(I) .CONTINUOUS PARAME1ER SEMIGROUPS OF POSITIVE MATRICES ... · (i) .continuous parame1er semigroups of positive matrices. (2) inequalities for p-functions. by alan george cornish

(I) .CONTINUOUS PARAME1ER SEMIGROUPS OF

POSITIVE MATRICES.

(2) INEQUALITIES FOR P-FUNCTIONS.

by

ALAN GEORGE CORNISH.

Abstract

In Section 1 of this thesis it is pointed out that if the matrix functions

(pij(t)), t > 0, enjoy only non-negativity, measurability and the semi-group

property they nevertheless exhibit many of the analytical properties of

transition functions of a Markov chain. It is shown (with certain exceptions)

that the functions are continuous for t > 0 and that they possess finite limits

as t O. They also possess the always-or-never-zero property for t > O.

On the other hand if the non-negative matrix functions (r..1-3(x)), A > 0,

satisfy only the resolvent equation they will, in fact, behave very much

like Laplace transforms of Markov transition functions. It is shown (again

with certain exceptions) that there are non-negative continuous functions

(p..(t)), t > 0 satisfying the semi-group equation whose Laplace transforms

are the functions ri (X) and that the limits lim Ar (A) exist. j X-->00 ij

Finally, two examples are given of semi-groups with the property that

one of the functions pij(t) can tend to +.0 arbitrarily fast as t

A discussion of p-functions and semi-p-functions follows in Section 2.

P-functions occur, for example, as diagonal transition probabilities pii in

continuous time parameter Markov chains, semi-p-functions as the corresponding

diagonal functions in matrix semigroups.

The variation of a p-function over a time-interval is studied; it is

1 shown that if a standard p-function exceeds 1 - e- for some t > 0 it is

necessarily greater thane in [0,t]. A corollary to this result is that a

measurable p-function that is almost everywhere zero may never exceed 1 - 1,

The class of semi-p-functions has a similar structure to the class of

p-functions. It is shown that a measurable semi-p-function is either zero

almost everywhere or is a constant multiple of a standard semi-p-function.

The well-known result.for p-functions is deduced as a corollary.

Preface

This thesis is in two independent sections. The introduction to

Section 1 is given in § 1.1 and the results of Section 1 have been

published in the Proceedings of the London Mathematical Society, (3)

23 (1971) 643-67. Section 2 is in two parts. Paragraphs 2.1 -2.3

deal with inequalities for p-functions, introduced in § 2.1, and the

results have been accepted for publication in the Journal of the

London Mathematical Society. Paragraphs 2.4 - 2.6 discuss semi-p-

functions, introduced in § 2.4, and the results have also been accepted

for publication in the Journal of the London Mathematical Society.

Acknowledgements.

I should like to record my gratitude to Professor G.E.H. Reuter for

his understanding, guidance and constant encouragement during the preparation

of this thesis.

. I am indebted to Professors J.F.C. Kingman and G.E.H. Reuter for

considerable simplification of § 2.3 and to the former for his proof of

the corollary to theorem 2.6 and.for allowing me to see his paper [17] before

publication.

4

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Contents

SECTION 1. CONTINUOUS PARAMETER SEMIGROUPS OF POSITIVE MATRICES.

PP

1.1. Introduction 6

1.2. Continuity of transition functions 7

1.3. Resolvents and semigroups 17

1.4. Examples 22

1.5. Ergodic limits 31

SECTION 2. INEQUALITIES FOR P-FUNCTIONS.

PP

2.1. P -functions 40

2.2. Some identities 42

2.3. Bounds for.p-functions 4s

2.4. Semi-p-functions 48

2.5. Continuity of measurable semi -p -fUnctions 49

2.6. Representation theorem 51

REFERENCES 57

SECTION I. CONTINUOUS PARAMETER

SEMIGROUPS OF POSITIVE MATRICES.

1.1. Introduction

Suppose that {(pii(t)), t > 0, i,j = 0,1,2, ...1 are the transition probability

functions of a continuous parameter Markov chain with a denumerable number

of states, i.e.

Plj (t) < c°, (1.1)

pii(s+t) = ?ik(s)pki(t), (1.2)

X pii(t) < 1. (1.3)

It is well known (see, e.g., [3]) that if all of the functions pii(t) are

measurable then they are all continuous for t > 0, the limits

u.. = lim p..(t) 13 t+0 13

exist for all i,j and the functions pii(t) are either identically zero or

never zero in (0,00).

We shall show in § 1.2 that these results hold even if we drop condition

(1.3) (except possibly for transitions from i to j when pii(t) and pjj(t)

are both identically zero; the exceptional cases are illustrated by

examples in § 1.4).

Now suppose that the functions {(r1 (x)), A > 0, i,j = 0,1,2, ...1

satisfy

CI r..13 (A) <

r.. (X) - r.. (p) + (A -11 ik) r (x)rkJ .(p) = 0, 3.3 k

y Ar..(A) 1.

Then it is known (see [20]) that there exist continuous functions

(1.3')

f(pii(t)), t > 0, i,j = 0,1,2, ..,1 satisfying (1.1), (1.2), (1.3), such

that

r.. ( A) = f e- Atp..(t)dt 13 0

and such that the limits lim xr. (x), lim p. (t) exist and are equal. A->o. t4-0 lj

We shall show in § 1.3 that, if we assume only (1.1') and (1.2'), we

can still obtain continuous inverse Laplace transforms pij(t) satisfying (1.1),

(1.2) and lim xr..(x) = lim p..(t) (these results possibly not holding for 13 13 t+0 pairs of states i,j with the property that rii(X) 0 and rjj(X) 0; the

exceptional behaviour is exhibited in examples in § 1.4). It will then

follow that, if we assume (1.1'), (1.2') and

xrij(x) 4 S. as X 4 co, (1.4')

then we can find continuous inverse Laplace transforms pij(t) satisfying (1.1),

(1.2) and

p--13 (t) S.. as t+0. 13 (1.4)

In § 1.5 we consider the question of whether, for non-negative matrices

with elements satisfying (1.1), (1.2), the ergodic limits (see [11])

log pij(t) lim t4.0

(when these exist) can take the values + .0. We show by means of two examples

that such situations can occur and moreover that p..(t) can tend to +

arbitrarily fast.

1.2. Continuity of transition functions

If the matrices {(pij(t)), t > 0, i,j = 0,1,2, ... } satisfy (1.1) and

(1.2) it is convenient to denote by the set of states i for which either

pii(t) 2 0 for all j or p..(t) 0 for all j. In particular, p11(t) = 0 if

i 6 1. It follows at once that, for any i,j, Pij(S+0 = 40pik(s)pki(t).

Also, if k ¢ , there exist i,j Osuch that pik(t) 0, pki(t) O. We

shall find that the set performs a similar role to that of the set F, described

in [3] § II.1, and the two sets are equal in - the Markov chain case. We now

prove the following:

THEOREM 1.1. Suppose that f(pij(t)), t > 0, i,j = 0,1,2, ... } satisfy ,

(1.1), (1.2). If all pij(t) are measurable then any pij(t) is continuous for

t > 0 unless both i,j pij(t) can be discontinuous only if pik(t) =

P3 .k (t) = 0 for all k,t > 0).

8

Proof. Suppose that pij(t) is not identically zero (otherwise the result is

trivially true) and suppose first that j . Then pip (to) > 0 for some k

and some to > 0. Since a measurable function is approximately continuous

almost everywhere (see, e.g., [8] II, p. 257) there exists at most a null

set of points at which any of the transition functions are not approximately

continuous. So, given t > 0, we can find a positive number t1 < to such that

all the transition functions are approximately continuous at t1 and t1 + t.

Now, from the semi-group property

pjz(to) = y p_;111(ti)plia(to-t1), m

we deduce that there is a state m such that pim(ti) > 0. Since pjm is

approximately continuous at t1 and pim is approximately continuous at t1 + t,

there exist constants c 0, d > 0 and a measurable set S ..c(-t1,...) such

that, for s e S,

0 < pim(ti+t+s) s c, pjm(t+s) d > 0, S has density 1 at 0. (1.5)

Using the semi-group property, for any k,

pkm(ti+t) pkj(t-s)pjm(ti+s) dpkj(t-s) for s e S, s < t. (1.6)

Choose 6 such that 0 < 6 < min(lt„lt) and such that m(Sn[0,26]) *5

(where m is Lebesgue measure). Then m(Sn[16,46]) i(S. Let

Sh = {s c Sn[18,4flisth e S} for - 16 < h < 16. Then

mSh 16 for - 16 < h < 16.

Now, for h e [-16,16], s e (16,46),

Ipij(t+h)-pij(01 < Ipik(s+h)-pik(s)Ipkj(t-s).

Integrate with respect to s over Sh; then

(1.7)

Ipii (t+h)-pij (t) 1h jshipik(s+h)-pik(s)Ipkj(t-s)ds

3 2pkm(yt) 6 (sd Ipik(s+h)Is(s+h)-pik(s)Is(s)Ids,

16 (1.8)

using (1.6), (1.7), where I is the indicatorrfunction of S. The series on

the right of the inequality (1.8) is dominated by

2pkm( 1

X Sd .(u)du+ Pik pik(u)du

Sn[0,26] 'Sn[16,36]

B 4

Sd I p.R (t +t+u)du Ta. 2Sc < SA[0,26]

using (1.5), and so converges uniformly in h e[-16,16]. Also (see, e.g.,

[8] I, p. 636) we have, for each k,

36 ik(s ik h-)-0

lim IP +h)Is(s+h)-p (s) IS (s)Ids = 0,

so that, using uniform convergence in (1.8),

1pij (t+h)-pij (t) 0 as h 0.

The result, if i /a, follows at once by considering the transpose of

(pij(t)), which also satisfies (1.1), (1.2).

REMARK. If i,j ej, pij(t) is lower semi-continuous since

pij(s+t) = kg. pik(s)pkj(t)

and the terms of the series are non-negative and continuous; however, Example1.2

of § 1.4 illustrates that pij may not be continuous for all t > 0.

We now generalize an algebraic result of Doob (see [3] Theorem 11.1.2).

THEOREM 1.2. Suppose that the matrix U = i,j = 0,1,2, ... satisfies

0 uij < uij = uikukj for all i,j.

Then the set of states may be partitioned into disjoint subsetskI,I,J,

with the following properties.

