I
..I
I
I
I
I
I
I
GENERALIZATION OF A THEOREM OF MARCINKlEWICZ
by
lL D. Miller
Birkbeck College, London and
University of North Carolina, Chapel Hill
Institute of Statistics Mimeo Series No.
423
February 1965
--I
I
I
I
I
I
I.
I
I
This research was supported by the office of
Naval Research under contract No. Nonr-855(09).
DEPARTMENT OF STATISTICS
UNIVERSITY OF NORTH CAROLINA
Chapel Hill, N. C.
I
..I
Generalization of a Theorem of Marcinkiewicz
by
I
I
I
I
I
I
--I
I
I
I
I
I
I.
I
I
H. D. Miller
Birkbeck College, London and
University of North Carolina, Chapel Hill
1.
Introduction.
Let F(x) be a distribution function, that is a nondecreasing,
right-continuous function satisfying F(-~)
= 0,
F(~)
= 1.
The Fourier-Stieltjes
transform
~
~(z) = J
(1.1)
e
izx
,
dF(x)
which always exists for real z, is the characteristic function of F(x).
We
shall be interested in cases where ~ ( z) exists for all complex z and under such
circumstances ~(z) is an entire function of z (Lukacs, (1960), p. 132).
One of
the problems connected with characteristic function is that of characterizing
them, i.e., given a function, can we say whether or not it is a characteristic
function.
Necessary and sufficient conditions are given by Bochner's theorem
(see e.g. Lukacs (1960) p. 62) but these are difficult to apply in individual
cases and so it seems worthwhile to seek characterizations of a more particular
kind.
If ~(z) is an entire function, then the moment generating function (m.g.f.),
(1.2)
M(t)
=J
e
tx
dF(x)
,
-~
is an entire function of t.
We prefer to work with the m.g.f. rather than the
characteristic function since this avoids frequent and slightly inconvenient
multiplications by 1.
I
In connection with the characterization of entire m.g.f.'s,Marcinkiewicz
(1938) proved a strong necessary condition, namely that an entire function of
finite order p > 2, the exponent of convergence of whose zeros is less than p,
can not be a m.g.f.
In particular this result implies that if p(t) is a
polynomial then exp{P(t)} is a m.g.f. if and only if p(t) = a t 2 + a t with
2
1
a ~ and a real. This latter result is usually known as the theorem of
2
l
°
Marcinkiewicz.
Lukacs (1960, p. 146) has extended this result to functions of
the form ckek{p(t)} where ek(z) is the k
by el(z)
constant.
= e Z,
ek(z)
= exp(ek_l(z)}
(k
th
iterated exponential function defined
= 2,
3, .•• ) and c k is a normalizing
Lukacs (1958) has also shown that the function
is a m.g.f. if and only if Al ~ 0, A2 ~
° and p(t) = a 2t 2 + alt with a 2 °
~
and a
real. Some further extensions of Marcinkiewiczts theorem have been
1
given by Christensen (1962), who shows, in particular, that for certain speci-
fied m.g.f.'s g(t), a function of the form
cannot be a m.g.f. if the degree of p(t) exceeds 2.
Some further generalizations are stated in the following section and
proved in SUbsequent sections.
In particular it is shown that if f(t) is any
nonconstant entire function, then f(P(t)} cannot be a m.g.f. if the degree of
p(t) exceeds 2 (Theorem
4).
We rely on certain elementary properties of m.g.f. 's.
Firstly the fUnction
M(t) defined by (1.2) is obViously real and positive when t is real.
M(t) is a strictly convex function of t when t is real unless M(t)
1960, p. 136).
Further if t = u + iv
(u,
2
v real) then
2
Further,
1 (LUkacs,
JI
I
I
I
I
I
I
~
I
I
I
I
I
I
J
I
I
I
..I
I
I
I
I
I
I
--I
I
I
I
I
I
I.
I
I
(1.4 )
jM(u+iv)1
=: M(u)
,
= rei~
ie
IM(re ) I ~
or, writing
t
(1.5)
M(r cos e) •
In establishing that certain functions are not m.g.f.'s we shall, in common
with previous authors, show that these functions contradict the elementary
inequality (1.4) or (1.5).
