,
UNIVERSITY OF NORTH CAROLnTA
Department of Statistics
Chapel Hill, N. C.
ON NECESSARY AND SUFFICIENT CONDITIONS FOR THE CONVERGENCE
OF THE RENEWAL DENSITY
by
Walter L. Smith
August, 1961
This research 'Was supported by the Office of
Naval Research under Contract No. Nonr-855(09)
for research in probability and statistics at
Chapel Hill. Reproduction in whole or in part
is permitted for any purpose of the United
States Government.
·e
Institute of Statistics
Mimeograph Series No. 305
ON NECESSARY AND SUFFICIImr CONDITIONS FOR THE CONVERGENCE
OF THE RENEWAL DENSITY
by
Walter L. Smith (1)
1.
Introduction.
If k(x)
write k*l(x)
is any non-negative function in
= k(x)
Ll(-oo, -kia), and if we
and
-kia
k*n(x)
=
J
k*(n-l) (x-z)
k (z)
d
Z
-co
for
n
= 2,
3, ... , then the function
k*n (x)
is defined almost everywhere (although it is possibly infinite for some,
and its asymptotic behavior has been an object of study since the earliest
days of rene'Wal theory.
iPS) that as
x
Interest 'WaS centered upon establishing (or deny-
-> +at,
A
£1' (xL7
->
J..t
-1
l
Prior to the important paper of Feller (1941) there appears to have been
some -controversy concerning this asymptotic behavior.
Feller provided
sufficient conditions under which the rene'Wal density would converge to
(1)
This research 'tma supported by the Office of Naval Research und.er
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 purposes of the
United states Government.
the desired litut.
These conditions have been modified and simplified
by Tgckl:ind (1948) and by Smith (1954, 19'5).
The simplest sufficient
conditions to date are those given by Smith (1955);
f(x) -> 0 as
x -> +
<Xl
;
(ii)
rex)
€
these are: (i)
L1+e ~or some
In the course of some work on the theory of dams
1Ye
e >
o.
have recently
discovered that if
<Xl
f(x)
=
J
e -(t + x) xt - 1
r{t)
x >0 ,
o
=
e
then III = 1
,
0
and 6. ff(xL7
-> 1, so that the renew.1 density
does converge to the appropriate limit.
Ll +e
for
e > O.
not necessary.
x <0 ,
However,
f(x)
is:in no class
Thus the sufficient conditions of Smith (1955) are
l'1e refer to Smith (1960) for a discussion
of this
example.
This paper is mainly concerned with establishing that a certain
set of conditions on f(x)
is both necessary and suffiCient f'or tbe
asymptotic convergence of the renew.1 density.
Before we state this
main theorem, however, we must introduce some notation.
Let U{x) =p {~x }be the Heavisideun1t funetioJ!l; i.e.) U(x)=l if x>o
andU(x}=O if x<0.
and be(X) = rex) U(x)
Thus
a(x)
For any e > 0
-
ce(x);
define
ce(x)
= f(x) u(x - e),
then define a(x) = f(x) - be(X) - ce(x).
vanishes for all positive x,
be(X)
vanishes outside the
interval (0, e), and ce(x) vanishes whenever x < e; for all x, however,
f(x) = a(x) + be(X) + ce(x).
Denote Fourier transforms thus
ct
+00
(Q)
=
J
J
00
e
iQx
ax
c 5 (x)
=
e
iQx
f(x)
ax
5
-~
We can now state,
Theorem 1.
variable
If f(x)
is the....;erobability density function of a randoIil
X with a mean value
=
III
e
X which is strictly positive
(though, possibly, infinite), then in order that 6. ff'(xL7
as
x
->
-1
where 1J.1
00,
'
is to be interpreted as zero if III
->
=
+00
Il -
l
l
,
it is necessary and sufficient that
(i)
and, for
fex) -> 0
sar.~ 5 > 0,
(ii) 6. rb" (x)
-
0
-
as
x
7 ->
->
as
0
+ooz....
x
-> tf:I,
(iii) cr (Q) bclon~s to SOlJ.e class L , whore p :.:JD,y depend upon 5.