(i) Either uij = 0 for all i or uji = 0 for all i, if j clu. (1.9)

(ii) There exist numbers {xi > 0, i 1161 such that, if

10 -

then

x. uij -1 v.. = 1.. for i, j x.

(1.10)

vii = SIJvj if i. e I,j e J, (1.11)

where {vi ,j ¢ 3'u} is a set of numbers satisfying

v. > 0, y v. = 1. (1.12) jeJ 3

Proof. Definel by equation (1.9). (Note that this is consistent with the

definition of for non-negative matrices which are independent of t.) If

U E 0 then all states j are in33-u and there is nothing to prove. So suppose

that uji > 0 for some i,j. Then )j- c (the complement of 't))is not empty,

because there exists k such 'thatuik ukj > 0. We consider henceforth only H_c

the reduced matrix "Jux Jo which satisfies

j i ui 0, u = u.,u, . = X u.kuki j LicK k 3 kJ/. U

for all i,j e

Also,

k e 1J- *there exist r,s such that urk > 0, uks > 0

*there exist t ,m such that u u k > 0, u km u > 0 Q ms there exist t ,mea uc such that ut.k > 0, ukm > 0.

(1.13)

Define 2- j a. = 3 1 + .0kj '

j ¢ U'

and note that, if

x. = u..a.131 U.

3' U' 3/

then, for each

0 < x. = X u..a.+ u..a. u..a.+ 2j < 1

i'd- 3-33 j%13-j3-3-j 3 • • 3 <1 5"U

If a = (a.) then x = Ua, so that Ux = U' a = Ua = x. Define (vij) by (10),

so that

vi 0, y v.. = 1, v.. = y v.,v,., j

jou 13 13 kj3.

u

Thus (v..) satisfies the conditions of Doob's theorem ([3] Theorem 11.1.2),

where it follows from (1.13) that (in Chung's notation) j F for every j,

so that (1.11), (1.12) hold.

COROLLARY. The only non-negative idempotent matrix with all diagonal

terms zero is the zero matrix.

Proof. If U 0 then 'a is not empty, so that from (1.10),(1.11),(1.12),

3 u3 .. > 0 for some j.

REMARK. Notice that the theorem and corollary will not necessarily

hold if we relax the conditions to U 0, U U2. We need consider only

the simple example

0 1 1 1 0 1 1 1 0

THEOREM 1.3. Suppose that f(pij(t)), t > 0, i, j = 0,1,2, ...) satisfy (1.1),

(1.2)andthatallpij(t) are measurable for t > O. Then, if i,j are not

both in the limits u.. = lim (t) exist and satisfy ij to Pli

uji 0, u.4 = X uikukj; uii > 0 if i a 1-1 ko-

p..(t) Xu. p .(t) p. (t)u . if i , (1.14) 13 k kJ kia

pij(t) = ipik(t)ukj k uikpkj(t) if j . (1.15)

In the notation of Theorem 1.2, may be partitioned into disjoint classes

I,J, ... and there exists a matrix (ITIJ),I,J e C, where Cis the index set

U =

and

- 12 -

for the classes distinct from satisfying

TTIJ

(t) 0, ITIJ

(s+t) = X KIIIK(s)II ( ), lim flij(t) =

too 6/J, (1.16)

where v.x.

p..13 (t) = Fl (t) if i I,j e J.

x. 1J 3

There exist functions II1J , i c J eG satisfying

riij(t) o, riij(s+t) = xrIiK(s)ram(t),

K

where v.

p.. (t) -/ = fi.1 (t) if ie,jc J.

)c- J 3

There exist functions17/ , I c G,j E satisfying

Ij(t) o,

Ij(s+t) = ix( s ) nic ( t )

where

pi j (t) = xilllj (t) if i e I,j e .

If i,j c , then

pi j (s+t) = IniK(s)rrici .

(1.17)

(1.18)

(1.19)

(1.20)

(1.21)

(1.22)

Proof. Suppose that i,j 4 and pij (t) 0. (If pij (t) E 0 then uij = 0.)

We can find states k ,m, and numbers s1 > 0, t1 > 0 such that IA i(si) > 0,

pim(t1) > 0. By Theorem 1.1, pz j is continuous at sl and pjm is

continuous at tl'

so there exist positive numbers c,d,6 such that

pti(si-t)

Thus, for any k and t

pa(si)

c, pjm(tl-t)

e (0,6] ,

pzi(si-t)pik(t)

d for all t c

cpik(t),

[0,6].

(1.23)

pkm(ti) pkj (t)pjm(t/-t) dPkj (t)

So, for each k, pik(t), pkj(t) are bounded in (0,6]. We may therefore

(by the Bolzano-Weierstrass theorem and the diagonal procedure) find a

- 13 -

sequence {tn) in (0,6] which decreases to zero such that the sequences

{pik(tn)}, {pkj (tn)} converge to finite limits uik,uki for each k.

Suppose that {tA) is another such sequence, so that {pik(tA)}, {pkj(tip}

converge to finite limits ulk, uy j for each k. Using (1.23),

1 1 Pik(s)P1-.&) .(tn ) ` k Pik(s) Pkm(t1) = pim(s+t,) < co,

so that the series on the left converges uniformly in n. Similarly, we

can show that 11 p (tn)pkj (s) converges uniformly in n. Since, from ik

Theorem 1.1, p..ij(s) is continuous for s > 0, we have (using, e.g.,

[8] II, §81)

pii(s) =114-co

i4(s+tn 11-->co ) = lim

vaX p.k(s)pkj(tn) = pik (s)ukj , (1.24)k

J

pij(s) = lim pi.;(tA+s) = lim X p.,(t)p,.(s)= k ix n xj k J

Note that, from (1.23),(1.24), the continuity of ptj , and Fatou's

lemma,

(1.25)

kX Pik(tn')uk* < X — P k(si)uk3 . < p - -J ko c k c tj(s+0) - c ( 1 < s ) P

so that the series on the left converges uniformly in n. Similarly,

we can show that lw ulkpkj(tn) converges uniformly in n. So, letting

s 0 along {tA} in (24) and along {tn} in (1.25), we have

u13 !. = lim k¢ n y p.k(v)uk, ulkukj,

(1.26) u

1. = lim y = 1 uiku kj* j n.4.0

Thus,from(1,216)u!--ui i.e. u. = lim p. (0 exists for all j' t+o

1,i / , and

u..0, u.. y uikuk . (1.27) 13 ko

- 14 -

Now we must have

umm > 0 for all m

(1.28)

• Otherwise suppose that m j and umm = O. The semi-group property

implies that there exists some i,j / such that pim(s) 0,

pmj(s) O. On the other hand, equation (1.27) and Theorem 1.2

imply that either ukm = 0 for all k /4 or umk = 0 for all k j ;

this would lead to the contradiction that either pim(s) E 0 from

(1.24) or pmi(s) E 0 from (1.25).

For i and any -56 / we have, by Fatou's lemma and the continuity

of pit,

> = lim pik(s)pkt(t) lim sup pit(s)ptt(t) + s+o k/J- s4,o

kWiLoiffif pik(s)pkt(t).

kft

So lim sup pit(s) < for all 2, Let t 0 and use Fatou's s+o •

lemma:

> lim inf pit(t) lim sup pits) - lim inf pit(s) utt + t+o s+o s4,0

X lim inf pik(s) k k s4o

(1.29)

Hence the series on the right is convergent. For any j multiply

by utj, sum over 4, and use (1.27):

X[

lim supP . (s) - lim inf p. (s)1u....u_. 3 = O. . 2.1,1 s+o it s+o it xx x

•

In particular, if 2, = j then u.. = lim p..(t) exists and equality .13 t+o 13

holds in (1.29), so that

u. y u. . if i 1, ko kJ

- 15 -

By considering the transpose of (pii(t)) we obtain similar results

for i j ef. The equality in (1.15) is obtained by letting

s+0 in

pii(t+s) = X Pik(t)Pkj(s), j

(1.30)

and using the continuity of p..ij and the uniform convergence of the

series on the right for s e (0,6] as in (1.24). The inequality in

(1.15) is obtained by letting t4-0 in (1.30) and using Fatou's lemma.

A similar argument proves (1.14). Equations (1.27),(1.28) imply

that we may apply Theorem 1.2 to the restriction of (u.) to

x'J.c. Using (1.9),(1.10),(1.11), the equalities in (1.14),(1.15)

become repectively vk p.. (t) = x. X. — (t) if i e I,

11K T ij ei

V. pli (t) =

x ik X p (t)xk if j e J, j keJ

so that, if i,j then pij(t)xj/xivi is independent of i e I by

(1.31) and independent of j e J by (1.32). We may therefore define

Hip) > 0 by (1.17) and then it follows that lim rhj(t) = t+0

Also, c_ p..(s+t)x.

J x.

laIJ(s+t) - 1-) - J

v.x. v.x. -s)1)k-(t) Pik( J 3 i 3 i k0

x. v x v.x , k - X X k „ t,$) (t) V.X.

Ke keK xk IK x.

= riIK(s) Ke (t),

proving (1.16). We note also from (1.32) that, for i c4,

pii(t)xj/vi is independent of j e J. We therefore define functions

i65,JeCby (1.19) and note that Ilij(t) ?, 0 and

(1.31)

(1.32)

- 16 -

p..(s+t)x. x.

= 1J v - Pik(s)Pki(t)

J 3 V x. v-1c x,v.

= y y II kJ V (t) Ke keK k

y nix(s)I7 (0, Kec,

proving (1.18). From (1.31) we see that, for j e , pij(t)pi is

independent of i e I. We therefore define functions riij(t) for

I c C, j c 3by (1.21) and note that Ilii(t) 0 and

pii(s+t) 1

nij (s+t) X-

1 - x. 114 Pik(s)Pkj (t)

x.vk

y y 1 fl(s)x (t) xi Ke g keK xk IK k K3

-- Y 1-111((s) TI-(t), KJ KY proving (1.20). Finally, if i,j e then

Vi P..(s+t) =1p. (s)p, .(t) =1 niK(s)x_ri .(t) ik KJ k KJ 13 k¢1- Ke keK k

= X rIiK(s)riKi(t). KeC

REMARK. If both i and j are in Examples 1.3, 1.4, and 1.5 of

§ 1.4 show that u.. may not exist. 1)

THEOREM 1.4. If f(p.i .(t)), t > 0, i,j = 0,1,2, ...} satisfy (1.1),

(1.2) and each p..(t) is measurable for t > 0, then each p..13 is either

identically zero or never zero in (0,w).

