Interpolating between the compound-geometric and compound-Poisson distributions

compound-Poisson distributions

Citation for published version (APA):

Harn, van, K., & Steutel, F. W. (1975). Interpolating between the compound-geometric and compound-Poisson distributions. (Memorandum COSOR; Vol. 7510). Technische Hogeschool Eindhoven.

Document status and date: Published: 01/01/1975

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STATISTICS AND OPERATIONS RESEARCH GROUP

Memorandum COSOR 75-10

Interpolating between the compound-geometric and compound-Poisson distributions

by

K. van Harn and F.W. Steutel

by

K. van Harn and F.W. Steutel

O. Summary

The starting point of this paper is the characterization of the compound-Poisson, i.e. the infinitely divisible lattice distributions (class C

1) and the subset of compound-geometric lattice distributions (class CO) by the non-negativity of recursively defined quantities (section 1). In section 2 we generalize these recursion relations by introducing sequences c (a) for _n a £ [0,1] and thus we obtain classes of distributions C t which we wish to

a

be increasing with a. This can be achieved by an appropriate choice of cn(a) (section 3). For the classes C we obtain other properties, which generalize

a

known. properties of the class C

I • Another subdivision of this class is given

in section 4, where we consider the compound-negative-binomial lattice dis-tributions.

1. Preliminaries

We only consider lattice distributions (p) 0' i.e. distributions on the n n2

nonnegative integers, with PO > O. The corresponding probability generating

function (pgf) "" P(z) :=

L

n=O n p z , n

is then unequal to zero on

Izl

< p for some p > O. If (p ) >0 (or P) is

infi-n infi-

n-nitely divisible (inf div), then P is unequal to zero on the closed unit disk

Izl

S I (cf. [4]).

For an inf div pgf P we have the following representation (see [1]).

Theorem 1.1. A pgf P is inf div if and only if there is a A > 0 and a pgf Q), such that

(I. 1 ) P(z)

=

exp[A(QI(z) - l)J •

Thus, the class of inf div distributions on the nonnegative integers coin-cides with the class of compound-Poisson distributions on the nonnegative integers. We recall the definition of such distributions.

Definition 1.2. A compound (lattice) distribution is a distribution with pgf P(z)

=

G(Q(z». where G and Q are pgf's. The corresponding random va-riable x can then be written as

+ ••• + x , -n

where ~'~1'~2"" are independent, while n has pgf G and each of the ~i has pgf Q.

Example 1.3.

- p

i) _n has a geometric (p) distribution: G(z)

=

~--.-- t so a

aompound-geome-- pz

tria distribution (p ) >0 has pgf

n n-(1.2) P ( z) = -:---~-i..,-~ 1 - P

- pQO(z) •

ii) n has a Poisson (A) distribution: G(z)

=

exp[A(z - J)J, so a aompound-Poisson distribution (p ) >0 has pgf

n

n-(1. 3) P(z)

=

exp[A(Ql(Z) - I)J •

Definition 1.4.

Co

is the set of all compound-geometric distributions.

C

1 is the set of all compound-Poisson distributions, which, according to theorem 1.1, is equal to the set of all inf div distributions. If P is a pgf corres-ponding to a distribution in a class C, we shall also say, that P E C.

Remark. To some of the generating functions, corresponding to distributions in

Co

or C_I, we add an index 0 or I, to be able to fit them in the more ge-neral notation of the following section. Further we denote the coefficient of zn of a generating function Ra(z) by rn(a), or r_n, if no confusion is possible, while the sequence (rn(a»n~O is simply denoted by rn(a).

We now formulate some more or less known theorems, which will be the start-ing point for our investigations in the next sections. For the sake of com-pleteness we write down the proofs in detail and in a few lemma's we give some properties of the appearing quantities. We start with the definition of absolute monotoniclty, and then consider the distributions (pn)n~O in

C) •

Definition 1.5. A function R on the complex numbers is called

absoZuteZy

mo-notone

(abs mon), if there are rn ~ 0 (n

=

0,1,2, ••• ) and p > 0, such that

00

R(z)

=

I

for Izi < P •

n=O

Theorem 1.6. A pgf P is inf div (P €

C

I) if and only if p' (z)

RI(z) := P(z) is absolutely monotone. Proof. Let P €

C

I• Then there are A >

0

and pgf QI' such that log P(z)

=

A(QI (z) - 1) •

Hence:

R, (21) = p' (z)/P(z) = [log P(z)]' = ,AQi (z) , which is a abs mon function (izi < 1).

Conversely, let P'(z)/P(z)

=

R1(Z) be abs mon, i.e.

00

P'(z)/P(z)

=

_R1(z)

=

for Iz 1 < p ,

with r (1) ~ 0 for all nand p > O. Integrating from 0 to z (1211 < p), we

n obtain ~ rn (1) zn+1 log P(z)

=

log PO + L n=O n + 1 121 1 < p , 00 r (I) or, if we define A := -log

Po

and Qt(z) :=

*

L

nn+]

n=O P(z) = exp[A(Ql(Z) - 1)], 1 Z I < p , with A > 0 and Q 1 abs mono n+I 21

Since P'(z) = P(z).RI(z), Izi < p, we have the following relations

(I.4) n = 0,1,2, ••• t

where r (1) ~ 0 for all n. It follows (see lemma ].7), that n

00 r (1)

\' n

L _ _ '!"<oo_

n=O n + 1

New Ql(z) is convergent for Izl S 1, so that exp[A(QI(z) - I)J is an analytic continuation of P(z) to the whole disk Izl S 1_ As P itself is analytic on

I z

I

S I, we have

P(z)

=

exp[A(Ql(Z) - )J, for

I

z I S I •

From P (1)

₌

I _{we get now Qt(l)}

₌

1 , _{so Q) is a pgf and P is a} compound-Poisson pgf, i.e. PEe 1 •

From Ql(l) = 1 we then obtain:

00 r ₍₁₎ 00 r ₍₁₎

I

n 1 ,

_L

n It _-log

Po

0

r

=

or:

₌

= _•

n=O n + 1 n=O n + 1

Remark. For the quantities

R),

It and Q} from the preceding proof we have the relations

( I .5)

=

_{AQj (z), A}

=

_-log

_{Po =}

z

J

R) (u)du •

o

Some properties of the recursions (1.4) are now summarized in the following lemma.

Lemma 1.7. The sequence r (I), defined by (1.4), and its generating function

n

RI have the following properties

i) R) has a radius of convergence p > 0, and p' (z)

RI (z)

=

P(z) for Izi < p • ii) If rn(l) ~ 0 for all n, then

(1.,6)

00 r (I)

\' n

L -~'r<oo, n=O n + )

so that RI(z) is convergent for Izi < I.

Conversely, if a sequence r (1) with r (I) ~ 0 for all n and satisfying (1.6)

n n

Proof. From (1.4) we obtain n por = {n + l)p I - \ ' p r n n+ L k n-k ' k=1 so that n Po

I

r _n

I

~ (n + 1 )p _n+ 1 + \' _L P _{k n-k '}

I

r I k=l or Hence, if we define N 1~I(x) :=

I

(x > 0, N E: E) : n=O N N n 2pol~1 (x) :s;

l

(n+ I)p IX + n

l l

Pk 1r _k1xn ~

n=O n+ n=O k=O n

00 _N N

I

n

l

l

Irn_klxn

~

~ _{(n+l)p IX +} Pk n=O n+ k=O n=k N N k+ :s; p'(x) +

l

P k

I

Irn1x n ~ p'(x) + p(x)I~1 (x) • k=O n=O

Now, if we choose a Xo > 0, such that P(x

O) < 2PO' then we have p' (x

O)

1~I(xo) :s; 2p _ P(x ) < 00 for all N ,

o

0

00

from which we can conclude that

L

Irnlx~ <

00,

n=O

radius of convergence p ~ xo. Taking generating mediately have

P'(z)

=

P(Z)Rt{Z) •

Let r ~ 0 for all n. Then we can write n

00

and R1(z)

=

L

r zn has a

n=O n

im-eo eo n ""

n~o ~

k~O

Pkrn- k

=

~ ~ r n-k l. _Pk l.