2.
Statement of results.
Essentially, Marcinkiewicz's result can be stated
if p(t) is a polynomial of degree m > 2 and if g(t) is an entire
as follows:
function of order p < m, then g(t) exp{p(t)) cannot be a m.g.f.
More generally,
we prove the following.
Theorem 1.
Let f( t) be a nonconstant entire function, p( t) a polynomial of
degree m > 2 ~ g(t) an entire function of order p < m.
~
g(t) f[exp{P(t)}]
cannot be a m.g.f.
.!f
Corollary.
f(t) is a nonconstant entire function and p(t) a polynomial of
degree greater than 2, ~ f[exp{P( t)}] cannot be a m.g.f.
Necessary and sufficient conditions for r[exp{p(t)}] to be a m.g.f. are
available if we restrict the class of entire functions as in the following
theorem.
Theorem 2.
f(l)
= 1,
If'
-
f
>
n-
f (t) =
r.(J)
f t n is a nonconstant entire function satisfying
n=e n
-
° (n = 0,
1, ••• ) and if p(t)
f[exp{P(t)}] is a m.g.f. if and only if p(t)
a
2
;::
= alt
= alt
m
+ ••• + a t , then
m
+ a 2t
2
-
With aI' a 2 real and
o.
It may be thought that the condition of nonnegativity on the coefficients
is a necessary condition for f[exp{P(t)}] to be a m.g.f. when p(t) = alt + a t 2
2
n
(a real, a > 0). (It clearly is necessary if a = 0.) That is is not necessary
2
2
l
1 2 , so
is shown by the simple example given by taking f(t) = 2t2 - t, P ( t) = '2t
f
3
I
J
that
f[exp(P(t))J
= 2et
2
,
which is the m.g.f. of
dF(x)
= (--L e "(~;4)
..
y;c
..l...
,J2;
2
e "(X ;2)) dx •
(It is easily verified that (djdx)F(x) ~ 0.)
However, we can, as in the following theorem, write down necessary and
sufficient conditions for f[exp(P(t)]J to be a m.g.f. without restrictions
on f(t).
But these conditions are rather obvious and at the same time diffi-
cult to apply to individual cases; i.e., it would be difficult to determine
whether a given entire function f(t) = ~ f t
n
n
satisfies the condition (2.1)
below.
Theorem 3.
If f(t) = ~en f t
-
n=a n
n
is a nonconstant entire function and if
m
p(t) = alt + ••• + amt , ~ f[exp(P(t)]J is a m.g.f. if and only if
p(t) = alt + a 2t
2
with al~' f n is real (n
= 0,
1, ••• ), f
o
~ 0, f(l)
=1
and
en
1
--
na
2
1
-
~ f n 2 (exp(_~)]yn ~ 0
n=l n
2
(2.1)
(0 < y::: 1)
or
(ii) a
2
= 0 and f
>0
n-
(n = 1, 2, ••• ).
One may also ask what may be said of functions of the type rip (t)] Where
f is an entire function.
Clearly, even if p(t) were of degree 2, f would have
to be rather special for f(P(t)] to be a m.g.f.
However, in the following
theorem we show that we may rule out all entire functions if the degree of
p(t) exceeds 2.
4
I
I
I
I
I
I
I
--I
I
I
I
I
I
.-I
I
I
..I
I
I
I
I
I
I
--I
I
I
I
I
I
I.
I
I
Theorem 4.
Let f (t) be a nonconstant entire function and p( t) a polynomial of
degree greater than 2.
~ f[P(t)) cannot be a m.g.f.
Finally, the following result generalizes that of Christensen (1962,
Theorem 3.1) and partially generalizes that of Lukacs (1958, p. 489
or 1960,
p. 158), in connection with (1.3).
Theorem 5.
Let g( t)
= E:_ oo ~tn,
gn ~ 0 (n
= 0, .!
1, ••• ), be regular and
oo
and let f(t) = En=e f ntn, f n -> 0 (n = 0, 1, ••• ),
m
be a nonconstant entire function. If p(t) = alt + ••• + & t and if a i8 real,
m
at
then g(e ) f[exp[P(t))] is & m.g.f. if and only if g(l) f(l) = 1 and
nonconstant for 0 < \t\ <
=;;;......;~..;.;;;.;;--
= alt
p(t)
3.