Proof. If we defineIlIJ(0) = IJ for I,J e C and use a theorem of

Chung ([2] Theorem 10.1), we see that the functions TILT are either

identically zero or never zero in p),-). Using the corollary to Chung's

theorem it follows that the functions also have the always -

or -never-zero property. From (1.17),(1.19),(1.21) it follows at once

1,7

- 17 -

that all pij have this property provided that not both i,j c 4 . If i,j c q and pij(ti) > 0 for some t1 > 0, then, for any t > 0, we

can find t2 such that 0 < t2< min(t,t1), and k / 4 such that

pik(t2)pkj(ti-t2) > O. Since k ¢ 4 , pkj is never zero so that pkj(t-t2) > O. Thus pij(t) > 0, so that the theorem holds for all pij.

Jurkat ([10]) has obtained further analytical properties of semi-

groups of the type (Flip)) and of functions of the type nip), i c 4

(and riii(t), j c'4 ). Using (1.17),(1.19),(1.21) the reader can

readily deduce corresponding results for our functions p..3.3(t) (when i,j

are not both in '4 ).

1.3 Resolvents and semi-groups

We now consider matrices {(rij(A)), A > 0,i,j = 0,1,2, —.1 satisfying

(1.1') and (1.2'). It can be shown by an induction argument that the

functions r..(A) are completely monotonic for A > 0 (see [19]) and hence

analytic (see, e.g., [22]). It follows that either r..13(A) = 0 for all

A > 0 or rij(A) > 0 for all A > O. For, if r (X o) = 0, then since

r. (A) is a non-increasing funcjion, rij (A) = 0 for all A A o' • this

inTliesthatrijN rij(A) is analytic.

With a slight abuse of notation we denote by S the set of states i

for which either rij(A) E 0 for all j or rji(A) = 0 for all j. If

E = {0,1,2, —.}, we write i E if there is a state j for which

rij(A) > 0 for one (and hence all) A > O. Similarly, we write E i

ifr.3 i

(A) > 0 for some j and (all) A > O. Thus i cS if and only if

E i ,v+ E.

THEOREM 1.5. Suppose that {rij(A)), A > 0, i,j = 0,1,2, ...1 satisfy

(1.1'),(1.2'). Then, if i,j are not both in , the limits u.. = limAn.(A) x

exist and there exist functionsPi (t), continuous on (0,..), such that

- 18 -

rii (A) is the Laplace transform of pii (t),pii (t)uij

t+ 0, and

0 < pii (t) < pi j (s+t) = 44

X p.k (s)pk . (t) s,t > 0. (1.33) 1 3

Proof. For fixed a > 0, any A > a and any i,j, (1.1') and (1.2') yield

(A-a) Xrik (A)rki (a)„.5 rij (a).

Multiplication by positive constants

2j

a. = , j = 0,1,2, ..., 1 +rk . (a)

1c .j

and summation over j give

(A-a) Xrik (A)yk <yi,A > a, i = 0,1,2, ..., k1.34)

where

y. ,Xa.r.1.(a), i = 0,1,2,

Now

y. = a.r.. (a) + X a.r.. (a) < a.r.. (a) + X < 1 13 3 13 j<j_ 3 13 ai

so that, if i E, then 0 < yi < We define

Arij (A) = r1. j ( +a) yi if E, j E, (1.35)

and note that r. (A) 0,

r.. A, A, ..(x) - r.. (p) + (X- 1p) y r. (A)r .(p) 13 1( k3

r. (A+a) - (114.00 [(A-Ea) - (11 4-a)] / r. 1kE lj ri j X+a)rkj (p+a)

and, using (1.34),

- 19 -

1 X A'1"-ii (A) ---[(X+a) - a] X r.-(A4a)Y. 1.

jE Y. j%4E 13

Collecting these together, we have, for all i E, j E,

% % % r..(A)>,0,n(A)-r..(1.04-(A-OZT..0(11)=0,D,.(x) 1.

ik kj rii lj

k%4- E

(1.36)

Reuter has shown ([20]) that (1.36) implies that there exist functions

% p..(t;a), continuous on (0,00), such that

r.. = f e-Xt ..(t;a)dt, A > 0,

P.(t;a) 0, (s+t;a) = y i5.k(s;c )ki(t;.), y i3. (t;0 ) lj .

3`''÷E

(1.37)

and the limitslimAr . (X), lim ..(t;a) exist and are equal. Putting X4°°

li ci t+0

p..(t;a) = P"..(t;a) eaty./y. if i E, jr\->E, and using (1.35),(1.36), 13

(1.37) we can easily check that, for all i E, j E, pij(t;a) is

continuous on (0,00) and satisfies

CO

r.. (x) = e-Xtp..(t;a)dt, > a,

0

;c0<,1)(s-q;c6)=X131-k(s;(t ;a) a)Pkj

kE }

(1.38)

and the limits lim Ar..(X), lim p..(t;a) exist and are equal. ij 13

t+o

If we now-repeat the above procedure with a different constants > 0,

say, in place of a > 0, we obtain (1.38) again with a replaced by (3.

- 20 -

Since pij(t;a), p1 (t;$) are continuous and have the same Laplace

transforms for A > max(a,O, by the uniqueness theorem for Laplace

transforms the functions are equal for all t> O. Thus for all i%-41,

jrv-)E there are continuous functions p..13(t), say, independent of a, such

that

c° r. (A) = e- Xtp..(t)dt, A > 0,

0

0 < p..(t) < p..(s+t) = p (s)p .(t) lim p..(t) = lim Ar..(A) 13 13 ik k3 13 t+o 13 A4-=,

(1.39)

By considering the transpose of (rij(A)), which also satisfies (1.1')

(1.2') we can construct similar functions p..13(t) when E#i and Ej

which satisfy (1.39) .(except that 4,,E is replaced by ) LE%+k'' Note

that, if i and j , the two constructions give identical functions

p..13(t) by the uniqueness theorem for Laplace transforms. Finally, if we

define

Pij (t) E 0 if i ,v1+ E or if E ti j,

we may replace the summation in (1.39) by lko and then -(1.39) holds

whenever i,j are not both in

COROLLARY. If {(r..(A)), A > 0, i,j = 0,1,2, ...} satisfy (1.1'), —

(1.2') and (1.4'), there exist functions pij(t), continuous on (0,0,,),

satisfying (1.1), (1.2) and (1.4) such that 13

r..13(A) =

-Atp..(t)dt.

Proof. Since Ar..11(A) 1 as A co, for all i, 5- is empty.

THEOREM 1.6. If i,j are both in and if rii(A) c..13 as A co, then

there exists a function pij(t) satisfying

0 < pii(t) + c°, pii(t) < + > 0, pii(s+t) = YPik(s)Pkj(t), s,t > 0,

and such that

rij(A) = cij + f e xtp..(t)dt, A > 0.

- 21 -

Proof. The function 40pik(s)pkj(t) is a function of s+t only. For,

if s + t = s1 + t1 and s > sl, say, then

kigpik(s)pki m = Ipic6(s1)pa(s-s )pki(t) = piz(si)p,j(t1). 44.

We therefore define

p..(t) = y pik (s)pkj (t-s) for any s e (0,t). 13

If p > A > 0 then, using B. Levi's theorem,

r.. (X) -r.. (u) OD CO

13 W

lj = r.1k (P)rkj (X) = f e-Pse-At P ik(s)Pkj(t)dtds - A kf/9 o o ki'3-

= I

I e-Pse-Atp..(s+t)dtds = I e- (i-X)sF(s)ds, o o 13

where F(s) = Ie-Aupij(u)du. Since F(s) is monotonic, it follows thatu=s

F(s) is finite and hence absolutely continuous in (0,=), and

F'(s) = -e-Aspij(s)

for a.a. s > 0. Integrating by parts,

> r..(x)-r..(11) = lim e-(P-X)aF(a) - re-Psp,.(s)ds a+o a 13

= lim I I(a,c.)(s)(e-As-(P-X)a-e-Ps)p..(s)ds. a+o o

It is easy to check that the integrand is non-negative and

t(e-X s _e-p 13 s)p..(s) as a+o and thus, by monotone convergence,

> fc° (e -X -p s-e s)p..(s)ds, 0

'so that pii(s) < + w a.a. s > 0. The result follows by letting u '00

and again using monotone convergence.

Ifi,jarebothirthen.limxrij (x) may not exist (§ 1.4, Examples 1.4,

00

A4-co

1.5) even if r..13(X) -+ 0 as A o. Examples can occur (§ 1.4, Example 1.1) in

- 22 -

which pij(t) takes the value + when pij(t) is everywhere finite'it

may not be continuous (51.4, Example 1.2)and lim p,.(t) may not exist t+o 13

(51.4, Example 13)even if lipuXr..(x) exists. However, if lim p..(t) x4. 13 t+o

exists so does limXrij . (X) and the two limits are equal. (This is jam

a standard Abelian theorem (see, e.g., [22]).)

1.4. Examples

Examples 1.1-1.5 are based upon the examples of Yuskevic and Orey

(asp. We first construct a (substochastic) semi-group, whose states

we shall denote as

0,1; (n,k), k = 1,2, ..., mn; n = 1,2, ...,

by means of a matrix Q = (qii) defined as follows

g(n,k),(n,k+1) = -g(n,k),(n,k) = in(n)' k = 1,2, —'111n-l'

= - g(n,mn),1 '

n 1(n,mn),(n,m ) =

m(n),

goo = an' gij = 0 otherwise,

n = 1,2, ...,

where the integers mn, positive numbers m(n) and positive constants an

such that n=1 an < 1 are to be specified later. All states are stable

and Q gives rise to a minimal process P(t) = (p1 (t)) < 1 (see, e.g.,

[3] 511.18) satisfying

p'ij(t) = i«licikpki(t) = ?ik(t)qkj, t > O. (1.40)

The Laplace transforms r..(X), calculated from the transformed equations 13

(XI - Q)R(x) = I = R(x)(XI - Q) (where R(X) = (rij(X)), are

- 23

r,

ort,$),(n,s+r)(A) - n r fl

[x+111(n)]r+1 r = 0,1, ..

m -s+1

(,) _ [111(n)] r X[X+m(n)]mn-

s+1 '

1 an[

m.(n)]

s-1 ,

roo- (X) = A+-1- r11(X) = 1

ro,(n,$)(x)

(x+1)[A+m(n)]s

1 co an[m(n)] rol(X) -X (X+ 1)

r.. = 0 otherwise, n=1

mn [X+m(n)

s = 1,2, ..., mn

Hence we can evaluate (pli(t)). We shall require only pll(t) = 1

and m

rt) - [ m.( 1)] n e-m(n)ssinn-1ds,

P(n,1),1`-) (ml 1).

n = 1,2, ... .

it) t 0,

_ [m(n)] n -m(n)t mn -1 13(n,1),1 -̀1

We now introduce new states a0 and. define the following functions

for t > O.

paa(t)Epf3 (t)Epia(t)Epo(t)Epl3a(t)Epao(t)Epis(t)E0,

Pa.(t)Ep' oy .(t)+q

op03 .(t)Ep'

03 .(t)+p

03 .(t), j 74 0,

•

pir3(t)Ep'il(t)+qpil(t)Eptil(t), i 1.

