UNIVERSITY OF NORTH CAROLINA
Department of Statistics
Chapel Hill, N. C.
Mathematical Sciences Directorate
Air Force Office of Scientific Research
Washington 25, D. C.
ON A SPECIAL CLASS OF RECURRENT EVENTS
by
"
M. P. Schutzenberger
University of North Carolina
June, 1961
Contract No. AF 49(638)-213
Qualified requestors may obtain copies of this report from the ASTIA
Document Service Center, Arlington Hall Station, Arlington 12, Virginia.
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,.e
Institute of Statistics
Mimeograph Series No. 291
ON A SPECIAL CLASS OF
RECURRENT EVENTS
M. P. SchUtzenberger
I.
Let
F be the set of all finite sequences (words) in the symbols xeX.
A recurrent event
(A,~)
on
F
is defined by the set A of all words at
the end of which it occurs and by the probability measure
feF,
~
~;
for any
is the measure of the set of all infinite sequences which begin
We call A the support of (A,~) and we denote by T(A,J.l) the mean
by f.
recurrence time of
(A,~).
If (B, ~,) is another recurrent event of
F,
(A n B, ~,) is aga in a
recurrent event and it results from the general theory of Feller (Cf 2.
Chap. VIII) that when
T(B,~') is finite the ratio
~
= T(B,J.l')/r{AnB,~')
is, in a certain sense, the limit of the conditional probability that a
random word
feF
belongs to A when it is known to belong to
B.
For
given A, it is in general possible to find infinitely many (B,J.l') with
finite
T(Bs~r) which are such that
~
= O.
The main point of this note is to verify several statements which,
together, imply the following property
If A is such that
T(AflB,~') is finite for every (B,~') with finite
T(B,~'), then, for every such (B,~'), ~-l is an integer at most equal to
a certain finite number
8* which depends only upon A.
Classical examples of this occurrence are the return to the origin in
random walks over a finite group (Cf 3) and in particular the recurrent
event which occur at the end of every word whose length is an integral
mUltiple of a fixed integer
In the section 2,
k.
we discuss some properties of a class of recurrent
events which we call birecurrent; in section 3, we verify the statements
mentioned above and in the section 4 we describe several examples of birecurrent supports.
2
II. We consider F as the free monoid (Cf 1. Chap. 1) generated by X;
the empty word
the words
e
is the neutral element of
f and f'
is the word
ftl made up of
(f') is called a left (right) factor of
different from
F and the product
f
ff'
of
followed by f'; f
ftl; a word is proper if it is
e.
Feller's condition (Cf 2. Chap. VIII) that the non empty subset A
of F
is the support of a recurrent event can be expressed as follows
U
if aeA and f€F, then, afeA if and only if feA.
r
This condition implies that A is a submonoid of F (i.e. that eeA and
A2C.A). We shall say that A is birecurrent if it satisfies
symmetric condition
if
U1
feF, then,
faeA
if and only if
It follows immediately that if (Ai)
recurrent (birecurrent)
the sets Ai; indeed,
of
r
Ul
aeA and
and, if, e.g.,
U and the
feA.
is any collection of supports of
events the same is true of the intersection C of
C is a submolloid because every Ai is a submonoid
a,afE C, the word
f
belongs to all the sets Ai(because
U)
and, consequently it belongs also to C.
r
In all this paper, A will denote a recurrent (or, eventually, bire·
current) support and we shall use the following standing notations:
A*
= the
set of all the proper words at the end of which the event whose
support is A occurs for the first time (and, for
B,
B*
any recurrent support
is defined similarly).
S ... F - A*F (= the complement in F of the right ideal A*F);
R ... F - FA*.
We state explicitly the following well known facts:
II. 1. Every feF
aEA
and SES
admits one and only one factorization
and at least one factorization
f
= ra'
with
f
= as
a'EA
with
and
r€R; if and only if A is birecurrent the second factorization is unique
3
for all
f€F.