+ a 2t
2
~ a , &2 real and a
l
Proof of Theorem 1.
Lermna A.
of R.
00
Let R be
&
2
~
o.
We require the following lemma.
large positive number and let ¢(R) be a bounded function
Let
(m ~
where a = a exp(U~ ) ~
--- m
m
m
o.
1)
Then the roots tk(R) of the equation p(t) = R + i¢(R)
satisfy
1:
exp[
t k ()
0
R - (.E..)m
(2k1r-(3 )i
m m}
(R
-> 00;
k = 1, ••• ,m)
m
Proof.
Clearly, the result is exact if p(t) = a t
m
m
and ¢(R)
is also intuitively clear in general, since p(t) - a t
m
large It I and R respectively.
m
iii
O.
The result
and R + i¢(R) - R for
However, a proof is easily obtained by means of
Rouch~fS theorem. Without loss of generality we may take a
m
= 1, for other-
wise we make a change of variable s = [a exp(if3 1m) h. The result is clear
m
m
m
if m = 1. Suppose m > 1 and let R = C (C > 0). Define
5
I
J
B(t)
For given
radius
€
€,
C.
0 <
€
<
=tm -
A( t) - Cm•
sin(~/m),
consider a circle with centre C
exp(2i~/m)
and
For t on this circle and for C large, it is easily seen that
and hence for C sufficiently large IB(t)/A(t)1 < 1, so that A(t) and A(t) + B(t)
have the same number of zeros inside the circle. But A(t) + B(t) = t m _ Cm
has exactly one zero in this circle, namely C exp (2i~/m) •
The corresponding
zero, t(C) say, of A(t) therefore satisfies It(C) - C exp(2i~/m)l<
€
C.
Hence
(c -> (0) •
t(C) - C exp (2i~/m)
The conclusion of the lemma therefore follows for k = 1 and similarly for
k
= 2,
••• , m.
We proceed now to the proof of Theorem 1.
modulus of fez) on the circle
I zl
= e
R
Let F(R) be the maximum
and suppose that this maximum is
attained at a point z = exp{R + i¢(R)) where 0 :: ¢(R) < 2~.
pet)
where a
m
equation
tR
m
j
E ajt
j~
= Qm exp(i~m ) ~ 0
pet) = R + i¢(R),
t
(3.1)
If
=
=~
Let
1:
(0 < ~
< 2~).
Let t R
so that by Lemma A,
-
.. (~)m exp(
Q
R
m
m
(2~-~ )i
m)
m
+ iVR (uR' vR real) then a.s
R
1:
2~-~
= tl(R)
be a root of the
(R -> ex».
R
.... 00
1
21(4'
(_)m cos(
m)
Q
m
m1
(cos(
fu
(cos (
= o(R )
6
m
m) 1=
2~~ )
0 )
m m ) = 0 ).
I
I
I
I
I
I
I
--I
I
I
I
I
I
.'I
I
I
..I
I
I
I
I
I
I
II
I
I
I
I
I
I
I.
I
I
Hence for large R it follows that
2rc-{3
(coSf3J 0, cos( m m)
=
0
(R)
(otherwise)
m
Icos { (2rc-f3 )/m) I
m
<l
0
•
Now for m ~ 3 and for any e satisfying 0 < 0 :::: 21f we have
follows that
f
Icos(e/m)1
< 1.
It
and hence
for all sufficiently large R and some fixed K > O.
Now if f(z) is not a linear function; i.e., f(z) ~ fa + f z, then the
1
function
-r1
(3.3 )
max
I z/
r
=
is ultimately a steadily increasing function of r.
This can be seen by
applying the maximum modulus principle to the function f(z)/z in the annulus
o<
r
l
< Izi < r for r' fixed and r increasing.
If f(z)
then the function (3.3) tends to a finite limit, namely
= fa
If11,
+ f z (f ~ 0),
l
1
as r ->
00.
