*A PhD dissertation under the direction of GORDON SIMONS.
This research was partially supported by the National Science Foundation under
Grant No. MCS75-07556 .
.e
DISTRIBUTION FUNCTIONS ON PARTIALLY ORDERED SPACES*
by
Kiang Liu
Department of Statistics
University of North Carolina at Chapel Hill
Institute of Statistics Mimeo Series No. 1086
August, 1976
.e
ACKNOWLEDGEMENTS
I wish to express .my sincere appreciation to my advisor Professor
G.D. Simons for suggesting this problem and for his patient guidance and
constant encouragement.
I will always remember with gratitude the personal
interest he has taken in me and my work.
I am also happy to thank Professor W.H. Graves for his enthusiastic
service as my joint advisor and for his perceptive criticism and advice
throughout this dissertation.
His suggestions made a substantial contribution
to the direction and progress of my research.
Also I wish to thank the other members of my doctoral committee,
Professors D.G. Kelly, I.M. Chakravarti, T. Brylawski and K. Petersen, for
their review of my work.
.e
Especially I would like to thank Professor T•
Brylawski for suggesting the approach which I use for deriving Mobius f1.mctions of history spaces.
I am also grateful to Professors N.L. Johnson,
M.R. Leadbetter, S. Cambanis and the faculty of the Department of Statistics
for the excellent instruction I have received here.
Finally I am particularly grateful to my wife Shuguey for her patience
and understanding throughout my graduate study and to my parents for their
continual encouragement and support.
-e
iii
TABLE OF CONTENTS
Chapter I. INTRODUCTION AND SUMMARY
1.1
Introduction
1.2
Summary
Chapter II.
•
(I
•
..
•
1
•
9
..
THE EXTENSION OF THE MOBIUS INVERSION FORMULA
2.1
Introduction . • •
2.2
Some Preliminaries
2.3
The Extended M~bius Inversion Formula
Chapter III.
•
.
.
. 10
· 13
• . • . 14
THE HISTORY SPACES
3.1
History Spaces and Their M~bius Functions
3.2
Basic Properties of HCn)
3.3
The Uniqueness of the Mass Function for B(n) • 30
3.4
An Inversion Formula for H(3)
. . 23
27
· 38
Chapter IV .. THE FINITE PARAMETER SPACES
4.1
a-lower Finite Spaces
4.2
Finite Parameter Spaces
4.3
A Class
AP PEND ICES
REFERENCES
of
44
• 47
52
Finite Parameter Spaces •
.
•
• • 71
· 81
e.
CHAPTER I
INTRODUCTION AND SUMMARY
1.1 Introducti.on
The consideration of distribution functions on
partially ordered spaces (sets) is motivated from a boundary
crossing problem arising in sequential analysis.
First of
all, let us consider the following statistical example.
Example 1.1.
Let U and U be two parallel lines with
l
z
positive slopes, and let L
l
and L be two parallel lines
2
with negative slopes as in Figure 1.1.
These four lines
divide the positive half plane into three regions, region a,
region b and region c, say.
Simons (1967)
co~sidered
a
sequential three hyp6thesis test as follows.
Let Xl' X ,
2
..• be a sequence of independent, normally
distributed random variables with unknown common mean p and
known common variance cr
Z
.
He want to test which one of
three simple hypothesis HI' HZ and II is true, where H. is
1
3
the hypothesis
S
n
=
n
Li=l X 1••
p
=
Jl • ,
.1
Jl I
< Jl
< Jl3'
Z
For n = 1,2, •.. , let
The test procedure can be described as
a) Accept HI if (n, S n ) touches region a before b or c,
h)
Accept HZ if (n, S ) touches region b before a or c,
n
c) Accept H if (n, S ) touches region c before a or b.
n
3
·e
Simons replaced S
n
by a Wiener Process Wet),
t
~
0 with
mean Jlt and variance crZt, and found approximate values for
-[J
f-----·-region b
Figure 1.1
e
Figure 1. 2
2
n
the operating characteristic functions such as Pp{accept HI).
The
co~responding
until(t,
Wet»~
Figure 1.1.
test procedure is to graph W{t) versus t
is on the boundary of one of the regions in
(Note that Wet) is almost surely continuous.)
We do not know much about where Wet) goes except that
it starts at the origin.
However, the history of the Wiener
paths can be described by words made from an alphabet of
three letters a, band c; for instance, the path which
touches region a before region b and region c, then touches
region b, then touches region c, and then it never touches
region a and b again can be described by the word abc.
(See Figure 1.1.)
Since it can be shown that Wet) touches a
finite number of regions with probability one, we can list
all possible Wiener paths as follows.
e - Wet) touches none of. the regions
a - Wet) touches region abut never touches the other two
regions
b - Wet) touches region b but never touches the other two
regions
c - Wet) touches region c but never touches the other two
regions
ab - Wet) touches regiona before b, then touches b, then
never touches a again, and it never touches region c
Let X be the collection of all such words. In other
3
words, X consists of all possible "words" of finite length
l~tters
made from an alphabet of three
a, band c, with
the property that no letter may follow itself immediately.
(Note that e can be considered as a w0rd with zero length.)
On X, a partial ordering" :;; " is defined that for x, x'
x
:;; x'
(See Figure 1.2.)
iff x' is a subword of x.
m(x)
?
X,
This
partially order.ed space is called a "history space".
spaces will be d~scussed ~ore generally later.
E:
History
A mass
0 is assigned to each word in X as the probability
that the history of Wet) is x.
M(x)
(1.1)
=
L
The function
x
m(y),
E
X
y:;;x
is said to be a distribution function.
M(x) has a probabilitistic meaning.
x
t
=
For each x
E
X,
For instance, when
a a ... a , M(x) is the probability that there exist n
n
l 2
1m e s 0 < t 1 < •••
in region aI' a 2 ,
< t n wit h
( t 1.' W( t 1) ) ~
... , and region a
n
••• ,
(t
n
,
Wet
»
.
n
respectively.
The. reason for introducing the functions m and M will
now be made clear.
The O. C.
quite easily in terms of m.
Pjl(accept HI)
Thus,
functions can be expressed
For instance,
mea) + m(ab) + ro(ac) + m(aba) + .•.
it is useful to be able to compute m.
However, in
practice, it is easier to compute M than it is to compute m.
(This will be made clearer with the next example.)
Consequently, one is led to ask whether the functional
4
values of M determine the functional values of m.
That is,
is there an inversion formula which expresses each m(x) in
terms of the set of values {M(y), y E xl ?
this dissertation will be directed toward
A large part of
a~swering
this
question and answering related questions.
Let us consider now another interesting example which
is the same as example 1.1 except that the middle region in
Figure 1.1 is dropped.
~xample
1.2.
crossing
In this example we consider the boundary
probl~rn
of Wet) for two regions, region a and
(See Figure 1.3.)
region b.
The corresponding partially
ordered space is shown in Figure 1.4.
m and M are defined
in a similar way as in example 1.1.
Doob
u
2
>
(1949) described how to compute M.
0 and 11 < 0, 1
2
U and L respectively.
Let u
1
> 0,
< 0 be the slopes and intercepts of
He obtained the formulas
if x
=
(ab)
n
and n is even
M(x) =
where
(ab) n denotes the word consisting of the first n
,
letters of the sequence abab ...
be obtained by the interchange of
5
When x =
(ba)
n , M(x) can
(u ' u ) and (11' Ii)'
l
2
o
L
Figure 1.3
b
ba
ah
bab
aba
baba
abab
Figure 1.4
6
_. e
He was interested in expressing a probability such as m(e)
in terms of M(x), x
E
X.
In particular, he found
2
_e- 2 [n (1112+u1u2)-:n(n-1)11u2-n(n+l)ul12]+
This formula in our notation
b~comes
00
m(e) = 1 -
(1.2)
(_1)n-1
L
n=l
L
M(x),
Ixl=n
where Ixl is the length of x.
Anderson (1960) obtained a formula for P(W(t) touches
regi.on a before region b).
This can be expressed in terms
of the masses asm(a) + m(ab) + m(aba) + . . . .
Anderson
evaluated the probability as
M(a) - M(ba) + M(aba) - M(baba} +
(1.3)
In fact an inversion formula (formula (1.4»
found for this
X
E
spa~e
which expresses
m(~)
can be
in terms of M(x),
X:
0<>
(1. 4)
m(x) = M(x) -
L
(_1)n-1
n=l
E
M(y)
lyl=lxl+n
Formula (1.2) and (1.3) can be derived easily from (1.4).
More generally, let X denote an arbitrary countable
7
partially ordered space and m(x)
~
0 denote a mass or
probability which is assigned to the point x, x
M, in terms of m, by (1.1).
E
Define
We ask the question whether
there exists an inversion formula (such as (1".4»
permits each m(x) to be expressed in
values {M(y), Y EX}.
X.
te~ms
which
of the set of
As we shall see the answer depends
on the structure of X.
Now, we would like to summarize and abstract our
p~oblems
as follows:
We begin with a partially ordered space X
countable.
wh~ch
is
1'Je shall confine all of our attention to locally
finite spaces (by which we mean that the number of points in
each interval [x, y]
= {z: x
~
z S y} is finite), because
it is for such spaces that a MBbius function is defined.
(Note: The spaces discussed in Example 1.1 and 1.2 are both
locally finite.)
X, a mass m(x)
~
0 1s assigned.
further assumed that M(x) = E "m(y)
"
ysx
<
00
To each point x
E
called a distribution function.
for each
Xe
It is
X, M is
Unlike the distribution
function defined on the real line, for certain spaces, M(x),
X
X need not determine m(x),
E
AI.)
II n
X uniquely.
(See Appendix"
We shall concern ourselves with the following questions:
(1)
f
XE
c. t ion,
When does the distribution function determine the mass
X
E
x't
(2) How are they found when they are determined?
8
The~e
problems have been solved by Simons (1975) for
what he calls a-lower finite spaces.
A brief discussion of
these spaces will be given in Chapter IV.
1.2 Summary
In Chapter II we will discuss a Mobius inversion formula
due to Gian-Carlo Rota and consider extensions of his Mobius
inverion formula.
Under certain
condition~,
these formulas
not only hold for nonnegative real valued measures
I
but also
hold for abstract measures.
In Chapter III, we will discuss history spaces.
M~bius
function will be computed.
to be a determining space.
The
A history space is proved
By this we mean that: each mass
function m is uniquely determined by its distribution
function M.
An inversion formula can be found for the space
that we have discussed in Example 1.1.
F~nally,
in Chapter IV we will briefly discuss a-lower
finite spaces and show how they lead to a consideration of
what we shall call finite parameter spaces.
A necessary an.d
sufficient condition, based on the Mobius function, will be
given for a space to be a finite parameter space.
Further-
more, a theory will be developed for certain types of finite
parameter spaces.
I For . eac h su b set A of X, define meA) = r X" AM(x), then
C
m forms a nonnegative real valued measure on X.
9
CHAPTER II
..
THE EXTENSION OF THE MOBIUS INVERSION FORMULA
2 '.1 Introduction
The method of M6bius inversion, in the form we shall
describe, was first systematized by Gian-Carlo Rota (1964),
Rota begins with a locally finite partially ordered space X.
Let A(x) denote the set of all real valued functions f(x,y),
defined for x, y ranging over X, and with the property that
f(x,y)
= 0 if x i y.
Three operations, addition, scalar
multiplication and product are defined in A(X), where
addition and scalar multiplication are defined as usual. The
product fg of two f.unctions f and g is defined as
fg(x,y) =
E
f(x,z)g(z,y).
x~z~y
One can easily verify that the three operations are welldefined and A(X) with these operations forms an algebra over
the real line.
of X.
A(X) is referred to as the incidence algebra
It has an identity 0, given by o(x,x)
=0 when x j
=
1 and 8(x,y)
y, the so-called Kronecker delta function.
The zeta function Z;; is defined as Z;;(x,y) - 1 if x
l; (x,y)
.-
0
othF~rwise
.
Clearly
Z;;
is contained in A(X).
inverse lJ.(x,y) is called the Mob ius function.
~
y,
Its
That an
inverse exists is inmediate from the following lemma.
Lemma 2.1.1.
A
if and only if f(x,x)
functi~n
~
f(x,y) of A(X) has an inverse
0, for all x
€
x.
We can see easily that p(x,x) = 1, for every x in X,
and that in general p(x,y) is integer-valued.