II. 1'.
Every proper
a
from A admits a unique factorization as a pro-
duct of elements from A*.
The two statements are qUite intuitive but a formal proof of them has
been given in
(5)j 11.1' shows that any bijection (i.e. one to one
mapping onto) of A* onto a set
A onto the free
Y can be extended to an isomorphism of
monoid generated by Y.
The following remark will be
used repeatedly in the course of this paper.
11.1". When A is birecurrent, if
s,s' € S (r,r'€R) are such that
is a right factor of
s' (r is a left factor of r
(fr,fr l €A) for some
f€F,
then
s =
Proof.
Sl
(r = r
l );
this implies
implies
= f's
for some
f'€ Aj because of
is proved; if sr € A
at € Aj as above, a
sf, s'f €A
=F
s' €S
sr €A with
sand
Sl.
sf, f'sf
- A*F
s'
and
€Aj
because of
U
l
,
11.1', this in turn
= es = s.
s€S and r€R; if
(e}, 11.1' shows that
V (e}.
Ul we can limit our-
r
fl€S and
f' = e and we have proved that
Let us assume now that
and that
U Md
self to the proof of the statement concerning
s'
)
if, furthermore, f€R (f€S), then sf € A*
Because of the perfect symmetry of
By hypothesis,
l
s
sr
cannot be a left factor of
sr
= aa'
= e,
with
the result
a€A* and
s and, consequently,
a'
is a right factor of rj but, by a symmetrical argument, this shows that
at
=e
and that consequently
sr
=a
€A*.
This concludes the proof of
11.1".
Let us assume now that A is
the
~
birecurrentj we denote by t8f (~f)
set of the right (left) factors of
f
that belong to
the set of the triples (r,a,s) which are such that
f
S (R)
= ras
r€R, a€A, S€Sj such a triple will be called an A-factorization
aod
8f will denote the number of distinct triples in the set
A-factorizations of
f.
and by
and that
of
~
f
of the
4
11.2. For any f,f' e F,
8ff'
>
max (8f, 8f') and
and only if for every left (right) factor
ff" (fllf') has a factorization
f"
where
!!22!.
ff" = sa
is a right (left) factor of
Let us consider any element
jection
O'g: Mg .......
which is such that
~g;
g'
(f"f' = ar') where
aeA and
geF and prove that there exists a bi-
for some
(determined by the conditions
of f' (of f) the product
a.
indeed, by 11.1
= rg l
g
f"
8ff' = 8f (=8f') if
= as,
to any re6Rg (Le., to any reR
gleF) it corresponds a unique
se~g
ogro, be-
aeA, seS) which we call
cause of the symmetry implied by the hypothesis that A is birecurrent
we can construct in a similar manner a mapping
0g
-1
; since, clearly, for any
that
0
g
re.6Rg we have
~g ~
og -10
the proof of 11.2.
which are such that
if and only if we do not have
f', i.e.
~g.
this shows
~f)
-~f
8ff' is equal
plus the number of proper
fr'e R; thus, 8ff t
ff" e R
g are in
We now revert to
By the above construction we know that
8f (i.e. to the number of elements in
r'e~'
r = r
g
is a bijection and also that the A-factorizations of
a one-to-one correspondence with the elements of
to
0
Mg which we call
~
8f with the equality
for some left factor
f" of
if and only if every such ff" satisfies the condition stated
in 11.2.
Because of the symmetry this concludes the proof.
For any feF, let us denote by af the smallest positive integer for
af
which f
eA with af, infinite if the only finite power of f that belongs to A is
fO
(= e, by definition).
11.3. A sufficient condition that the recurrent support A is birecurrent
is that af
is finite for all
feF; reciprocally if A is a birecurrent
support, then, for any feF, af is at most equal to the supremum
m
of 8f
over all the positive powers of f.
Proof.