The functions paj(t), pis(t) are finite and non-negative (see, e.g.,

[3] p. 135) and satisfy for s,t > 0

paj(s+t) = poak(s)pkj(t), pil3(s+t) = ?ik(s)pk(t).

m -s

Inn

co m

pas (t)= X anpi(n 1) 1(t) = X an

n-1 " n=1 m 1 n- •

e-m(n)t

tmn-1

.. (1.42)

- 24 -

The Laplace transforms of these functions exist and are given by

r aa (X)Er (X)Er. (x)Er .(X)

.ErP,04 MET- ao(x)Er113(X)E 0,

1a

-1

a,(n,$) [aX [n(L(r1))1iS

r , 00=(X+1)ro,(n,$)(A) - m s = 1,2, ...,

(1.41)

111-s+1_

r(n,$), (A) 'xr

(x) s III 01,$),1 m,-s+1 '

= 1,2, ...,

[X+m(n)] "

[am(n)]

an[m(n)] n

rof3

(x)=xrol

(x) - c--vT)1 n-1 m

n [A+m(n)]

m . a n [n(n)] h

r a 1 (x)=(X+1)ro,i

(X) = X m ,

n-1 n

Now, for s,t > 0,

YPa • k(s)Pk(t) Vo = {P'ok(s)+Pok(s)/Pikl(t)

= k o n

iYanP(n' 1)' k(s)1Pikl(t)' using (1.40),

= Xan P(n 1) k(s)Pikl(t) n "

= Xanp'(11,1),i(s+t)

In order that the semi-group property shall hold for transitions from

a to 13, we must define

Using B. Levi's theorem, we can show that the Laplace transform ro(x)

of pap,(t) exists and

- 25 -

m 0. a [m(n)] n

ras (X) = X n

n=1[x+m.(n)]mn < co. (1.43)

Since r (X) < 0,, we have past) < a.e. It is not hard to check that

all of the functions defined by (1.41),(1.43) satisfy the resolvent

equation. We now have

(r..(x)) satisfies (1.1'),(1.2'), (p..13(t)) satisfies (1.2),

0 pig (t)

where the state space now includes a,s c . Using the monotone

convergence of the series in (1.43) for 0< X < .0 and noting that each

term tends to zero as X .± 0., we have lim r (X) = 0. In examples X->-co

1.1-1.5 P(t) and R(A) are as defined above. The behaviour of pa (t)

and rc3 (x) is regulated in each example by the values of the parameters

m,m(n),an

Example 1.1. R(X) satisfies (1.1'),(1.2') and is the Laplace transform

of P(t) which satisfies (1.2) but p„(1) =

This is the example of Yuskevic. Put mn = n,m(n) = n-1. For any

t > 0, we have, from (1.42),

a Ca-1)n e-(n-1)ttn-1 af3(t) = n

n=1 (n-1):

y an( (tel-1)n

-le

= Xa f (t), ") (11-1)427(n-1)) n=1 (a-1): n=ln n

say.

Now- tel-t increases from 0 to 1 as t increases from 0 to 1, but decreases

to 0 as t increases from 1 to .0, so using Stirling's formula,

fn(1) .0 as n 4- 00, fn(t) 4- 0 as n 4- 0. if t 1.

- 26 -

Thus there exists a sequence {nk} + such that fn (1) >, 2k, k = 1,2, ... .

k Define

2-k 2 if n = nk'

0 if n nk'

so that r=1 an = 1 and p

a (1) >, y

k=1• f . . If t 1, (t) < 1 for

n n.

all n >,,N, say, so that

N-1 pa (t) anfn(t) + y a

n <

n=1 n=N

Since anfn(t) 0 as t 0 and rn=lanfn(t) < co for t e OD,to], to < 1,

by monotone convergence, lim p as

= O. Thus limXr ,,(X) = O. t+o a' A.->0. a'

Example 1.2. R(x) satisfies (1.1 1 ),(1.2') and is the Laplace transform

of P(t) which satisfies (1.1),(1.2) but has lim sup pap,(t) = + t+to

lim inf pats(t) < £ for some to > 0 and given e > O. toto

Orey's example with m(n) = (mn-1)/(t0+2-n) will suffice to give

lim sup p (t) > lim inf p (t). However, we can control p (t) better t+t

o t4-t0

by putting m(n) = (mn-1)/(to+bn), where 0 < bn < ... From (1.42),

m o-m.(n)ttmn-1 pa (t) = a

n1 n (n_1

m -1 -(m -1) (m -1) n e n ✓)/(27r(m-1)) = n

n=1 (mn-1)!

an

=

mn-1

to 1+b 21 n 1- j

t 1-t/(to+bn) t e o n

Co

= X h(t), say. n=1 hl

(t),

The terms behave in a similar manner to those in Example 1.1 inasmuch

as (for fixed t,n and variable mn) hn(to+bn) co as ml co whilst

hn(t) 0 as mn -1, co if t # to+bn. Choose any real numbers an > 0 such

that 41.1an < 1. Given e > 0, for n = 1,2, ... define bn = 4-n and

define mn so that

bn

' hn(to+bn) > n, hn(t) < — for t < to 2 + — and t to+2bn. 2n

We then have

CO

pcS(to+bn) hn(to+bn) n, p(to+2bn) < X = e, n = 1,2, ... , k=1 2

so that lim inf po(t) < c, lim sup pas(t) = + Clearly po(t) is finite t0 t+ to

for all t > 0; for,if V 1bn < t to + 2bn,n = 1,2, ... ,

Co

past)

1 + hn(to+bn) < k=1 2'

and, if t > 2b1 or t to,

CO

pa (t) X < k=1 2

Since hn(t) 4. 0 for 0 < t to, then applying monotone convergence to

yn=1

hn(t) we have lim p cy3(t) = 0. Thus lim Xr eiti3(x) = 0. t+o X'

Note that we can never have P(t) satisfying (1.1),(1.2) with lim inf paci(t)=0 toto

- 28 -

for some to > 0 and a,(3 e such that pa (t) O. (For if pa(t) t 0

then by Theorem 1.4, pa(t) > 0 for all t > 0. So

po(to+t) pak(t,)Pkk(t)P4(to-t1) > 0 for some k j , t1 < to.

Letting t 0, lim inf pul3(to+t) pak(ti)ukkpk (to-tl) > 0, since t+to

u > 0 by Theorem 1.3.)


P(t) which satisfies (1.1),(1.2) but has lim sup past) = + co, t+o

lim inf p ,(t) = 0, even though lim Xr (A) exists. t+o a A4-.0 as

Again, Orey's example with m(n) = (mn-1)/2-11, mn 2, suffices to give

lim sup pct (t) >lim inf p(t). however, it allows us no control over t+o t+o

the terms of the series for Xr (2,a(X)' so instead we define m(n) = (m-1)/bn

where 0 < bn < ... Just as in Example 1.2,

.., an Ou mn-le-(mn-1) 1/(2“

m n-1)) pas (t).

n=

y

1 n .n

[t el-tibn] mn-1 h (t), say. b 2-ff n=1 n

Again (for fixed t,n, variable mn) as mn co, hn(bn) co whilst hn(t) 0

for t bn.

Now, from (1.43),

anX anX

Xr a' M.

n =1 [(a/m(n))+1] n1 [(Abni(n-1))+1rmn

- n= 1

gn(X), say;

(m 1-1)!

- 29 -

gn(X) increases from 0 at X = 0 to a maximum at A = 1/bn and then decreases

to 0 as A -+ co; gn(l/bn) + an/ebn as mn > co. For n = 1,2, , define

bn = 4-n, an = bn/2n (so Xn_lan < 1). Define m so that

hn(bn) n, hn(t) < 4-n for t < lbn' and t 2bn'

and note that, for t < bn, k > e, hn(t/k)/hn(t) < (e/k)n-1 < e/k (since

et bn > 1 and ,-t/bn < e). So for n = 1,2, .

(4-n) hn(4-n) n,

p (2.4-n) = n2

h r 2.4r-r-1

1) + hr(2.4-n) < e(n-2) 4 .-r o as n

r=1 4 ( n / r=n-1 4n-1

r=n-1

so that lim inf p a' ,(t) = 0, lim sup past) = + Arguing just as in

to t+o

Example 1.2 we see that p03(t) < co for all t > 0. Finally, z =1g.n(X) is .Co

uniformly convergent for A > 0, since gn(x) . 1/(e2n) and 2,n=12-n < co.

So lim Xra„(X) = /11%1 gn(A) = 0.


of P(t) which satisfies (1.1),(1.2), P(t) continuous for t > 0, but

lim sup xral3(X) = + co, lim inf Xr a'

(X) = 0. A400 k+c°

As in Example 1.3 put m(n) = (mn-1)/bn, where 0 < bn < c.). Note that

gn(l/bn) an/4bn for all ma 3 2 and gn(X) < anX for all A > 0. Choose

co such that 0 < co < 1. Define an,bn,cn (in this order) for n = 1,2, ...

as follows.

Choose an such that 0 < 1/cn_i. < an < cn-1/2n. Then gn(x) < 1/2n for all A

Choose bn such that 0 < bn < cn-1 and bn < an/(4n). Then gn(l/bn) n.

Choose mn such that hn(t) < 1/2n for all t cn-1.

• •

- 30 -

Choose cn such that 0 < en < bn and Xw k=lgk(x)..< 1/2n for all 'X 1/ cn.

