INSTITUTE OF STATISTICS
BOX 5457
STATE COLLEGE STATION
RALEIGH. NORTH CAROLINA
..
UNIVERSITY OF NORTH CAROLINA
De':,a,rtrnent
~)f
C~apcl
Statistics
Hill
N. C.
SONE REt!ARKS ON A DISTRIBUTIOfl ~CURRIHG IN NEURAL STUDIES
"
\':fl ter L.
Februar~,
SIni tb
1964
This research ~ias sUPP'Jrteu b;y the Offie e of lJa\lal .Sesearcr:
under contract No. N.::nr-855(09) for research in ~)r6bability
anu statistics at the University of NortD Carolina Chapel
Hill) N. C. Reproduetior. in ','hole 'Jr in part is permitted
for any purpose of the United States Government.
lastituteo.f'tf,ti sties
1ltirneo Ser~ 10. 381
, '1'
r",'~
SOME REMABKS ON A DISTRIBUTION OCCURRING IN NEURAL STUDIES
•
l
by
Walter L. Smith
University of North Carolina
= = = ;: = ;: =
Introduction.
does
= = = = = = = = = = = = = = = = = = = = = ::: = = = ~ =
Suppose that
X is a non-negative random variable (which
not vanish With probability one) such that
finite and, for real positive
s) put
~(
s/\
~ log (1 + X) is
(! e- sX
G
=
For any
c > 0
we shall show there exists a distribution function G(x») of a non-negative
random variable, such that
s
(1)
G*(s)
=
J
e- sx dG(X)
=e
-cj o
1 - cp(z)
z
dz
•
o
This distribution function G(x)
arises in a variety of contexts.
The
author obtained it many years ago in some unpublished work on the initiation of nerve pulses.
It has also arisen in studies of a certain recording
apparatus (Takacs, 1955) and of the "present value" of a renewal process
(Dall 'Aglio, 1963). More recently it 'Was derived in a colloquium at
University College, London) by Dr. J.
whether G(x)
Keilson, who raised the question of
is absolutely continuous and, if so, of how the corresponditJg
probability density function behaves near the origin.
It is the object of
the present paper to prove the following theorem of several parts.
~his research 'WaS supported by the Office of Naval Research under contract
No. Nonr-855(09) for research in probability and statistics at the University
of North Carolina, Chapel Hill, N. C. Reproduction in whole or in part is
permitted for.'any purpose of the United States Government.
2
•
F(x)
If
Theorem 1.
is the distribution functi on of
j
log (1 + x) dF(x)
(1.1)
function
Equation
a(x)
(1)
defines an absolutely contin~~u!. distribution
i
with a probability density function
continuous on the open interval
(0,
creasing function Dl(x) such that
Jl
(;)
(1.2)
~).
a(x)
= XC
g(x), say, which .is
There is a strictly de-
Dl(X), and Dl(O+) is finite
(in-.addition to (2»
if and only if
F(x)
dx
<
x
o
in which case
~
,>
1
f
!
,
~
o
then:
I
<
and we assume that
X
r(l+c)
e
Furthermore, if
F(T) = 0
-c
e
Jo
for same
1 - e- x _ F(x)
dx
x
T> 0
then Dl(X)
is constant in
(0, T).
!
i
l,
(1.;)
There isa,strictly decreasing convex function D (x) such that
2
[1 - a(x)J= XC D (X). If, for some 0 ~Y < 1 ,
2
x
(4)
J [1 - F(y)] dy'" xY L( x) ,
o
where L(x) is a function of slow growth, then
,
[1 - a(x)] ,.,
as
as
x
x ->
->
,
~
~
If, however,
x
(6)
J
o
[1 - F(Y)] dy'"
x L(x) ,
as
x ->
~
,
3
•
then
[1 - G(x)] ....
-where
M (x)
L(z)
z
J
=
x
and M(x)
(1.4)
x _>
as
c M(x)
,
dz
is also a function of slow growth.
The continuous probability density functj:on g(x)
xCI-C) g(x)
= d(x),
say,
if (3) holds, in which case
Moreover
=c
d(O+)
c Y L(x)
x 2-y
g(x) ....
is such that
a strictly decreasing function of x, and x g(x)
is a function of bounded variation.