1"
---- .. -... _..
,
-.,Furtherr:lore, if .theso conditions hold for some 5 > 0, then they hold for
_
every
5 >0
.~LS
-
..
--"-,,,-~
-
II
for whieh
6
oj
l'
(x)
d
x <
1.
This is the first time that necessary and sufficient conditions for
the convergence of the renewal density have been established.
It will be
seen that they are substantially less restrictive than the sufficient
conditions of Smith (1955); for example, no restraint is placed on the
behavior of
f(x) for negative
Condition (ii)
x.
of Theorem 1 appears to 'beg the question' and does
not seem easy to verify in any particular case; however, the following
theorem enables one to recognize quickly a wide class of functions which
sstist,y, this condition.
4
Theorem 2.
Suppose that, for some
8
> 0,
8
J
b 8 (x)
ax <
1
o
functions which vanish identically outside the interval (0, 8)
and are
such that
(i)
W{ (Q)
(ii) W (x)
2
e
L
p
for some
;: O (x-M)
then we :may conclude
>
p
0;
for some M > OJ
tha~
6. Lb {xt7
8
->
0,
as
x
-> co
•
From this theorem we shall deduce a simple corollary 'Which states
that 6. Lb (xt7 -> 0
8
in the interval (0, 8).
as
x
- > co if f(x) is monotone decreasing
The motivation behind the conditions of Theorem I can be appreciated intuitively.
Only singularities of f(x)
in the ~ interval
(0, .) should affect the behavior of 6. Lf(x)_7 for very large
Singularities, if any, of f(x) at the
for f*n{x),
ties of
origin should reLJain at the origin
n > 1, and not be displaced to larger values of x.
for large values of x,
f(x)
6.. Lf(xL7
at the origin.
x.
Thus,
should not be affected by singulari-
Needless to say, the 'counter-example'
from the theory of dams, which we quoted above, satisfies the new set
of conditions.
The question arises as to whether there exist probability density
f'tmctions which fail to satisfy any of the conditions of Theorem 1.
shall show that indeed there are.
In addition we shall show that if
We
5
condition
(iii) of Theorem 1
r
large number
such that 6.
is not satisfied then tbere exists no
f
f(xL7
cerning condition (ii) of Theorem 1
Theorem 3.
'We
If there is some interval
D. [f(x)
7
D. Lbe(xL7
is bounded for
x
> r. Con-
shall prove
«(x,
~), 0 ~
a < ~ , within which
is bound.ed, then, for all sufficiently small
->
0
as
x
-> 00
B
> 0,
•
Thus, when condition (ii) of Theorem 1 fails to hold for any
0 > 0,
the rene'trlal density is unbounded in every interval, however small, to
the right of the origin.
On the other hand, a Corollary will show that
if D. Lb (xL7 tends to zero as
5
tially rapidly.
a
Let
-->
00
be the class of density :l:imctions
conditions of Theorem 1.
properties of
Theorem 4.
x
k
J .
Let
then it does so exponen-
f(x)
which satisfy the
It is of some interest to discover closure
In this connection we shall prove
1'1(x), f (x)
2
tive constants, p + q = 1;
belong to
~ ex)
let
0' ;
let p,
q be non-nega-
be a bounded non-negative function
such that
+00
+00
J¢
(x) 1'1(x)
ax > 0
-00
;
J
x
¢ (x) 1'1(x) ax
> 0
-00
Then the following probability density functions also belong to
+00
(iii)
J
-00
1'1(x - z)
1'2 (z)
dz
I"'i.
v
6
Finally we shall discuss one or two Iilinor
behavior of A Lf(xL7 as
IDatters ,
such as the
-> .. co , and the implications of a
x
recent paJ?er by Feller and Orey (1961).
2.
Preliminaries.
In all that follows
f(x)
'Will always stand for the probability
density function of a random variable
ini'inite) mean value.
explicitly invoked.