-n ""

k=O n=k n 1 - Po "" _l. \' P _n+1 n=O eo co r eo r

=

l

_Pk

l

n + ]n + k <!: POl n:-1 '

k""O n=O n=O

so that 00 r - P

l

-.!!.. s

0 < co , n=O n+J Po eo d eo r I ( ) \' n zn+ t l.'t < I. As R z = -d l.~) z n=O n r

and

I

_n+i

n z n+l convergent for I zl n=O

follows that R

1(z)

is

convergent for Izl < 1. Conversely, let rn <!: 0 for all

eo r

n and a:=

L

n:-I < 00. Choose Po E (O,IJ and define Pn' n <!: 1 by

n=O n (n + l)p 1 == \' P r n+ l. k n-k' k=O n=O,I,2, ••••

It follows that P_n <!: 0 for all n. Now define qn := Pn/PO' n ~ Ot then also

n=0,1,2, ••• ,

with qo

=

1. If ~ is the counting measure on {0,1,2, ••• } and fn(k) k <!: 0, n <!: 0, then fn(k) S fO(k) for all nand

f

fO d~ = a <

00,

so

the dominated-convergence theorem

r k : == '!'"k-+--1-+-n- , that by eo rk

I

lim

I

k + 1 + = lim f (k)d~(k) 1J:'7<X> k=O n ~ n lim f (k)d~(k) = 0 • n

So we can choose nO and aO < 1, such that

00 r k V

l

S a O • n<!:nO k=O k + 1 + n eo

We now estimate

L

qn as follows. For N > nO we have n=O

As a O < N+I

I

n=l q = n 1, we have N+l

I

_qn S n=t N

I

qn+l

=

n=O N

I

n=O n + N "" r k

I

qn

I

k + 1 n=O k=O S a + n for N > _nO n -1 0 00 a

I

and so

I

n

_o

-1

L

n=O 1 - a_{O n=O} qn' _n=O qn

""

=: c < 00

.

q S n

Since qn

=

_{Pn/PO' it follows that c} ~ I and n~o _Pn

=

POc, so that (pn)n~O

becomes a probability distribution, if we take PO == l/c.

0

Remark. The r _n (1) may be expressed explicitly in the P : from the first n _n equations of (1.4) by Cramer's rule we obtain

Po

0

PI PI Po ZPZ (I.7) _r n-l (l) -n det _3P3 = _{Po P} 2 p] PO. Po

.

_np Pn- l

.

(I ~ ..

.

C PI n

From theorem 1.6 and lemma 1.7 one easily derives the following characteriza-tion of inf div lattice distribucharacteriza-tions, first given by Katti [3J.

Theorem 1.8. A distribution (p ) >0 is inf div if and only if the quantities n n-r _n (I), n == 0,1, ••• defined by n (n + 1)p I"

I

Pkrn_k(I) n+ k=O are nonnegative.

Proof. Let (p ) >0 be inf dive According to lemma 1.7i) the generating func-n func-

n-tion R_t of the r

n(!) exists and is equal to P'(z)/P(z) for Izi < p. As this function is abs mon, by theorem 1.6 we have: r (1) ~ 0 for all n. Conversely,

n

let rn(l) ~ 0 for all n. Then (lemma 1.7ii» the generating function R) of the r _n (I) exists and is equal to P'(z)/P(z) for Izl < 1, so that P'(z)/P(z) is abs mono The inf div of P follows now from theorem 1.6.

0

We now turn to the distributions in CO' and derive some analogous results.

Theorem 1.9. A pgf P is compound-geometric (P E CO) if and only if

P(z) - PO

RO (z):= zp (z)

is absolutely monotone.

Proof. Let P E CO' Then there are p E (0,1) and pgf QO' such that

I-p

P(z) - 1 - PQo(z) • Hence:

which is an abs mon function (Izl ~ 1).

Conversely, let RO be abs mon, i.e.

n r (O)z ,

n for I zl < p,

with rn(O) ~ 0 for all nand p > O. As P(z) - Po

=

zP(z)RO(Z)' we have

Po

P (z)

=

-:----=--~ _{- zRO(z) ,}

I zl

< p 1

or, if we define p

:=

I - Po and QO(z) :=

p

zRO(z):

_ I-p

with p € (0,1) and Q

O abs mono Since

P(Z) - Po

z

=

P(z).RO(z),

I zl

< p ,

we have the following relations

(I .8) n = 0,1,2, ••• ,

where r (0) 2 0 for all n, from which we can derive (see lemma 1.10), that

n

I

n=O

r (0) < 1 •

n

Now RO(z), and hence QO(z), is convergent for Izi ~ 1, and, as

I

n=O

r (0) < 1,

n I ,

1 -

t .

~--~~ __ ~s an analytic continuation of P(z) to the closed unit disk

1 - pQo z)

Izi ~ 1. As P itself is analytic for Izi ~ 1, we have

From P (1) I-p P (z ) = -:---"!!"-~""-':­ - pQO(z) , = 1 we now get Q O (1) metric pgf: P € CO' From Q

O (I) = I we then obtain:

for

I

z

I

~ 1 • = 1 , so Q O is a pgf 00

~

RO(!)

=

I, or:

I

n=O and P is a compound-geo-r (0) = p = I _{- po·} n

Remark. The following relations now hold for the quantities R

O' p and QO

from the preceding proof

or if we take qo =

o ,

(1.9)

00

D

RO(z)

=

~

PQO(z), p = 1 - Po

=

I

_{rn(O) =RO(!)' Qo(z)}

=t

zRa(z) • n=O

Lemma 1.10. The sequence r (0), defined by (1.8), and its generating

func-n

tion RO have the following properties

i) RO has a radius of convergence p > 0, and

(1.10)

p(z) - PO RO(z) = zP(z)

ii) If r (0) ~ 0 for all n, then

n (l.lI)

I

n=O r (0) < 1 , n for 1 z 1 < p •

so that RO(z) is convergent for

Izl

~ I.

Conversely, if a sequence r _n (0) with r _n (0) ~ 0 for all n and satisfying

(1.11)

is

given, then (1.8) defines a unique probability distribution (pn)n~O.

Proof. Exactly as in the proof of lemma 1.7 we can show, that RO has a radius of convergence p > 0 (replace the factor n+ t by 1). Then, taking generating functions, from (1.8) we immediately obtain

Let r (0) ~

o

_{for all n. Then we can write} n

co co _n co 00 co

1 _{- Po =}

I

_{Pn+l =}

I

I

_{Pkr -k (0) =}

I

_Pk

I

r (0) =

L

r (0)

n=O n=O k=O n k=O n=O n n=O n

so that 00

I

n=O 00 r (0)

=

n 1 - Po < I ,

and RO(z)

=

I

r (O)zn is convergent for n=O n

Conversely, let r ~ 0 for all n and a

:=

n

1 z 1 ~ J. 00

I

rn < 1. Choose Po € (O,lJ,

de-n=O

,

fine P

n' n ~ I by (1.8) (then Pn ~ 0 then also

for all n) and define qn := PnlpO' n ~ 0,

We can write now 00

I

qn+1

=

n=O Hence, as a < I, 00

I

qn+1 = n=O 00 00

I

qk

L

k=O n=k 00 r n-k

=

a(1 + I

~

a ,or

L

qn

=

~---a-

=: c < 00 • n=O 00 00

Since qn

=

Pn/PO' it follows that c ~ 1 and

L

n=O a probability distribution, if we take PO

=

~

•

P _n

=

POc, so (p ) >0 becomes _n

n-o

Remark. The determinant representation (cf. (I. 7) for the reO) now becomes

n Po

0

PI PI Po P2 (1.12) _{r n- 1 (0)} = _Po -n det _{P2 PI Po} P3

.