(8)
IX)
d(O+) is finite if and only
If (4) should hold, then
Dl(O+).
,
as
x
_>
(X)
,
as
x ---:;.
IX)
while, if :(6) should hold, then
g(x) .... c L(x)
x
(9)
(1.5)
A ~ 0, \I ~ 0, and for all sUfficiently large x ,
If, for some A > 0,
1 - F(x)
<
e- Ax x\l
A
r(\I+l)
then, as
x ...>
IX)
,
\I
1
g(x) =
°
_e_xp_[ A._x_+_ _\l_~_I-:(::-A_C_)::",,\I_+_I_x_\l_+_l_} } •
{
,
1:
(\1+1)
2 \1+2
x
=
° ( e-Ax xAC-l)
v=
°
To prove Theorem 1 we find it necessary to establish the following
~
three theorems concerning a more general class of density functions.
Theorem 2.
If a(x) >
I
° and
a~x) ax =
l
co
Io
a(x)
l+x
a(x)
ax,
co ,
<;>
and if we write,
for
tR s
=
°,
J e-sx
o
then there is a probability density function
t
co
=
co
~
6 (X), say, on (0,
------------.IS aO(z)a
6~ (s)
ax <
e-sxb.a(x)
ax =
e
such that
dz
"0
where the contour integral in the exponent is taken along a
Theorem 3.
co)
strai~ht
line.
In the notation of Theorem 2, if a(x) is ~~ontinuous and of
bounded variation then v7e may take
6 (x) as continuous and of bounded
a
yariation in a~y interval not containing the origin. Moreover, if
a(x) < A e·~x for some A > 0, ~ > 0, then 6 (x) = O(e-~x xA- 1 ) •
a
Theorem 4.
If- a
ex) and a (x) both satisfy the conditions of Theorem 2
l
2
and if a ex) ~ a (x) for all x and
2
l
Jl alex) -a2 (x)
o
x
,
.
•
,L
ax <
then, in an obvious extension of notation
for all x > 0,
and
co
,
,
ax
e
for almost all x.
In part of our argument we make use of the continutty theorem for
Laplace-Stieltjes transforms.
There does not Seem to be any convenient
reference for this useful theorem (although its use occurs in the literature
from time to time).
'He therefore append a short proof in an appendix.
Proof of Theorem 2.
Write, for fixed
0> 0 ,
co
=
and define
go (x)
=
,
0
for
x
<0
,
a(x)
, for x> 0
x I
o
is a probability density function. Suppose that Zl' Z2' Z3'···
=
Then go (x)
is an infinite sequence of independent random variables, each governed by
the density function
go (x) • Suppose M is an integer-valued random
variable, independent of the
[Z},
such that for
n
r
= 0,
1, 2, •••
. -I
e
P[M = r}
=
Define a random variable Y
=0
otherwise.
o(Io)r
r!
if M = 0, and Y
= Zl
+ Z2 + ••• + Z-M
Then it is an easy matter to see that Y has a distribution
function Go(x), say, where
(in the notation already suggested in the enunciation of Theorem 2 we have
written
g~(s)
for the ordinary laplace transform of go(x».
Thus we have
6
(10)
(1 _ e- sx )
x
= - S.5
decreases to zero we see from
(10)
a(x)
dx
and Beppo Levi's theorem that
---> G*o(s), say, where
00
.)Eo
(11)
log G(s)
o
= -
Jo
dx
In view of our hypothesis about
the right of
(11)
a(x}
it is clear that the integral on
is absolutely convergent.
Also, from Lebesgue's theorem
G* (s) -...> 1 as s decreases
o
It follows therefore, from the continuity
on dominated convergence, we can deduce that
through real values to zero.
theorem for Laplace-Stieltjes transforms, that there is a distribution
function G (x)
o
over
(0, co)
such that
00
Furthermore 1 by Fubini 1 s Theorem ,
00
Jo
-sx,
(
~l_._,~e_·,..!.
x
00
a(x) dx
=
lJ
Jo J0
( ) dzdx
e -zx ax
Thus the theorem will be proved if we ShO'Vl that
continuous.