X 'With strictly J?ositive (posSibly
This assumption 'Yill be implicit even 'When not
Note that there is no restriction to strictly non-
negative random variables.
We shall alVlaYs denote
the understanding, sometimes tacit, that
as zero if III
= 00.
III
-1
( ; X by Ill' with
is to be interpreted
Our arguments will involve several manipulations
of multiple, especially double, series.
lations without comment.
vle shall indulge in these manipu-
The justification will alVlaYs be the non-
negatiVity of the terms involved.
Nearly every statement, relation, Dr
equation, in this paJ?er is really onJy true (or proven) for almost all
x;
to avoid tiresome repetition, however, vTe shall always omit mention of
this 'almost ever,yWhere t qualification.
We shall be prima.rily concerned 'With non-negative functions in
L
l
(-co, +00) and it is convenient to give this class the nal:le
shall always 'Write hex)
for the renewl density function
~.
f).
We
Lf(xL7;
m, n, r, sand N 'Will denote integers (frequently they are dUI:mJY
variables in summation); subject to the preceding exceptions, however,
we use lower case Roman letters for non-negative functions of the variable
x, and usually these functions
If a
and b
-0)
a (x - z) b (z)
\
jV .
are non-negative functions then the convolution
+co
J
are in
d
z
7
'Will almys be defined (though possibly infinite).
As
such convolutions
will occur very often in what follows we shall denote then siIrrply as
a*b or a*b(x),
whichever is the more convenient in a given context.
Notice that a*b
=b*a.,
and that, whenever the Fourier transforms eXist,
(a*b)t = a+ b+ .
If k
is any non-negative function
and, more generally,
shall write k *2 for k*k
"We
k*n for k*k*(n-l).
introduction, if we define k *1
=k
Thus, as explained in the
then we shall write
00
E
n=l
It is clear that the function 6. .L'!7 is defined alLlost everywhere, since
all the convolutions k *n are non-negative.
Ilk II
We shall also write
for the integral
+00
J
ax
k (x)
-co
3. The basic lemmas. In this section we prove the basic lemas on which
our main argument
Lemma L
II
(i)
If a
II = II
awb
a 1/
(ii)
and
II
Proof.
6.
1/
"b
is in? and
= II
is in
two ~-functions.
then a*b
is in
~
and
II •
a 1/ / (l - 1/ a
The fact that a*b
fact that a*b
integration.
'()()
and b are both i n '
If a
.L!:.7
(i)
depends.
~
That
is
II
a"
< 1 then
6.
.L~7
is in
'P
II)·
non-negative is trivial; and the
is a familiar property of the convolution of
"a*b
II
= " a"
"b 1/
can be shown by direct
8
(1i)
Since a
is non-negative it is trivial that a *n is like..
and hence that b. ["~7 is non-negative; more-
n = 2, 3, .•. ,
'Wise, for
over, this non-negativity justifies term-by-term integration, thus
II
But (i)
b.
co
I =
["~7
demonstrates that
II
E
n=l
1/ a *xl
a'*n
II
/I
=
"a lIn and so Lemma
1 (ii)
follows upon summing a geometric series.
We now define a subclass 1.) of
role in our arguments.
a(x)
1)
€
fJ
P 'Which 'Will play an important
The function a(x)
->
and a(x)
° as
JJ
€
x"""::>
00.
if and only if
Concerning this class
we have the following lemma which shows that.1J is closed under
additions, convolutions, and multiplication by non-negative constants.
Lemma 2.
A k
e
(i)
Ifk
and A is any non-negative number then
Xl .
(ii)
k
~J
€
If k
and k
l
are both members of
2
JJ
then both
l + k 2 and ki*k2 are members of 1) .
Proof.
kl
* k2
The only non-trivial part of this lemma is the fact that
€
1)
if both k
l
e: ~1
* k 2 e:
l
can find a large
we observe first that k
any e:
> 0,
lTe
and k
and k (X) < e: for all x >~.