If we express
the
we have
Mobius function as the left inverse of
p(x,y)
=
L
p(x,z),
~,
x < y fixed.
x~z<y
If we express it as the right inverse of
].l(x,y)
L
].l{z,y),
~,
we have
x < y fixed.
x<zS:y
These formulas are used frequently in a recursive manner to
obtain
~
explicitly.
THEOREM 2.1.2
(Rota).
Let m(x) and M(x) 2 be two real valued functions defined
on a locally finite partially ordered space X with the
property that
(2.1)
M(x)
=
L m(y),
y:O;x
X
E
X.
Suppose m satisfies the following condition.
Con.d'[t..<.oYl. A:
Then for x
(2.2)
E
An element
X
o
exists such that m(x)
= 0 unless
X,
m(x)
=
I
].l(y,x)M(y).
yS:x
Equation (2.2) is commonly referred to as the Mobius inversion
2
Elsewhere we use m and M to denote the mass function and
the distribution function respectively.
11.
A
~
formula.
a finite
Observe that the. sum in (2.2)
numbe~Qf non~zero
c~nsistsof
terms since X
It is clear that when X is finite,
~s
locally
at most
fin~te.
(2.2) holds for any mass
function m and its corresponding distribution function M.
It provides an inversion formula which expresses m(x) in
terms of M(x), x
Howeve~,
~
X.
Hence X is a determining space.
when X is infinite, Condition A might not be satisfied
by some mass functions; the number of non-zero summands in
(2.2) could become infinite and the meaning of the sum
would be potentially ambiguous.
A is too restrictive.
We have found that Condition
It can be relaxed to some extent.
For example, the assumption of the absolute convergence of
the summand would be adequate to make (2.2) correct.
this condition is still too strong.
But
It suffers from the
fact that it involves M and not just the structure of X.
Also it largely ignores the partial ordering.
We would now like to consider extensions of the Mobius
inversion formula given by (2.2) and would like to obtain
alternative conditions, depending on the structure of X
only, under which the sum in (2.2) has a valid interpretation
(even with an infinite number of non-zero summands) for any
m and M defined on X.
Furthermore, under certain conditions,
the extended Mobius inversion formula holds not only for
real valued measures but also for abstract measures which are
defined in the next section.
12
In this section, we will state some results, given by
Graves and Molnar in 1974. which
wil~
b~
used to derive
Theorem 2.2.1 below.
Let R be a a-field of subsets of a non-empty set X.
~.easur~
An (abstract)
on R with values in a locally convex
Hausdorff topological vector space (l.c.s.) W over the
real field· R is a map V:
1)
R -+ W sa tis fying
veep) = 0,
= VeAl + V(B)
2) V(AuB) + V(AnB)
for all A,B
3) for every increasing sequence Al
A
C
2
C
R,
E
A ..• of
3
sets from R, V(A.)
converges . to V(uA.).
.
1
1.
One can see that if W
= R :
the real line, V becomes
a signed measure in the usual sense.
Let S(R) denote the collection of all simple functions
defined on X.
R.
Clearly S(R) forms a vector space over
Let X be a map from R to S(R) such that for each A
X , the characteristic function of A.
A
any measure V: R-+ W, define a map
€
R, X(A)~
For any l.c.s. Wand
V: S(R) -+ W as follows:
n
L
V (f)
0.. V (A .
1.
1.
i=l
) = ffdV
f
€
S(R),
where
n
L
Cl,X
i=l
1.
f =
A•
1.
is the unique representation of f
coefficients
0.
1
'
• • .,
0.
n
in terms of non-zero
and dis j oint s et s AI'
""
An
€
It can be verified that V is the unique linear map which
13
R.
satisfies V =
Let
T
VoX.
be -a topology
def~ned
on S(R) such that a base
for the neighborhoods of 0 consists of sets of the form
..... -1
-
-"'-1
VI (N1)n ... nV r
(N r ) where for each i
=
1 ~ ... , r.
W.
1.
l.c.s., N. is a neighborhood of 0 in W. and V.: R
1.
1.
1.
+
is
a
Wi is
a measure. T forms a locally convex topology on S(R) and the
fol10w~.
following theorem immediately
THEOREM 2.2.1
(Graves).
(1) X is a measure.
(2) For ariy l.c.s. Wand any
measure V: R + W,
there exists a unique continuous linear
map V from (S(R),
T)
to W satisfying V =
iox.
If X is countable, X ~{xi}~=l say~ and R
power set of X, S(R) and
~
=
P(X) the
can be given explicitly as
follm,.Tf3 :
S(R) can be identified with the set of all sequences
(a.),
a.
1.
1.
e
R,
which take on only finitely many values.
any sequence (n.), n. e R, vith n.
1.
lim.
1. +00
1.
1.
n.1. = 00, let D(n.)
= {(a.)
1.
1.
~
>
0 for all i and
1.
a countable number of points.
lao1_ I
S(R)
The D(n.)I s form a local base of 0 in
Let X be a locally finite
(~(R),
pRrtla~ly
Let
For
F=
<
n.,
1.
for all i}.
T).
ordered set with
{F} be the collection
of finite subsets of X~ and let D ~ F b~ any directed set
under set-theoretic inclusion.
L
=
r.
We shall write
f(x)
xeX
14
(D)
if liru
FE
DE
XE
~
~
f(x)
j.'
L, where f
is a function defined on X
with values in a l.c.s •.
For x
E
X, let Vl(x) = {[y,x]
y
~
x}
•
Lemma 2.3.1. If X has the property that for each x,y eX,
exis~s
there
a point z for which z
~
~
x and z
y,
then Vl(x),
with set-theoretic inclusion, forms a directed set.
P r..9 0 f :
5 e e A pp end i x
A2 •
THEOREM 2.3.2.
(1)
for each x,y
(2)
for each x
If X satisfies
X,
E
eX,
there exists a z for which z::<; x and z ::<;y,
there exists a t
< x such that
x
s~t
X
,u~
s, u
1s}~
K <co
X
then
p(t,x)X..-
(2.3)
t
in (5(R),
..-
T), where t
=
{y e X
y : : ; x} and R is the power
set of X.
Proof:
Let x e X.
For each s, s ::::; x, let
It is easy to see that fS'x is a simple function on X and
[{fs,x :
[s,x]
E
V (x)}; Vl(x)] formR a net in S(R).
Suppose that
to 0 in (5 (R),
.[),
1
[{f
s x
'
:
[s,x]
E
V1(x)}; V (x)]
1
then
lim
r
p(t,x)Xt = Xix} ,
[s,x]eV (x) te[s,x]
1
15
converges
i . e.
-E
lJ ( t , x )Xt
(VI
t~x
Vl(x)] converges to 0 in (S(R), T).
hood of 0 in (S(R), T), and let X
and lim.~-+oo 11.~ =
00
,
that
ni
So
u 1 ,u 2 ' ... ,u N·
S
~
K '
x
ex» .
LetU(I1.) be a neighbor~
=
00
{uili=l .
Since K
x
<
00
there exists a positive integer N such
for i > N.
Choose So such that So
For all s, s
E
S
~
t
x
and
So
lJ(t,x)
tE[ui,x]n[s,x]
for u. < x and
~
u. ~ sand u.
~
o
~
~ s
otherwise
( ) of or s ~·so.
. < 11 • .f or a 11'~ an d fS'.x € Un.
If s,x(u.)·1
~
~
~
s x
That [{f) :ls,x] E V1(x)}; Vl(x)] converges to 0 in (S(R),.)
h ence
immed~atly
f611ows.
Co~olla~y2.3.3.
If X satisfies the conditions in
Theorem 2.3.2, then for any l.c.s. Wand any measure V: R + W,
(2.4)
V({x})
=
+
lJ(t,x)V(t)
E
(VI (x»
t~x
Proof:
V: R
+
By theorem 2.2.1.(2), for any
W,
l.c.s~
Wand any measure
'" from
there exists a unique continuous linear map V
(S(R), T) to W satisfying V
=
VoX. Hence,
V({x}) -- V(X{xl)
= V(
. lim
E
l!(t,x)Xt)
[s,x]EV (x) tds,x]
l
16
=
lim
[s,x]EVl(x)
=
=
L
tE[S,X]
lim
[s,x]EV (x)
1
L
L
tE[S,X]
lJ(t,x)V(t)
lJ(t,x)V(X+)
t
+
.
lJ(t,x)V(t)
(V
t:s;x
Co~olla~y
e
~
lim
L
V(
lJ(t,x)X+)
t
tE[S,X]
[s,x]EVl(x)
2.3.4.
1
(x».
If X satisfies the conditions in
Theorem 2.3.2, then X is a determining space.
In particular,
for any mass function m and its corresponding distribution
function M
m(x) =
(2.5)
for all x
Proof:
in
R.
E
L
lJ(t,x)M(t)
t:S;x
X.
Let V(A) = L
m(y).
yEA
V is a measure on X with values
By Corollary 2.3.3
m(x) =
E
lJ(t,x)M(t)
tEX
(V
1
(x»,
for x E X.
Since R is Hausdorff and the limit is unique, it follows
that X is a determining space.
Ex~mple
2.1.
Let X be t.he partially ordered set shown in
Figure 2.1.
17
(1,2)
...
(0,1 )
Figure 2.1
ll«i',j'),(O,l»
-1
if i'=3k+l for k=O,l, ..•
1
if i'=3k+2 for k=O,l, •••
or i'=O, j'=l
o
if i'=3k+3for k=O,l, .••
=
-1
if
i'-~=3k+l
for k=O,l, ... ,
and jfj'
ll«i',j'),(i,j»
=
1
if i'-i=3k+2 or 3k for k=O,l, ••. ,
and j=j'
o
otherwise
In this space, all suprema in Theorem 2.3.2. (2) are
bounded by 2.
for x
€
Thus X is a determining space and (2.5) holds
X.
It can be shown that if condition (1) in Theorem 2.3.2
holds and (2.5) holds for any mass function m and its
corresponding distribution M, then condition (2) in Theorem
2.3.2 must hold.
This fact is proved in the Appendix A3.
In generali V (x) i$ not always a directed set.
1
18
We may
want to consider another set which always forms a
set if X is locally
For x
E
X~
fin~te.
let V (x) be the collection of all finite
2
It is easy to
unions of intervals with right enrl point, x.
V2 (x)
show that
directed
for~s
with set-theoretic inclusion
a directed
set.
THEOREM 2. 3. 5.
X
E
If X has
the property that, for each
X,
~up{1
p(t,x)
E
I:
F
tEFn[u,x]
EV 2 (X),u
K
E X}
X
<
ro,
then
p(t,x)X+-
(2.6)
(V
t.
Let x E X.
For each F E VZ(x), let
E
p(t,x)Xt tEF
It is easy to see that fF,x
forms a net in S(R).
show tha t
(S(R),
(x»
T), where R is the power set of X.
in (S(R),
Proof:
2
X{x}
E S(R) and
•
[{fF,X:FEV2(x)}; VZ(x)]
To show (2.6) is true, we orily need to
F
[{f , x.: F E V (x) }; O (x)] converges to 0 in
2
2
T).
Let U(ni) be a neighborhood of 0 in (S(R),
X =
{u.}~
I.
.1. 1=
Since K
X
<
00
and lim._~oo n
- U',N' ] [u ., xl
, ].=.
1
0 _"
0 (x),.
EZ
n.1
~
K
For all F
~
~
a positive integer N such that
F
1 ~
19
x
i
==
00
'
T), and let
there exists
N.
Let
VZ(x) and F
~'FO
for i
>
.e
1:
t EF n [11 i ' xJ
lJ(t,x)
o
hence
IfF,x(u ) I < fl_
i
i
-for F
::>
F '
O
That [ { f
otherwise ,
for all i .
Fx
~
Therefore, fF,x
U(n ),
i
V (x)] converges to 0
F E-V (X)};
:
E
2
2
in (S(R), T) immediately follows.
COJtO.t.ta.JLlj
2.3.6.
If X satisfies the conditions in
Theorem 2.3.S:then for any 1.c.s. Wand any measure V: R + W,
r
(2.7)
+
lJ(t,x)V(t)
t~x
Proof:
By Theorem 2.2.1.(2), for any 1.c.s. Wand any
measure V: R
+
W, there exists a unique continuous 1iriear
map ~ from (S(R), T) to W satisfying V =
iox.
,..,
V({x})
= V (X{x})
,..,
= V(
t
lim
v(t,x)Xt)
F€V (x) tEF
2
,...,
=
lim V( r
lJ(t,x)Xt)
tEF
FEV (X)
2
=
lim
r. lJ(t,x)V(Xt)
FEVZ(X) tEF
=
lim
r lJ(t,x)V(t)
F'EV 2 (x) tEl"
=
+
r
+
lJ(t,x)V(t)
t~x
zo
Hence,
COfLolfalLlj 2.3.7.