So we have Z.lan < 1, (1/bn)rot (1/bn) (l/bn) n, and

n (1/cn)rc4 (1/cn) = gk(l/cn) + gk(1/cn)

k=1 k=n+1

< 1/2n + X gk(1/ck_i) < (1/2n) + X (1/2k) 0 as n k=n+1

Hence lim inf xraf3(x) = 0, lim sup Xr(x) = + co (and hence we must have A4-0

p(113(t) limit as t 0). Now p a0 (t) is finite and continuous for t > 0.

For, if bn < t, hk(t) < hk(bk) if k n, and hk(t) < 1/2k if k > n. Now

Xi(=ihk(bk) + Xk=n+11/2k < co, so that 1;„,hk(t) is a uniformly convergent

series of continuous functions on [bn,co) and therefore p03.(t) is

continuous on [bn,00) for every n.


P(t) which satisfies (1.1), (1.2), P(t) continuous for t > 0, but

Xr(i a (X) 4- co as X

This is a simpler example which still gives Xrao(X)44. limit as A

Put m(n) = l/bn, mn = 1. From (1.43),

... a X Aral (a) = X an+1 - X gn(X), say,

n=1 n n=1

where ga(x) fi an/bn as A 00. From (1.42),

pe(t) = X (an/bn)e-t/bn

" n=1 7

and so by monotone convergence lim pot = lim Xr(A) =n=1an/bn. If too X-+.0

k=n+1

00

- 31 -

we put an = bn = 2-n then the limits are + 00. For any 6 > 0, pot (t)

is the sum of a series of continuous functions, uniformly convergent in

[6,00). So pot0(0 is finite and continuous for t > 0.

Example 1.6. R(A) satisfies (1.1'), (1.2') with ra. (X) c > 0 as

A 4. co. The simplest example is

0 c I. R(X) =

0 0

In general, if R(A) satisfies (1.1'), (1.2') with states a,f3 c!-1 such

that r (A) 0 as A -4- co, on putting

raf3(x) ras + c,

% ij r(x)E rij (X) unless i = a and j = 0,

we obtain a new process ?(A) satisfying (1.1'), (1.2') with lim r A) = c. A400 a'

„,(

1.5. Ergodic limits

We turn now to the question of the behaviour of

log p..(t)

t as t

when P(t) E (pij(t)) satisfies (1.1), (1.2). This behaviour is discussed

by Kingman in paragraph 7 of [11] under the additional assumption (1.4).

We note that in view of our Theorem 1.4, (1.4) could be replaced by the

weaker condition

pii(t) > 0 for t > 0 and all i

(i.e.)j* is empty), which is the condition that Kingman imposes on Markov

processes in his paragraph 2.

The following examples show that it is possible to construct P(t) wiith

StateS i,j for which

logpij(t) + cc. as t 00

- 32 -

and that pii(t) + co arbitrarily fast. In Example 1.7 all states lie

in different irreducible classes (i.e. there are no pairs i,j with i y j

for which both pii(t) > 0 and pii(t) > 0) whilst in Example 1.8 all

states lie in the same irreducible class.

Example 1.7. Suppose that

0 0 a3 a4 • 0 •

0 0 0 0 • • •

Q 0 b3 3 0 • • •

0 b4 0 4 • • •

• • •

7 where ak > 0, bk > 0. Jurkat ([9]) has pointed out that, if there exists

P(t) satisfying (1.1). (1.2), (1.4) such that P'(0) = Q, there will

necessarily be a minimal process. We shall choose values for the constants

ak,bk so that there is a minimal process P(t) with the property that,

given .any function f(t) as t co,

P 1 =1122 = °' P12 = c°' P(t) < p

12(t) 3 f(t) for all t sufficiently

large,

where log p.1 .(t) • l Pij im

It is convenient to use the construction of Reuter and Ledermann ([21])

for P(t). Suppose that Qr is the matrix obtained by truncating Q after

the first r rows and columns and let xij ,x. (11) be the elements in the ij

ith row and jth column of Qr,(Qr)n respectively. Then, for 1 i r,2 < j r,

xi i(1144:1 = x. (n+1) = bkxik

(n) 12 k=3

xij(11+1) = aixii(n) jx..

13 (n) = ix. (n)

Solving these equations we have

t+co t

- 33 -

xij ;--- a-j (n) -n-1 , 2 < j < r,

(n) .n xii =1 2 <i<r ,

r rro x1.2 - Xakbkkn-2 n 2,x12 = 0, k=3

x12 1 (n) .-1a-1 b 2 2,

Pij(t) = a(ejt-1)/j, j > 2,

pi2(t) = bi(eit-1)/i, i > 2,

00

P12(t) = y akbk(ekt_ 1-kt)/k2, k 3

pii(t) = 0, for all other values of i,j.

Thus, each state is in an irreducible class by itself. Noting that

2 lu Au e2 eu-1-u < lu2eu for u 0, we have

00. At2 X akbke

kt/2‹p12(t)<1't2 1 akbkekt,

k=3 k=3

Now suppose that f(t) coand choose yk t co such that

n=1 n:

-34-

1k 4 lm for k = 0,1, ...,[f(m+1)] + 1, m = 1,2, ... .

Putting akbk = e-ky k, we obtain

P12(t) le X ek(t-yk) <

k=3

since yk t+1 for k sufficiently large. On the other hand, if

t e [m,m+l), m = 2,3, ...

p12 (t) > k:yklm

1 [f(m+1)] + 1 f(t),

and clearly p12 = m'

Example 1.8. Suppose that

0 a2 a3 a4

b2 2 0 0

Q b3 0 3 0

b4 0 0 4

• • •

where ak > 0, bk > 0, M = max(1,I =2akbk), M < (1.44)

We shall again use the construction of Reuter and Ledermann to obtain

a minimal P(t) such that P'(0) = Q, with

log co '111 ' lim — +

and we shall show that the constants ak,bk may be chosen so that, if

f(t) + co as t co, then pli(t) f(t) for all t sufficiently large. There

is just one irreducible class in this example (since, for all

i,j,ajbi > 0 implies that pij(t) > 0), so it follows from the corollary

- 35 -

to Theorem 1 of [11] that pi, sup k = + k

Suppose that xi,x2, xr is any row of Qr and denote by

x1 (n) 'x2 (n) '

xr(n) the corresponding row of (Q

r)n. Thus

r x1 (n+1)

b-x-(n), j2 33

(1.45)

(n+1) x. a-x

1(n)+jx-(n), 1 < j < r,

Suppose first that xi(n),x2 n), xr

(n) is the first row of (Q

r)n.

Then, from (1.45),

x1

0, = 0 3 = a. '

r x1(2) a b )(- (2) 1 k k 3

a3 k=2

r x (3) = Xabk, x.(3) = a. .

k=2 Xakb+a-j2.

k=2 3 1 k k k 3 •

(1.46)

For the terms where n > 3 we verify by induction that

x1(n) 2n-3Mn-3 akb

k' k=2 (1.47)

(n) n- -n- n--1 x. :a Pr- -+a X a b kn-31

i • -k=2k k -

Observe that, for any integer p > 0,

r j akbk

kP -15 ak bk

kP +jIakbkg k=2 k<j k>j

- 36 -

j at.hi,j13 + akbk0+1

k=2 " k=2

< mjp+1 i akvp+1. (1.48)

k=2

Note, from (1.46), that (1.47) holds for n = 3. Then, using (1.44),

(1.45), (1.48) and the induction hypothesis (1.47), we have

r x (n+1) = b.x.(n) 2n-3Mn-3 { a.b.jn-1+ ( a.b.) ( akbkkn-3)} 1

j=2 3 3 j=2 3 3 j=2 3 3 k=2-

r r

2n-3.W-3 a.b.jn-l+Mb kn-3 j=2 3 3 k=2

ak k

m f < 2n

-1 j=2 3 '

x

(n+1) = Ca) (n) - +jx.

J

z 2 33{a. 1 akhirkn-2+a.jn+a.j at.b.tyn-'

3k=2 3 k=2

2n-3mn-3 i a. a b kn-2+a.in+a. [ min-2+ a b kn-21 }

3k=2 k k J J k=2 k k

21-1-3W-3 i 2a< i a kb kr1-2+2Ma - jn 3k=2 J

in-21,e1-2 f a. jn4.a. a b kn-2

3 3k=2 k k •

So, by induction, (1.47) holds for n 3. Now, from (1.46), (1.47),

t2

tnx

(n)

P11 (r) = l+tx1+ xl

(2) +

1

2 n=3 n!

-37

1-2

'. n=3 k=2

2 r a b = 1 Nt

k=

7

2 8k14 k

(2Mkt)11 2 L 3k 2 n=3

n!

2 r a b 1 + Mt + 1 7 k k 2Mkt

2 --y- (e -1-2Mkt). 8m- k=2 k

Letting r -4- 0. and again using eu

- 1 - u < lu2eu, u 0, we have

Mt 2

. t n2m n-3-n-3 r

1 + -7- + n! akuh k0,

M 2 + t2 ci a b e2Mkt. p11(t) 1 -7- 4M k=2 k k

(1.49)

Also,

131j(r)(t) = tx. + .2 t nx (n)

x.(2) i

2 3 n. n=3

J r a.jt2 tu211- 3Mr1-3ain-1

tn2n-3mn-3a.

akbkk

a.jt2

a.e2MUt

a. r .akb

t+ + J k 2Mkt a.. e 1 -2Mkt

2 3.-

8M 8M3 k=L 2 k3

M2k2t2 )

using (1.46),(1.47). Letting r . and using eu-1 -u-u2/2! < -16u3eu, u 0,

a.jt2 a.e2Mit a.t3 .

plj

(t) ajt + 3

2 + + kke2te

8M3j 6 k=2

(1.50)

Now, from (1.45),

x.3 (11-1-1)

3 ?...jx-(11), 1 < rp n % 1 1

so that

(n) .n-1 x a., n 1.