• Po Pn- 1 PI Pn

As an analogue to theorem 1.8 we now have from theorem 1.9 and lemma 1.10 (see also [4J):

Theorem 1.11. A distribution (p ) >0 ~s compound-geometric if and only if n

n-the quantities r (0), n _n

=

0,1, ••• defined by

are nonnegative.

Two other classes of lattice distributions, which are known to be inf div, are the log-convex and the completely monotone distributions. For the latter we give a useful characterization in theorem 1.14 (see [4J).

Definition 1.12. A distribution (p ) >0 1S called log-convex if

n

n-n ~ I •

Definition 1.13. A distribution (p ) >0 is called completely monotone if

n

n-n ~ 0 and k ~ 0 •

A

is the set of all completely monotone lattice distributions.

Theorem 1.14. (p ) >0 is completely monotone if and only if there is a finite n

n-measure ~ on [0,1) such that

or, equivalently, (pn)n~O is a mixture of geometric distributions: 1

Pn

=

I

(I - p)p dF(p) , n

o

with F is a distribution function on [O,IJ.

We conclude this section with a theorem concerning the relation between the classes

A,

B, Co

and

C}

of (inf div) distributions (see [2J, [4J and [5J).

Theorem 1.15.

A

c

B

c

Co

c

C

I, or: the following implications hold:

(pn)n~O completely monotone ~

- (p) _{n n~O} compoundgeometric

-Proof.

A

c

B:

Let (pn)~O be completely monotone. Then (theorem 1.14) 1

I

(1 - p)p dF(p), n n ~

a ,

a

with F a distribution function on [O,IJ. If ~ is a random variable with dis-tribution function F, x := (I

-~) ~~Hn+l)

and

1..

:= (I

-~) ~~Hn-1),

then,

us~ng Schwarz's inequality,we have

so that (p ) >0 is log-convex. n

n-B c CO: Let (p ) >0 be log-convex. If PI = 0, then it follows by induction

2 n n_

from Pn ~ _{Pn+)Pn-I' that Pn} =

a

for all n ~ I, which is a trivial case. So PI > 0, but then it follows by induction from P

~

p2 IIp 2' that p >

a

n n- n- n

for all n. Now we can write P > k >

---

_Pk-

_t k=l, •••

,n,

and so

(1.13) k= I, •••

,n.

We want to use theorem 1.11 and prove the nonnegativity of the r (0) by in-_n duction from (1.13): rO(O) = P1/pO>0. Let rk(O) ~O for k~n-l, then it follows

P _n par _n (0)

so r (0) ~ 0, and (theorem 1.11): (p ) >0 € CO.

n n

n-Co c

C

I: We can simply prove this by showing that a compound-geometric distri-bution is also compound-Poisson.

I-p

If P(z) = I _ PQO(z) €

CO'

then

(1.14) A := -loge) -p) > 0, Q) (z) :=

log(I - pQo (z» 1 ""

logO - p) =

!

n~1

n

(pQO(Z» n

is abs mon, with QI(I) = I, while

An other method uses the theorems 1.6 and 1.9: If P € CO' then

P(z)

with RO abs mon, from which we derive

Rl(Z)

=

P'(z)/P(z)

=

[log P(z)]'

=

[-log(1 - zRO(z»]'

=

[zRO(z)] , I

=

1 - zRO(z)

=

p;

P(z)[zRO(z)]' ,

which is an abs mon function, so: P € C I' 0

Remark. The relations between R

t and RO' derived in the preceding proof, (1.15) [ZRO(z)J'

=

1 - zRO(Z) t and (1.16) RI(Z)

=

P(z)[ZR (z)]t PO

a

'

give by equating the coefficients of sequences r (I) and r (0) ([zRO(z)]'

n n

zn the following relations between the

00

=

I

(n + l)r (O)zn) n=O n (1.17) and (1.18) r (1) = n (n + l)r n (0) + 1 n r (I) -

I

(k + l)rk(O)P n_k • n

Po

k=O

From both relations we immediately obtain: If r (0) ~ 0 for all n, then n

(1.19) r (I) ~ (n + I)r (0),

n n n = 0,1,2, ••••

In view of the theorems 1.8 and 1.11 we can formulate the following

Corollary 1.16. If, for a given distribution (p ) >0' the r (0), n ~ 0,

de-n de-n- n

fined by

n

P_n+1

=

I

Pkr _{_k(O),}

are all nonnegative, then the r (I), n ~ 0, defined by

n

(n + I)Pn+1 = n

=

0,1,2, •••

are all nonnegative, too.

2. Interpolating between 1 and n + I

In the preceding section we have given characterizations of the distributions

(pn)n~O in

Co

and

C]

by the nonnegativity of the recursively defined quanti-ties r (0) and r (1) (n _{n n} ~ 0) (see theorem 1.8 and 1.11). This suggests the possibility of dividing the class of distributions

C)\C

O (cf. corollary 1.16) in a number of increasing subclasses of distributions, characterized in a similar way.

Let us introduce therefore sequences (cn(a»n~O ' defined for a € [O,IJ, with

the following properties

(2.1)

c _n (0)

=

1 and c _n (1)

=

n + 1 for all n ~ 0 ,

c (a) is nondecreasing 1n n for each a, and in a for each n • n

For such a sequence cn(a), and a lattice distribution (pn)n~O' define the sequence (r (a» _{n n_} >0 by

(2.2) n

=

0,1,2, ••• ,

with generating function

R (z) :

=

a 00

r

n=O n r _n (a)z •

For a € [O,)J define the following class of distributions

v

>0 r (a) ~ O} ,

n- n

or, in terms of generating functions,

C

_a

:=

{pgf P

I

R is abs mon} • _a

Of course, for a

=

0 we get the recursion

(1.8)

with the rn(O), RO and

Co

from section 1, and for a

= ]

the recursion (1.4) with the rn(l), R) and C₁ from that section.

Examples of c (a) we can obtain by an obvious interpolation between c (0) :::: 1

n n

and c (1) :::: n+ I, or, if C (z) is the generating function of the sequence

n a -I -2

cn(a), between CO(z) = (I - z) and C

I (z) :::: (I - z)

Example 2.1. (the properties (2.1) are easily seen to hold)

i) C (z) a -I -I 00 kn H) C (z) a (l - z) (I - az)

:::: I

a z , so m=O 2 n n+l c (a) = 1 +a +a + ••• +a :::: - a n -a 00 00 00 n

iii) C (z) :::: _{(1 - z)} -) _{(1 - az)} -a ₌

I

_z m

I

_{( )(-az) =} -a k

I

I

_{( k )} -a _(-a) k n _{z ,}

a

m=O k=O k ncO k=O

n

Ca)(_a)k n (a+~-I )ak c (a) =

_I

::::

I

n

k=O k k=O

iv)

c

(z) = (l-z) [J-(I-a)zJ::::(l-z) -2 - 1 - 2 +az(l-z)

=

a 00 co

I

n

I

(n + I) z n so c (a) 1 + an •

=

z + az

,

:::: n n=O ncO v) _{cn(a) = (n +} 1 )a

.

vi) c (a) = ( 1 + an) a

.

n n k vii) c (a) =

I

a k a n k=O ,n n with a k ,n ~ 0, aO ,n :::: 1 and

I

k=1 a

=

n. k,n

In the following lemma we summarize the main properties of the sequence rn(a) and its generating function R (cf. lemma 1.7 and 1.10).

a

Lemma 2.2. For every a € [O,lJ and a fixed sequence c (a) we have

n

i) Ra has a radius of convergence p > 0, and

(2.3)

(2.4)

0>

R (z)

=

p(z)-l

I

c (a)p Izn,

a n=O n n+ for

I

z

I

< P •

so that R _a (z) is convergent for Izi < 1. If, furthermore, lim c (a) =: c(a) < 0>, then

n n-+<>o (2.5) iii)

L

n=O r (a) < c(a) , n

so that Ra(z) is convergent for

Izi

~ I.