(12)
G (x) is absolutely
o
To this end, we differentiate (11) and find that
d
- -d
s
*
G (s)
0
7
~(x)
Hence, if
is defined by
x
~(x)
it is a consequence of
= Jo
(12)
a(x - z) d G (z)
o
that
co
" -sx
(
Jo e x d G x)
J
=
0
e-
sx
f(x) dx
o
•
From (13) we infer that, except for a possible discontinuity at the origin,
G (x)
is absolutely continuous "\lith a density function
o
.fu2
x
Finally, v1e rule out the possibility of a point mass of probability at the
origin by observing that its l-reight must equal (taking the limit through
real values)
co
-j
lim
e
s ->
=
e
o
a(x)
dx
x
co
Which is zero, by our hypothesis that the integral
d.iverges.
Thus the theorem is proved.
Proof of Theorem 3.
for
fj
a
(x)
The continuity and bounded variation properties claimed
are easy consequences of the representation
x
fj
a
ex)
=
~ Jo
a(x - z) d G (z)
o
8
From this equation 'We· also see that if' a(x) < A e-11 x then
-Ax
(14)
Therefore
x
~ log
t
e1'1z 6a (z) d.z
<
-Ax
and so
log
Thus
A log
:It:
x
J e1'1Z 6a (z)
d.z
<
o
and hence, from
(14) again, 'We have
1
eT1x 6 (x)
a
< A xA- 1
J e'P;Z 6a (z) dz
,
o
'Which completes the proof of the theorem.
Proof of Theorem
4. We extend the notation used in the earlier proofs,
with the aid of suffices, in an obvious way.
Except for the convergence
of the integral
j1
a 1 (x) ;
a 2 (x)
dx
o
the non-negative function a 1 (x) - a (x) satisfies all the conditions im2
posed upon a(x) in Theorem 2. Thus we can say there is a distribution
function
H(x), say, of a non-negative random variable, such that
s
-j
e
(16)
9
(a~(z)
a~(z)}
-
0
dz
= H*(s)
•
Because of the convergence of (15) it Will be seen from the proof of
Theorem 2 that the function
H(x)
will have a discontinuity at the origin,
but will otherwise be absolutely continuous.
From (16) it follows that
x
Go1(x)
S
=
Go2 (x - z) d H(z)
o
and therefore
as claimed.
Let us now write
x
g[2](x)
5
J g&(x-z) g5(Z) dz
=
0
,
and, for n>2
x
g[n}(x)
0
i
=
g~n-l](x_z)
g5(z)
dz
Then, from the proof of Theorem 2, it is evident that Gol(x)
-I
at the origin of amount
e
51 , but is absolutely continuous otherwise,
and, for almost all x > 0 ,
=
has a jump
E
n=l
g[n] ex)
51
•
10
But, by our hypothesis,
for all x.
Thus, if 0
Hence
<a <~ ,
5 decrease to zero we find that
If we now let
,
that is
~
J
Ct
-J
al (x) dx;: e
lI
0
Since the last inequality holds for arbitrary a and
~
(> 0), the
final contention of Theorem 4 is proved.
Proof of Theorem 1, Part (1.1).
It is clear that the integral
Jl 1; F(x)
dx
o
diverges, and an integration by parts will show that
00
00
J log (1
+ x) ClF(x)
=
o
Thus, if we put a(x) =
conditions of Theorem 2.
c{~
- F(x»)
Jo
1 - F(x)
1 + x
dx
•
then this function satisfies all the
Upon noting that
11
co
S
zx
e-
{l -
1 - ep(z)
F(x)} dx
=
ep(z)
dz
Z
o
we can therefore infer that
o
-c
Z
o
e
i~
1 -
S
indeed, the Laplace-Stieltjes transform of an absolutely continuous
distribution function G(x).
Furthermore, if we write
function corresponding to G(x)
g(x)
for a density
then we may put
x
g(x)
= ~
So {l - F(x - z)} g(z)
dz
x
=
c G(x)
x
c
x
The distribution function G{x)
x
S
S
G(x - z) dF(z)
o
is continuous and therefore
G(x - z) dF(z)
o
is also a continuous function of x.
Equation
(17)
therefore shows
to be continuous as claimed.