If X satisfies the conditions in
Theorem 2.3.5, then X is a determining space.
In
part~~ular.
for any mass functionm and its corresponding distribution
function M,
E
m(x) =
p(t,x)M(t)
t~x
for all x
Proof:
E
X.
EYE~ m(y).
Let VeAl
values in R.
V is a measure on X with
By Corollary 2.3.6,
Hausdorff, the limit is unique.
(2.8) holds.
Since R is
It follows that X is a
determining space.
Example 2.2.
Let X be the partially ordered set shown in
Figure 2.2.
(0,1).
Figure 2.2
(_l)i'-i
V.«i',j'),
(i,j))
if (i',j')
~
(i,j)
0
=
{
otherwise
It can be proved that all suprema in Theorem 2.3.5 (2) are
21..
~.
bounded by 3.
h~lds.
for x
Then X is a determining space and (2.8)
€
X.
22
CHAPTER III
THE HISTORY SPACES
3.1 History
~aces
and Their Mobius Functions
The spaces that we have discussed in example 1.1 and
1.2 involving two and three letters respectively are ones of
special interest.
In this chapter, instead of considering
these two special cases, we would like to consider a class
of spaces that are generated by a finite number of letters.
Ven~n{t~on
3.1.1.
subword of x.
Let G be a set of ndifferent letters and
Then H(n) is said to be a history space.
It is easy to see that history spaces are locally finite.
Hence, the M~bius function can be d~fined on these spaces.
Before we get into the details, we would like to
introduce some notation which will be used frequently in
this chapter.
Some NotatJ...on
(1) Usually X,3 Y and Z represent words in HCn), small letters
represent letters, and Sand T represent subsets of R(n).
I ...
3Elsewhere we use X to denote the partially ordered space.
(2 )
Ix.1 denote the length of the word X. (leI = 0.)
(3)
Let X and Y be two words.
e
If the last letter of X and
the first letter of Z are not the same, then define
Y-Z
YZ~
and Y-e = e·Y
=
Y.
For example, aba·ca.= abaca.
(4) Let Y - Y1 Y2 •• 'Y E H (n) .
k
=
Y.
1.
.Y
1.
Y1 Y2" ·Y i
if 1
~
Y
if i
> k
e
otherwise
Yi Yi + 1 •· 'Y k
if 1
~
e
if i
> k
Y
otherwise
Y . . . . Y·
1.
J
if 1
.Y
if 1 s i
s k
Y.
if i s 1
:S:
e
otherwise
1
. Y.
1
J
J
(5) For a =f: b,
j
(6)
~
i
i
i
~
k
~
k
~
j
j
:S:
k
::;
j
:S:
k
e
(ab). denotes the word consisting of the first
J .
letters of ababab ..••
For example,
(ab)6
= abobab, and (ab)O = e.
For' Z
E
H(n), Z- {Y: Y ~
(7) For S c H(n), S
=
U
ZES
(ab)5
= ababa,
zL
Z, m(S) = E
m(Z), where m(Z)
ZES
is the mass of Z.
The following lemma will be used in deriving the
function of H(n).
24
M~bius
Le.!7Jma.3.1 .2. For any pair of w.ords X,Y, with X > Y,.
there exists a 1-1 function f
If
(l )
= JZ I
(Z) J
[Y,X]
± 1 for -Z
+
[Y~X]
satis~ying
:
[Y, X] •
E
(2) If f(Z) = zt, then f{Zt) = Z.
(Note: An example is given in Appendix A5.)
Proof:
Let
words for
X
=
~hich
x x
l 2 ... x n and y= YlY2."Yn+kbe two fixed
Y
=
X, and let Z
<
zlz2 ••• zr be an arbitrary
The word zt = feZ), which we
word in the interval [Y,X].
shall define, is obtained by deleting an appropriate letter
from Z which is common to
or by inserting into
Y
appropriate letter appearing in Y.
the smallest index u
s
>
Z
an
To this end, set
0 such that Z
u
=
Y ,
u
=
o
if no such u exists,
and
t
= max{u
5
r
u
Z
Z' - f (Z) -
{
= Yu } + 1.
: Z
•
s-l s+l
if s :f 0
Z
if s
= o.
It is easily seen that Z' is a well defined word, that
Z' E [t,X] and that condition (1) holds.
It remains to show that condition (2) holds. To this end,
let s' and t
l
be defined for
zt
as sand t are defined for Z.
Then e2} easily follows from two facts about the indices s,
t, s' and t
t :
25
a)
If s = 0, then s·
b)
If s
f
p~.then t '
~
t and
s~
f O.
~
sand
s'
0..
First of all, we need to show that
x t-l = Z t - l '
Suppose Xt - 1 j
~hen
s= 0,
Z
I ' let u be the first index
t-
k > 0 such that Xk - l = Zk-l and x k j
zk' and let v = max{k:
(Note: if there exists no such k,
then we let v = 1.)
That is a
Since 1
contrad~ction.
~
u
~
t-l and u X
~
u+lZ, s=vjO.
Therefore Xt - l = Zt_l'
Since Y
t
can
be deleted from Z' and the remaining word Z is in [Y,X], we
can easily see that 1
is to show that 1
~
s'
~
s'
< t
~
t.
The remainder of the proof
is impossible.
,
Suppose that 1
z
~
s'-l • s'+l Zt-l "y"
t t Z.
By definition of
s' < t.
Since X
t-
1 = Z t - 1 and
Sl
Sf,
~
X
t-l,
>
this
Then one
of the following cases must be true: either'
SIX
> s'+lZ or
= Zv and v+ IX > tZ for some s' ·S v < t-l. The first case
implies s = s' j 0. which is a contradiction.
The second
Y
t
case implies that one of the letters zv+l,.",Zt-l can be
deleted from Z and the remaining word is still in [Y,X].
to.)
(That is : s ~ v+l
It is also a contradiction.
If s j 0, Z' = Zs- I" s +l Z ,
t.'
=
s.
Since t '
>
By definition of
8', it is easy to see that st = O.
Otherwise s = s' which is a contradiction.
26
t~,
..
THEOREM 3.1.3.
The Mobius function of the history space
H(n) only takes on the values -1, 0 and 1.
__
In particular,
when Y s X,
{(-OI)IYI-IXI
ll(Y,X)
otherwise.
Proof:
(_1)1
We only need to prove tha t when Y < X, II (Y ,X) =
Y
I-IXI.
By Lemma 3.1.2, there exists a 1-1 function f
= IZI+l or IZI-l and
from [Y,X] to [Y,X) satisfying If(Z) I
f(f(Z»
and Z
=
~
Z, for all Z
fey)}
=
€
[Y,X).
: Y < Z
{feZ)
S
Y
Thus we have {Z
X and Z
~
fey)}
<
Z
:S
X
.
The remainder of the proof uses induction based on the
total number of elements in [Y,.X} and the above facts.
induct~on
-e
The
step is
Il(Y,X) =
L
Y < Z ::;; X
Y < Z :;;: X
=
(-1) IZI-IXI
L
ll(Z,X) =
_(_l)lf(Y)I-IXI_ (1/2)[
(_l)IZI-IX!+
L
Y<Z::;;X,Z~f(Y)
+
=
E
(-1) IZI-IX']
Y<Z::;;X,Z#f(Y)
-(-1) If(Y) I+I-IXI~(I/2)[
(-1) \ZI-IXI+
L
Y<Z::;X,Z~f(Y)
+
I
(-1) If(Z)!-IXI]
Y<Z::;;X,Z~f(Y)
3.2 Basic Properties of H(n)
In this section we would like to
27
dis~uss
some basic
Let A, B, C, D be
properties Qf history spaces.
of H(n) and let X,
y~
sub~ets
Z be words in H(n).
sequences of subsets of H(n) satisfying
(1) Ai n Bi .~ Ai +1 U Bi + 1
(2) meA.~ n B.)
~
~
i
0, 1,
=
as i
0
-)-
00,
then
00
(3.1)
( -1 )
Proof~
i
{ m (A . ) +m (B • ) }
~
.
.
~
First of all, we use induction based On k to
prov~
that
k
i
E (-1) {m(A.)+m(B.)} +
(3.2)
. 0
~
~=
~
(
k+l.
)
-l)m(A nB
k
When k= 0, since m is a measure, m(AOuB )
O
m(AOoB )'
O
Assume that (3.2) is true for k
= n.
=
k
m(AO)+m(B )O
Then,
~
1
(-1 ) i fm (A . ) +m (B . )} + ( -1 ) n + m(A nB )
i=1
~
~
11
n
Hence,
=
n
i
..
n+l
I (-1) {m(A.)+m(B )} + (-1)
m(A + uB + )
n 1 n 1
i
i=O
~
=
n+l
i
n+2
E (-1) {m(A.)+m(B.)} + (-1)
m{A n + 1 nB n + 1 ).
i=O
~
~
(3.2) is true for all k = 0,1,2, ••.•
28
Then (3.1)
e-
f Q110wing from condi.tion (2) upon let ti.ngk +
00 • .
PRQPOS1TION 3.2.2.
a)
If AnB c CuD, then XoAoYn.W-:.Y c X1C,YUX;n:y' •.
b)
If AnB
Cvil, then XoA'Yn.X.BoY
Proof: a) Let Z
€
Z.1. o.1. +l Z 1., - 1
Z, where i
1
.Z ~ Y}.
2
=
Zi
·1.'
l
i1
+lZ, -1"
1.
2
~1Z.
.
1
'
1.
cun.
Z
•
1.
€
2
-
1.
Z
1
1
=
min.{j: ZJ' ~ X} and 1
+lZ,
1. 2 -
1
0
wr~tten
2
Z E X'A'Yand Z
•
1. 2
X.B.Y, there exist a P
€
X-:-C-;Y:u.X;}).Y.
The word Z can be
X.A·Y X.B.Y.
Since Z
J
1.
0
2
1
;=
€
A and a Q
=
max{j:
=
Zi •
1
B such
€
2
1 is an element in AnB 3nd hence is an element in
Since Z.
1.
X and. Z
1.
2
~
1
---X·c·Yux·n·Y.
~
Y,
it follows that
---- ---
Thus X'A'YnX'B'Y cX·C·YUX·D·Y.
b) Because of a), we only need to show that if AnB
then X·A·YnX·B·Y
~
Similarly, define i
then. +1Z. _]
1.
1.
1
2
€
X·c·Yux·n·Y.
= m1n{j:
1
~UD
;=
~
AnB.
Let Z
Z.
J
~
€
J
Q.
€
A and a Q
E
Thus, Z
This implies Z
€
By the aid of Proposition 3.2.2.b), we
29
CUD,
max{j:.Z ~ X},
2
There exist a P
~
~
X·c·Yux·n·Y.
X} and i
P and.1. +l Z 1.' -1
1
2
X·A-Y and Z ~ XoQ.y c X-B·Y.
Proof:
as
X·A-ynX·B·i,
~n1y
need to
B
P rove. that a 1 na Z
-
0-
._~--
a
:0:
na
2
is
-----~
1
and X
~
a
2
imply either X 5 a a
or X 5 a a ·
l 2
2 l
X'a
C
Proof:
1
c
Also alna
a18ZUaZ81 simply because the ineq.u a1 iZ
trivial.
ties X
a
C
Because of Proposition 3.2.2. D),
prove aloynYoa Z c aloYoa2UYoazoYlyl'
define ii
=
max{j
: qj
=
Yly
w~
Let Q
1
0a oozu
2
only need to
E
a1oynyoaZ'
I} and i 2 = max{j : qj = a Z }.
Therefore, Q
E
a 1 .y °a 2 uy °a Z °Y IyI'
These four propositions apply to spaces other than
history spaces.
0
For instance,
they apply to "free monoids",
which are like history space but have no restriction on the
repetitions of letters.
The proofs go through with no
changes.
3.3 The Upigueness of the Mass Function for H(n)
In this section we shall prove that each mass function
m defined on a history space is uniquely determined by its
corresponding distribution function M.
That
~s,
we shallo
prove the following theorem:
THEOREM 3.3.1.
H ( n ) i s a d e t e r min in g spa ceo.
(n = 2 , 3 , • • . . )
Since the details of its proof become rather involved,
we shall begin by explaining the basic ideas:
30
Let X be an arbitrary fixed word in H(n).
m(X) = M(X) - m(X The proof consists in showing
lower idea1
4
X-{X}
th~t
Then
{X}).
the value of m on the
~s uniquely determined by the values
assumed by the distribution function M on H(n).