Conversely, every sequence r (a) with r (a) ~ 0 for all n, and

satisfy-n n

ing (2.4), or, if lim c (a) =: c(a) < 0>, (2.5), by (2.2) defines a

pro-n

n-+<>o

bability distribution (pn)n~O € Ca'

Proof. We can prove i) and the first part of ii) in the same way as in lemma 1.7 (with the factor n+ 1 replaced by c (a». For the second part of ii) we

n

can write, since c (a) is nondecreasing in n,

n

so

1 - PO

=

L

_Pn+1

₌

n=O

0> n 0> 0>

I

c (a)

L

Pkrn_k(a)

~ ~

L

Pk

L

rn(a) ,

n=O n k=O c\aJ k=O n=O

L

rn (a) ~ (I - Po)c(a) < c(a) • n=O

Following again the proof of lemma 1.7, we see (using again the monotonicity of c (a», that for iii) it is sufficient to prove

n

(2.6)

0> rk(a)

lim

L

< I •

n-+<>o k=O ck+n(a)

"" rk(lX) 00 _rk(lX) lim

I

C k (lX) .,.

_I

_lim Ck+n(lX) =

o ,

n-roo k=O +n k=O yt+<X>

so that (2.6) holds. I f lim C (lX)

n

=

C(lX) < 00

,

then yt+<X> 00 _rk(lX) 00 _rk(lX) lim

I

Ck+n(lX)

=

I

c(lX)

,

n-+<» k=O k=O

and this is less than I, i f (2.5) holds.

One would like to have, for all lX and 8 € [0,1]

(2.7)

We now investigate for sequences C (lX), satisfying (2.1), to what extend n

this implication holds.

Lemma 2.3. For all n ~ 0 and all lX,S € [O,lJ

(2.8) Proof. Define A (z) := lX 00 (convergent for JzJ < I) ,

then it follows from lemma 2.2i): AlX(z)

=

P(z)RlX(z) and AS(z)

=

P(Z)RS(Z), from which

Equating the coefficients of zn, we obtain (2.8).

Theorem 2.4. For all lX € [O,lJ

This is an immediate consequence of the following: If r (0) ~ 0 for all n, then

n

i) For all a € [O,IJ and all n

(2.9)

o

ii)If

(2.10)

then

and

cn+1(a)

c (a) is nondecreasing in a for all n ,

n

r (a) n

c (a) is nondecreasing in a for all n ,

n

r (a) is nondecreasing in a for all n • _n

Proof. Let r (0) ~

a

for all n (or: (p ) >0 € CO), According to (2.8) (with

n n

n-S

=

0, and hence ckCS)

=

I) we can write

n-l

=

_{cn (a)Pn+1 -}

L

Pk+l rn_1_k(a)

=

k=O

so that

po[r (a) - c (a)r _{n n n} (0)]

=

=

which is nonnegative, as rn(O) ~ 0 for all nand cn(a) is nondecreasing in n. We conclude

r (a) ~ c (a)r (0),

n n n n

=

0,1,2,.,. ,

so that all r (a) are nonnegative, and (p ) >0 €

C

N ,

n ( ) n n- u

C i a _n+

Furthermore, let c (a) be nondecreasing in a for all n. Then we have for

n

S > a and k ::;; n

cnca) n-l ct+1(S) _{n-I c t +1 (a)} ck (a) =

R.~k

cg, (13) ;:::

.t~k

cg, (a) or

c (a) n

(2. II ) for a < S, k

=

O,I, ••• ,n • From (2.2) and (2.8) we obtain for a < S

poe c (a)r _{n n} (S) - c _n (S)r _n (a)]

=

which is nonnegative on account of (2.11). It follows that c (a)r (S) ~ c (S)r (a),

n n n n for S > a and all n ,

r (a)

n

which implies that c (a) is nondecreasing in a for all n. Consequently

n c (8) r (8) ~ n() r (a) ~ r (a), n c a n n n for S > a ,

because c (a) is nondecreasing in a. So we finally have, that r (a) is

non-n n

decreasing in a for all n.

0

Remark. One easily verifies, that condition (2.10) is satisfied in the follow-ing cases n+J c (a)

=

(a+n) I - a n n ' I-a a , I + an and (n + I) •

We now consider distributions (p ) >0 in

C

with a > O. It turns out that for

n n- a

many choices of c (a) we do not have the desired property (2.7). We shall give _n some simple, necessary conditions for (2.7), from which we obtain counter-examples for almost all our choices of the sequence cn(a). To this end in the following lemma we introduce a special distribution.

Lemma 2.5. For a

O E (0,1), the for all n ~ I by (2.2) defines which

(2. 12)

sequence rn(a

O) with rO(aO)

= )

and rn(a O) = 0 a probability distribution

(p )

'0 E

C

t for

n n", a

O

Proof. The sequence rn(a

_o)

satisfies the conditions of lemma 2.2iii) and hence defines a distribution (~p) € C by the recursion, from which we

n n~O 0.

0 obtain: cn(aO)P

n+1

=

Pn ' and from this (2.12) immediately follows.

0

Lemma 2.6.

i) If there is a 0.

0 € (0,1) such that

(2.13)

then for r 2(a) corresponding to (pn)n~O we have: r 2(1) < O. ii) If there is a 0.

0 € (0,1) such that

(2.14)

then for r 2(a) corresponding to (pn)n~O we have: r 2(0.0) = 0 and ri(a

_o)

< 0 (here c_{1 and c2 are supposed to be differentiable).}

So, in both cases we do not have (2.7).

Proof. For the rn(a) corresponding to

(p )

>0 we have on account of (2.12)

n

n-from which we successively obtain

I

= '( )

c 1(0.) - I , c] 0.

It follows, that, if (2.13) holds, then

and consequently, (pn)~O in C

ao' but not in Ct' If (2.14) holds, then

whence (pn)n~O in C

ao' but not in

e

a for a E (aO,aO + e), with £ > 0

suffi-ciently small.

0

In the following lemma, for the different choices of cn(a) (cf. lemma 2.1) we list values of aO' for which (2.13) and (2.14) hold.

Lemma 2.7.

i) c (a)

n

=

(a:n): satisfies (2.13) for aO >

13 -

1 (~0.732), and (2.14) for a_O >

H15-

1) (~0.618).

n+l - a c (a) =

-"":""---n I-a satisfies neither (2.13) nor (2.14).

ii)

iii) c (a)

=

n

n

I

(a+~-l)ak:

satisfies (2.13) and (2.t4) for a_O > 1 - £, with

k=O

£ positive and sufficiently small.

iv) c (a)

₌

) _{+ an:}

n satisfies (2.13) and (2.14) for aO > ~

.

cn(a)

=

(u'+ 1) : a satisfies (2.13) for a 1 - £ (£ positive and

suffi-O >

2 2 ciently small) and (2.14) for a

O > log log 3.

v)

c _n (a)

₌

( 1 + an)a: satisfies (2.14) for a_O > 1 - £ (e: positive and

suf-vi)

ficiently small) •

We further have (cf. example 2.1ii) and vii»:

Lemma 2.8. If c

J is linear and c2 quadratic in at then in order to exclude (2.13) and (2.14) for all a

O E (0,1) it is necessary and sufficient that for all a E (0,1)

Proof. As c (0) = I and C (I)

n n n + I, cl and c2 already have the following form

with b

1

o.

Further, from the requirement cI(a) ~ c₂(a) for all a E (0,1) we

I

obtainib ~ I _ a for all a € (0,), so: b ~ I.

Now suppose, that (2.13) does not hold for all a

O E (0,1). Then it follows

> 2a - 1

that b - a(2 _ a) for all a E (0,1), which implies: b ~ I. So we obtain (2.15)

as the only possibility for c

1 and cZf which gives no counter-examples

accord-ing to lemma 2.7ii).