Proof of Part
(1.2). From (17) we see that
x
d
dx G(x)
=
XC G(x)
-
fo
xC
G(x-z) dF(z)
so that
x
(18)
c
"" -
x
(l+c)
S
o
G(x-z) dF(z)
g(x)
12
~he right-hand side of
{18}
is negatiVe. therefore
x" C G(x) :::; D (X),
1
say, is a decreasing function as was to be proved.
Suppose that
creases to zero.
increases to a finite limit
Dl(X)
A, say, as
x
de_
Then, by a fam11iar Abelian theorem for laplace..Stieltjes
~e c!"{s) --> A r(l
transforms (Hidder, 1941, p. 181)
(through real values).
+ c) as s_>
tm
Therefore
s
clog s
S
c
,
log [A r(l + e)]
o
that is
s
c
as
.f
s --->
.
co
s
C
---> log [A r(l
d.z
+ c)]
But, by Fubini's theorem ,
cp(z)
Jl
Jo. l."zcp(z)
1
cp{z)
d.z
z
z
dz
=
s
o:l
l
J0
o:l
J
:;;
e- zx F(x) dx dz
e -x
.. e-sx
F(x) dx
x
0
ide can thence deduce from Beppo Levits theorem that
00
o:l
J
1
cp(z)
z
dz
e"x F(x)
x
= J
I1hen
A
<
00
0
From all this we may conclude that
F(x)/x belongs to
dx
Ll(O,
is finite if and only if
A
1), as is claimed in this part of the theorem.
happens to be finite
vle
see that
1
o:l
log [A r(1 + c)]:;;
c
J
o
dx
..
r
c
Jo
1 .. cp(z)
z
dz
r ,
13
But
co
J (1;
=
x
e- ) [1 - F(x)] dx
,
o
and so
co
log [A r(l + c)]
=
S ~ - e-: -
- c
F(x)
dx,
o
which proves the value for D (0+ ) •
1
This completes the proof of Part
Proof of Part (1.3).
~
(19)
[x·
c
also note that should F(x) = 0
c
x· G(x)
(18),
for all x < T then, by
vIe
is constant for all x < T.
(1.2).
x > 0 ,
From (17) we have, for
t1 - G(x)}] =
-
x
(l+C)
t1 - K(x)}
where K(X) is the absolutely continuous distribution function
x
(20)
From
K(x)
=
S
F(x - z) dG(z)
o
(19) it is apparent that x- c {1 • G(x)} = D2 (X) ,say, is a de-
creasing function with an increasing derivative; in particular,
D (X)
2
is convex.
Now suppose that for some
0
~
Y:5
1
and some function of slow
growth L(x)
x
J [1 - F(y)]dy....
y
x
L(x)
,
as
x ->
co
•
o
Then, by a slightly more complicated Abelian theorem than the one we have
already used (Doetsch,
1950, p.
), we have that
14
.1...,:,
r(l + y) L(~)
cp(s)
s
as
s -> 0+
that as
sY
through real values.
However, we can discover from (1)
s --> 0+
*
1 - G(s)
-
~
s
1 - p(z)
c
s
s
z
r(l+y)
dz
dz.
Before we can proceed we must discover the asymptotic behavior of the
integral on the right. By an obvious change of variable we have
0:>
s
L(z·l)
1
L(u/s) du •
1
dz =
s 0
zY
sY
u2-y
-
- J
I
Now Karamata (1930) has shO"l'1!l that for a given function of slow growth
L(x)
as
there is necessarily a function
x ->
00
p(x)
such that
p(x) ---> 1
and
L(x)
_ p(x)
x
From this fact it is an easy deduction that for arbitrary
L(u/s)
L(s·l)
o <
for all sufficiently large
€
> 0
<
u and all sufficiently small
s.
if Y < 1, we can appeal to dominated convergence to infer that
Hm
s -> 0+
J
1
du
2-y
u
=
1
(l-y)
Therefore,
L(8· 1 )
-
1
8
sY(l-y)
Hence
*
1 • G (s)
c
8
as
s -> 0+.