In
all cases, however, it follows that if R > R' and if R is sufficiently large,
then
F(R) >
e
(3.4)
.
~.e.,
R
F~R~
F
R ) >
F(R')
c~,
e
C
for a fixed c (0 < c :::: 1).
e R-R' ,
We may take c = 1 if f(z) is nonlinear, but we
must take 0 < C < 1 if f(z) is linear.
7
I
J
It therefore follows that for all sufficiently large R,
r[exp[P( t ))]
R
r[exp[p(~))J
> c exp [R - ~[P(uR))J
>e
the last inequality following from
K1R
(3.2).
We now turn to the function get).
= r ne
ia
= 1,2,
(K > 0),
l
Suppose get) has an infinity of zeros,
< r < •••
If €(O < € < m-p) is given
l - 2 then outside the circles with centre T and radius r -2m we have, according to
n
n
T
n
n (n
••• ), where r
a theorem of Borel (Cartwright,
1956, p. 22), that
Further, since get) is of order p, we have
log
Ig( t) I
< It
I p+€
If get) has no zeros, or a finite number of zeros, then
for all It I sufficiently large and
(3.7) also holds.
Now define
M(t)
= get)
f[exp(p(t)JJ.
Then
8
(3.6) holds a fortiori
I
I
I
I
I
I
I
--I
I
I
I
I
I
.-I
I
I
..I
I
I
I
I
I
I
ft
I
I
I
I
I
I
I.
I
I
log
Consider the sequence of values peT ) (n ~ 1, 2, ••• ).
n
bounded above as n ->
If ~(P(T )] is
n
then for R sufficiently large, all the points t R
are outside the circles with centre Tn and radius r n-2m • We may therefore
00,
apply the inequality (3.6) to (3.8).
p +
€
Using also (3.5) and remembering that
< m we obtain
log
>0
for R sufficiently large, in virtue of (3.6) and (3.1).
If ~tp(T )) is not
n
bounded above, let R < R < ••• , R -> CD, denote all the positive values
l - 2 n
of ~ (p ( 'I"n)] and let <Tl' <T2' ... denote the corresponding members of the
sequence (T ].
n
radius
I<Tn I-2m .
Let t I, til be any two points in the circle with centre
<T
n
and
For all <Tn sufficiently large we have
and we can therefore find a constant
~
such that
Hence if R > 0 lies outside the intervals
Rn - K~~-m-l
n
-~un
' Rn + K2u~-m-l
(n . "
1 2
••• )
-2m
then t R lies outside the circles with centre Tn and radius ITnl
• The sum
) is 2K E <Tn-m-l which is finite since
of the lengths of the intervals (
3.10
2
9
I
Hence we can let R -->
m+l exceeds the order p of g(t).
00
outside the
intervals (3.10) and so again we obtain the inequality (3.9).
We have thus
(1.4) and M(t) cannot be a m.g.f.
contradicted
4. Proof of Theorems 2 and 3. We need the following result which seems
natural enough but a simple proof has eluded the author.
Lemma B.
00
Let f ( z) = r.
-
n=o
f zn be a nonconstant entire function and
n
ml
p(t) = a t + a It - + ••• + alt (a ~ 0) a polynomial of degree m > 1m
mm
-- -If f[exp(P(t)}] is real for all real t, then the coefficients f (n = 0, 1, ••• )
m
n
and a
-
n
(n
Let a
B(t)
= ~k
= 1, •••
k
= bk
m) are all real.
+ iC
k
bkt , C(t)
respectively.
k
where a , b are real (k
k
k
= r.k
k
ckt.
= 1, ••• m)
and define
Let p and q be the degrees of B(t), C(t)
Then m = max(p, q) and
p(t)
= B(t)
H(t) = f[exp{P(t)}]
=
00
r. f
n=o
n
exp(nB(t) + i n C(t)}.
For real t, H(t) = 'if{'t'), where the bar denotes complex conjugate.
Hence, for
real t
00
00
(4.2)
r. f
n=o
exp[n(B(t) + iC(t)}] = r. f n exp[n(B(t) - i C(t)}],
n
n=o
but since both sides of (4.2) define entire functions of t, the relation (4.2)
holds over the whole t-plane.