0

a

It is a pity, that cn(a) = (n + I)a and cn(a)

=

(a:n)(~

r{an+ 1) , n

~

00) do not have the required properties, because they seem to give a "better" sub-division of

C]\C

_{O' than e.g.}

c (a) = I + a + ••• + an

=

n

n+I - a

I - a. ('" I _ a. ' n ~ 00) •

However, since the latter yields no counter-examples and is the most obvious in view of lemma 2.8, we shall consider it in more detail in the next section. If we drop the monotonicity of cOCa), then we can consider sequences cn(a)

a 1 n+l-a. -1

like (n + a) , and r(I _ a) ( n+l) • The latter we obtain by means of

"frac-tional differentiation": a-I

=

z from which 00 \' r(n + I) l. r(n - a + ]) n=I n-a z = 00 \' (n + I)! n l. f(n + 2 - a) Z , n=O (n + I)! cnCa.)

=

f(n + 2 - a) = n+l-a -I

r

(I - a) ( n+ I )

We shall not persue this in this report.

We conclude this section with a property of

C ,

which holds for every choice a

Theorem 2.9. If (p) €

C

t then also (q(Y» €

C

for all Y € [O.IJ,

n n~O a n n~O a •

with

n

q_n (y) .-_.-

_PTYT

Y P _n' n

=

0,1,2, ••••

Proof. Let (p ) >0 € C t so rn(a)

~

0 for all n. Define the sequence r(Y)(a)

n n- a n by n c (a)q(Y)

=

n n+l

L

q (Y)rCY)(a) k n-k t n = 0,1,2, •••• k=O On the other side we can write

n+1 n+l n

c (a)q(Y) - Y c (a)p Y

~

r (a)

=

n n+1 -

PTYJ

n n+l

=

PTYJ

L Pk n-k so that r(Y)(a)

=

n k=O

=

n \' (y) n-k+l ( ) L qk Y r n-_k a k=O n+l ( ) 0 Y r _n a ~ for all n •

It follows by definition, that (qCY» €

C •

n n~O a

3. Case c (a) _n

=

I + a + a 2 + ••• + a n

In this section we investigate the recursion relations (2.2) with n+l

- a

1 - a n

=

0,1,2, ••••

o

So, the quantities r (a), R (z) and the class C , defined in section 2, now _{n a a} correspond to this choice of c (a).

n

First, we formulate lemma 2.2 more precisely.

Lemma 3.1. For all a € [0,1) we have

i) Ra has a radius of convergence p > 0, and

(3.1) R (z) = 1 I

a - a z { 1

P (az) }

00

(3.2)

I

n=O

r (0.) <

~--n - 0. •

so that Ro.(Z) is convergent for Izi S I.

iii) Conversely, every sequence r (a) with r (0.) <:: 0 for all n, and

satisfy-n n

ing (3.2). by (2.2) defines a unique probability distribution. (p ) "'0 E C • n n", 0.

Proof. For Iz\ < 1 we can write

00 00 \' n 1 \' n t.. c (o.)p IZ

=

1 {t.. Pn+lz n=O n n+ - 0. n=O 00 \' n+l n} t.. 0. Pn+)z == n=O P(Z) - Po p(o.z) - P = "':"'-_ { O}

=

- 0. z Z

= "':"'--

_{- 0.} Z {P(z) - P(o.z)}

Now (3.1) immediately follows from (2.3). As lim c (0.)

=

~---

,

the rest is

n 1 - 0.

a reformulation of lemma 2.2. n-+<x>

o

Remark. In lemma 3.1 we only consider 0. < t. The case 0.

=

I, which is

essen-tially different, has been treated in lemma 1.7. It is a limiting case in the following sense: lim R (z) == o.tl 0. lim ":'1---0.- z o.t) p' (z)

=

~P~("'zp)

{I _ P(az)}

= )

lim p(z) - P(o.z)

=

P(z)

PTZr

at) z - o.z

Since (3.1) gives an explicit expression of R in P, we can also prove

theo-0.

rem 2.4 using generating functions.

Lemma 3.2. For all 0. € [0,1) we have

(3.3) and

(3.4)

RO(Z) - o.RO(o.z) Ro. (z)

=

"'!"')---a 1 - o.zRO (o.z)

R (z) 0.

p(o.z) RO(z) - o.RO(o.z)

= ----

or, in terms of the coefficients, (3.5) and (3.6) r (a) n n-I k+l = c (a)r (0) +

I

a rk(0)rn_1_k(a) n n k=O I n n-k r (a)

= --

l

ck(a)rk(O)a P-k n Po k=O n 1 Po Po

Proof. As RO(z)

=

z

{I -

PTZJ}'

we have P(z) = I _ ZRO(Z) • Substituting this

in (3.1) we obtain (3.3):

I 1 - ZRO(Z) RO(z) - aRO(az)

R (z) =~--{I - }

=~-a - a z 1 - azRO (az) - (l 1 - azRO (az)

from which, using (3.1) once more, (3.4) follows. The relations (3.5) and

(3.6) now follow from (3.3) and (3.4) by equating the coefficients of zn.

0

From this lemma we immediately obtain the abs mon of Ra from that of R_O' and so: Co c C

a, while the inequality (2.9) follows from (3.5) or (3.6). Further-more, letting a

+

1 in lemma 3.2, we get the relations (1.15), ••• ,(1.18).

The following lemma expresses P in terms of R • a

Lemma 3.3. For all a € [0,1) we have (with the p from lemma 3.1)

(3.7) I f P E: (3.8) 00 p(z) = Po

n

k=O

C , then (3.7) holds for Izi ::;; 1 and we a 00 _{1 - (I -}

(l)a~

(ak) P(z) =

n

a k k

,

k=O 1 - (I - a)a zR (a z) a for Izi can write for Izi ::;; 1 _•

Proof. According to (3.1) for Izl < p we can write

P(z)

=

[I - (1 - a)zR (z)]-IP(az) , a

from which we obtain for every n € E

P(z) or

n-l IT k=O k k [J - (J - a)a zR (a z)J a

Po

This tends to PTZ) , for n +

00,

from which (3.7) follows. If P E Ca' then

on account of lemma 3.1 the derivation above is valid for

Izi

$ I. Taking

z == I in (3.7) we get

k k

p == IT [1- (1 - a)a R (a )J ,

°

k=O a

and (3.8) follows from (3.7).

o

Lemma 3.3 gives us a characterization of C • For a =

°

and 1 we already have a

one: Co is the set of compound-geometric distributions and C) the set of com-pound-Poisson, or inf div distributions. C now appears here as the rather

a

special set of infinite products of compound-geometric pgf's, given by (3.8).

Condensing the notation a little, we have:

Lemma 3.4. For a E (0,1): P E C if and only if there is p E (0,1) and pgf

a

Q with Q(O)

=

0, such that

(3.9) P(z)

=

00 k

IT 1 - pg(a ) _{k •}

k=O 1 - pQ(a z)

Proof. Let P E C • Then P has the

a zR (z)

representation (3.8), which becomes (3.9), a

if we define Q(z) := R (1) and pgf, while p € (0,1) onaaccount

p := (1 - a)R (I). As R is abs mon, Q is a

a a

of (3.2).

Conversely, if P has the representation (3.9), then we have for the corres-ponding R a R (z) == ~~-a - a z = " ' ! " ' - -

1..

{I - (l - pQ (z»} = - a z ()O k IT 1 - pQ(a z) }

=

k+l k=O 1 - pQ(a z) p Q(z) - a z which, as Q(O)

=

0, is an abs mon function. So: P € C •

a

o

Theorem 3.5. For all a € [O,lJ we have: C

a C C10 Or: every pgf P with an ab-solutely monotone R is inf dive

Proof. Let P E Ca with a € (0,1). Then P has the representation (3.9), or (3.10) co P(z)

=

IT k=O I - 'lJ' k - 'lJ' 8 (z) , k k k k k

with 'lJ'k

:=

pQ(a ) E (0,1) and 8

k(z) := Q(a z)!Q(a ) a pgf. 80, P is an

infi-nite product of compound-geometric pgf's, which is inf dive

o

Remark. We can also prove theorem 3.5 using the following relation between RI and Ra: from (3.7) we obtain

so (3.11)

00

P'(z) d d k k

R1 (z) = P{z)

= -

log P(z) =- - log IT [I - (l -a.)a. zR (a z)J =

dz dz k~O a

co

= -

I

.Ld loge I - (1 - a.)a.kzR (a.kz) J =

I

k=O z a. k=O co [zR (akz)J' R1 (z) = (l -a)

I

ak

----a.-.,k~-"';"k-

• k=O 1 - (l - a)a zR (a z) a (l-a.)a.k[zR (a.kz)J' a. k k ' I - (I - a)a zR (a z) a

Now, if Ra is abs mon, then it follows from (3.11), that R] is abs mon, too.