1950, p. 511)
r(l + y) L(~·l)
,
(l-y) sy
From a Tauberian theorem for Laplace transforms (Doetsch,
we can then deduce that
x
J (I ..
c L(x) xY
G(y)}dy -
,
(1 .. YJ
o
x ->
as
00
•
Furthennore, from (20) ,
,
so that,
as
s
-> 0+ ,
1 • K* (8)
S
-
1 .. ce(s)
+
s
*
1 .. G(s)
s
c
r(l+y) L(8- 1)
(l-y)sY
Hence, by another Tauberian argument ,
x
Jr
(1 ..
{I .. K(;y-)}dy -
c)L(x) xY
(1 - y)
o
If we multiply (19) by
Y+
x(l - G(x)}
as
x ->
00
)l+C) and integrate by parts we find that
x
(21)
,
= (1
+ c)
J (1
o
x
- G(y)}dy .. c
J
0
[1 .. K(y)} dy
•
16
From the asymptotic results we have obtained it follows from (21) that
1 - G(x)
c y L(x)
(l-y) xl-V
N
In the case y = 1
,
as
x
->
<Xl
•
we cannot employ the dominated convergence
ment and the results come out somewhat differently •
argu~
Let us define
<Xl
M(x)
L(z)
z
dz
Jx L(az)
z
dz
Sx
=
Then for any fixed a>O
<Xl
M(ax)
=
and hence, for an arbitrary
(1 - e)
S
x
L~Z)
dz
€
> 0 and all sufficiently large x,
L~Z)
S
< M(a x) < (1 +€)
x
It is obvious therefore that M(a x)
M(x)
N
is consequently a function of slow growth.
as
x
dz
-->
<Xl
and that
M(x)
We thus obtain for this case
1 - G*(s)
s .
1
as
s
->
0+
,
and so, via the iauberian theorem,
x
Jo {I - G(y)}dy
N
C
We shall show in a moment that
x M(x)
,
as
L(x)jM(x) -> 0
then follows, as before, that
*(s)
1 - K
+
s
c Mes';'l}
s
x ->
as
<Xl
x ->
<Xl.
It
17
and so ,
x
S (1 - K(Y)}dy -
c
x M(x)
•
o
From (21) we can then deduce that
(1 - G(x)}
C
N
,
M(x)
as
x
->
ex>
,
which was to be proved.
To see that
L(x)jM(x) -> 0
as
x ->
ex>
we observe that for 6.
arbitrarily large and positive
x6.
>
M(x)
Jx
L(z)
z
dz
6.
=
Thus
r
J.
L(ux) du
u
6.
M(x)
>
1m
S1
(L(UX)}
udu
L(u)
and so, by a dominated convergence which can be justified much as before ,
lim inf
x -> ex>
M(~J
L{X)
6.
>
S
1
-du
u
This establishes the correctness of our assertion.
Proof of Part (1.4).
(22)
so that
x g(x)
x g(x)
By
(17) and (20) we have
= c G(x) - c K(x)
is of bounded variation as claimed.
If we differentiate
this Jast equation (and write k(x) for the, necessarily continuous, densit,y
function associated with K(x»
we find that
18
= -c k(X)
x g'(x) + (1 - c) g(x)
which implies that
~ [X(l-C) g(x)] =
c k(x)
c
x
Therefore x(l-C) g(x) = d(x), say, where
function.
=
If d ( 0+ )
andtherefore G(x) >
00
d(x)
then, given any large
xCfj./c ,
is a strictly decreasing
we have
fj.
g( x ) >
for all sufficiently small x.
x
-(l-c)
Hence
only if D (0+) = 00 • On the other hand, if d( 0+ )
l
c
it is clear that, for small x, G(x) _ d(O+) x Ie, so that
d( 0+) =
fj.
is finite
00
= c Dl(O+) and, incidentally, Dl(O+) is sean to be finite.
d(O+)
To complete the proof of this part we need the following
x
Lemma 1.
If
I
{l - F(y)Jdy ,.. xYL(x)
as
x.....:>
00
,
where
o
and L(x)
is a function of slow growth, then
sY cpl(S)
- L(s·l)
..
Proof.
o:s Y ~ 1
We note that, as
x
x --.:>
00
as
"I),
s -> 0+
,
x
~
J{
I
o
1
x1+"1
- > (1.. y) r (1 +
1
[1. F(Z)]dz} dy-
xl +y
0
Io yY L(y) dy
,.. L(x)
"{+l
by Theoreme 1 of Karamata (1930, p. 40).