Suppose first that B(t)
(4.2) that for all z
=O.
Then putting z = exp(iC(t)} we obtain from
F0,
10
I
I
I
I
I
I
I
--I
+ i C(t) •
Let
(4.1)
J
I
I
I
I
I
.-I
I
I
..I
I
I
I
I
I
I
--I
I
I
I
I
I
I.
I
I
CD
E
f
n::o
n
n
z
00
=
E -f z -n
n::o
n
=0
n
(n = 1, 2, ••• ) so that f(z) = constant, contrary to our hypothesis.
Since a Laurent expansion is unique, it follows that f O
= f O'
f
Suppose we can find a path L extending infinity in the t-plane such that
as t ->
00
along L,
~(B(t) + i C(t)}- ~(B(t) - i C(t)}
with both sides of (4.3) tending to
-00.
of (4.2) tend to zero and we obtain f
The exponential terms on both sides
o = f o so that f O is real, possibly zero.
The relat,1on (4.2) now holds with the summations starting at n = 1.
f
in the first nonvanishing coefficient after fOe
k
exp[k(B(t) + i C(t)}] we have,
Suppose
Dividing through by
00
f
k
+
f exp[(n-l)(B(t) + i C(t)}]
n=k+l n
E
00
= f
(4.4)
If we now let t ->
00
k exp(-2ki C(t)}+
E
n=k+l
f n exp[n(B(t)- i C(t)} -k {B(t)+ i C(t)}].
along L all terms inside the summation signs in (4.4)
tend to zero and we have
Since C(t) is a polynomial with zero constant term, it follows that C(t)
As before, therefore, f
n
is real for all n.
=O.
It remains to show that the path
L exists.
We choose L from among those curves in the t-plane on which 3 {C(t)}
11
= o.
I
We have, for t
J
= re i8 ,
C(t)
= c qt q
+ ••• + c t
a iq8
= cqr~
l
+ ••• + clre i8 ,
and
Each of the rays 8 = 8
n
3(C(t)J
= O.
= n(n/q)
= 0,
(n
We choose n so that b
1, ••• ) is an asymptote to a curve
cos P8
p
< 0 and then take L as the
n
curve J(C(t)] = 0 which is asymptotic to the ray 8
= 8.
n
Then, as t --> 00
along L,
!R{B(t) + i C(t)) - b
p
rP
cos P8
,
n
(r -> 00)
!R{B(t) - i C(t)) .... b
p
rP
cos P8
and L therefore satisfies our requirements.
e = 8n
2
such that cos (p n
2
or 8 =
1
Lemma B.
en
l
n/q) > O.
according as b
p
2
We observe that since q::= 1 and
such that cos (p nln/q) < 0 and an
p ::= 1, we can always find an integer n
integer n
,
n
l
> 0 or b < O.
p
This completes the proof of
real), then f[exp{P(t))] is the m.g.f. of a
lattice distribution, while if p(t)
= alt
+ a t
2
2
(a , a real, a > 0), then
l
2
2
f[exp{P(t))] is the m.g.f. of an infinite mixture of normal distributions
together with a discrete probability f
O
at the origin.
TO prove the necessity, we observe from Theorem 1 that if f[exp(P(t))]
is to be a m.g.f. at all, then p(t) can only be of the form p(t)
and from Lemma B, the coefficients a
l
til
I
We choose L asymptotic to the ray
Turning to Theorem 2, we see that the sufficiency of the condition is
clear since if p(t) = alt (a
I
I
I
I
I
I
I
and a
12
2
must be real.
= alt
2
+ a t ,
2
Further we cannot
I
I
I
I
I
.-I
I
I
..I
I
I
I
I
I
I
II
I
I
I
I
I
I
I.
I
I
have a
2
< 0 for in this case exp {p{t)} and, therefo~f[exp{p{t)}] would
as t ->
be bounded
.:!:
which is impossible for a convex function.
00,
The
theorem is therefore proved.