By hypothesis,
A satisfies
8'f
U and, in order to show that it is
r
birecurrent, it will be enough to show that if a and fa belong to A
5
then f
also belongs to A;
positive finite
a, {fa )m-1 eA
we have
m;
and
let us assume that
{af)m= a{fa)m-1feA and, because of
Ur , this implies
Let now A be birecurrent and
(o ~ n
any fn
~
{af)me A for some
fe A.
f
This proves the first part of
such that
o'f
o'f) admits an A-factorization
is finite; by 11.1,
fn = (e,an,sn) and,
sn it corresponds one A-factorization of f o'f ;
8'f
since, by definition, 8f
< 8'f, we must have s n = sm (= s, say) with
by 11.2, to each such
o~
m,n
8'f and, e.g., m < n.
~
n
= as
n-ma'
= a.
Thus,
f
and fm = a's with a,a'eA
Because of U , this last
1
n-m
f
belong to A and, since 0 < n-m ~ 8'f, by con-
and, after cancelling s, we obtain
relation shows that
f
struction, the result is entirely proved.
Let us assume now that A is birecurrent and that
8f = 8f2
< (X). We consider the set
f
is such that
K (containing at least
f2) defined
by
K =
II.4.
(f'
e
fFf
8f' = flf J.
0: K ~G
There exists a group G, a subgroup H of G and a mapping
that have the following properties:
o is an epimorphism (i.e. homomorphism onto) and G is finite;
0-lH = KnA and the index of
H in G is at most
Proof. According to lI.2, the hypothesis
of a bijection
unique
a*:
6Sf ~ 6Rf
any keK.
K, we have 6Rk
Thus, recalling the definition of
0
se t:JBf
sr € A; trivially, a*e
we can associate to any k € K a bijection
~=a*
of = 8f2 implies the existence
defined for each
re 6Rf which is such that
11.2 and the very definition of
of.
= 6Rf
and
by
= e.
~Sk
the
a*s =
Also, by
= t:JBf for
of given in the proof of 11.2,
~
6Rf -=16Rf
defined by
Ok.
Let us now verify that for any k,k' €K we have
deed, i f (r,a,s)
€
~k'
=~
~,
;
in-
& and (r' ,a' ,s') e&' we shall have (r,a" ,s') e&k'
6
for some aN eA
if and only if
Because of
sr'
€
the hypothesis that
shows that the set (o~)
finite, the elements
is finite, this construction
cr
is an epimorphism.
~
Let us observe now that
~
of
is a group G and that the mapping
keK
which sends every keK octo
that is, if and only i f
A and the identity is verified.
k belongs to A if and only if
lets
e
(e,k,e)€6k,
invariant; again, because G is
keK which have this last property map onto a sub-
group H of
G and, clearly,
the index of
H in G is at most equal to the number of elements in
(i.e., to of)
is contained in A. The fact that
11.4 we state
If A is such that the supremum
11.4'.
finite and if
of
representation of
Proof.
= 0*,
G
.~*
of
then the representation
over the cosets of
oft over all
ft €F is
is isomorphic to the
~
H.
The property stated amounts to the statement that
f~)
is transi-
or, in an equivalent fashion to the fact that for every
=
exists at least one keK which is such that
such that
~
is a standard result from group theory.
As a corollary of
tive
cr-IH
k = as with aE A.
For proving this, let
5,
se~f
there
i.e. which is
(r,al,s)
€
6f; by
11.3 we know that there exists finite positive integers m and m' which
m mt
are such that f m € A and r m' € A; thus the product f mr mt-lf = frats
t
admits the factorization ails with a" = fmrm al E A and it belongs to K
since, under the hypothesis that
of
is maximal,
K is identical to fFf.
The next statement is not needed for the verification of the main property;
its aim is to show that the representation described in section
covers all the birecurrent supports with finite
4 below
0* ( = sup of, by defini-
tion) •
11.5.
If A is a birecurrent support with finite
monoid M and an epimorphism
r: F
~
0* there exist a
M which are such that
7
1 -1 1 A = A, and that
~.