3 2 n=3 n! n=3 n! k=2

Using (1.45) again

- 38 -

r x1(11+1) j2 a.b.jn-1 , n 1,

= 3 )

SO

n , t x1 (n) n r n-2 (r)( = v L Pll >, 1+ X Ya.b. j

n=2 n! n=2 n! j=2 J 3

r a.b. n-n r a-b. jt .

j2 / 33 / t 3 - ).23 (e -1-3t).

j=2 n=2 n! j=2 )

Letting r .÷ co and again using eu-1-u 4u2 elu , we have

,2

P11(t) y a.b.elit, 4 j=2

(1.51)

Suppose that we have chosen the coefficients akbk so that the series

in (1.49) and (1.50) converge for t > 0. We can then show that

p.. (t) < c for all i,j. 13

For the inequality

p1j(r)(2t) pli(r)(t)pij(r)(t), i 1,

gives,letting r co,

-39-

P- p1 (2t)

Plitt)

since ai > 0 implies that pli(t) > 0. So P(t) < co provided that we can

find values of ak,bk satisfying (1.44) such that the series in (1.49)

converges. To show this and to show that pi/(t) can be made to tend

to + co.arbitrarily fast we proceed exactly as in Example 1.7 choosing

Yk,ak,bk just as before. Note that we do have M <

-40-

SECTION 2. INEQUALITIES FOR P-FUNCTIONS.

2.1 p-Functions

Some important problems of the theory of Markov chains with

a continuous time parameter and denumerable state space [3] have been

solved recently (e.g. [15]) by the study of a class 5 of functions

called Eltanc.Lions [12,16]. A real-valued function p on (o,&) is a p-function if, for n 1, it satisfies the inequalities

F(ti,t2, to ; p) 0,

G(ti,t2, to ; p)

where, for 0 = to < t1 < < tn, we define

(2.1)

(2.2)

F(ti,t2, , l)r • =

11 p(t. -t- ), • 3 3i

t ; L n = 11 +

n r=0 0j0<il<...<jr+1 1 . 0

G(ti,t2,...,tn ;p) = 1 - y F(ti,t2,...,tr;p). r=1

(2.3)

(2.4)

If n = 1,2,3, for example, the expressions (2.3),(2.4) become

F(t1 ; p) = p(t/), F(t1,t2 ;p) = p(t2) - p(ti)p(t2-t1),

F(ti,t2,t3;p) = p(t3) -p(t1)p(t3-t1)-p(t2)p(t3-t2)+p(t1)p(t2-ti)p(t3-t2),

G(t1;p) = 1-p(t1), G(t1,t2;p) = 1-p(t1)-p(t2) + p(ti)p(t2-t1).

G(ti,t2,t3;p) = 1 - p(ti) - p(t2) - p(t3) + p(t1)p(t2-t1) + p(t1)p(t3-t/)

+ p(t2)p(t

3-t2) p(t

1)p(t

2-t

1)p(t

3-t2).

A p-function p belongs to the class 6) of standard p-functions if

lim p(t) = 1, . t+o

and in this case it is customary to define p(o) = 1.

-41.-

One of the mysteries of p-function theory (and, in particular,

of Markov chain theory, since Markov transition functions pii(t) are

p-functions [12]) is as follows. If p c P, t > o, p(t) = M, m = inf

{p(s): 0 < s < t}, 0 m < NE< 1, what are the possible values of the

pairs (M,m)? Some partial results are already known. Davidson [4,

Prop 4.1] and Blackwell and Freedman [1] showed independently that

- m + m2

if m M.<

3 4 if m

by using (2.1),(2.2) for n = 1,2. In a posthumous paper [5] Davidson

2 proved that the critical value of M does not exceed 7 and Griffeath [6]

has recently sharpened this result to

3n2 - 2m + 3 4 if m>, 1

M 2 1 if m .

These results were obtained by using (2.1),(2.2) for n = 1,2,3. Davidson

(see e.g. [16]) using an inequality of Bloomfield also proved that

M .1 e M < (2.5)

1 - e1 1 if m < -6 .

These inequalities point out those values (M,m) which a p-function

cannot assume.; something is also known about the values which are accessible.

Davidson has given an example (see [4, example 4.1]) which illustrates

that any pair (M,m) satisfying

- 42 -

M < e-(1-m), 0 < m < 1, (2.6)

is assumed by some standard p-function. Only the region between (2.5)

and (2.6) is now uncharted; if this region were accessible certain

questions relating to Markov groups [23] would be answered.

Our investigation makes a corresponding contribution to the solution

of an (almost) equivalent problem posed by Kingman in [14]. A consequence

[4, Example 4.3] of Davidson's example is that if ej is given the topology

of pointwise convergence then Q is not a closed subset of . It is a

simple corollary to our result that those p-functions inewhich are almost •

everywhere zero may never take values above 1 - 2, (This class of functions

is not empty: indeed, Davidson's Example 4.3 is such a function).

2.2. Some identities

It is convenient to list here a number of elementary, but useful

algebraic results for the functions F(ti,t2, to ; p),

G(ti,t2, to ; p), which we shall require later.

PROPOSITION 2.1. If p is a real-valued function on and if F,G

are defined as in (2.3),(2.4), then for n 2,

F(si,s2, sn ; p) = F(si,s2, sn_2,sn ; p) - F(sn-sn_l ; p)F(si,s2,

sn-1 ; p), (2.7)

F(si,s2, sn ; p) = F(sn-sn_i,sn sn_2, sn-si,sn ; p),(2.8)

F(si,s2 sn ; p) = F(s2,s3, sn ; p) - F(s1;p)F(s2-s1,s3-sl,

sn-Y,P), (2.9)

n-1 F(sn;p) =F(sr;p)F(sr1.1-sr,sr4.2-sr,...,sn-sr;p)+F(si,s2,...,sn,p).

r1

(2.10)

- 43 -

G(si,s2, sn ; p) = G(sps2, sn_2,sn;p) -

G(sn-sn_l;p)F(si,s2,...,sn_1;p) (2.11)

G(si,s2,...,sn;p) = G(s2,s3,...,sn;p)-F(s1;p)G(s2-sl,s3-sl ,

...,sn-sl;P), n-1

G(sn;p) = r1 F(s

r;p)G(sr+i-sr,sr+2-sr,...,sn-sr;p) +

=1

G(si,s2,...,sn;p),

n-1 n-1 X G(sr+1-sr'sr+1-sr-1'...,sr+1-s1;p) = G(sr+1-sr'sr+2-sr' r=1 r=1

...,sn-sr;p),

(where s1 may be zero in (2.14).)

Proof. From (2.3),

(2.12)

(2.13)

(2.14)

n-2

F(si,s2,...,s;p) = (-1)r p(s. )p(s.; -s4 ) p(s -s ) n r=o o<ji<...<jr<n-1 31 J2 J1 n jr

n-1 r-1

p(sn-sn_1) X (-1)r n p(s. -s4 ) r=1 o=j0<j1<...<4=n-1 i=o 3i+1 :i

= F(si,s2,...,sn_2,sn;p)

n-2 r

- F(sn- sn-1'131) (-1)r Y II p(s. -s_;.)

r=o o=j0<ji<...<4.1.1=n4i=o Ji+1 )1

= F(s1,s2,...,sn_2,sn;p) - F(sn-sn-1;p)F(si,s2,...,sn_1;p),

(noting that p(t) = F(t;p)) which proves (2.7). Also,

F(sn-sn-1' sn-sn_2,...,sn-si,sn;p)

- 44 -

n-1 = y (-1) X p [(s-s4 )-(sn-s. )111(s -s. )-(s-sj ) r=o °=jo<il<"'<jr+1=n n Jr 3r+1 n 3r-1 n

p [

(s -s. )-(s -s. )1 n

o n 31

n-1 = (-1)r

p(s. -s. )p(s. -s. )...p(s. -s. )

r=o °=jo`il<'''`jr+1=n j1 3° 32 31 3r+1 3r

=

proving (2.8). The remaining identities (2.9) - (2.14) may now be deduced

from (2.7) and (2.8).

F(s1,s2,...,sn;p) = F(sn-sn_psn-sn_2,...,sn-spsn;p)

= F(sn-sn-l'sn-sn-2,...,sn-s2,sn;p)-F[sn-(sn-si);OF(s

n-sn-1'

sn-sn-2'"

s n-s .p)

= F(s2,s3,...,sn;p) - F(si;p)F(s2-si,s3-si,...,sn-si;p),

proving (2.9). Identity (2.10) follows from (2.9):

n-1 X F(sr;p)F(sr+i-sr, sr+2-sr,... sn-sr ; p)

r=1

n-1 = y {F(sr+i,sr+2,...,s n.n)-FC, ,sr,sr+1,...,sn;p)}

= F(sn;p)-F(s1,s2,...,s ;

Introducing the functions G, n-2

G(s1,s2,...,sn_2,sn;p) = 1 - X F(si,s2,...,sr;p) - F(sl,s2,...,sn_2,sn;p) r1

n-2 = 1 - X F(si,s2,...,sr;p) -{F(s1,s2,...,sn;p)

r=1

+ F(sn-sn_l;p)F(sl,s2 ,.. .,sn_l ;p) }

• • •

r=1

•

= G(s1,s2 ,...,sn;p) + G(sn-sn_l;p)F(s1,s2 '""sn-1.43)

-45-

which proves (2.11). Identity (2.12) follows from (2.9):

G(sl,s2,...,sn;p)

n = 1 - F(s1;p) - X IF(s,,s3,...,sr;p)-17(s1;p)F(s2-sl,s3-si,...,sr-si;p)1

r=2 4

= G(s2,s3,...,sn;p) - F(s1;p)G(s2-si-s3-si,...,sn-si;p)

and (2.13) follows from (2.12):

n-1 y F(sr;p)G(sr.„-sr, sr4.2-sr,...,sn-sr;p)

r=1

n-1 ° X {G(sr4.1,sr4.2,...,sn;p)-G(sr,sr41,...,sn;p)}

r=1

= G(sn;p) - G(si,s2,...,sn;p).

For (2.14) we use (2.8):

n-1 rX 1G(sr+1-sr'sr+1-sr-1 =

n-1 =

rX 1 q {1 - y 1F(sr+1-sr'

sr+1

-sr-1"—'sr-1-1-se)} ==

n-1 n-1 r =X1 -y XF(s

q+;-sq,s(14.2-sq,...,sr-scesr+i-sq;p)

r=1 r=1 q=1

n-1 n-1 = X {1 - X F(sq+i-sq,sq+2-sq,...,sr+1-sq;p)} q=1 r=q

n-1 = X 1

G(sq+1 -sq' sq zq' ...,sn-sq;p).

Noting that (2.14) still holds if si = 0 the proof is now complete.

2.3 Bounds for p-functions

Suppose M,p are numbers such that, for some p e9 and some 0 < s < t,

M = P(t), p = P(s),

where we assume, without loss of generality, that p < 1. Define

t = inf { u o : p(u) = p },

so that, since p is continuous and standard,

0 < T S < t, p(r) = p, p(u) > p if u < T .