In proving Theorem
3 we see from Theorem 1 and Lemma B that for f[exp{P(t»)]
to be a m.g.f., it is necessary that p(t) = alt + a t 2 (a , a real) and that
2
2
l
f be real (n = 0, 1, ••• ). By the argument at the end of the previous paran
graph it is also necessary that a
where f
n
(n
= 0,
1, ••• ), a
M(t) =
J
2
and a
l
~
Now let
O.
2 are real and a 2 > O.
We clearly have
00
e
tx
dF(x)
-00
where
(4.6)
dF(x)
= f O dH{x)
J (41tna
z
+ [
2
fn
00
n=l
2
)
(x - na l )
exp {4na
)] dx,
2
H(x) being the unit step function with a jump at x = O. We thus have
alx
1
exp (2)
00
-'2
dF(x) = fodH(X) + [
~Z
f n {exp
(4.7)
J{41ta )
n=l
n
2
2
where y = exp (-x /4a ).
2
We now see that for M(t) as defined by (4.5) to be
a m.g.f. it is necessary and sufficient that f
O
~ 0, f{l) = 1 and that the sum
on the right hand side of (4.7) be nonnegative for 0 < y ~ 1.
If a
5.
2
= 0, the result is obvious and so Theorem 3 is proved.
Proof of Theorem 4.
We may assume without loss of generality that the
coefficient of the highest power of t in p(t) is unity.
p(t) = amtm + am_ltm-l + ... + a O ( am
..L
F
13
For if
0) and if f ( z) = E fnz n then
I
= tm +
m l
(a l/a )t - + ••• + (aO/a ),
mm
m
n
and then f(P(t)} = fl(Pl(t)}, where fl(Z) = E f n a:z is an entire function.
Accordingly, let
f(P(t)}
=E fn
an (Pl(t)}n where Pl(t)
m
p(t)
= tm
+ a _
ml
t m-l + ••• + a
O
(m > 2).
i
Consider the complex number Re ¢, where the argument ¢ may depend on R
but is always defined to be in the interval '](/2 ::: ¢ < 5'](/2.
We consider the
roots of the equation
p(t)
= Re i ¢ .
We assert that for given
€,
0 <
€
'](
< 2m' there is always a root t
R of
this equation which satisfies
1
m
ItRI - R
(R -+ 00
)
1
We observe that if p(t) == t
m
and
we take
ItRI
R = ¢/m,
t~en t
satisfies (5.1). lIn general, if we consider a circle CR with centre
R
-m
-6
m
R exp(i¢/m) and radius R (0 < 6 <.!), then for all sufficiently large R, the
circle C lies within the angle
R
'](
5'](
2m - € < arg t < 2m
+ € •
It is now easily verified that for t on CR,
1
IP(t) - t
m
I =0
1--
(R m)
14
(R-> 00).
I
I
I
I
I
I
I
--I
m
= R and arg t
m
J
I
I
I
I
I
..I
I
I
..I
I
I
I
I
I
I
lit
I
I
I
I
I
I
I.
I
I
1
'
Hence, since 1 - 6 > 1 - m'
it follows from Rouche's
theorem that for all
m
sufficiently large R, p(t) - R ei~ and t - R ei~ have the same number of
zeros inside CR.
t m - R e i~ has a zero at t
•
S~nce
= R11m
e i~/m , the centre
of C , it follows that P ( t) - R e i~ has at least one zero, say t = t , inside
R
R
CR.
angle
It immediately follows that for all sufficiently large R, t
R
lies in the
(5.2) and that
which gives the result (5.1).
Now consider the function
M(t) = r{p(t)}.
If M(t) is to be a m.g.f. then clearly f(t) cannot be a polynomial, for if
f(t) were a polynomial, then M(t) would also be and a polynomial cannot satisfy
the inequality (1.4).
We suppose therefore that f(t) has an essential
singularity at infinity.
I z/ = R,
If F(R) is the maximum modulus of f(t) on the circle
then F(R)/R is ultimately a strictly increasing function of R.
Hence
for all sufficiently large R , R with R < R we have
2
l
l
2
Suppose that
If( z) I
attains its maximum on
is defined to be in the interval 1(/2
and so that t
R
~~
I zl
< 51(/2.
satisfies (5.1). Let ~ = ~tR.