The heart
of the proof is based upon a transitive relationship which"
we describe now.
For each subset S of H(n) we associate the collection
C(S) of all lower ideals
Z
H(n).
E
s·z
which can be formed with
The collection C({e}) is the collection of all
principal lower ideals {Z
Z e H(n)}.
Thus the distribution
function M determines, in a trivial way, the values of m on
C({e}).
We shall say that a subset Sl of H(n) determines
a·second subset S2 of H(n) and write Sl
+
S2
if the values
of m On C (Sl) and the distribution fmiction M jointly determine the values of m on C(S2).
Sl
+
Expressed more concretely,
52 if for any two finite measures
ill
l
and m2 on H(n),
the conditions
A e C(Sl)
U
C({e})
entail
Obviously, the relationship represented by " + " is transitive.
4 A set S is said to be a lower ~tdeal if S = S.
Furthermore, S is said to be a principal lower ideal if S=X for some
XeH(n) ..
31
Now, let S be the set of all words Z immediately below
i.~.,
X,
of wards Z < X for which
JZ
I
IXJ+1.
=
that the lower ideal X-{X} is a member of C(S).
Observe
Thus, in
order to show Theorem 3.3.1, it suffices to show that
{e} ""* 8.
SUPPOS2 now, for definiteness,
i
IX
=
~ O.
I
that X = x x ... x where
l 2
i
The argument to be employed in showing that
{e} ""* B tonsistB
in defining a chain of sets So' 8 , 52"'"
1
8 + with the following properties:
i 1
(1)
So = {e} and Si+l = S.
(2)
Each set S. contains words only of length j
(3)
So
The
chain of sets is defined by backward induction: The set
J
-)-
Sl ->- S 2
So and 8 i + l
j
E
...
-+
S.
1
-+
8i
3
i+l).
.~
+l ·
j+l
as the set of words Z e H(n) for which
Sj+l (l ~ j
~ i).
For example, suppose G ={a, b, c} and X = abo
8
j
~
are defined by (1), and the set Sj is defined
in terms of S
Z 'x
-)-
(0
S = {bab, cab, aeb, aba, abc},
8
2
=
Then
{ba, ea, Be},
The proof of Theorem 3.3.1 requires two kinds of
arguments, one to show So ->- 8 , and a second to show 5.
1
J
5'+1 for 1
.J
~
j
~
i.
(The latter is unnecessary when X
in which case i = 0.)
=
-+
e,
Both kinds of arguments are immediate
consequences of a lemma.
Therefore we sha11digress for a
moment and discuss some preliminary results that are required
for deriving these two arguments.
32
e
For any word X
denote the
i~
c~llection
H(n) and any subsetAof
of words
wh~ch
G~
let TAX
can he created by
inserting into the word X a single letter from the set A.
O'or instance., if X =ah and A =' {~,
ach, aba, abc}.)
cL
then TAX = {cah,
When A is empty, TAX is defined as the
empty set.
Suppose X = xlxZ .•. x
i
is a word of positive length,
S is a nonempty subset of H(n), and a is a letter in
distinct from the last letter of X.
e
The word X can he
expressed uniquely in one of two forms:
I.
X = X ahab •.• ab.
u
II.
X =
X
u
(x :/:a and x 1-h if u>O)
u
u
babab ..• ab .
(Note: b is assumed to be the last letter of X.)
For the definition which follows, let X' be the word formed
from X by inserting the letter b or a into X immediately
after X
u
according as I or II holds.
We shall say that the
triplet (X, S, a) satisfies Condition R if
(1) S
TeX,
c
(2) X·a
(3) X'
E
Eo
S,
and
·S.
Now, let us assume that the triplet (X, S, a) satisfies
Condition R.
For j = 0,1, .•• , let D(j) denote the set
S'.b.(ab)j u X.TA(ab)j' where S' is the set {Z : Z·b
and A is the set {c : X·c e Sand c :/: a}.
e:
S}
Some properties
of D(j) can be described as follows:
Observe
that each word in D(j) is formed by inserting
33
a letter () fG into X· (a b) . • therefore
J'
(.1.1)
D(j)
Since X'
E
c
X.(ab) ..
J
for
J ::
0, 1, ...
S, the following relation holds.
X.(ab)j+2 < X'. (ab)j
S' ob(ab)j c D(j).
E
It implies
(3.4)
for j
=
0,1, ...
It is not hard to s,ee that each word in S'ob'(ab)j+1 is
less than some word in S' obo(ab)j and, each word in XoTA(ab)j+l:
is less than some word in XoTA(ab) ..
,
(3.5)
D(j+l)
c
Thus,
J'
D (j)
for j = 0, 1 •.•.
(3,6)
In fact the converse of (3.6) is also true.
prove the case when j
is even.
(When j
We now
is odd, the proof
Is similar to what we shall describe.)
Since Xo(ab)j+lnD(j)
-------
_ _--
can be expressed as UQED(j) Xo(ab)j+ln Q, we only need to
..
show that for each Q
X-:-( a b ) j ;2 uiiU+Tf.
E
-,
D(j), Xo(ab)j+ln Q is a subset of
Let Y = X (a b ) j .
0
Ea c h
W0
r d Q in D (j) can
be expressed uniquely as one of the two forms:
ILY·c.
IfQ is of the first form l by Le"mma 3.2.4,
-
"---.---.
e.
X' (a b ) j + 1 n Q :::: Yk' k Y'.anY k • c • k Y .
cYk'kY'a.b u Yk -c'. k Y·a
.
:::: X.(ab)j+2 u Q·a
c X'Cab)j+2 u D(j+l).
If Q is of the second form, by Lemma 3.2.3,
X.(ab)j+l n Q :::: Y·a n Y·c
== Y·a.c U Y.c.a
c i-(ab)j+2 U D(j+l).
Therefore, the converse of (3.6) is true.
That is to say,
for j==O,1,2, ....
(3.7)
Suppose Z is a word in HCn), for which S'Z is
The set S·z can be expressed as X'a.Zu D(O)·Z.
d~fined.
By formula
(3.7) and Lemma 3.2.2,
for j=O,l, . . . .
as j
+
00.
Furthermor.e, 1l1(){:(abTj~in]r(rf'Z):<;;MCX'(ab). )+0
J
Thus, by Lemma 3.2.1, we obtain the following lemma.
35
Le.mma. 3. 3. 2 .
with last
lette~
Suppose X is a word of
(3.8)
~nd
b, S is a subset ofR(n),
efG distinct from b.
Condition R,
pos~t~ve
If the triplet
length
a is a
letter
(X, S. a) satisfies
then
m(X.i)
for all Z for which S.Z is defined.
Le.mma. 3.3.3_. Let X be a word of positive length and let
S be a~ubset of TGX.
Suppose that the triplet
(X, S, a)
satisfies Condition R,
for each a with X·a e S.
Then S'
Proof:
When {a : X'a e
Therefore 8'
Sf·b.
+
S}
= .,
the setS can be expressed as
S.
Now; let us assume that {a : X'a
denote the number of letters in {a
When N(S)
E
T•
5}
and let N(S)
: X·a e S}.
1, the set 5 can be expressed as S'-bu{X.a}.
Therefore, for j
triplet
S.
+
0,1, ... , D(j)
=.,
=
S'·b· (ab) .•
J
Si.nce the
(X, S, a) satisfies· Condition R, by Lemma 3.3.2,
co
m(S'Z)
L (-l)j{m(X' (ab) "+l'Z)
j =0
J
+ m(D(j).Z)}
co
=
Si
It C
e
L (-1) j { m (X. (a b) "+ l' Z)
j=--=O
J
X--:- (ab) j + 1 - Z
e
+
m( S
C ({ e}) and S'·b· (a b) j •
i •
b. (ah) " • Z)} •
J
i
E
C ( S ' ), S' +
S.
The remainder of the proof uses induction based on N(S),
Let N(S) = k.
X-a
z , ... ,
The set S can be expressed as S'.b u {x.a ,
l
x-ak}'
Since (X, S, a ) satisfies Condition R,
1
by Lemma 3.3.2.
36
4It
.e
CQ
ttl(S "Z) =I: ( ... I)j{m(x·(ab) . • Z)
j=O
1
J+1
Observe
t hat D (J
.) =.5' b ~ (a 1 b) j
0
=
induction assumption,
=
=
5' + D(j), for j
-e
D' (j)
+
D(j).
+
=
k-1.
By the
It is easy to see
1, and D'(O) = 5'.
~
0,1, ....
C04olla4r 3.3.4.
Proof:
Furthermore, N(D(j»
D(j-l), for j
length one, {e}
i s a subset
Z, •.. ,k, the triplet (X" (a 1 b)j' D(j), ai'
satisfies Condition R.
that D'(j)
u X' TA
(al b ) j
The set {a : Xo (a 1 b) j o a€ D(j)} ={a " ' . '
Z
of TG(X. (alb)j).
ak}' and fo~ i
+ rn(b'(j).'Z)}.
That is to say 5'
Therefore,
+
S.
For any setS consisting of words of
S.
Suppose S = {a}, for some a in G.
Obviously,· { e} + S.
The remainder of the proof uses induction based on the
number of words in S.
Let S.=
{al~
a
2 , •.. , ak} .Sincem is
* measure
and
with the aid of Lemma 3.2.3,
k.
m(SoZ) = m( uaioZ)
i=l
k
k
= m(a 1 o Z) + m( u a.oZ) - m(a1oZ n
i=2
1
u aioZ)
i=2
k
=
m(a 1 o Z) + m({a 2 , ... ,a k }oZ) -me u a 1 Z n aioZ»
0
i=2
=
m (a 1· Z)
+
m ({ a 2 ' ••• , a } • Z)
k
37
+
-m ({ a a ,a a ,.
1 Z 2 1.
0.' a l a k ,aka l }.
Z).·
The set {a l a 2 , a 2 a 1 ,
... , a 8 , aka } is a subset of T
l
1 k
8
G l
,
= 2, ••• , k, the. triplet (aI' {ala2.aial, ••• ,alak'
and (or i
aka }, ail satisfies Condition R.
l
.• ~, a } +
k
Thus' {a ,
2
{ala2,82al, •.. ,alak,akalLNow, by the induction assumption
Ie}
{8 Z , .•. ,a }.
k
+
We can conclude that Ie}
Proof of Theorem 3.3.1:
Sj
+
S j +1
for 1
~ j
~
It remains to show that So + 51 and
Sj+l
=
TGX j
-
Since So = Ie}, and 8
i.
of words of length one,
Corollary 3.3.4.
5.
+
So
+
a
consists
Sl immediately follows from
It is easy to see that Si+l
{X j + l }, for
1
~ j
~
i-I.
=
TGX and
Therefore, the
triplet (X., 5.+ , a) satisfies Condition R, for each a with
J
J 1
X.oa
J
E:
S'+l
J
(1 ~ j
~ L)
.
By Lemma 3.3.3, 5.
J
+
8'+1' for
J
-
lS;j~i.
Remark:
A similar procedure c.an be applied to prove that
a free monoid is a determining space.
3.4 An Inversion Formula for H(3).
t~o
The history space generated by
1.4) satisfies Theorem 2.3.1. (1).
letters (see Figure
From Theorem 3.1.3, we
know that the M~bius function is defined as
\ley, X) = (_l)IYI-IXI,
y
~
X.
All suprema in Theorem 2.3.1.(2) are bounded by 1.
Therefore,
Corollary 2.3.4 holds and (2.5) provides an inversion
formula to express m in terms of M for H(2).
is clear and is of a simple form.
38
This formula
One would expect that
e
for mu 1 a ( 2 • 5) a I soh old s for H (3) .
not the case.
..
wesh~ll
ror H(3),
Un for t u.n ate 1 y, t his is
find an inversion formula
to compute the values of m from M.
However, the formula
that we shall describe is not as simple as formula (2.5).
lette~s.
In H(3), e consists uf three
cto represent these
assume that x
thre~
= b.
lXI
We use
different letters.
When x
IX
~,
band
We shall
\ ~ b, a similar proof
holds.
Lemma 3.4.1. If Q
Proof:
X-c, then Qe ~eX -
<
Since X-c = {X-c} u Te(X-c),we only need to show
that if Q e Te(X-c), then Q e TeX then Q
Q e Tee ,
X
{X-c}.
+e
E
{
{X-c}.
If X = e and
be, a c , c b , c a} c { b, a} = Tee· - {c}.