M admits minimal ideals.
f€F and denote by Crt) the set of all
Let us consider any
f'€F
which satisfy the following condition:
for any f , f € F, f ff € A if and only if f f'f € A.
2
2
l 2
l
l
The relation f' € (rt) is reflexive and transitive and it is well known
that it is compatible with the multiplicative structure of
a congruence);
thus we can identify each set Crf) with an element
a certain quotient monoid
flf f
€ A with
2
trivially,
1
F (i.e., it is
-1
M of
Since
of
f € A if and only if
A is the union of the sets (1a) , a€A ~nd
f =f = e,
l 2
1
F.
1f
A = A.
Let us now take and element
quantity; according
to
f
11.2
which is such that
of = 0*, a finite
the maximal character of
Of
implies that
for every f
r
the product flf has a left factor flr € A for some
l
thus, because of the symmetry, any relation flf f € A implies
2
€ ~f;
flr, sf
2
(r,a,s) € ~.
€ A with
It follows immediately that for any two
1k = 1k' is equivalent to the relation
11.4.
Thus, 1K
k,k'
dk = dk'
is isomorphic to a group
section of a right and of a left ideal of
€
K = fFf, the relation
in the notations of
and since
K is the inter-
F, this shows that
M admits
minimal ideals.
We now revert to the preparation of the proof
and we
consider
A, a birecurrent support,
C = A n B; we assume that
of the main property
B a recurrent support and
C does not reduce to
e
and that consequently
C* (the set of the proper words at the end of which the events whose
supports are
A and
B respectively occur together for the first time)
is not empty.
II .6.
with
Any element
f
l
€
B -C*B
belongs to F -C*F
f
and
from F -C*F
f
2
€
F-B*F;
has a unique factorization
f= f f 2
l
reciprocally any such product
flf~
8
any l'
Because of II.l
Proof.
has a unique factorization
= 1'11'2
t
with
€ F-B*F· since C is a recurrent support contained in
'
B any product = ft 1" with 1" e: B and 1" e: F - B*F belongs to F-C*F
1 2
2
1
i f and only if fi belongs to B - C*B and this concludes the proof.
1'1 € B and
l'
2
As mentioned in II.l', there exists and isomorphism
where
Q is the free monoid generated by
P of C by
verified that the image
according to our hypothesis
submonoid of
-1
qe:Q
-1
As above we define a
s
p,
e: S
since
of
s)
with
P*
= ~C*.
P-factorizations of
For any b
Proof.
Let
and
indeed,
P
r
€
l
is surely a
p,p',
q = Fps
and that
F e: R =
All the remarks made in
Q, and we define
Q - QP*, pe:P,
II.2 apply here
5q as the number
q.
8'
B,
when,
U
-1 p,~ -1 p',~ -1 pqp' e: A imply, e.g.,
P is a birecurrent support in
II.7.
U
r
q e: A, by U and, finally q € P = ~(A '" B).
l
P-factorization of an element q€Q as a triple
which is such that
= Q-P*Q
~
~b ~
5b.
be any element of R and define
~* r
as the (uniquely
determined) element re:R which is such that
(r,a,e) e:6b for some
We show that the Iestriction of the mapping
~*
an injection (i.e., is one to one into);
e.g.
r'
=r
q'
for some
the following relations
with a'e:A; rat
because of
= rab'
~
-1_
r
= ~ -1 q'
= ~*r' = r,
€ B; consequently,
= ~b'
r' =r q'
= e because of the relation
= r and our contention is proved.
The remark II.7
-r,r-,
ra € B with a e: A; ~ -1
this shows that q'
finally q'
ThUS, r'
=
~*r
ae:A.
to any set ~q
indeed, if
q'€Q; thus, if
with b'
Ur' b' e: A·,
Q
~
and it is easily
satisfies
~
A is birecurrent;
( because
qp e: A, by Ur , then ~
(r,
= ~B*
Q and it is enough to verify that the relations
pqp' e: Pimply
~
Q*
: B
~
(q€Q) is
e: 6Rq we have
say, we have
r' = r
a'
belongs to
a' € B
= ab'
and,
P and that
€ R.
is also proved since we have shown that for any
be:B there exists an injection
iR ~b .., 6Rb •
9
= sup
II.8.