But an integration by parts shows
x
1
1+"1
x
S
y{l .. F(y)} dy
o
x
=
Jo {l .. F(y)}dy
-
l~
x "I
x
y
1{~f[l-f(z)]dz }dy
0
19
and hence we have
x
J
y {1 - F(y)}dy
y xy+1 L(x)
y+l
N
o
The Laplace transform of xtl - F(x»)
1 - ep(s)
2
,
as
x
-> co
is
+
s
s
and so, by the Abelian theorem for Laplace transforms, as
1 - ep(s)
s
2
-
cpr(s)
+
•
s
y r(y
f
s
->
0+ ,
2) L(s·l)
(y + 1) s Y + 1
But
-
1 - ep(s)
s
x
from the hypothesis
S
r (1
+ y) L( s -1)
,
sY
{1. F(y)}dy - xY L(x) • Thus
o
sy cpt(s)
L(s·l)
->
(1 - y) r( 1 + y)
as claimed.
Returning to the proof of Part
(1.4), let us define
x
~(x) = So
g(x - z) z d F(z)
Then
rex) ax
= .
cpr(s) G*(5)
and so, under the conditions of Lemma 1 ,
as
Therefore, by the Tauberian theorem we have been using
5
,
-> 0+
20
x
1
x'Y L(x)
J r(y) dy
-> (1.. 'Y) ,
as
x
-> 00
•
0
If we convolute both sides of (22) with F(x) we find
(23)
x k(x)
where H(x)
-
=
rex)
c K(x)
-
c H(x)
is the distribution function
x
H(x)
=
J K(x .. z) dF(z)
o
On integrating (23) we obtain
x
(24)
S
-x{l - K{x)} +
x
S
(1 - K(y»)dy -
o
r(y)
dy
0
x
=
c
J (K(y) - H(y)}
dy
•
o
From (21), (22), and (24) we then find
x2 g(x)
c
x
= (1
+ c)
~ {G(Y)" K(y)} dy
f
r(y) dy
o
x
..
c
J {K(y) - H(y)}
dy
•
o
The function
(G(y) - K(Y)}
is non-negative and its Laplace transform
is easily seen to be
1 - cp(s)
s
reI
G*(s)
+ y) L(s-l)
sy
as
s
->
0+
•
21
x
Thus
I (G(y) - K(y»)dy""
,
xY L(x)
as
x
-> 00
•
o
Similarly, the non-negative function
and so
(K(y) - Hey))
has La.place transform.
xY L(x) ,
as
x
l'o
(K(y) - H(y») dy...
x ->
00
,
also.
We now have enough asymptotic results to deduce from (25) that
x
2
g(x)
c
...,
i.e.
Y
Y x L(X)
2- y L(x~
2.'1(
x
This completes the proof of this part.
...,
g(x)
Proof of Part 1.5.
Lemma 2.
If
flex),
as
x
->00
We begin first with
f 2 (x), f3 (x)
that
are bounded integrable functions such
x
ll(x)
and if, for
,
=
So f 2 (x -
z) f3 (z) dz
x > 0 ,
,
22
t' or some
Proof.
d
> '" > 0,
J.l
A
> e,
~ ~
0, 1
>
0:
>
0, then
For any m > 0
r
axl
0:
Ax
e
(m +
xl
Axo:
e
(m + x)'P
}
=
f
~
Ao:
1 - 0:
X
l
J1
(m + x)
so that we can always choose m large enough to make
Axo:
f(x)
=
_e_........(m + x)~
an increasing function of x >
stants
o. . Having
chosen m we can then find con-
N , N , such that
l
2
<
for all x.
Thus
x
f 1 (x) :5
Nl N2
Jo
e-J.lX + J.lZ - ~~ + A~
•
0:
(m + zl
x
Jo
in view of the fact that
f(x)
increases.
e -(J.l - "')(x - z)
dz,
Thus the lemma is proved.
That part of (1.5) concerning the case
Theorem 3.