For any integer n >, 3, consider sequences

- ct-

-46-

0 < t1

< t2

<

with

to-1 =T,t = t.

Using the convention to = 0, (2.13) and (2.14),

n-1 M = 1 - p(tr)G(tr+i-tr,tri.2-tr, , tn

-tr;p)

r=0

n-1 - 1 - p G(tr+/-tr,tr+2-tr,

•

tn-tr;P) r=0

n-1 = 1 - p G(tr+i-tr,tr+/-tr_l, tr+/-to ;p)

r=o n-2

- 1 - p G(tr+i-tr,tri4-tr_1' tr+i-to;p).

r=o

Now choose t t 1, 2, ..., tn_2 so that

tr

= inf {u 3 0 : p(u) =r1,

(2.15)

where (3n-1

= p. Since p is continuous and standard,

0 < t1 < t2 < < to-2

<tn-1'

p(tr) = f3r, r = 1,2, ..., n-1 (2.16)

We now require

PROPOSITION 2.2 If, for n r 3, ti,t2 , tn_2 satisfy (2J61 then for

r = 0,1, ..., n-2,

1 - 4 G(tr+1-trtr+1-tr-1' tr+1-to;p)

Proof. From (2.10),

r p(tro.) =

k p(tk)F(tki_l-tk,tk.4.2-tk, tr+i-tk ;p), =o

so that

Rr+1 RkF(t t t t = .." tr+ftk;P)

k=o

Sr F(tiol-tk,tk4.2-tk, tr.4.1-tk;p), k=o

and using (2.8),

r .); F(tio_1-tk,tk+2-tk , tr+i-tk;p)

k=o

(2.17)

-47-

= y tr+1-tk;11)' k=o

Hence r

tr+l-to;p) = 1 - k_y F(tr+/-tr,tr+/-tr_1,...,tr+l-to =o

as required.

Thus (2.15),(2.17) show that

1

M < 1 - p (n -1)(1-p7i7-1

We now have the following

(2.18)

THEOREM 2.3 If p e6", t > 0, M = p(t), p = p(s), se[0,t] then

M < 1 + plop . (2.19)

Proof If p c8 , (2.19) holds for s e (0,t) by letting n -+ .... in (2.18),

and for s e [0,t] by continuity of p. Since the topology of 5 is the

topology of pointwise convergence, p E Q if there exist pk e 4 such that

pk(t)

COROLLARY

then

since

p(t) for all t > 0. Hence (2.19) also holds for p

1. If p e(?, t > 0, M = p(t), m = inf {p(s) : 0 s t},

l+mlogm ifm —e (2.20)

1 . if m 1 - (2.21)

{

Inequality (2.20) is an immediate consequence of (2.19). If

p is continuous and standard, there is some s e [0,t] for

M <

Proof.

1 m < T

which p = 1- , and (2.21) follows from (2.19).

COROLLARY 2. If p e6" is almost everywhere zero, then

p(t) 1 - for all t > O.

Proof. If p ei;and p(t) > 1 - for some t > 0 then there exist

-48-

pk ce such that pk(t) p(t) and pk(t) > 1 - 2, Thus, by Corollary 1,

pk(s) > 1 — for s E [0,t] and every k and hence p(s) > e [0,t].

2.4. Semi-p-functions

A real-valued function p on (p,c.) which enjoys property (2.1) but not

necessarily (2.2) is called a semi-p-function [17].

Suppose, for example, that the matrix functions

{(pii(t)), t > 0, i,j = 1,2,3,...I satisfy

CO

0 < 1 pi. (t) < co, p..(s+t) = X p.

1k(s)pk.(t) (2.22)

k=1 3

An easy calculation [17] shows that the diagonal functions pii are semi-

p-functions. On the other hand, if these functions also satisfy

CO

y p.. (t) = 1 (2.23) j=1 13

then the functions p..13 are transition functions of a continuous time

parameter Markov chain, and a corresponding calculation [17] confirms that

the transition functions pii are actually p-functions. As we have seen

already, Markov transition functions defined by (2.22),(2.23) share many

common properties with the matrix semigroups of functions defined only

by (2.22). More generally, Kingman [17] has shown that p-functions and

semi-p-functions exhibit similar behaviour.

In the following paragraphs we shall point out a similarity in

structure between the classes of (Lebesgue) measurable semi-p-functions

and measurable p-functions. We show that if p is a measurable semi-p-

function then either

(i) p is almost everywhere zero, or

(ii) p is of the form upo, where 0 < u < 1 and po is a standard semi-

p-function, i.e.

lim po(t) = 1. too

This corresponds to the well known result for p-functions which is proved

by a rather different method in [13].

2.5 Continuity of measurable semi-p-functions

The following theorem is motivated by the known results for Markov

chains [3], Theorem 11.1, and by Theorem 1.1 of this thesis.

THEOREM 2.4. A measurable semi-p-function is either continuous on (0,.)

or zero almost everywhere.

Proof. Suppose p is a measurable semi-p-function and let S = {t > 0 ; p(t) > 0}

have positive measure. Given x > 0, we shall show that p is continuous

at x. Writing t1 = s, t2 = s+t, t3 = s+t+u, the inequalities (2.1) for

the cases n = 1,2,3, become

p(s) 0

(2.24)

p(s+t) p(s)p(t)

'(2.25)

p(s+t+u) + p(s)p(t)p(u) p(s)p(t+u) + p(s+t)p(u). (2.26)

Now (2.25) implies that S is a semi-module and that -log p(s) is finite

and subadditive on S. So by theorems of Hille and Phillips [7, Theorems

7.3.2, 7.4.2] there exists a c(x,...) such that (a,-)IS and p is bounded

on all closed subintervals of (2a,..). Given c > 0, define for n=1,2,3,...,

( 1

Sn= fyc(3a,4a):p(y_x) ena $ p(y-,x) < 6 ems, cenxl. (2.27)

' p2(y-x)

00

Since U S = (3a,4a) there exists a positive integer N such that m(SN) > 0 n=1 n

(where m is Lebesgue measure). Now define

p1(t) =

e-NtP(t),

t > 0,

and note that p1 is also a semi-p-function satisfying (2.24),(2.25),(2.26)

in particular. Thus, if y c SN,

y c (3a,4a), 1 < e5Na, Pl(Y) p1(2y-x)

t.Y , Pi(Y-x) < 6 ' 2 P (Y-x)

< E.

i

If hc(-x,x) and y c SN we have, using (2.25),

(2.28)

- 50 -

Pi (Y4) 1 Pl(x+11)-pl(x) Pl(x) < pl (y-x

, Pi (Y) [P1(Y+h)-131(Y)] /31(y-x)

(2.29)

and applying (2.26) for pl with s = u = y -x, t = x+h, and using

(2.25) again,

2p/(y+h) p1(2y-x+h)

Pl(x+h) -131(x) %' Pl(x) Pl'Y-) x Pi (Y-x)

2 . 1 p1(2y-x) ( [P1(Y+11) Pi(Y)] 2 [Pi (2Y-x+h) -Pi (2Y-x) ] 2

pl Y-x) Pi (Y-x) Pi (Y-x)

(2.30)

Thus, using (2.28),(2.29) and (2.30) imply that for all h c (-x,x)

and all y c SN

1111(x+h) -P1(x)k 2e5Na

1111(Y+h)-P1(Y)14e10Na

IP1(2Y-x+h)-p1 (2y-x)I + E.

(2.31)

Integration of (2.31) with respect to y over SN gives

5Na 11)1(x+h)-Pi(x)1

2e 1Pi(Y+h)-131(Y)IdY m(SN)'SN

el0Na +1p,(2y-x+h)-pi(2y-x)Idy +

m '1■1) SN 4a 10Na 8a-x

I Pi (Y+h) (Y) I dY e2m 6a-x

I Pi (Y+h) (Y) I dY 6 3a

2e5Na

m(SN)

2e10Na (8a

IP1(Y+h)-Pi(Y)IdY 6 M ( N) .3a

(2.32)

Since p1 is bounded on (3a-x, 8a+x) and measurable, letting h40. in (2.32)

and using a familiar result ([8], I. n.636),

- 51 -

8a 2_10Na

lim sup Ipi(x+h)-pl(x)! lim sup Ip/(y+h) -p/(y)Idy z = E. ho (SN)

Remembering that e>o is arbitrary, it follows that pl and hence p, is

continuous for all x>o.

REMARK. We cannot hope to prove continuity without using (2.26)

or a higher inequality of (2.1) since examples are known of discontinuous,

measurable, finite, subadditive functions [7, p 246].

2.6 Representation theorem.

We first prove some formulae for the functions F(ti,t2,...,tn;p)

and G(t1,t2,...,tn;p) which we shall require later.

PROPOSITION 2.5. Suppose that p is a continuous semi-p-function on (0,w)

and u = lim inf p(t). Then 0< u 1 and for m 3 1, t 0, t+o

lim sup F(tl,t2,...,tm,t;p) = (1-u)F(ti,t2,...,tm;p), t+t m

(2.33)

lim inf lim inf lim inf lim inf F(t1,t2,...,tm,si,s2,...,st,t;p) s1m

s s2+s1 s3+ s2 t t-1

=

- ak F(t-tm;p)F(ti,t2,•••,tm;p), (2.34)

Q whereat = X (1-u) s, k = 0,1,2,... . (2.35)

s=o

If, moreover, p is a p-function,

lim inf lim inf lim inf lim inf G(t1,t2,...,tm,s,,s2,...,sk,t;p) s1+tm s2"1 s34-s2 est-1

= G(ti,t2,...,tm_u,t;p) - akG(t-tm;p)F(ti,t2,...,tm;p). (2.36)

3a

- 52 -

Proof. Letting s+o in (2.25) and using the continuity of p we have

p(t) ; p(t) lim sup p(s), s+o

which implies that either p(t) E o or o <, lim sup p(s) < 1; so s4,0

0 4 U < 1. From (2.7),

F(ti,t2,...,tm,t;p) = F(tpt2,...,tm 1,t;p) F(t-tm;p)F(t1,t2,...,tm;p),

and (2.33) follows using the continuity of F(ti,t2,...,tm_i,t;p) in

t e (tm_1,=) and letting t+tm.