If
0* (
then
8* (
= sup
Proof.
that
of) is finite and if
8q) is a divisor of
B contains an element
5*j
for at least one b€B,
8*.
we have
B
f
which is such that
and we use the notations of
By construction the image
n a-l(H
theory the index
G (i.e., of
8*).
G'
by
G'} = A n Bn Kj
5'* of
H ~G'
the role of
in
define a bisection
o~*
G'
is a divisor
8'*
~*
-1
: 6Rf
II.4 and II.4'
II.8 is proved.
a
H in
with ~(Bn K)
~(B (\ K)
H of
--
(\~*6R~f -? 6R~f
a.
for
in
-+ G which
We recall the
~**.
which is such that
- - 1
~~f
a natural fashion an epimorphism ~**:
H by
of that of
used in II.7 and we observe that we can
is the identity mapping of
is the inverse image of
G and
is in fact equal to 5*;
is the index of the subgroup
~*
and
thus, by a standard result of group
We prove now that
definition of the mapping
= 8*
B n K is a sUbgroup of
K and we obtain an epimorphism
is such that 8*
8f
II.4 and II.4'.
a of
this we repeat the construction of
-1
= 8*
Under these hypothesis, we may assume without loss of generality
8 ~f =
~*
8b
onto itself;
~*
induces in
G' ~a and, trivially,
Thus
8*
is equal to
8'*
H" G'
and
10
III. We keep the notations already introduced
is arecurrent event; according to Feller,
~e
M:
o
Mr :
We
=1
if
and for any fEF, ~
aEA and rEF then,
shall say that
~
~f
= l:(lJfx:
=~
~
and we assume that
satisfies the two conditions:
XEX).
~.
is a positive product measure if
for any f, f' E F, and, in this case,
F~
~f'
= IJf
1Jf'
>0
Mr is trivially satisfied.
We denote by If I the length of the element f
F'
(A,~)
and for any subset
of F we use the following notations:
=
{fE F' : If I:::; n}
It follows that
~'~
~'=
;
1
if
lim l: (1Jf : fE F').
n
n...,oo
F'
is such that any fEF
left factor which belongs to F';
has at most one
this condition is satisfied in particu-
lar by any subset of A* and, according to Feller's definition, we shall
(A,~)
say that
is persistent if
and only if
~*
= 1. The next two
statements are proved at the imitation of Feller.
(A,~)
111.1. For any recurrent event
Proof.
Let us introduce for any
sE8
we have T(A,~)
the notations
= ~.
8(s) = 8
f)
sF and
A*(s) = A*" sF. We verify the identities.
(IIL1).
for all m~/s/:
(111.1').
for all m ~ 1 : (l-~*) + (~* - ~:) = ~m - ~m-l·
Indeed, (111.1) is an immediate
o~
~s-~:+l(s) =~m+l(s)-~m(s);
consequence of Mo and of the fact that
(s) U 8m(s) X and 8m+l (s) U A~l (s)
is the special case of (111.1) for s = e.
the sets
From this second identity we deduce that if
lim (~m- ~m-l)
m..,oo
~*
= 1
implies
= O.
~s
~*
are identical ; (111.1')
=1
we have
ThUS, a fortiori (from the first identity) that
= ~*(s).
We now sum the second identity fromm=l
to m=n; after rearranging terms, we obtain:
11
~
n :::: (n+l) (l-~*)
n + t ( /a/ ~: a€ A*)
n
This shows that if (A*,~) is not persistent ~ is infinite and we assume
(III.l").
