A .)
= 1 ,
u. dll
= 0,
Let N be a positive integer,
be sets in AO ' and aI' •••
,~
be nonnegative numbers such that
N
(5.4)
j=l
I
(5.5)
Then p(x)
=
N
j=l
a.
J
JA.
r a.IA (x)
j=1
1
J
i
= 1,
•.• ,k •
1
J
, where
j
I A denotes the indicator function of the
set A, is the probability density with respect to
Po(~)
).
and therefore in
P.
~
of a distribution in
Hence conditions (5.4) and (5.5) imply
ERR A T ASH E E T
The following paragraph was omitted from page 29. It should follow the
third displayed equation and precede the paragraph which begins: "Let
AI' ... ,An be any n sets ... "
We first assume that
0
lB. ' U d~/~(Bj)'
)
is in the interior of the convex hull of
U(~)
B ,
,Bk+
in AO of positive ~-measure
l
l
is an inner point of the polytype whose vertices are
Then there exist
such that
0
k + 1 sets
j = I, ... ,k + 1 .
•
-29N
N
L
(5.6)
L
a.
jl=l
a. G(A. ,
JI
In
JI
j =1
n
,A. ) =
In
o,
where
g d~n .
Al x ••• XA
n
(The existence of the integrals in (5.6) is guaranteed by the assumption that
J
(5.7)
Jg d
pn
exists for
P
P .)
€
Conditions (5.4) and (5.5) also imply that the origin
0 of
Rk
is in
the convex hull of the set
•
Let
any k
AI' ... ,An be any n
of the sets
sets in AO and let
Then i f
numbers such that conditions (5.4), (5.5) with N = n
condition (5.6)
with
N=n + k
A +l ,
n
are nonnegative
+
k are satisfied,
is satisfied.
We now use the argument in the proof of Theorem lA, with
and
g(x l , '"
,xn)
replaced by
,An+k be
a., AO' A., J U d~
J
J
A.
Pj"X, Xj' U(x )
j
and G(A , .. , ,An)
I
J
respectively, to infer that there exist symmetric real-valued functions
n- l
such that, for any
HI' ... ,~ on A0
k
n
L L
i=l j=l
(5.8)
t.
J
The matrix
(5.9)
.~
is nonsingular.
By Lemma 3A, with X replaced by
o'
A
we have for
-30-
G(A ,
l
(5.10)
...
n-l
,An) =
where, with Vi (A) =
(5.11)
Tn,m(Al'
m=O
fA v. d1-l ,
vex) =
1.
... ,An)
...
( -1 )n-m-l T
n,m (AI'
L
u-1-1 l
,An) ,
U (x)
k
k
I
= Lm n-m . L
,
1. -1
C
{"<II ,
\:t '..
j
;
i n-m=1
.'
)V. (A.
)
B. , •.. , B.
1
1
1
J
+
m
1
1
n-m
l
Hence if we define, for
(5.12)
t
n ,. m(x 1 , ..• ,xn)
I
_ (A.).
n m n
v.
_ (x.),
I
n m n
1
(xl'
= Im,n-m
k
k
i1
2
=1
.
I =1
i.
:~
n-m
,B.
g(X. ,
Jl
(5.13)
V.
1
g(x l , ••• ,xm' AI' ... ,An _m)
= fA
)v. (X
n-m
1
1 j m+1
)
1.
d1-l(xm+l )
1
then
(5.14)
f {g (xl' ••.
C
for all sets C
= Al
'Xn )
-
n-l
~ (_1)n-m-1 tn,m ( xl' ... ,x ) ~
n
l.
n . 1-1 =
m=O
x ••• x An
0
in A~.
Let w(x l ,
,xn ) denote the integrand in (5.14). The integral
r
n
n
J(C) = J W d1-l is zero for C € Ao
Let B be a set in AO
By a stanc
dard argument, J(E n Bn ) = 0 for all E in the a-field A(~) and hence
on Bn •
•
-31-
Now
.
and let
let
Be
P be a distribution in P, let p be a version of
= {x:
p(x) > e}.
I
(1 +
Thus
Be
AO for all
1
d~ ~
e
f
B
e
on
Hence w(x l '
UE>OB~ '. a se~ P~-rne~s':lre one~f, that.. · "
r.;n,.
(5.15)
€
g(x
> O.
€
=
l , • \..' ,xn)
a.e. (p n) , for all
P
n-l
~
rn=O
P
€
,
Then
lu.l)p
i=l
dP/d~
(_l)n-rn-l t
(xl""
n,m
,xn)
It is easily seen (compare the proof of Theorem
lA) that the sum in (5.15) can be written in the form (5.2) where each function
h.
1
is symmetric and A(n-l)-measurable.
We now show that the functions
condition (5.1) is satisfied.
strictly positive numbers
Let
With
AI'
k+l
~ A.
j=l J
h.1
can be so chosen that the integrability
as above, there exist
k+l
~ A. = 1 and
such that
j=l J
B , •.• ,B +
k l
l
iAk+l
f
B.
J
=0
u d~/~(B.)
J
.
Po be the distribution whose density with respect to
PO(x)
Then Po is in
PO(~)
nonsingular and the
r
j=l
A. I B (x)/~(B.) .
J
j
J
and hence in
A.
J
P.