We prove (2.34) by a simple induction on t. If t = 0, there are no

limiting operations and (2.34) reduces to (2.7). Assume that (2.34)

holds for some k 0 limiting operations; then by the induction hypothesis,

(2.33) and (2.7),

lim inf {lim inf lim inf lim inf F(t t . .,t s s ...,s 4- t•p)} l' 2' m' l' 2' 21"

= lim inf {F(t1,t2;...,tm,t;p) - at F(t-s1;p)F(t1,t2,...,tm,s1;p)-1 s1m

= F(t1,t2,...,tmt;p) - ak F(t-tm;p)(1-u)F(t1,t2,...,tm;p)

= {F(ti,t2,...,tmi,t;p)-F(t-tm,p)F(t1,...,tm;p)}

ak F(t-tm;p)(1-u)F(ti,t2,...,tm;p)

= - ut+1F(t-tm;p)F(t1,t2,...,tm;p).

Thus (2.34) holds for t + 1 limiting operations and hence for all t o

by induction.

If p is a continuous p-function we prove (2.36) similarly by induction

on t. If t = 0, there are no limiting operations and (2.36) reduces to

(2.11). Assume that (2.36) holds for some t 0 limiting operations and

then by the induction hypothesis, (2.33) and (2.11),

s1+ in s2+s1 s34, s2 st+1

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lim inf Ulm inf lim inf lim inf s14,t.m 4,st s34,s2 st+1

= lim inf {G(tl,t2,...,tm,t;p) - azG(t-s1;p)F(t1,t2,...,tm,s1;p)} s14,tm

= - crtG(t-tm;p)(1-u)F(ti,t2,...,tm;p)

= { G(ti t2 , ,tnh t ;p) -G (t-tm;p) F (t i , t 2 , tra;r3) }

- otG(t-tm;p)(1-u)F(t1,t2 ,...,tm;p)

= G(t1,t2,...,tm_1't;p) -crt+1G(t-tm;p)F(t1,t2,...,tm;p).

Thus (2.36) holds for t + 1 limiting operations and hence for all t by

induction and the proposition is proved.

REMARK. Obviously a dual result holds by interchanging the roles of

lim inf, lim sup. We shall find, however, that lim p(t) exists for too

continuous semi-p-functions and then we may replace lim inf and lim sup

by lim in (2.33),(2.34);(2.36).

We now prove our main theorem on semi-p-functions.

THEOREM 2.6. .Suppose p(t) is a measurable semi-p-function. Then either

(i) p may be expressed in the form

p(t) = upo(t), o o.

Proof. Suppose p(t) > o on a set of positive measure so that p is

continuous for t>oby Theorem 2.4. Let us write u = lim inf p(t). As t+o

in Proposition 2.5 we note that o lim sup p(t) 1, so if u = 1 then t+o

limp(t) exists and p(t) is a standard semi-p-function. t+o

Suppose therefore that o u < 1 and define az by (2.35). Our first

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aim is to show that a p is a semi-p-function for each t = 0,1,2,... .

We therefore check that the inequalities (2.1) hold for the functions

cs p. For o < tl,

1 — F(t .a p) = p(t1 ) >' o.

For o < t1 < t2, choose variables sps2,...,st so that o < t1 < s1 <..

• < st < t2 and then let st st_1+ ... s/ t/ in o < F(tpsps2, .

•

st,t2;p). Using (2.34) (with t = t2, m = 1) we obtain

0 <lira inf lira inf • • • lira inf. F(tpsi,s2,• • • ,st,,t2;p) s1+t1 52+s1 yst_l

= F(t2;p) - ak F(t2-t1;p)F(t/;p) (2.38)

1 F(tl, ;(5g).

We cannot have u = o, otherwise (2.35) would imply at + co as t.+0.,

contradicting (2.38) for suitable tl,t2. Thus o o) we have

lira sup p(t2-t1) . u, t2A1

i.e. lira p(t) exists and equals u (and henceforth we may replace lira inf, t+o

lira sup, by lira in (2.33),(2.34),(2.36). ). We now perform a rather

complicated induction; we show that if

o<tl<si<s2<...<st<t2<st+1‹...<s2t<t3<...<s(n_2)t<tn_1<s(n-2)2,4.1<..

...<s(n_wtn, then for n = 1,2,... the operation of taking the (n-1).

-55-

limits

urn lira 000 lim lira *00 lim ... lim

S

1

+t

1

s24,s1 s(n-2)t+14,tn-1 s(n-1)2,+s(n-l)t-1 st+s t-1 st+14,t2

(which we shall denote by limn) in the expression

F E F(t s „ s , s n 1' 1- L- , k,t2,sk+1, s(n_lot,tn;p) ;• 0

produces

0 lim Fn a 1 = F(t t n l' 2'''''tn;GtP)* R

(2.39)

We have, in fact, already shown (2.39) for n = 1,2. Assume now that

(2.39) holds for some n 1; using (2.34),

liM lim

lim F s(n-l)t+1n s(n-l)t+2(n-l)t+1

snt+sn2,-1

= F(ti,S1,4 ,St,t2,• • • ,S(11......1)24 ,tri+1;P) ...atF(tri+ITtn;P)Fn•

Hence, taking the remaining (n-1)2, limits and using the induction hypothesis

and (2.7),

1 p t t 0< lim F = .; _p) n+1 n+1 ' n+i t

atF(tn+1-tn;p){6 F(t1,t2 ,...,t23;601)}

1 = at {F(tl,t2,...,tn...1,tn+1;atrO F(tnil:t11;atP)F(tpt2,...ptn ;atP)}

= F(t1,t2,...,t1144 ;aip).

Thus (2.39) now. holds for n+1 and hence for all n = 1,2,... by induction,-

and ap is' a semi-p-function for each t = 0,1,2,... . Now if po is

defined by (2.37),

aQ 1 rr.. ÷11, EAL-1,t2,...,tn;a9p) -+ F(ti, ...,tnypo), as

-56-

implying that po is a standard semi-p-function and the theorem is proved.

We may use Theorem 2.6 as an alternative means of establishing the

corresponding theorem for measurable p-functions [13].

COROLLARY. Suppose p(t) is a measurable p-function. Then either

(i) p may be expressed in the form

p(t) = u po(t), 0 O.

Proof. (Kingman) By Theorem 2.6 we have only to show that po is,a

p-function. First note that po(t) < 1 for all t > O. (For po(t) is

bounded above by u, andpo(s) > 1 would imply that po(ns) w as n 4-

For any h > 0, the sequence {po(nh), n = 1,2,...} is a generalised renewal

sequence bounded above by unity and so [17, Prop.1] is actually a renewal

sequence. Thus [12, Prop. 6] since po is continuous and standard, po is

a p-function.

REMARK. It is possible to check directly from the inequalities (2.1),

(2.2) that po is a p-function by proving that G(t1,t2,...,tn;6zp) 0

using (2.36) and a similar induction to that used with (2.34) in the

theorem to show that F(t1,t2,...,tn;atp) >, O.

Finally, note that p(t) = 0 almost everywhere can occur, even for

p-functions; moreover, non-measurable semi-p-functions can also exist.

We refer the reader to [13] and [3, p 290].

- 57 -

REFERENCES

1. D. BLACKWELL & D. FREEDMAN, On the local behaviour of Markov

transition probabilities', Ann. Math. Statist. 39 (1968) 2123-27.

2. K.L. CHUNG, 'On the boundary theory for Markov chains,' Acta

Math. 110 (1963) 19-77.

3. , Markov chains with stationary transition probabilities.

2nd. edn (Springer - Verlag, 1967).

4. R. DAVIDSON, 'Arithmetic and other properties of certain Delphic

semigroups II', Z. Wahrscheinlichkeitstheorie und Verw. Gebiete

10 (1968) 146-72.

5. , 'Smith's phenomenon and jump p -functions,' (Stochastic

geometry and stochastic analysis, ed E.F. Harding & D.G. Kendall,

Wiley, London, to appear).

6. D. GRIFFEATH, 'The maximal oscillation problem for regenerative

phenomena', (to appear).

7. E. HILLE & R.S. PHILLIPS, Functional Analysis and Semi-groups

(American Mathematical Society Colloquium Publications, 1957).

8. E.W. HOBSON, The theory of functions of a real variable, Vols I

and II, 3rd edn. (Cambridge, 1927, reprinted by Dover Publications

Inc. 1957).

9. W.B. JURKAT, 'On semi-groups of positive matrices', Scripta Math.

24 (1959) 123-31, 207-18.

10. , 'On the analytic structure of semi-groups of positive

matrices', Math. Z. 73 (1960) 346-65.

11. J.F.C. KINGMAN, 'The exponential decay of Markov transition

probabilities', Proc. London Math. Soc. (3) 13 (1963) 337-58.

12. 'The stochastic theory of regenerative events', Z.

Wahrscheinlichkeitstheorie und Verw. Gebiete 2 (1964) 180-224.

13. , 'On measurable p-functions', Z. Wahrscheinlichkeitstheorie

und Verw Gebiete 11 (1968) 1-8.

14. , 'A class of positive - definite functions', Problems in

Analysis, (Princeton University Press, 1970).

- 58 -

15. , 'Markov transition probabilities V', Z. Wahrscheinlich-

keitstheorie und Verw. Gebiete 17 (1971) 89-103.

16. , Regenerative phenomena (Wiley, London 1972).

17. , 'Semi-p-functions', (To appear).

18. S. OBEY, 'Non-differentiability of absolute probabilities of Markov

chains' Quart. J. Math. (Oxford) (2) 13 (1962) 252-54.

19. G.E.H. REUTER, 'Nul solutions of the Kolmogorov differential

equations', Mathematika 14 (1967) 56-61.

20. , 'Note on resolvents of denumerable submarkovian processes',

Z. Wahrscheinlichkeitstheorie und Verw. Gebiete 9 (1967) 16-19.

21. and W. LEDERMANN, 'On the differential equations for the

transition probabilities of Markov processes with enumerably many

state', Proc. Cambridge Philos. Soc. 49 (1953) 247-62.

22. D.V. WIDDER, The Laplace transform, (Princeton University Press, 1941).

23. D. WILLIAMS, 'On operator semigroups and Markov groups', Z.

Wahrscheinlichkeitstheorie und Verw. Gebiete 13 (1969) 280-85.

Documents

(I) .CONTINUOUS PARAME1ER SEMIGROUPS OF POSITIVE MATRICES ... · (i) .continuous parame1er semigroups of positive matrices. (2) inequalities for p-functions. by alan george cornish