~*
now that
= 1. Under this hypothesis,
T(A,~)
is defined as
lim
t
n~ CD
(/a/~
: a€A*)
n , and since ~*:::: 1 implies that (n+l ,) (l-A*)
n
:::: t ( (n+l) IJB. : a € A*-A~), we can write for all
n
t( /a/ IJB. : a€A*) < ~
< t( /a/ IJB. : a€A*-A*)
n n n
For any
s
~
e
111.2.
T(A,~):::: ~
as (e)
when
which are such that
~
S€S, 1 ::::
is so.
s:::: e and, when
sf€A*.
a product measure and
and, for all
when this last
(A,~)
persistent,
~*(s).
Under these hypothesis all the notions are perfectly symmetrical;
thus, the
identity (111.1") shows that
Since any a€A*(s)
and since
(111.2)
f€F
If A is birecurrent,
T(A,~)
is infinite when
let us define R*(s)
as the set of those
we have
Proof.
S€S
~
t( /a/IJB. : a€A*)
n
~:::: T(A,~)
This concludes the proof since it shows that
quantity is finite and that
+
~
~
n ::::
~
n and, as a special case,
has a unique factorization
a:::: sf with f€R*(s)
is a product measure, we have for all
m > /s/ the identity
~ A: (s)
::::
~s ~:_/s/s).
ThUS, because of the formula (111.1) we have in any case R(s):::: ~*(s)/~s
<
1 with the equality sign when
111.3
If A is birecurrent and
(A,~)
~
is persistent since, then,
a product measure,
Proof. We use the notations of the section
T(A,~)::::
~s
8*.
II and we recall the following
facts:
1) According to ILl",
2)
for the same reason,
and
R*(s')
R*(s)
if
is a subset of R;
s,s' € 6Sf for some
are disjoints.
f€F, the sets R*(s)
12
3)
if
0*
is finite and
corresponds one
se ASf
the union of the sets
Now to the proon
then, by 11.2, to every reR
which is such that
R*(s)
over all
~
of
is equal to
= 8*
~
f'eF
is infinite;
the inequality
=1
~*(s)
R.
we have the inequali-
and, trivially, the result will follow by
(A,~)
The second inequality is vacuously true when
since, then,
8f'
=
when
(A,~)
~ (~*(s)
and since the sets
is
111.2.
is not persistent
persistent we have for any
: se ASf') ~ ~
R*(s}
there
sr€ A*; thus, in this case,
s€6Sf
We shall show that if
$ of $
~
ties
= 5*
of
since, then,
are pairwise disjoint.
Thus the
second inequality is always true.
If now
of
= 0*,
E (~*(s) : se ASf)
3 above that
we know by
= ~;
since in any case as we have seen in the proof of 111.2, we have
~*(s)
$ 1, it follows that
111.4.
(B',~)
If
Let
fied that
B=
(B,~)
and the result is proved.
is a recurrent event and if A is birecurrent we have:
T(A" B' ,~) = '5* T(B' ,~)
Proof.
~ ~ 8*
(beB':
where
~
8'*
>0
is defined below.
) and
C
=A f)
it· is easily
B ;
vf.r1-
is again a recurrent event and that according to 111.1
we have
T(A () B' ,~)
= T(A n B,I-1) =
,~) = T(B
T(B'
,~)
I-1(F - C*F)
~(F - B*F).
=
We keep the notation used in the proof of 11.6 and II.?
and we observe
that, by taking into account 11.6 and the condition
on
remark 111.4
For
v~b
is equivalent to the relation
~,
for all
beB; because of M
r
v on
T(P,v ) = v(Q - P*Q)
But, by definition,
Q.
= 8*
~,
Because of
p
the
•
Q by the relation
and of the definition of
a positive product measure and, since we know that
(p, v) is a recurrent event on
r
~(B-C*B)
proving this identity we define a measure
=
M
= ~C
B, v is
is birecurrent,
III.l a.nd III. 3
= -*
0
v(Q-P*Q)
=~
(B-C*B)
= ~(B
-C*B) and the result is
13
proved
III.,.