~
=°
is already covered by
We shall therefore assume from here on that
C > 0,
there are constants
1 -
~
> 0, and suppose
> 0 such that
~
,
F(x) <
x > 6
•
Define
c Max
Then
e(x) >
F(x)}
c{l -
sufficiently small x.
(
for all
e(x) -
J
o
and we can deduce from Theorem
g(x)
x
J
and
e(x) = c{l
F(x)}
-
for all
Therefore
I
(26)
I
II - F(x),
=
c{l x
F(x)1
dx
<
<Xl
4 that
,
0(6 (x»
e
for almost all x.
Define
Ci(X)
=
~(x)
=
C e-/I.x x~
r(v+l)
and
Then
~(x)
-> 0
Moreover, since
Theorem 3 that
and
~(x)
~(x)
=
e(x) - Ci(X)
° for all
= O(e-ilx )
x>
6.
Hence
for arbit.rarily large
~
~
(x)
r),
is defined.
it follows from
24
(:. (x)
(27)
=
'T
O( e -:l x )
for 11 arbitrarily large. He als 0 note that
x
(28)
(:. (x) =
(:. (x - z) b. (z) dz
'T
cr
e
~
For typographic ease, let us
v~ite
]; (~)
2 \>+2
x
Then, in view of (26), (27), (28), and Lemma 2, we shall have proved
Theorem (1.5) if we show that
one of estimating
b. (x)
say that
(29)
Our task thus becomes
From
all that we have proved so far we can
.
cr
Acr(X)
= O(e- Yx )
for any Y < 'A..
Thus we can deduce from
of Widder (1941, p. 66) that
b. (x)
cr
=
1
lim
21fi
T ->c:o
CC
-y+iT sx +
\>(s+'A.)\>
e
J
-y-iT
Let us put
h(B)
= sx
+
Cc
Then
Cc
so that
= O(~(x».
is continuous and locally of bounded variation; also,
cr
from Theorem 3,
Theorem 7.3
b. (x).
b.cr(x)
hr(s)
=0
where
d.s
•
25
s
= -/\
+
= -/\
+
6(X) ,
say
,
and this is a point on the real axis a little to the right of the point
s
= -/\.
Choose an arbitrarily small
As the real parameter t
>
0 •
runs from
= -/\
s
€
+
-€
o(x) _ t 2 + it
runs along a small parabolic arc C ,say.
s
on
NOvl
CC(\I +1)
h"{s)
Thus, for all
to +E 6(x) the point
6(X)
=
C
(s+/\) (\1+2)
we have
,
where K is some constant which does not depend on 6 and E, provided
l
,
they are both small. Therefore, if s is any point on
h"(s)
where
=
s* is some point on (: between
sand -/\ + 6.
Hence, on
s:
i
26
where
< K2
Ipl(t)I
not depend on
(j{ 0;( s)
or
E
= -Ax
K being some further constant which does
2
On
we thus have
€ /)
,
r:
/).
Cc
-+
+ x/)+
\,)0'J
= -Ax
Cc
+ x/)+
\,)
where
Ip2(t) I < K
3
Hence, noting that
€
\,)-
/)
0 , for some constant
Cc
(1 + 2€ o)e
If we substitute for
K not depending on
3
€
or
= (i - 2t) dt, we have
ds
-Ax+xo+ -
~
CC(\,)+1)t 2
2 0(\,)+2)
\,)0
/)
\,)
+e:/)
S
e
in terms of x
in the last inequality, we dis-
cover
S
eh(s)ds
=
O(*(x»
,
as
x ->
(X)
•
~
Let T be a large pos i tive number)
11 a small one, and le t
£.
(T)
be the line mapped out by
as
t
runs from
0 to T.
Notice that
£(T)
is a straight line segment
sloping away from the imaginary axis and linking up with one end of
t: .