If
and Q e TeC, then Q must be of one of the following
forms:
I.
for some j ,
Q = X-Ij
j+l X'C
0
:0;
j
:0;
I X I,
II. Q = X-c-a,
III. Q = X-c·b.
When I holds, Q < X.-l-.X
J
Q < X-a
TeX -
E
{Xoc}.
TGX - {X-c}.
E
J
When I I holds,
When·III holds, let i =max{j:, xj;=a}
so that X ~ Xi-(cb)IXI_i. or X = Xi-(bc)IXI_io
imply that Q < X - i + X-c-b
2
i
PROPOSITIQN 3.4.2_
TeX -
E
If X
+ e,
{Xoc}.
Both cases
Thus,
Q~
-.-._-TGX-{X-C}.
then
00
= .L
J=O
( -1 ) j { M (X (ab) • +1 • Z)
0
J
39
+
m ( (T e X I X I -1 - { X } ) - (b a) j
+1 - Z) }
•
Pro Q f :
Let S = TeX - {X ~ c },
I t i s not ha r d to a e e th at
By Lemma 3~3,2~
(X, S, a) satisfies Condition R.
00
4. (-l)j{m(X.(ab) '~l~Z) +
j=O
J
m(S'Z) '"
where D (j)
=
Recall that.S
S I . b' (ab) j
l
X. T (ab) j .
A
Y'b e S }= TeXIXI_l~{X}and A
= {y
~b
dee, X·d e S, d
U.
m(D(j)~Z»),
and d I a}
=..
There~ore,
= a·Z
u b •Z•
=
{d;
Thus formula (3.9) is obtained.
When X = e,
a·Z n b·Z
(TeX
-
{X'c}) ·Z
Since
= {ab,ba}'Z and triplet (b, {ab,ba}, a) satisfies
Condition R, by Lemma 3.3,2, We obtai.n
00
(3.10)
,
L (-l)J{M«ab) '+l·Z)+M«ba) '+l'Z)},
J
.
J
j =0
PROPOSITION 3.4.3.
(3.11)
Proof:
m(X·c)
=
For X
E
H(3),
M(X) - m(X) - m(TeX - {X·c}).
Observe that the principal lower ideal X can be
expressed as {X} u TeX. By Lemma 3.4.1, TeX
TGX - {X·c}.
=
{X·c.} u
Thus, we obtain (3.10).
Finally, we would like to find m(e).
lab, ba, ac, cal
:=
Tea,
m(e) = M(e) -
mea
u b
u
40
c)
Since i- nib ·c}
:=
= M(e) - M(a) - m(b u c) + m(TGa).
The triplet (a, TGa J b) satisfies Condition R
by
Le~ma
1
Therefore~
3.3.2,
co
E (-l)j{M(a.(ba)'+l) + m(D(j»},
j=O
J
where D(j) = {b,c}.a.(ba)jU a.T{c}(ba)j= TG(ab)j+I~{(ab)j+2}'
Combining the above formulas, we obtain
co
(3.12)
m(e ) = M( e ) -
Formula (3.9),
.e
L
•.
j =0
(3.10),
( -1 ) J
{M « a b) . +1 ) +m ( TG(ab) . - { (ab) . +1 }) } •
J
J
(3.11.) and (3.12) can be used in a
recursive manner to obtain values of m from M.
(3.11)
pro~ides
J
Formula
a way to express m(X.c) in terms of M(X),
m(X) and m(TGX - {X.c}).
The value of m(TGX - {X·c}) can
be computed by using (3.9) and (3.10).
Using (3.l2),. (3.9),
and (3.10) we can find the value of m(e).
Let M(X) = 1/(3 IX !), for X
Example 3.1.
to find m(X).
E
H(3).
We want
First of all, we use induction based on IXI
to prove that
0/ 4) I X I +1
I X 1+ I Z I +.1
3 -2
m
« TGX
- {X. c }) • Z) =
3
When IXI = 0,
co
L
( -1 ) j { M ( (a b ) j
j=O
41
+1 • Z +
M ( (b a) j
+1 .Z) }
00
l:
j::::O
(.,.1) j
-3i-Z·t +J' +'1
+
-T"IZ~~c-+~'~+-l
3
J
2
3 1ZJ + 1
3
.4
::::
---l=li3/4) JXI+l
3!Xj+IZ!+1
2
---;rZl~
•
Ixi = k.
Assume that the statement is true for
IXI
=
k+l.
Then
<Xl
=
,
k (-l)J{M(Xo(ab)'+l' Z ) + m«TGXjXI_l-{X})o(ba)J'+l°Z)}
j=O
J
00
,
}~ (-1) J [
=
~
1
3 k + 1 + I Z I + j + 1-
l
j=O
4-2 (3/4)
k~~_
=
+ (
3- 2
( 3 ' k +1
-4)
1
) 3k+ 1 -I- Iz I +j
+1
}
e.
3_2C3/4)k+l+l
00
L (-l)j
3k+l+IZI+1
By
Now, let
.3k+l+IZI+l
j=O
3-2 (3/4)IX 1+1
(.3.12)
m(e)
3IXI+IZI+1
and (3.13),
:::: 1 -
00
,
k
(-1)J
1
------
3 j +1
j=O
_. 1 - -
1
(1 + 3 )
3
4
3
(-1)j
L --;-j=O
3J
00
I
_
_ _ ..L._ _
3-2(3/4,j+l
'r
3 j +1
2
+ ---
4
:
_l-l),j
j=O
4J
£.
242
5
:::: 1 - - . - - + - - . - '.3
4
4
5
42
e·
By (3.11 ) and (3.13)" for: X
m(X)
I
= 3}xJ+1
=
=
2
4]XI
-
f
,
e
-
m (X I X J -1)
m (X
3 -2 GJ1!LL! xl
3 xj
j xl-I)
2
2
4'Xj
4 1X1 - I
.. .
+
_2_(I+(_L) I X I)
5
4
2
5
(±)+
if Ixi is odd
=
2
5
=
.e
_2_
5
~(I~(-!-)IXI)
. 5
if I X I is even
IX I
C_L)
4
Therefore, m(X) = .;
Remark:
4
(-t-) lXI,
for all X
€
H(3).
When n ? 4, a similar procedure can be applied to
find an inversion formula for H(n).
However, the formulas
become very tedious, and we will not discuss them.
43
CHAPTER IV
THE FINITE PARAMETER SPACES
The space discussed in AppendixA1 is not a
space.
determin~ng
However, it can be shown that for this space, each
expr~ssed
m(x) can be
as a linear combination of "M(x), x E X,
and a parameter m(O,l).
determined by M(x), x
In other words, each m(x) is uniquely
X, and m(O,l).
E
In this chapter, we
would like to consider a necessary and sufficient condition
based on the Mobius function for a space to be a finite
parameter space.
By a finite parameter space X, we simply
mean that each m(x) can be expressed as a linear combination
of M(x), x
i
~
E
X, and a finite number of parameters m(a ),
i
1, 2, .•• , n (with coefficients independent of m).
First
of all, let us consider a particular class of one parameter
spaces.
4.1 a-lower finite spaces
Simons (1975) considered a class of spaces which he
referred to as a-lower finite spaces.
Definition 4.1.1.
A partially ordered space X is called lower
finite if for each x
E
X, the set {y: y
~
x} contains only a
finite number of points.
pefinition
4.1.~.
A partially ordered space X is called 0-
lower finite if there exists a sequence of partitions
x
=:
A
k
+ B satisfyi,ng
k
(1) A
> B
k
k
(i.e;, each point. of A
k
exceeds each. point
of B ),
k
(2) A
t X as k
k
(3) each A is
k
+
00,
lowe~
finite.
Suppose X = A + B + C with A > C > B, where A, B or C
may be the empty set.
Define whenever the number of summands
is finite:
~
ll(y, A)
y
ll(B, x) = -
-:;'z
ll(Y, z),
YEA' ,
ll(z, x),
X
<A
~
E
B1
,
B<z~x
ll(B, A) = -1 +
~
ll(y, x).
B<x,y<A
The a-lower finite spaces can be classified into three
different types:
X is lower finite.
Type I:
Type
X is not lower finite and (B + , A ) = 0, for
II~
k l
k
an infinite number of k.
TypeI!I: X is not lower finite and (B + , A )
k
k l
=
0, for
a finite number of k.
Type I and type II a-lower finite spaces will not be
discussed here, because the set {y: ll(y,X) ~ O} is finite,
for each x
E
X.
Therefore, all type I and type II spaces
are determining spaces and formula (2.2) provides an inversion
formula to express m(x)
in terms of M(x), x
45
E
X.
Type III spaces are the Ones of special interest,
since the mass function m is not necessarily uniquely
determined by its distribution function M.
X
E
A function 6(x),
X, is introduced to give a necessary and sufficient condi-
tions for a type III a-lower finite space to be
space.
For x
E
~
determining
X, let
The function S is well-defined in the sense that the
values of Sex) does not depend on how one chooses k. Define
if y
E
AI'
ify
E
B .
1
The function v has the property that, for each x
number of points in {y: v(y,x)
~
O} is finite.
X, the
~
With the aid
of this property, the following theorem is obtained.
THEOREM 4.1.1.
(Simons)
Let X be a type III a-lower finite space. Let M be a
distribution function corresponding to the mass function m(x),
X
E
X, and let Y
= m(B l ).
(1) M and Y together determine m.
."-
E v(y, x)M(y) + y6(x),
yEX
m(x)
(4.1)
In particular,
X
E
X.
(2) X ~s a determining space if and only if E<
x~xO
for some
X
o
E
X.
If X is a determining space, then
46
18(x)
\=00,
y - lind
4 )v(Y. x)}1(y)J
I:
/
.
B <x <AI yEX.
k-+ oo
k
is uniquely
L
Ja(x)J}
Bk<X<A 1
.:
determined~
Let a be a minimal point of AI"
Observe that the
parameter m(B ) can be expressed as M(a) - mea) and (4.1)
1
becomes
(4.2)
E v(y,x)M(y) + a(x)M(a) - m(a)a(x).
m(x)
yEX
That is to say each m(x) can be expressed as a linear combination of M(x), x
Th~refore,
E
X and a parameter: a mass of some point.
a type I I I a-lower finite space is a one parameter
space.
4.2 Finite Parameter Spaces
The M~bius function defined on a type I I I a-lower finite
space X has the following representation:
p(y, x) = v(y, x) +
(4.3)
~(y,
where, for each x e X, the number of y
is finite.
a)B(x),
E
X such that v(y,x)fQ
We have found that equatidns (4.2) and (4.3) are
closely related to each other in a way that has rtothingto do
with the a-lower finite partial ordering.
THEOREM
4.2.1. If for each mass function m defined on a
partially ordered space X,
m(x) -
L v(y,x)M(y) +
YeX
47
n
L m(a.)B.(x),
i=l
J.
J..
for x
E
X,
then
(4.5)
}l
(Y ,x)
;;::
V
(y , x)
+
n
~
l-l (y, a . )
i=1 .
1..
6. (x) ,
for x,Y
E:
X~
1.
where
functions v and 6 , i=1,2, ••• ,n, are independent
i
th~
(1)
of m,
(2) for each x
of y
E:
X, there exists only a finite number
E:
X such that v(y,x).j O.
Conversely, if (4.5) holds, then (4.4) holds for all mass
functions defined on X.
Suppose (4.4) holds for all mass functions defined
Proof:
on X.
Let gx(y) = ll(Y,x) - v(y,x) -
let A(x)
=.
{y: gx(y) :f OJ.
E~=lll(y,ai)ei(x), and
In order to show that (4.5)
holds for all x and y, it suffices to show that A(x) =
for all x
E
~,
X.
Suppose A(x ) :f
O
4>
, for some
X
o
E
X.
Since the set
{y: v(y,x ) 1: O} is finite and A(x ) is a subset of
O
O
+
+
n
{y: v(y,x ) :f O} u X
o u u.1= 1 a.1 , there exists at least one
O
Let YO be a max1mal point of A(x O).
if x
=
YO'
otherwise,
.f
1.
X
>
-
YO'
otherwise.
The functions m and M form a mass function and,its distribu48
tion function respectively_
For each i=1,2, ... ,n,
the set {y:
a subset of [Yo,a 1.J, therefore,
L:.
YE
j..l(y,.ai)M(y)
t-
O}is
X1jl(y,a.)M(y)J'<oo.
1.
.
By Fubini's theorem, we have
E jJ(y, a.)M(y)
1
YEX
=
m(z)
E 1I (y , a 1. )
1:
yeX
z ~y
=
m(z){ L
jJ(y, a.) }
1
y?z
1:
ZEX
=
m(z){
L
z~a
=
m
.