Before we investigate, whether the implication (3.12)

holds generally, we prove, that the special distribution (pn)n~O €

C

ao (see lemma 2.5), from which we obtained counter-examples for many choices of cn(a), belongs to every

C

_e

with

e

~ a.

_O'

Lemma 3.6. For the r (a) corresponding to the distribution

(p )

>0 € C from

n n n- a._O

lemma 2.5 we have

(3.13) n

=

O,t, ••• , and a. € [O,t) •

From this it follows that (3.14)

and (3.15)

00

'"

Proof. On account of (3.7) for p(z)

:=

I

p zn we have, as R

(z)-n=O n aa

,

from which for a E [a,l) we obtain

' " 00 1 - (I -a )akz

R", (z) == ~_1.. {I _ P (a z)}

=

-:-~

1..

{I _ II 0 0 }

"'" - a z 'i?(z) - a z k=a 1 - (1 -

aa)a~a1!

Hence, for m ~ J

1..u

m-l k R _{m(z) ==}

-

II [I - (I

-

_{tla)aOz] }}

,

m z k==O a O - a 0

which is a polynomial in z of degree (m - l) t so that rn(aa) m

=

o

for n ~ m, or

(3. t 6)

However, with (2.12),

+. •• ,

which is a polynomial in a of degree n. Hence, (3.13) immediately follows from (3.16). From (3.13) we easily obtain (3.14) and prove (3.15) by proving the equivalent inequalities

(3.17) c (a) n II (8 - aa) k ~

n k=1

using mathematical induction. For n == I:

n k

c (8) II (a - a O)' n k==l

Suppose that (3.17) holds for a fixed n, then

n+l k n+1 k n+l k c+l(a) IT (13 - 0. 0)

=

C (a) IT (13 - 0.0) + IT a(S - 0.0) ~ n k=l n k=1 k=1 n k n+1 k ~ (S - an+l)c (13) IT (a - 0. 0) + IT 13 (a - 0.0) ~

a

n k=J _k=t n+1 k n+J k ~ c (13) IT (a - aO) + Sn+l IT (a - 0. 0) = n k=1 k=l

The result of the preceding lemma gives uS hope, that (3.12) does indeed hold. Therefore, we look for a convenient expression for RI3 in terms of Ra.

Lemma 3.7. For a,S € [0,1) the following relations between Ra and RS hold

(3.18) and (3.19) I - (1 - S)zRS(z) 1 - (I - a)zR (z) a 1 - (1 - S)azRS(az)

=

1 - (I - a)l3zR (Sz) a RQ (z) ==

""1-~

1..

{I - IT k k 1 - (1 - a)a zR (a z) k a k } 00 p S z k=O 1 - (1 _ _{0.)0. SzR (a Sz)} a P(az) Proof. According to (3.1) we have: J - (I - a)zR (z) = ~~- , so

a P(z)

[I - (I - a)zRa(z)][1 - (I - S)azRS(az)]

₌

P(az) _{P(z) • P(az) =} P(Saz)

_ P(Sz)

- P(z) • P(Sz) -P(aSz) - [I - (I - S)zRQ(z)][1 - (I - a)l3 zRro(Sz)]

p '"

which gives (3.18). By iteration of (3.18) we obtain for every n ~ 1

o

n-l 1 - (I -a)akzR (akz)

a n n 1 - (I -13)zR

S(Z) == _k=O IT - - - -... _{- (I - a)a SzR (a I3z)} k--=.:..-k-.[ I - (I - S)a ZRS(a z)] • a

Since the last factor tends to 1, if n +

00,

(3.19) follows.

o

Though (3.19) expresses RS proof of the abs mon of RS' fore, we return to relation

explicitly in R , it does not provide an easy a

if we know that R is abs man and i f S ~ a. There-a

Lemma 3.8. For a,B ~ [0,1) we have (3.20)

or, in terms of the coefficients, (3.21)

with

(3.22) o (a,B)

:=

~ l (B k+1 - a n-k+1 )rk(a)rn_k(S).

n k=O

Proof. Writing out (3.18) we obtain (3.20), from which, by equating the ff ' . f n+1 (3 21)

coe ~Clents 0 Z t • follows.

Lemma 3.9. If P £

C ,

then for all n ~ 1 a

(3.23) C (a)r (B) ~ C (S)r (a),

n n n n

lIn

for S £ [a , I J •

Proof. Let P £

C •

According to (3.21), (3.23) is equivalent to

a

(3.24) o _n-l(a,S) ~ 0 for

B

~ [a I

In

,IJ.

o

I ( ) ( ) () () Of S S [ ",lIn+I,IJ

For n

= :

_{00 a,a}

=

a - a rO a rO 8 ~ or B ~ a. uppose £ ~

and 0k(a,B) ~

a

for k = 1,2, ••• ,n-l. Then it follows from (3.21), that rk(B) ~ 0 for k

=

1,2, ••• ,n, so that with (3.22)

n+1 n

o (a,a) ~ (a - a)

I

rk(a)r _k(8) ~ 0 •

n k=O n

Thus, (3.24) has been proved by induction.

Remark. Using (3.21) and (3.22), we can prove something more than (3.23), namely

c (a)r (8) ~ c (S)r (a), for 8 ~ a, if n

=

0,1, ••• ,5,

n n n n a 3/n-2

for ~ ~ a , if n ~ 6 •

IJ

In the following lemma we formulate some relations, from which another par-tial result follows.

Lemma 3.10.

i) If a - Sy with a,y € [O,IJ, then

(3.25)

ii) For all 13 € [O,IJ

(3.26) iii) For (3.27) iv) IfP (3.28) (1 + S)R 2(z)

=

Ra(Z) + SRa(Sz) - 13(1 - 8)zR S(Z)RS(Sz) • 13

all 13 € [0, I ] and all n ~ 0

2 _S n k

oneS ,13)

=

+ 13 (I - an+2)

2

S rk(S)rn_k(S) •

k=O

€ C 2 with 8 € [O,IJ, then

S

2 2

c (13 )r (a) ~ c (8)r (13), n == 0,1,2, ••••

n n n n

v) For aIlS € [O,lJ C 2 c Cst

a

Proof. If a - Sy, then, on account of (3.19), we can write

== k+1 k+1 00 1 - (1 - a) a zR ( a z) == [1 - (I -a)zR (z)J II a == a k-O 1 - (1 - a)akSzR (ak8z) a

=

[ l -k k 00 I - (1 - a)a 8zR (a 8z) I (1 - a) zR (z) J { II k a k } - ==

a k-O 1- (1 -a)a y8zR (a ySz)

a

- [I - {I -a)zR {z)J[l -_a (1 -y)azR {Sz)J-1 ,

y

from which (3.25) follows. Taking y

=

a, and mUltiplying out, we obtain (3.26),

Hence. with (3.21), I - Sn+2 2 n k

=

{(1 +8)r 1(8) +8(1-8)

I

8 rk(S)rn_kCS)} -1 - -13 2 n+ k=O 1 - on+2 2 0 2 n k .., (0 ) .., (1 _ on+ ) \' 0 (0) (0) r n+J IJ

= T"'+i3

..,

l.. I-' r k .., rn- k .., 1 - 13 k=O

which is (3.27). I f furthermore. P €: C 2' then all r (8 2) are nonnegative

a

n

and from (3.21) and (3.27) by induction we obtain for n

=

0,1.2 •••• ,

which gives (3.28). From this we see, that all r (8) are nonnegative, too,

n

Corollary

3.11.