/) •
27
On the line
::- (T)
+ -
Cc
\)r\)
vThere
r
2
= 62[(1
-
€
222
2
&) + e }
Thus
1
\I
at h( s)
< -Ax
\)+1
-
+
xV+I (C)\>'+1
c
- 'IIxt
'I
+
\)
,say
V
,
where
\)
2
v=x/)-e: 2 &x+
-Cc \I
xV+l
- \1+1
\)
\)r
1
(Cc.)\)+l
& in terms of x we find
On substituting for
\I
x (\)+IT w , say ,
where
1
--\I
w
2
=
1 - e /) +
= - (21
Hence there is a
R
h(s)
<
v1 [(1 2
+ /) - 0 ) e
2
222
e/»
-K
on
£: (T)
T
J e-n,xt dt
0
we have
I J eh(s)ds I
~
.1",
e2
\I
ds = (1 - lI)dt
<
<
\)
(4
\I
1
-..-Ax + (\1+1) x\l+l (Cc)\l+l ( 1 -
Thus, noting that
\)+1
+ 0 e )
> 0 such that w <
K
2 2
+ e }
-'Tlx
1
,
and we see that on
-,2
~)
\1+1
,
-
llxt
and that
.
,Z (T)
28
Hence)
x _>
as
,
m
and this result is uniform in T.
Lastly, consider the straight line segment
~ (T), say, which is
parallel to the real axis and mapped out by
s
as
t
=
runs from 0
(-A + C - e
to Tl T.
2
52 - t)
+
iCE 0 + T)
.t} (T) we have
On
for some K which is independent of T provided it is sufficiently large,
4
and of
e and
0
provided they are both
IJ eh(S)ds I
.~ (T)
-AX +
<
-x
=
o(1\1(x) )
e
small~'
2
xo -
€
,
as
Thus
2
0 X
x ->
m
,
uniformly in T..
In ( 29) we may suppose that
correctly if e
is small enough.
=A+
y
e
22
Combining
noting especially the uniformity of
0 - 0, for then
0"
(x)
;:
<
A
(29), (;0), and (31), and
(30) and (;1) With respect to T,
we can now easily prove
/:,
y
O(V(x»,
This completes the proof of the theorem.
as
x ->
m
•
29
Appendix
.... -a
Let
(Fn (x)}
be an infinite sequence of distribution functions of
non-negative random variables and, for real
=
J
s
~
0, let
e- sx d.F (x)
n
n
= 1, 2, •••
0-
be the corresponding Laplace-Stieltjes transforms.
Suppose F(x)
is
a further distribution function of a non-negative random variable and
that Fn (x) -> F(x)
as
n ->
0)
,
at every continuity point of F(x).
Then, by dominated convergence,
I
as
e-
SX
Fn(X) clx - >
n -> co, for every fixed real
I
e-
F(x) clx
Hence F*(s) -> F*(s)
n
s > O.
*
n -> co, for every s ~ 0 (for Fn(O)
SX
= F*(0) = 1
On the other hand, suppose Fn(s)
* ...> ~(s),
as
n --> co; suppose further that
origin.
~(s)
as
for all n).
for every real
s ~ 0 ,
is continuous to the right at the
By the usual 'Helly-Bray compactness argument there is a bounded
non-decreasing function M(x), say, and a subsequence
{F (x)} such that
nm
Moreover, we can
Fn (x) -> M(x) at every continuity point of M(x).
m
take M(x) = 0 for x < o. By the dominated convergence argument already
used we see F*
(s) ---> M*. (s) and so M* (s)
n
*
But Fn(O)
=1 m
for all n and so
*
~ (0)
= 1.
= i *(s)
for all real s > O.
However, i2 *(s) is continuous
to the right at the origin and hence M*(0+)=1. ,This proves that M(x) is
a distribution function, and indeed, the unique distribution with LaplaceStieltjes transform M*(s)
= i2(s).
By familiar reasoning it now follows
that Fn(X) -> M(x) at every continuity point of M(x).
30
REFERENCES
G. Dall'Aglio (1963), Present Value of a Renewal Process, Institute
of Statistics Mimeo Series No. 366, Chapel Hill, N. C.
G. Doetsch (1950), Handbuch der Laplace-Trans-fornation} Vol. I, Verlag
Birkhauser, Basel.
J. Karamata (1930), Sur un mode de croissance reguliere des fonctions,
Mathematics (Cluj), ~ , 38-53.
L. Takacs (1955), On stochastic processes connected with certain physical recording apparatuses, Acta Math. Hun~, §, 363-380.
D. V. Widder (1941), The Laplace Transform.
Princeton University Press.