1
= LyeX
a. ) }
1
(a,) .
1
Similarly, the set {y:
thus, m(x O)
il(y,
1:
z~y ~ai
jl(y,xO)M(y)
C
.
O} is a subset of [YD,xOJ,
e-
Substituting m(x O) = LyeX
il(y'XO)M(y).
lI(y,x )M(y) and m(a , )
1
O
t-
Ey E XjJ(y,a.)M(y)
in (4.4), we obtain
1
n
E jl(y'XO)M(y)
yeX
=
v(y,xO)M(y)+ 1: { L jl(y,ai)M(y)}Bi(x )'
O
yEX
i=l yEX
L
This implies
(4.6)
I
g
yE X
Xo
(y)M(y) =
Observe that when y 'A(x )' gx (y)
O
and y
t-
o
Yo' M(y) = O.
I
yeX
o.
= 0
and whe~ y
e A(x )
O
Thus,
gx (y)M(y)
0
This riontradicts
(4.6).
That is to say, A(x)
=t
for x e X.
.e
49
\---:j
Conversely, suppose (4.5) holds for all ~, Y
E
X.
Th~n
E. v(y,x)M(y)
YEX
r.
(4.7)
m{-z}
Z 5.y
Ii
=
E m(z) { I: (J.l(y,x)
ZEX
y~Z
E J.l(y,a.)f.L(x»} •
. 1
'J.
J.
J.=
Since
n
E (J.l(y,x) -
y~Z
r.
J.l(y,a.)/3-.(x»
i=l
J.
J.
n
=
E (E
J.l(y,a i »)8 (x)
i
i=l y~z
E
y~Z
=
1
if z = x an¢( x
-13.]. (x)
if z = a . and x
J.
1-13. (e .. )
if z - a
0
otherwise,
J.
:f a.J.
J.
i
:f- a i
= x
equation (4.4) follows.
THEOREM 4.2. 2.
..
Suppose the Mobius function of space X
has the representation as shown in (4.5).
d~termining.space
c
1
'
cZ'
(1)
(3)
••• ,
c
n
if and only if there exist no constants
such that
t'Y5.xlI:~=l
n
Then X is a
Ei=l cil3i
ci13i(y)l<
00
for x
E
X,
for x
E
X,
is not identical to z~ro.
50
Suppose the.re eXi.at constantsc , c2~
1
.Eroof:
(l)~
satisfy
negativ~
tlve and
Then
6+ and 6
(3) ~
(2) and
+
t3
Let
and
part of the function
dif£er~n~
are two
.-.
6
denote the posi-
Eni =l c. S.
....
1.
1.
respecti~~ly.
mass functions which give
rise to the same distribution function.
not a determining
••. ~ Crt which
Therefore, X is
space~
Conversely, suppose the mass funttions m
1
and m
2
give
From (4.4), we
rise to the same distribution function.
obtain
flm (x)
n
E flm (a . )
~
•
J..=1
1.
S. (x) ,
where l1m=m -m •
2
1
1.
Since flm is not identical to zero,
Th~s
identical to zero.'
condition (3) holds forc.=flm(a.).
1.
.
1.
Furtherm.ore,
n
E
I
E flm (a . ) S. (y)
y:;;x i=l
I
E
=
I flm (y) I
< 2 M(x) . <
co,
y:;;x
1.
J..
and
n
E
y:;;x
E flm (a . ) S. (y) =
• =]..
J..
Thus, conditions
J..
(1) and
functions v and
s.J..
= O.
(2) follow.
Theroems tl.2.1 and 4.2.2
finite parameter spaces.
I flm(y) = M(x) - M(x)
y-:;,x
J..
give some general results about
Of course, we know that the
depend in some concrete wayan the
structure of the space X.
Howe~er,
at our current level of
generality, it does not appear to be possible to find
explicit formulas for these functions.
51
Recall that these
functions can be defined explicitly for a
finite space.
~ype
III a-lower
We shall now consider a "class of spaces for
which the M~bius functions have representations as shown in
(4.5) and the func tion v and, B. can be defined in an analogous
1.
way as they are defined for
a
type III a-lower finite space.
4.3 A class of Finite Parameter Spaces
In this section we shall consider a class of
space~
which has the property that each mass function can be
uniquely determined by its distribution function and n
parameters, where n = 0, 1 or 2.
Exa~p1e
4.1.
Let X be the space shown in Figure 4.1.
Figure 4.1
For each x
E
X, let
if x = (n,l)
(n=1,2, ... )
otherwise,
and "let
B (x)
2
=
.
(_1)n-1
i.f x =
(n,1)
{ (_1)n(n_1) otherwise
(n=1,2, .•• )
It can be shown that the M~bius function has the following
representation:
52
- - - - ..
--------------
The fune tion v satisfies
the required property tha.t
~
for
x to X, Iy: v(y,x)
f
O} js a finite set.
Theroem 4.2.1, m(x), x
E
X is uniquely determined by M(x),
2.a~h
x
£
Therefore by
X and the parameters m(l,l) and m(l,2).
Let X be the space shown in Figure 4.2.
Figure 4.2
For n=l,2, ..• , let s(n,j)
that for x,y
= (_l)n-l.
It can be shown
X,
€
II
(y,x) =
Similarly, for x
E
V
(y,x) +
II
(y, (1,4»S (x).
X, m(x) is uniquely determined by its
distribution function and a parameter m(l,4).
Both spaces are not
a-low~r
finite spaces.
However,
they have some properties which are analogous to the
properties of a-lower finite spaces.
We describe tbese
now: .
(4.8)
There exists a sequence of partitions
53
X=
Ak+B +
k
~on-empty)
+Ck+D k (all sets are
(1) A
k
Ck+D k , Uk
>
>
satisfying
C and nooeof the polnta o~
k
Bk relate to any point of D ,
k
(2) Ak+B
k is lower. finite and Ak+B k t X as k
(3) the mi It i mal poi. n t s
0
+
oo~
f A k and Bk .don 0 t r e 1 ate
to each other,
(4) the maximal points of C andD do not relate
k
k
to each other.
In example 4.1, for k=1,2, ..• , let B =·{(k,l)},
k
Ak = {(n,j): 1 ~ n ~ k, j=1,2,3} - B , D = {(k+I~3)}, and
k
k
Ck
{(n,j): n
~
k+l, j=I,2,3} - D .
k
partitions X = Ak+Bk+Ck+D
k
Then the sequence of
satisfies (4.8).
In example 4.2, for k=I,2, •.. , ·let
= { { (n , ~ )
: n :
2k ,
2. k: 1 ,
j=3,4}
{(n,J). n - 2k, 2krl, j=I,2}
if k is odd
if k is even,
B = {(I,l),(l,2)} and B + = D ,
l
k
k l
A
k
C
k
= {(n,j): n
~
2k-l, j=l,2,3,4} - B ,
k
= {(n,j): n
~
2k,
j=1,2~3,4}
- D .
k
Then, the sequence of partitions X : Ak+Bk+Ck+D
k
satisfies
(4.8).
Now we shall develop a theory appropriate to spaces
which have property (4.8).
..
(For the remainder of this
section, we assume that the space X satisfies (4.8).)
We shall begin by poining out some immediate consequences of
condition (4.8).
It is easy to see that X has to be locally
54
finite.
From condition 4.8
ot
k ~l~ no point
(l)~
L~t
(4».
conse~uence
X(a
Yl and Y2 be two
It is clear that P(Yl'x) =
maximal point of Ck (or Uk)'
E
know that for each
B exceeds any point of A and no point
k
k
ofC k exceeds any point of D ·
k
P(Y2'x). for x
WP
pf condition 4'.8 (1) and
Similarly, let Y and Y4 be two minimal points of
3
Ak (or Bk ).
For each x
bf 4.8 (1) and (3».
E
X, P(x'Y3) = P(x'Y4)
In the rest of
thi~
(a consequence
section, we shall
use a , b , c and d to denot.e a minimal point of A ,
k
k
k
k
k
a minimal point of B , a maximal point of C and a maximal
k
k
point of D respectively.
k
PROPOSITION 4.3.1.
(4.9)
Proof:
We shall divide the proof into four parts as follows:
I:
x
€
B ,
k
Y
E
Dk ·
II:
x
E
A ,
k
Y
E
Dk ·
III: x
E
B ,
k
Y
E
Ck ·
IV:
E
A ,
k
Y
E
Ck ·
1.
=
O.
x
When x
Therefore,
B and Y
k
E
D ,
k
E
(4 .. 9 ) hold.s.
1
II.
When x
E
A and Y
k
E
D , p(y,b )' = O.
k
k
In order to
prove (4.9), it suffices to show that p(y,x) = -v(y,ak)p(dk,x).
If y is a maximal point of D , then p(y,a ) = -1.
Therefore,
k
k
Since for each y e D ,
k
[y,x]
n t
k
= • , we can use induction based on the number of
55
"
In partic.ular,
lJ(Z,x)
L
y<z~x
lJ(Z,x) -
=
-
=
(1+
r
lJ(z,ak))p(dk~x)
y<z<a
III.
p (z, x)
k
When x € B , Y € C , p(dk,x)
k
k
suffices to show p(y,x) a maximal point of C •
k
=
-p(y,bk)lJ(ck~x).
Then v(y,b )
k
=
-1.
Similarly, the remainder of the proof uses
on the number of points in [y,x].
v(y,x)
here.)
=
0
=
-p(y,bk)p(ck,x).
O.
Therefore, i t
Suppose y is
Thus,
induct~on
based
(Recall that when y
\l(Z,x)
y<z~x
E
z~x,
lJ(z,x);"
z€B
k
56
D ,
k
Thus, induction can be used
We have
l;
E
1:
\l(z,x)
y<z,z€Ck+D k
IV.
In II we proved that formula (4.9) holds for x € A
k
and y€ D "
k
Now we can use induction based on the number of
points in [y,x] to prove that formula (4.9) holds for y
Let y bea maximal point of C .
k
Since p(y,b )
k
~
-1
=
E
C .
k
p(y,a ),
k
we have
p(y,x)
Finally, it thus follows from the induction assumption that
p(y,x)
= -E
p(z,x)
y<z~x
= -
E
P(2;,X)-
z~x,zEAk+Bk
(1+
E
E
p(z,x)
y<z,zECk+D
k
p(z,bk»p(ck,x) + {(l~
y<z<b k
&
y<z<a k
57
~(y,ak»+
.
e
-(1+
L
lJ(y,bk»}}!(dk?x)
Yk <z <b k
PROPOSITION 4.3.2.
(4.10)
det
€
A
k
+
B •
k
lJ (ck+l'x)
( 11(d
Proof:
Let x, y
k + 1 ,x)
From formula (4.9), we obtain
.e
Since
-(l1(Ck+l,ak)-l1(Ck+l,bk»)
- (lJ (d
k +l ' a k ) -11( d k + l ' b k ) )
formula (4.10) follows.
58
Then
Now spaces which satisfy cortdition (4.8) can be
classified into three
Type I:
di£f~rent
Fot each k
on k)
~
types:
1, there exists a q
suchtha~
~
1 (depending
the matrix
Type II: X is riot a type I space and the matrix
(4.11)
is singular for infinitely many k.
Type Ill: The matrix (4.11) is singular for only
finitely many k.
It follows from Proposition 4.3.2 that for q
~
1,
Thus the distinction between type I and type III and the
distinction between type II and type III are independent of
the particular sequence of partitions.
Furthermo~e,
follows £rom Proposition 4.3.1 that if there exist i
j
~ 1 such that ~(ck+i,ak)
~(dk+i,bk)= 0,
then for y
=
E
p{dk+i~ak)
=
~(ck+j,bk)
it
~
=
Cmax(i,j)+k + Dmax(i,j)+k'
59
1 and
p(y,a )
k
~
p(y,b )
k
~
O.
Therefore the distinction between
thepart~cular
type I and type I I is independent of
sequence
of partitions.
Since type II and type III spaces are unaltered by
adding or deleting partitions, we can restate them as follows:
X is not a type I space and the
Type II:
is singular for k
(4.11)
matr~x
1.
~
Type III: The matrix (4.11) is non-singular for k
THEOREM 4.3.3.
A type I
~
1.
space is a determining space
and formula (2.2) is an inversion formula to express the
mass function in terms of its distribution function.
Proof:
In order to prove this theorem, it suffices to show
that the number of
non-z~ro
summands in (2.2) is finite.