If
5*
(B',~)
is finite and
persistent for some measure
~
which
satisfies the condition that for every f€ F at least one element from
FfF
Proof.
Because of the conditions satisfied by
an element f
= b's'
f
is a divisor of
has positive measure, then 8*
which is such that
with b'€ Band
= 5*
s' € F-B*F.
~(B*
follows from II1.1 that
5f
- s'F)
5*.
and
~
and that
= ~';
~>
(B,~)
Because
5* we can find
0;
we have
is persistent, it
since this last 'quantity is
positive, there exists at least one element b€ B*nsIF. Finally, because of 11.2 we have
5b'b
= 5* with bIb €B. Thus, we can apply
II.8
and the result is proved.
The next statement is intended to give a characterization of the birecurrent supports in terms of their intersection with other recurrent
event;
by E we mean any fixed birecurrent support which 1s such that
T(E,~*)
~*;
is finite for one positive product measure
(E',~)
as usual and we say that
E*
belongs to the family «E»
is defined
if the two
following conditions are met:
(E',~')
is a recurrent event on F;
there exists a finite integer m which is such that any element from
EI*
is the product of m words from E*.
It is trivial that under these hypothesis
E'
itself is a birecurrent support (With F*
= X)
family «E»
m
Since F
a simple example of a
is the family of the birecurrent events
(F(m)' ~m) where
is the set of all words whose length is a mUltiple of m and where
F(m)
~
is birecurrent.
is a suitable measure.
II1. 6 •
If the recurrent support A is such that
sistent for every
Proof.
(E',~')
€ «E», then,
If af
is per-
A is a birecurrent support.
This is a simple epplication of II.3
of this remark.
(A (\ EI, ~' )
is finite for all
f,
and we use the notations
then we know by 11.3 that
14
A is birecurrent; thus we may suppose that A and f
is infinite and we show that (An
able (E' ,~,).
elements from E*;
~'
and
is not persistent tor some suit-
Indeed, by the second part of 11.3 we know that
for some ftnite positive m;
of m'
Er,~r)
are such that af
f m admits a factorization as a product
thus
we take E' defined by the condition E'*=E*m'
defined by the condition that
any other f' € E'*.
fm €E
~'f
m
= 1 and that
The conditions Mo and Mr
(AnE' ,~,) cannot be persistent since A () E'
for
recalled at the beginning
T(E',~')
of this section are obvtously satisfied and
=0
~'f'
is finite.
reduces to
Finally,
1et and this
ends
the proof.
Clearly, the conditions of 111.6 are satisfied if A is such that
T(A
n B,~) <
that for any
ID
(B,~)
with finite
T(B,~).
(Other algebraic characterizations of the birecurrent supports have
been discussed in
"Publications scientifiques de
serie Mathematique, tome VI.
If A is birecurrent and if
measure,
~,
d' Alger
1959 p. 8'-90").
The next statement is a simple application of
111.7.
l'Universite'
8*
11.2.
is ftnite, then, for any product
the distribution of the recurrence time of
(A,~)
has moments
of every order.
Proof.
Let A'
= la€ A:
IJB.
by 11.7 we know that every f€F
> 01;
trivially, A' is birecurrent and,
has at most
8* A'-factorizations;
since the distribution of the recurrence times of
(A,~)
and
same, there is no loss of generality in assuming that A
~
(A',~)
= AV,
are the
i.e., that
is positive.
Since
8*
is finite there exists an element
11.2 has the property that for any proper
factorization
tion
S = F-A*F
f€F
which because of
s€ S the product
sf = ar with a€ A*A; thus, for any integer
allows us to write the inequality
sf
n
has a
the defini-
15
I.L
A*
(n+1) If I
..A *
<
- t¥'"n/f I -
(1 _ ..~)n+l •
~
Consequently the distribution of the
lal
(8£ A*), i.e.
of the
recurrence time of A*, is dominated by an exponential distribution and
this proves the result.