For all a €:

[O,IJ

we have

C

c

C

1 c

C

1 c ••• c

C

c ••• c

C

1 •

a a2 a4 2-m

a

D

We have now obtained in a rather laborious way a few results (theorem 3.5, lemma 3.9 and corollary 3.11), which indicate, that (3.12) is true. Indeed, we can prove (3.12) by a method, inspired by still another proof of

IIC

_a

cCI",

which follows from the next theorem and lemma.

Theorem 3.12.

1.' ) If P €

CIt

then for all a €

[O,IJ

P(z) P(az) is abs mon and P(az) P(a)P(z) €

Cl

o ii) For all a € [0,1) we have the following characterization of

C :

a (3.29) P _€

C

_{a ~} P(a)P(z) _{P(az) €}

C

_O'

Proof. If P €

CI'

then P has the representation (1.1), from which it follows

that

(3.30)

p~(!;fz) =

exp[A(Q(Z) - Q(az) + Q(a) - I)] •

Now let P € C • First we see from a

(3.31) P(a)P(z) _{P(az) ::} pea)

1 - (1 - a)zR (z) , a

that P(a)P(z) . _{P(az) 1.S a pg.} f If _{we e 1ne p:=} d f' 1 _- P() _{a an} d Q( ) _{z:= j _} 1 -_pea) a _Z R ( ) _{a Z ,} then it follows that p € [0,1) and Q is a pgf (as R (1) == 1 -

pea»~.

Hence

a I-a

P(a)P(z) I - ~

P(az)

=

1 - pQ{z) €

Co •

However, this and its converse, i.e. (3.29), we can immediately prove using h f P(a)P(z) the following relation between the RO corresponding to t e pg P(az) (de-noted by R6a

»

and Ra:

(3.32) _RO (a) _{(z) ==}

z

I _{{I - P(a)P(z)JP(az)}} pea)

₌

z

1 _{{I -} P(az) _{P(z) }}

₌

( _{I - a Ra} ) () _{Z •}

Thus R(a) is abs mon if and only if R is abs mon, which is, by definition,

_{, a}

_a equivalent to (3.29).

Lemma 3.13. The following relation between R) and Ra'holds

(3.33) or (3.34) [zR (z)]' a

=

I - (1 - a)zR (z) • a In terms of the coefficients

(3.35) c (a)r (1)

n n

Proof. On account of (3.1) we can write

[ P(az)], P(az)P'(z) P'(az)

(l -a)[zRa(Z)]' == - P(z) == p(z)2 - a P(z) ==

== P(az) {R (z) - aRI(az)} ,

pez) I

o

from which (3.33) follows. Using (3.1) once more we obtain (3.34), from which, equating the coefficients of zn, we get (3.35) by using

(3.36) and (3.36)'

~

_{a {R1}(Z) - _{aR 1(az)}}

=

L

n-O n c (a)r (1) z , n n [zR (Z)JI ... a

L

n=O (n+ l)r (a)zn • n

Corollary 3.14. For all a € [O,IJ we have C

a c CI. Furthermore, if P € Ca, then

(3.37)

o

Proof. Let P € Ca. Then Ra is abs mon, so that the abs mon of Rl follows from

(3.33) and (3.29), or immediately from (3.34). By induction we obtain the

in-equality (3.37) from (3.35).

o

We now try to find a relation between Ra and RS similar to the relation (3.33) between Ra and R). On the one hand we can write for (J - a)[zR

a (z) ] t the limit

zR(zl - SzR.(Sz)

(I -a)lim a a ... lim

~

:

~

{Ra(z) - SRa(SZ)} ,

st) z - SZ Stl

and on the other hand, according to (3.33), P(az) {R (Z) - aR

1(az)} ... lim P(az) {RQ(z) - aRQ(az)} •

P(z) 1 Stt P(Bz) ~ ~

Equating the two expressions, we may hope, that the following relation holds

1 -

a

1 8 {R (z) - SR

(Sz)}

- a a

or, symmetrizing in a and S,

(3.38) PC I -_az a ) {R (z) - SR (Sz)} _{a a}

The truth of (3.38) is expressed in the following lemma.

Lemma 3.15. The following relation between Ra and RS holds (3.39)

Proof. According to (3.1) the left-hand side of (3.38) can be written as

~(;z~

{RS(z) - aRS(az)}

=

zP(Bz) {(I -

:t:») -

(1 -

:t:~»)}

=

1 {p(aSz) 1 }

=

Z

P(az)P(6z) -

PTZJ '

which is symmetric in a and

S,

and hence equal to the right-hand side of

(3.38). Rewriting (3.38) gives (3.39).

o

Now, using relation (3.39), we can easily prove the desired property (3.12).

Theorem 3.16. For all a,S € [O,IJ the following implication holds

Proof. The case a ~ 1,

a

=

1 has been treated in corollary 3.14. Now let a < 8 <

follows lation

(3.40)

1 and P € Ca' Then also P € CI, and on account of lemma 3.12i) it

P(z) P(8z)

that , and therefore P( ) , is abs mono Considering the

re-p(i

z) az (3.39) and noting that

1 -

B

{Ra(Z) - BRa(Bz)}

=

L

n=O

n c _n (B)r _n (a)z ,

p(az)

we obtain the abs mon of RB from that of Ra and P(az) • Hence P €

CS'

o

Using lemma 3.15 we prove inequalities, stronger than r _n (B) ~ 0 (n ~ 0), and of which (1.19), (2.9), (3.15), (3.23), (3.28) and (3.37) are special cases.

Corollary 3.17. If P € C

a, then we have for all a € [a,l]

(3.41) n

=

0,1,2, ••• ,

and so: every r is nondecreasing on [a,I].

n

Proof. As we saw in the proof of theorem 3.16, if P € C and a _a < at we can ~ite

ro

p(az) t n

P(az)

=

n~O sn(a,a)z ,

coeffi-cients of zn, it follows from (3.39) and (3.40), that

c (a)r (13)

=

n n

from which (using

So ...

I and sn ~ 0) we obtain (3.41). The last assertion of this corollary is obtained from (3.41) by the fact, that cn(a) is

nondecreas-ing in a.

o

From theorem 3.16 we obtain an extension of corollary 1.16, as follows.

Corollary 3.18. If the r (a), n ~ 0, defined by n n+J n - a

~

Pkrn_k(a) , n

=

0,1,2, ••• Pn+l

=

I-a k=O

with a E: [O,lJ, are all nonnegative, then the r (13), n ~ 0, defined by n

_ I3n+1 n

Pn+1

=

I

Pkrn_k(13), n

=

0,1,2, ••• I - 13

k==O

with 13 € [a,IJ, are all nonnegative, too. We then, in fact, have

n+J _ I3n+1

- a _{r (13) ~ r (a), 0,1,2, •••}

1 - 13 n

=

.

I-a n n

Theorem 3.16 says, that for a ~ 13 the abs mon of RI3 follows from that of Ra' As on account of lemma 3.1iii) an abs mon R may be any abs mon function f

a

with f(l) < 1 ,with lemma 3.7 we can formulate the following assertion - a

about abs mon functions.

Corollary 3.19. If ~ is an abs mon function with ~(O)

=

a

and ~(l) < I, then the function~, for

a

~ a ~

B

~ 1 defined by

or by

I - ~(z) I - ~(z) J-~(az)'" J-cp(Sz)'

~(z) ... 1

In theorem 2.9 we proved the following implication (now formulated in terms of generating functions)

(3.42) P C _ P(yz) C

€ a P(y) € at for y € [O,IJ •

If we denote the generating function of the r (a) corresponding to the pgf

(y) P(yz) (y). n

P (z):= P{y) by Ra (1n the sequel we shall use such a notation without

further explication). then with (3.1) we can write

Hence (3.43)

(I - a) zR (Y)(z)

a

p(y) (az) P(yaz)

... 1 - ... 1 - ... (1 - a)yzR (yz) •

p(y)(z) P(yz) a

R(y)(z) ... yR (yz) ,

a a

from which (3.42) immediately follows.