There exists a q
p(ck+q,a k )
+B +
k q
1 such
~
=
and b
k
E
Ak+q+B k+ q , it follows from (4.9) that
Since y
we have y
Therefore,
E
Ck+D .
k
~
Ck + q +D k.+q '
Again, it follows from (4.9) that
the number 6£ non-zero summands in fo~mula (2.2)
if fini.te.
Now, let us consider type II spaces.
a type II space, there exists a k
60
~
Observe that for
1 such that
for q
)l(dk+q,B )
k
~
for i
It follows from
1.
:?-
= 0, for some"q
~
that i f )l(ck+q,a )
k
1, then )l(ck
=
.
.,a ) = )l(d
,2)
+q+l
k
k +q+i
k
= )l(dk+q,B k ) = 0, for
1, then )l(ck+q+i,b k ) = )l(dk+q+i,b k ) = 0 for i ~ 1.
1.
some q ~
(4.9)
Similarly if )l(ck+q,b )
k
Therefore, one of the following two cases is true:
Case 1:
Case 2:
For each q
~
)l(ck+q,a k )
is not zero.
For
eac~
q
~
1, at least one of )l(dk+q,a ) and
k
1, at least one of )l(dk+q,b ) and
k
is not zero.
We shall only discuss case 1, since case 2 follows
similarly.
We assume without loss of genarality that for
each q >. 1, at least one of )l(dq+l;a ) and )l(c + ,a ) is not
q 1
1
1
zero.
For each x
seX)
{
E
Ak+B , define
k
"(ck+l,x)/"(ck+l,al)
if )l(ck+l~al) ~ 0,
ll(dk+l,x)/)l(dk+l,al)
if 1l(d + ,a ) ~ O.
k 1
l
PROPOSITION 4.3.4.
The function
S
is well defined in
the sense that
(1) if 1l(dk+1,a1)~ 0 and ll(c k + l ,a l ) ~ 0, then
)l(ck+I,x)/ll(ck+I,al)
= ll(dk+L,x)/ll(dk+l,aI)'
(2) Sex) does not depend on the choice of k.
Proof:
-----
Condition (1)
and the fact that the
is an
~atrix
immediate consequence of (4.10)
(4.11) is singular for k
61
~
1.
.e
Now, we want to show that condition (2) holds. Let q?-l.
Suppose p(ck+q+1,a l )
=
~
O.
By Proposition 4.3.1,
S(x).
s (x) •
PROPOSITION 4.3.5.
Proof:
Let X be a type I I space.
If x
E:
By proposition 4.3.1, we have
Define v(y,x)
=
whenever x
€
Ak+B
whenever y
~
x and y
number ofy
€
X.
k
~(y,x)-~(y,al)S(x).
and y
~
E
C + +D + .
k 1
k I
Then v(y,x)
= 0
Likewise, v(y,x) = 0
-1· Thus v(y,x) # 0 for only a finite
Therefore, the following theorem is an
immediate consequence of Theorem 4.3.1.
62
THEOREM 4.3.6.
Let X be a type II space and let m and
dist~ibution
M be a mass function and its
tively.
Then M and meal) uniquely
function respec-
determin~
m.
In parti-
cular,
E
m(x) =
(4.12)
v(y,x)M(y) + m(al)B(x),
X.
X
E
l
to b .)
l
yEX
(Note: When we are in case 2, ·we simply
THEOREM 4.3.7.
a
A type II space X is a determining space
=
if and only if Ix<y P(x)
-
ch~nge
0
00, for some Yo
E
X.
If X is a
determining space, then
(4.13)
=
r.
1 im
k+ oo
J
ck~x~al
Proof:
The "if" part is an
4.2.2.
Since S(a ) = 1,
1
II (y , x) M (y)
yEX
II
E
I B( x) I .
ck~x~al
immediate result of Theoreni
B is not identically zero.
Thus
in order to show the "only if" part, it suffices to show
r.
Y
(x)
=
<
13
E
Ak+B k ,
x-y
0, for y
E
f3 (x)
x~y,xfak
E
X.
Fix y.
Then for each k such that
=
XE'"Ak+B k
Without loss of generality, let us assume ll(ck+l,a )
l
Then
63
~
O.
E
Sex)
x::;y, x/a
k
xEAk+R k
=
=
0 - 1 + 1
=
and since L x<y xJa xEA +B lI(d
x),
T
k'
k k ~ k'
0,
it follows that
sex)
L
x~y,xlak'
Letting k +
xEAk+B
o.
k
o.
we obtain E <Sex) =
x-y
00,
=
Finally, we want to show that if X is a determining
space, then (4.13) h61ds.
Y
E
X.
Suppose E < IS(x) I = 00, for some
x-y
It is easy to see that E <
x-a
IB(x) I =
00
From (4.12),
1
we obtain
Im(a )
l
IS(x) I -
E
ck~x~a1
ck~x~al
m(x)
E
ck~x~al
Since limk+ooE
< <
E
ck-x-a l
~
I
E v (y , x) M( y)
II
~
yEX
M(a ) <
l
00
•
IS(x)j= E <
IS(x)1 =
x-a l
00,
(4.13) follows.
The space that we discussed in Example 4.21s a type II
space.
(The basic facts of this space are 'listed in Appendix
64
A4.)
Since Iy~(1,4)IB(x)1
=
00, X is a determining space.
The parameter m(1,4) cari be computed by substituting
a
1
e.
= (1,4) in formula (4.13).
Let X be a type III space.
Recall that for k
~
1,
By Proposition 4.3.2 and the above assumption we have that
for q
~
1,
and
With the aid of Prop ositi. on 4.3.2, one can see easily that
65
e
0_
the vaJues of BI(x) and C (x) do not depend on how one
2
chooses k.
Therefore 6
1
and B are well defined.
2
Proof:
Proposition 4.3.1, we can express p(y,a ), p(y,b ) and p(y,x)
1
1
as follows:
p(y,a ) = uJl(ck,a ) + v~l(dk,al)'
l
l
p(y,b )
l
p(y,x)
= Ull(ck,b l ) + vp(dk,b l ),
= up(ck,x) + vp(dk,x).
I{P(ck,bl)p(dk,al)-p(dk,bl)p(ck,al)}
66
· .
.
={u~(ck,x)+v~(dk,x)}~{~(ck,bl)~(dk,al)-~(dk,bl)~(ck~al)}
l1(y,x).
=
v(y,x) = 0 whenever y
E
Ck+D k and x
v(y,x) = 0 whenever y -I- aI' y -I- b l
each x
E
X,
E
Ak+B k ·
Likewise,
and y f, x. TherefOre, for
t'here exists only a finite number of y
that v(y,x) 'f O.
X such
E
e
Thus the following theorem is an'immediate
consequence of Theorem 4.2.1.
THEOREM
4.3.8.
Let X be a type III space.
Let m and M
be a mass function and its distribution function respectively.
M, meal) and m(b ) uniquely determine m(x), x E X.
l
In
particular,
(4.14)
m(x} =
E
yEX
for x
E:
v(y,x)M(y)+m(a )B (x)+m(b )B (X),
2
l
l
l
X.
THEOREM 4.3.9.
A type III space X is a determining
space if and only if for any pair of real numbers u
which are not both zero, there
exists an
67
X
o
E
l
and
X such that
U
z
e
.e
E
x~xo
IU1B1(x) + u 2 B2 (x)1
=
00.
The "if" is a irllUlcdiate consequence of Theorem 4.2.2.
Proof:
Thus, He only need to
there exist Uland u
2
sho~17
the "only if"
part.
Suppose
which are not both zero such that
/.:1 u l B1 (x) + u 2 B2 (x)]
<
for all y € X.
00,
x~y
Since for y
E
L
A
<
J.
Ak+B ,
k
x-y,xrak,xE k+Bk
p(c ,x)::E <
. 1 . p ( d ,x) = 0,
k
x-y,xrak,xEAk+Bk
k
it:
follows that E <.1.
A +B Bl(x)=E <.1.
A +B BZ(x)=O.
x-y,xrak,xE k k
x-y,xrak,x€ k· k
Therefore,
E <.1.
A +B {u l Bl (x)+u 2 BZ (x)} = 0.
x-y,xrak,xe k
k
.
We obtain for y
E
Let k
-~
X, the identity EX~y{uIBl(x)+uZBZ(x)}=O.
Since Bl(a l ) = 1 = B2 (b l ) and Bl(b l )
is not identically zero.
=
°=
BZ(a ), ulBl+u28Z
l
Thus, by Theorem 4.Z.2, X is not
a determining space.
Observe that when X is a determining space,
exist Xl € X and Xl € X such that E
x~xl
EX:<;;X
I B2 (x) I
l
00.
This implies that z
IBl(x)
x~al
I =
IBl(x)
there
00
and
I =
00
and
00.
Theoretically, when X is a determining
Ex<a 18 2 (x) I
-- 1
space, for any distribution function M, there exists a
unique pair of non-negative real numbers meal) and m(b
r ) for
which
m(x)
=
E v(y,x)M(y) + m(al)Bl(x) + m(b )S2(x)
l
~
yeX
and E
x-;;y
m(x)
M(y),
for y E X.
68
Unfortunately,
i t is not
0,
00.
ve r y ea f; y t a fin d the val U e s
0
f m ( a 1) and m ( b 1) in g en era 1.
HOHeve:c, for some type III spaces,
they can be found
explicitly.
Co~olla~~
a)
If lim
4.3.10.
. {E
k -+00 c!fx$a 1
18 1 (x) I IE
.
ck$x$a l
113 2 (x)j}=
0,
then X
is a determining space and
lim{
L
11: v(y,x)M(y)11
L
ISZ(x)l}
k-+oo ck$x$a YEX
ck$x$a
·'
l
1
lim{
E l L v(y,x)M(y)+m(bl)SZ(x)\
k-+oo ck$x$a YEX
l
I
b)
If lim
. {E
k. -+00 ck$x:o;a
l
I Sz
(x)
L
Ck$X$U
ISI(x)I}.
l
I IE ck$x$a
.
I Sl(x) I}=
0,
then X
l
is a determining space and
=lim{
L
k-+oo ck$x$a
lim{
r
k-+oock$x$a
I
Lv(y,x)M(y)1!
YEX
l
I
v(y,x)M(y)+m(a1)Sl(x)
c
$x$a
k
real numbers u
u2I3Z(x)' :::
space.
00.
1
l
113 1 (x) I/Z::c =:::x$a
and
k
U
z
L
ck$x$a
I
IS 2 (x)I}.
l
since b) follows similarly.
J
SZ(x) I} ::: 0,
then for any
1
whicb are not both zero,
EX$al'u1Sl(x}+
Thus, by Theorem 4.3.9, X is a determining
From (4.14), we have for each k,
69
'
.
We only need to prove a),
If limk-+oo{E
l
YEX
l
I
Proof:
Il3 (x)l}
l
L
ck$x$a
e L.
.e
~
4
m(x) <
<Xl
ck:::;;x~al
and
I m( a 1 )
(4 . 20)
L
ck~x:::;;al
<
I 13 1 (x) I -
L
ck:::;;x~al
IE" (y , x ) M( Y) .+m ( b 1 ) e2 (x) I I
YEX
.
00
Since
(4.15) follows from
"e
(4.19).
Since lim k + oo Lc <x<aISl(x)!
k-- - 1
=
Lx<a
- 1
IB 2 (x) I
=
00,
(4.16) follows form (4.20).
The space that we have discussed in Example 4.1 is a
type I I I space.
in Appendix A4.)
(The basic facts about this space are listed
Since lim k + oo {L(k,1):::;;xS(1,2)
L(k,1)s~s(l,2)1132(x)l} = 0,
more,
Il3 l (x) II
X is a determining space.
Further-
the parameters m(l,l) and m(1,2) can be found by suing
(4.15) and (4.16) respectively.
70
APPENDIX
AI. A space which is not a determining
spac~.
Figure Al
Let X be the space shown in Figure Al.
ml(n,l) =m (n,2) = m (n,3) = 2
l
l
other ml(x) = 0,
~nd
-n-l
Let
ID
l
(0,1)=1/2;
, for n=2,4,6, ••• ; all.
let m (n,l)
2
for n=1,3,5, •.. ; all other m (x)
2
O.
Both m and
l
ID
2
are
mass functions which giv£ rise to the same distribution
function M, where M(O,I) = 1; M(n,l) = M(n,2)
n
~
1.
= M(n,3) = 2 -n ,
Hence X is not a determining space.
Actually X is a type III a-lower finite space.
basic facts of X are listed in Table AI.
71
The
Table Al
r=-
.~=
-
-
{x: x>(n,l)}
A
n
..