Since the matrix
are strictly positive, the matrix
Up = (f
o
=
k+l
Bl
U
is
~
d PO'
'00
,
f
Bk
U
d po)
U~
in (5.9) is
-32-
is also nonsingular.
The conditions of Lemma 3B with v
= Po
are satisfied,
except that the representation (2.4) of g(x , .•• ,xn) holds a.e. (p(n)) .
l
Hence (2.15), with v = Po ' holds a.e. (p(n)). By Lemma 4, each
h.(x l , ••• ,x
1) can be chosen as a finite linear combination of terms of
nthe form (2.19) with v = PO. The pn-l-integrability of these functions
1
follows in the same way as in the proof of Theorem lA, making use of the convexity of P.
Now suppose that the origin 0 is a boundary point of the convex hull
of U (v).
i t bi
all
P
t
~
Then there are real numbers b l , ••• ,bk ' not all zero, such that
"i du = 0
P.
follows from the others.
lA.
f
A < AO ' andftheTefOTe i t b i
"i'dP = 0
Thus one of the conditions
u i dP = 0, i = 1, ... ,k ,
fOT all
for
The rest of the proof is similar to that for Theorem
-33-
6.
Proof of Theorem 28.
suppose that
u l ' ••• ,uk
Let the conditions of Theorem 18 be 'satisfied, and
g is bounded while every nontrivial linear combination of
is P-unbounded.
We must show that
g(x l , ... ,xn)
=0
a.e.
(p(n)) •
We again assume that
cl
= ... = ck = 0
•
First it will be shown that there exists a measure v
is (i) equivalent to the family
f Iu i I dv <
Since the family
on
i = l, •.. ,k .
00,
P is dominated by a a-finite measure, it contains a
sequence PI' P2 , '"
=0
)
d.
k
= I
for all
Jlu·1
which
P, (ii) finite, and (iii) satisfies
countable equivalent subset (Halmos and Savage (1949), Lemma 7).
P.(A)
(X,A)
of distributions in P be equivalent to
j
implies
d P.,
.
=0
peA)
= 2-)(1
for all
P in
Let the
P (so that
P). Let
00
i=l
1
J
are strictly positive and
J
+
d.)-l,
J
v
= L
b. P.• The numbers b).
j=l J J
I b. is finite. Hence v is a finite measure
k J
00
00
j
equivalent to P. Also, i~l
!uildv = j~l bjd j < j!l 2- < 00 , so that v
satisfies conditions (i), (ii), (iii).
b.
J
f
Since v
is equivalent to
P, we have that if u is a nontrivial
linear combination of ul ' ••• ,uk then v'(!u(X) I > c) ;l! 0 for all real
Let A+ denote the class of sets A in A such that v(A);l! O. For
A € A+
define the set functions
U.(A)
1
=
f
A
Ul '
u. dvlv(A) ,
1
c.
i
= 1,
,k.
,.
Then every nontrivial linear combination of Ul ' .•. ,Uk
is unbounded on A+
_J
-34-
Hence there exist k set
•
Bl , .•• ,B k in A+
such that the matrix
uv =
is nonsingular.
(f
By Theorem 18, the conditions of Lemma 3B (last paragraph) are satisfied.
Igi dvn is finite since g is bounded.) Hence the representation (2.15)
of g(x 1, .•. ,xn ) holds a.e. (~(n)). Let AI' ... ,An be n sets in A+ .
Integrating both sides of (2.15) over the product set Al x ••• x An in iA~
with respect to vn , we obtain
(6.1)
n-1
(A l' ... , An )
Gt(A l ' ... , An ) = L~ (_1)n-m-l Tt
m=O
n,m
where
n
g dvn/ TI v(A.) ,
;=1
k
i
L
J
t
G (A. ,
, B.
~
J1
n-m=1
t
V.
~n-m
t
V.(A)
= v(B.)
~
~
fA v.~
dv/v(A),
)
n-m
(A.),
In
l
Vex) = U-" u(x) •
v
The representation (6.1) of the set function Gt(A l , .•. ,An) is strictly
analogous to the representation (4.1) of g(x l , ... ,xn) .. Since g is
bounded, Gt is bounded on A~ , and the
(A) are unbounded on A+ ~ Thus
the proof of Theorem 2A implies that Gt (A1 , .•. ,An) = 0 on A~. Therefore
Vi
-35-
.
•
for all cylinder sets C = Al x ••• x An
a.e. (vn ) , and thus
in An.
Hence g(x l , •.•
r)
,X
=0
a.e. ( pen) ).
REFERENCES
Completeness of order statistics.
Canad. J.
[1]
Fraser, D. A. S. (1953).
Math. ~, 42-45.
[2]
Halmos, Paul R. (1946). The theory of unbiased estimation.
Statist. 12, 34-43.
[3]
Halmos, Paul R. and Savage, L. J. (1949). Application of the RadonNikodym theorem to the theory of sufficient statistics. Ann. Math.
Statist. 20, 225-241.
[4]
Hoeffding, Wassily (1956). The role of assumptions in statistical
decisions. Proc. Third Berkeley Symp. on Mathematical Statistics
and Probability, Berkeley and Los Angeles, University of California
Press, vol. I, 105-114.
[5]
Lehmann, E. L. (1959).
Wiley.
Testing Statistical Hypotheses.
Ann. Math.
New York, John