By means of similar methods we can obtain other properties of Ca' We formulate them in the following theorems.

Theorem 3.20. A pgf p. which is the limit of a sequence pg£ts P £ C ,

be-n a

longs to C (i.e. C is closed under weak convergence).

a a

Proof. Let P £ C for n

~

1 so R(n)(z) ...

n a ' a

But then we have, if P(z) ... lim P (z).

n n-+<= R (z) =

-:--_1.

{l _ P(az)} ... a - a z P(z) ... lim R(n)(Z) ; n-+<= a

which is abs mon, too, so: P f C •

a

Theorem 3.21. For all a £ [O,IJ we have:

1 P_n (az) _ a

z

{I - ~ (z) } is abs mono n P (az) _ a

~

{I - lim pn(z) } = n-+<= n (3.44) P _€ C _{a -} P(y)P(z) _{P(yz) £}

C

_a' f _{or y €} [ 0 ) _,a

€ CO' for y € [atlJ •

Proof. Define p(y)(z)

:=

P~1~!5z)

t for y E [O,lJ. If P € C

I, then on account

of theorem 3.12i) we know that p(y) E

C

I, too, for all y. Now let a € [0,1)

and P E

C •

Then we can write

a

(J-a)zR(y)(Z)

=

I _p(y)(az) a I _ P(az) P(yz) =

a p(y)(z) p(yaz) P(z)

P(yz) {P(ayz) p(az)} P(yz) ( ) { () ( ) }

= P(ayz) P(yz) - P(z) = P(ayz) ] - a z Ra Z -YRa yz ,

so that

(3.45) R(y)(z) - P(yz) fR (z) - yR (yz)} •

a P(ayz) a a

Using (3.39) we can rewrite this

(3.46) R(y)(z) _a = I -_{1 -} y P(az) {R ( ) _{a P(yaz) y z - a} R (a )} _{y.z •}

As according to theorem 3.12i), ptyz» is abs mon, using (3.45) we obtain for )' ayz

all y E [O,lJ the abs mon of R(y from that of R • Hence p(Y)

a a

y € [0,1]. However, P €

C

for y ~ a, so on account of (3.29)

that p(Y) €

Co

for all y

!

[a,IJ.

Theorem 3.22. For all a € [O,IJ we have:

€

C

for all a

'W'e can conclude

o

(3.47) P E

C

~ n-l IT P( k )

Yk

Z €

C ,

a k-O P(y ) Y

lIn for n ~ 1, y E [a ,1].

Proof. Take a fixed n ~ 1 and define

n-I k

p(y)(z) := IT P(y z) k=O p(yk)

for y E [0,1] •

If P €

C

I' then p(y) as product of pgf's in

C)

belongs to

C

I ' too, for all y. Now let a € [0,1) and P € C • For

a

€ [0,1) we can write

a or, with

e

=

y, (1 - y)zR(y)(z) y p(ynz) "" (1 n) R ()

=

1 - P(z) - y z n z y

Hence

R(Y)(z) = c l(Y)R (z) •

Y n- n

Y

(3.48)

As R n is abs mon for Y

~

a 1/n, it follows from (3.48), that

~Y)

is abs mon

for

~ ~

a

lIn.

Hence P (y) € C for Y

~

a I

In.

0

Y

Remark. Inview of theorem 3.16 it is evident, that theorem 3.22 is equivalent

to the assertion that for all Y € [O,IJ:

(3.49)

We can easily prove (3.49) using lemma 3.4: If P € C , then there are

yn

P € (0,1) and pgf Q with Q(O)

=

0, such that

But then co nR. P(z)

=

IT 1 - pQ(y )

nl

•

R.=O I - pQ(y z) n~1 PCykz) k k=O p(y ) Still another proof of

n

(3.49) we obtain from the characterization (3.29): If

P € C n' then P(y )P(z)

Y p(ynz)

n-l k

€ CO. But for p(y)(z)

:=

IT P(YkZ) we have

k=O P(y )

Corollary 3.23. For all n ~ I and all a € [O,IJ we have

(3.50) n-I k IT 1 - pQ(a ) c

C

k

'"

a ' k"O 1 - pQ (a z) with p € [0,1) and Q a pgf. I-p Proof. Take P(z)

=

I _ pQ(z) € Co in (3.47).

o

Theorem 3.24. For all ~ € [O,IJ we have:

(3.51) for Y E [0, 1] •

Proof. If P €

C

I, then, according to theorem 1.1, pY €

C

1 for all Y ~ O. Now let ~ € [0,1) and P € C • Then we can write

~

(y) pY (~z)

(I - ~)zR (z)::: 1 - :::

~ pY(z)

so that

!..

[(I - ~)ZR(Y) (z)J __ (P(~z»Y-l d [P(~z)]

=

dZ ~ - Y P(z)

rz

P(z)

Hence (3.52)

As on account of theorem 3.12i)

p~(~;jz)

€ C

I, and so

(p~(~;jz»]-Y

€ C1 for Y € [0,1], it follows that (P(z) )I-Y is abs mon for Y € [O,IJ. Now, using

P(~z) ( )

(3.52), we obtain the abs mon of R~Y from that of R~, and we conclude: pY €

C

~.

o

use characterization (3.29) for the proof of theorem Remark. If we wish to

3.24, then we have to p € [0. 1) and A a pgf,

1 -

!

prove (3.51) for ~

=

O. If P(z) ::: 1 _ pA z) € CO' with then we can write [p (z) JY

=

I

~ ~Btz)

, with

q

:=

1 - (] - p)Y , B(z) :=

1.

{I - (t -pA(z»Y} =

1.

{t -

1:

q q n=O <X>

::: 1.

~

r

(n-l-Y)( A(z»n q l.. n n-l p , n=t

Now let P E: C a.. Then, according to (3.29), P(a.)P(z) E: Cat so that

P(a.z) P(a.)Yp(z)Y

= (P(a. )P(Z»Y _{Eo Co '}

P(a.z)Y P (a. z)

too, for Y E: [O,lJ. Using (3.29) once more, we conclude that pY E: Ca. for

Y E: [0,1].

We conclude the properties of the classes C with the assertion that u

a. a<1

is dense in C

1 in the following sense: Theorem 3.25. If P E:

C

I , then there is an increasing sequence an' with lim a. '"' 1, and there are pgf's P E C ,such that

n+<'O n n _{a. n}

P(z)

=

lim P

(~),

n+<'O n for

I zl

~ I •

Proof. Let P E: CI' Then there is a X > 0 and a pgf Q, such that

Take a. :=

n

(3.53)

P(z) = exp[X(Q(z) - I)J •

I

- ~ • If for n > A we define the pgf's P_n by

P (z) n n n-I

=

IT k=O k 1 - A!n.Q(a ) n

k

'

I - A!n.Q(a. z) n C q

then, on account of corollary 3.23, we know that P nEe a. • As P (1) = P n (I) = I ,

n

it is sufficient to prove the convergence of P (z) to P(z) for z E: [0.1). So,

n

take a fixed z E: [0,1) and write

with n-I = IT k=O {J + A[Q(z) - IJ

=

n

Hence (3.54) For k < n we estimate k ~ zQ' (z) • (t - (l ) t n

because (t - (lk) z < 1;; k < z and Q is convex. As for k < n

n n.

-

(l~

= I - (1 - ; , ) k

~

1 - (I - k ; ' )

~

*

n n

we finally have for some C(z) > 0 and all k < n (3.55)

n I n

As lim Q(z)

=

Q(l) = I and lim (l

=

lim (I -~)

=

I, we further have for

ztl n~ n n~ n

all k < n

Using this and (3.55) we conclude from (3.54) that

(3.56)

Now it follows that

and

n-I

p (z)

~

IT {I +

A[Q(Z~

- IJ + Ifk(n)\}

~

n k=O

~ {I + A[Q(Z) - IJ + e(n)}n + exp[A(Q(Z) - I)J,