_-----
II « n , j ) , (n ' , j ' ) )
_(_2)n-n'-1
for n', <n
1
for n""n' ,j == j ,
0
otherwise
,
ll(Bn,(n',j'»
_(_2)n-n'-1
ll«n,j),A ,)
n
_(_2)n-n
ll(B ,A ,)
n
n
_(_2)n-n
B (n, j )
.... (- 2)
l:x~ (0, 1) I B (x) I
for n' <n
,
for n'::;n
,
for
n'~n
n
4
---_.
v«n,j), (n'
,j'»
for n""n'=O,j=j'==l
1
(_2)n-n'-1
for n'>n::::1
1/2
for n'=n::::1,j=j'
-1/2
for n'=n::::l,j,&j'
otherwise
0
72
A ')"
~
.
Le,mma 2. 3. 1 • If X has the property that for
X,
~
there exists a point z for which z
V1 (x),
x and z
ea~h ~~y
~
€
y, then
with set-theoretic inclusion~ forms a directed set.
Proof:
[Yz'x] and [Y3'x] belong to
and
[y ,x]
2
Let
~
[y ,x]
3
[Yl~x] and
exists a point Y
3
~
~
V1 (x)
such that
~
then [Yl,x]
forms a directed set f6r each x
73
€
3
[Yz'x]
[Y ,x].
3
[YZ,x] belong to V1(x).
X for whichY
[Yl~x] ~
S Y
l
X.
Since there
and Y S Y '
Z
3
A3.
PROPOSITION.
Let X be a partially ordered space.
If
X satisfies
~
(1) for each x,y e X there exists a z for which z
(2)
y,
~
and z
x
for each mass function m and its distribution
function M,
m(x) =
~
L
(t,x)M(t)
t~x
for all x e X,
then for each x e X, there exists a t
supil
~(t,x)1
L
x
s~t
te [s,xln [u,xl
Proof:
Suppose there exists an x
sup{1
E
x
, uf,s,
u~s}
<
00.
X such that for all
~(t,x)1
L
< x such that
.
s~t
te [s,xln [u,x]
x
, uts,
u~s}
=
t
X
<x,
00.
We want to show that it contradicts condition (2).
For s
~
x,
let v+(s)=maX{L t
and let v - (s)=min{L t
A = {s :
s~x,
v+(s)
[
ES,X
~
E
[ s,x 1n [ u,x l~1 (t,x): uts,
1n [ u,x l~(t,x):uts, u1 s }.
Iv_(s)l} and B
=
{s :
s~x,
u~s}
Let
v+(s)</v_(s)I}.
Now we can divide the situation into two cases:
1)
There exists a YO
that {z:
2)
For all
z~yo'
y~x
~
x and a positive integer K such
v+(z»K} n A is a finite set.
and for K
~
1, {z:
z~y,
v+(z) > K}. n A
consists of an infinite number of points.
Suppose case 1) holds.
Observe that for y ~ YO' '{z: z~y,
v+(z) > K} n A is a finite set.
74
Since for y
~
x,
sup{v+(z), I ,,_(Z)]: Z$y}== sup{Jt;
u~Z, u1z} ==
ro, we have
Ch a 0 s e sIs u c h t hat
Let u
1
be
a·
s 1.
.[..
t E
~or Y s YO'
~ YO'
s1
Z,X
]
n [. u, x ]ll(t,X): Z $ Y ,
..
ini{v_(z)~ ZSy,ZE~} ::;
-00.
Band (l I 3 ) v_ ( s 1. ) . < - 1 .
E
point for which t;
[
] .[.
]·~(t,x) = v_(sl).
tE s,x n u ,x
.
1
For n > 1,
choo~e
n
and (1/3 )v (s )
-
s n such that s n < s n-·1' s n <. u n-. l '..
-no
<
n
Let u
S
n
E
B
be a point for which
n
x) = v (8 ).Nowwe have two sequences
x]n[u x~(t
]'·
n
n'
n'
oc
.
00
{un}n==l and {sn}n=l.
Define
EtE[S
if ·u.::; u n , .
m(u)
otherwise.
Then m is a mass function defined on X.
For n
lution function corresponding to m.
r,
tE [s
~(t,x)M(t)
n
E
=
,x]
tE[S
=
Let M be the distri1,
~
E m(u)}
~(t,x){
n
,x]
uSt
E m(u){
I
~(t,x)}
u$x
tE[S ,x]n[u,x]
n
00
=
i
E (1/3 ){
E
p(t,x}}
1=1
tE[S ,x]n[u.,x]
n
When i
< n,
.
When 1
>
S
E
t
E
S
n
r
S
J
L
tE
.
J.
IS
n
S
E
n
.
p(t,x)::; E . I · ] p ( t , x ) ::;
. t
rS
(1/3
J.
n'
i
x
U j.
E
,x
o.
B, we have
,x ]. n r· u.,x ] ~ (t , x)
00
i=l
.
r
n ' x J n u l ' x]
Furthermore, EtE
E
Thus
U..
n, since
J.
I s
.
].nru x. ·]p(t,x)
.
n'
){
E
tE[S
n
,x]n[u.,x]
J..
75
Jv
(s )
-n
L.
v - (s n ).
~ t,x)}
Then we obtain
-
~t
ThUG,
co n d i
l
[_ S
E
ion
n
(1!2)(l/3 )v _. ( 8n) < -n/2.
,y.
~
]p(t,x)M(t)
-~
wh~ch
as n + 00,
contradicts
n
(2) •
Suppose case 2) holds.
~
For y
x and K
~
1~·
{z:
z~y,
v+(z) > K}
n A consists of an infinite number of points.
Therefore,
for y
6
1
x,
S
SUp{v+(z): zsy and
such that 51 e A and
(l/3)~+(sl) >
Z
1.
Let u
for ~hich E [_
] [
]p(t,x) ~ V+(Sl).
te sl'x n ul,x
s n su.ch that s n <
Let u
N0W,
S
n-
l'
S
n <
U
be a point for which I
n
sn
n-·.l '
t E
[
S
n
A}
E
=
Choose
00.
be a point
l
For n > 1, choose
n
A and (1/3 )v+(s n »n.
~
] [_
]p(t,x) = v+(s ).
,y.n
u ,x
n
n
have two sequences{un}~=l and {Sn}~=l' Define
we
if
U ::::
U
n'
otherwise.
Let M be the distribution function corresponding to m.
.For n
1,
?
n
E
j..t(t,x)M(t)
E
te[s
,x]
te[s ,x]
n
=
E m(u) {
uSx
00
E (1/3
i=1
When i
<
n,
s
n
p(t,x){ E m(u)}
uSt
Thus
76
i
ll(t,x)}
E
tees
n
,x]n[u,x]
){
lJ(t,x)}.
I:
te[s ,x)n[u.,x]
n
1
..
pet x)
. I tE[S ,x]n[u.,x]'
n
\.Jhen i
> n,
=
I
1
since sn e A,
teru. ,x]'"l'(t , x)
IIte[s ,x]n[u.,x]p(t,x)!
.
.
n
00
r.
Thus,I
We obtain,
n
i
r.
(1/3 ){
i:.=l
>
< v+(sn)·
1
Furthermore; It [.
"]n.[ u,x
.. ]p(t,x) = v+(~).
E S,X
n
·n
= o.
1
lJ (t ,x)}
"te[s n,x]n[u.,x]
·
1
n
/2.
[
]p(t,x)M(t)
te s , x
-+
00
as n
n
condition (2).
77
-+
00
which contradicts
A4. Basic facts about the spaces in Example 4.1 and 4.2.
=
~~
Example
X
l~
--
-
Example 4.2
.1
.
f.--
~
{(n,j) :1~n~k,j==1,2,3}-Bk
B
k
{(k, 1) }.
{(n,j):n~2k-l,j=1,2,3,4}-
B
k
{(1,1), (1,2)} i f k==l
Dk- 1
C
k
{(n,j):n;;::k+1,j=l,2,3}-D
D
k
{(k+1,3)}
i f k;;::2
{(n,j) :n;;::2k,j==1,2,3,4}-D
k
{(n,j):n==2k,2k+1,j==3,4}
if k is odd
{(n,j) :n=2k,2k+1,j=I,2}
if k is even
.
1---
a
(k,2)
k
(2k-l,4)
i f k is odd
(2k-1,1)
i f k is even
(2k-l,1)
if k is odd
(2k-l,4)
i f k is even
.
~.
b
k
(k,l)
-
f----
ck
(k+1,2)
(2k,1)
if k is odd
(2k,4)
if k is even
(2k,4)
i f k is odd
(2k,1)
i f k is even
d
k
(k+1,3)
L-.
78
k
(c.ontinued)
x
Example
Example 4.1
1••
2
-.-.--.----..c..-+-------,-------·---I-----------~l
j.l «n,j), (n'
,j'»
(-1) n~n
,
(_l)n-n
if n>n', j =1,2,
';'=1
n-n'
if (n,j)
(n t
o
or if n>n', j=3,
j'=2 or 3
(-1)
,
$
,
j ,)
otherwise
(n-n') if n>n',
j=l or 2,
j'=2 or 3
1
if n=n',j=j'
o
otherwise
t-----------1'-----·----·-----------+-------------1
j.l(c ,b )J.l(d ,a )o
k
k
1
1
J.l(d , b )J.l(c ,a )
k 1
k 1
---------'---.-'-------I-------------J
---,--.
s(n,j)
--.~-
8 (n,.;)
1
-----.-r__.-.--------------1f__------------_t
o
i f j=l
(_1.) n-1
_____.. __
..
._.~_+_------.---.------1f.--------------
(_1)n-1
(~1)
..
. f J=
. 2, 3
).'
._ _.
n
i f j=l
(n-I) if
j=2,~
-..c..._ _-I-
"-I-
v( (n, j ) , (n' , j , ) )
' 2,
(_l) n '+n-~' ).'f n=n ' >1,]==
--I
(_1)n'+n-1
if nil,
(n,j )/'(11' ~j ')
j'=l
or if n=n' >1,
j=3,j 1=2
or· if
(n,j)=(1,4) ,
(n',j ')~(1,4)
or if n=n'=l,
j=2,j '=3
o
otherW1se
or if n' >n>l,
J=1,2,j'=1
L-.
._-..I..
...J-
.
79
_
(continued)
F'---------
-
X
Example
If .1
Example 4.2
(":'l)n'+n-l or i f n'>n>l,
j=3,j '=2,3
(-1) n' +n-l (n-n') i f n'>n>l,
j=1,2,
j'=2,3
l'
i f n==n' >1,j=2,
j'=2
or i f n=n'=l,
j=3,j '=3
0
otherwise
80
funeti.onde~ined
A5. The 1-1
Obae~ve
that [a 1 acbac]
onja,' ,aebac].
={a~
ab, ac, ha, ca,
~bc,
aba,
e.
aea, acb, bac, cae, cab, abae, acae, acbc, acba , ebac,acbae}.
For
th~s
particular interval, the 1-1 function f discrissed
in Lemma 3.1.2' can be given explieitely as follows:
f(ac) = a,
f(a) = ae,
f(ab)
aeb,
=:
f(acb) = ab,
f(ba) = aba,
f(aba)
= Qa,
aca,
f(aca)
=:
f(ca)
=:
ca,
:f(abc)
=:
acbc,
f(acbe) = abc,
f (hac)
=
abae,
f(abac)
=:
f(cae)
== aeac,
f(aeae)
= eac,
f(cba) == aeba,
f(cbac)
=
acbae,
bac,
f(acba) == eba,
f(acbac) = cbae.
81
e
.
REFERENCES
Anderson, T.W. (1960). "A modification of the sequential
probability ratio rest to reduce the sample size."
Ann. Math. Stati~t. 31, 165-197.
Doob, J.L. (1949). "Heuristic approach to the KolmogorovSmirnov theorems." Ann. Math. Stati~t. 20, 393-403.
Graves, W.H. and Molnar, S.M.
(1974). "A universal ~easure."
Rota, G.e. (1964). "On the foundations of combinatorial
theory I.
Theory of Mobius functions. II Z. Wa.hlt~c.he.in­
llc.hke.i.t4the.oltie. and Ve.ltw. Ge.bie.te. 2, 340-368.
Simons, G.D. (1967). "A sequent.ial three hypothesis test
fOT determining the mean of a normal population
with know variance." Ann. Math. S.tati.6. 38,1365-1375.
Simons, G.D. (1974). "Generalized cumulative distribution·
functions: II.
The a-lower finite case." Anna.l~ On
Pltobabllity 3, 492-502.
·e
82