15
= R at a point R
Then
Choose t
R
ei~
where
so that p(t )
R
~
= R ei~
I
J
I
I
I
I
I
I
I
Now in virtue of (5.1) we have, for all R sufficiently large,
n (0
Hence there exists R > 0 and
O
<
~
< 1) such that
,
so that on applying
(5.4) to (5.5) we obtain
til
>
for R > R •
O
1
,
It follows from (1.4) that M(t) cannot be a m.g.f. and Theorem
4 is therefore proved.
6.
Proof of Theorem 5.
The sufficiency part of the Theorem 5 is clear.
For
g(eat)/g(l) is the m.g.f. of a lattice distribution and f[exp(P(t)J]/f(l) is
the m.g.f. of a lattice distribution if a
distributions if a
2
2
=0
or if a mixture of normal
> 0, with possibly a discrete probability at the origin.
To prove the necessity part of the theorem, suppose that
(6.1)
M(t)
= g(e at )
is a m.g.f., where p(t)
= a,t
.'.
f[exp[P(t)J]
m
+ ••. + a t •
m
Then M(t) is real for real t
and since g(e at ) is real for real t so also is f[exp{P(t)J].
16
Hence by Lemma B,
I
I
I
I
I
I
.-I
I
I
..I
I
I
I
I
I
I
the coefficients a , ••• , am must be real.
l
Suppose m > 3 and a > O. I f ~ is real and positive then
-
m
positive strictly increasing function of
given
~,
~
for all sufficiently
p(~)
is a
large~.
For
consider the equation
p(t)
= p(~).
By Lemma A, there is a root of this equation, say t
= t~
, which satisfies
1
t lr
':>
-
[pa)
a
iii exp
m
(21£i)
m
(6.2)
(~
Hence as ~ ->
Since p(~) - a
CD,
m
-> (X)).
we have !Rt~ - ~cos (21£/m) (m
I:
4), lRt~
=0
(~) (m = 4).
~m, it follows that
P
I
I
I
I
I
I
..I
I
for all sufficiently large
of ~ and
~,
say
~
>
~O.
Now
3t~
liS - S sin (21£/m); hence we may choose
~l
is a continuous function
> ~O in order that 3 t~
1
is an integral multiple of 21£/a.
(6.4)
g[exp(at
It then follows that
s1 )} = g[exp(affit~ 1 )}
•
Hence
r[exp[p(~l) }]
=----~--
f[exp[p(lRt
sl
)}]
Now since f(x) is nonconstant and has nonnegative coefficients f
we have f(x') > f(x') if x' > Xl> O.
It therefore follows from
17
n
(n
= 0,
1, ••• ),
(6.3) that
I
J
> 1 ,
which contradicts the inequality (1.4).
case a
m
< O.
A similar argument deals with the
It follows that if M(t) as defined by (6.1) is to be a m.g.f.
then we must have m ~ 2, i.e., pet) = a t + a t
1
2
Finally if a
2
< 0 then on letting t ->
00
2
with a
1
and a
(n
= 0,
+ a t
2
2
1, ••• ).
at
real.
along the imaginary axis
through integral mUltiples of 2rri/a we find that M( t) ->
the periodicity of g(e
2
00
on account of
) and the nonnegativity of the coefficients f
n
This again contradicts (1.4) and so we must have pet)
with a , a real and a ~ O.
2
1
2
= a 1t
This completes the proof ot the theorem.
References
[lJ
Cartwright, l.fary L. (1956).
Press, London.
Integral Functions.
[2 J
Christensen, Inge Futtrup (1962). "Some further extensions of a theorem
of Marcinkiewicz," Pacific J. Math., 12, 59-67.
[3 J
Lukacs, E. (1958). "Some extensions of a theorem of Marcinkiewic z,"
Pacific J. Math., ~, 487-501.
[4]
Lukacs, E. (1960).
[5 ]
" , de 1a 10i de Gauss,"
Marcinkiewicz, J. (1938). "Sur une propriete
Math. Zeits., 44, 612-618.
Characteristic Functions.
Cambridge University
Griffin, London.
I
I
I
I
I
I
I
til
I
I
I
I
I
I
,J
18
I
I