2
it follo"YTs that
x/2
=
'P
£
1; .
2
by Lemna. 1 (i).
-00
< e:
II
kl
II
Furthermore, given
such that both k (x) < e:
l
If, therefore, we choose x > 2 Xo
xo(c)
+00
J
To prove this point
+ e:
II
k2
II
9
PlaJnJ.y" this 1Inplies that k
l
* k (x)
2
--> 0 as x -> co and the
lemzna. is proved.
a.{x)" b 8(X)" c 8 (x)
The mutilations
in our introduction.
then there is a
Proof.
8*
JJ
e
and A.
> 0 such toot
Since 6. Lb8-7 e 1)
A fb8(xL7
*n
ba (x) <
€
have been defined
vIe next prove a lemma involving
If' 6. Lb8-7
Lemma. ,.
of f(x)
6.
b (X).
8
is any real non-negative number
LA. b8*-7
e ~ •
there must be an xo(e)
< e for all x> xo ' In particular"
for all x > X and all n.
But
o
x/2
such that
this 1Inplies that
x
J
o
x
=
I
2
b*n ( x - z) b *n
8
8
( z)
d Z ,
x2
*2n
b 8 (x) ~ 2 e
We choose 8*
6. Lb8*_7 . ~
II
*n
b8
"
=
2
~ 8 such that 4 A.2 ~ b8*
6. Lb8-7
and hence 6. fb 8*_7
e
~ b8 "
<
2 e
(i)n
and so
00
I:
n=l
<
e;
•
•
II < 1. Then" necessarily"
e IJ'J also. Thus the pre-
ceding argument on b 8 applies equally well to b8*.
all x > 2x (e) J
o
A.2n b:;n (x)
n
Therefore" for
10
S:lJ.uilar1y, one can show that
co
2n-1
b;i2n-1)
A
L:
<
(x)
n=l
x, and a combination of the last two in-
for all sufficiently large
equalities proves the lemma.
With the aid of' Lemma 3
we are able to prove the fo11ow.Lng 1ennna
which we shall use directly in the proof of our main theorem.
A
["'a + bfft;-7
Proof.
o<
A
E:l"J
If A ["'bo-7
Lemma 4.
Note that
o < 0* :5
0,
0*
> 0 such that
JJ
E
< 1 -
then there is a
1/
II
1/ < 1 so that
a
II .
a
i'Te
Iaay choose a
A
Then, by Lemma 3, we can find a
such that
0*,
such that
co
L:
n=l
'be1ongs to
OJ
A ["'b
i
For each n > 0
un (x)
=
o*-7 I
of course, ale 0 belongs to 1)
let us define the
non-ne~tive
•
function
00
L:
m=o
Evidently,
/I
=
a"
+
(1 - II
belongs to
identically for
x
~
0,
~
II
*
bo
1/
,
a" )n+1
•
Furthermore, since a(x)
vanishes
it is apparent fram the definition of u (x)
n
11
x > 5*.
that u (x) vanishes identically for
n
Next define
00
u(x)
=
Z
n=l
An u (x)
n
,
where A is the nuoberchoOOll above.
vanishes identically for
II
u
x > 5*.
("
II
and so" by our choice of A,
therefore show that both 6.
U E:
Moreover"
a 1/ + "
( 1 _
a
II
*-7
5
/I
II)
)n+i
u
and u
appeal to Lennna 2 (ii) to infer that u* 6.
:J')
*
b5
L , and hence
l
-1 b
LA
Plainly u(x) is non-negative and
LA- l
')J .
E:
belong to
b
*-7
5
We have
j'J
and may
also belongs to
• But
since the summation in this inequality involves only the Udiagonal" terms
that arise in the calculation of u * 6.
and, hence"
U
o +
L")...-l b~7.
Thus
z~ un * b:
z~ un * b: belong to ~J • However,
00
Z
m=o
=
z
Z
k=o :r+s=k
=~
k =0
(a + b
)*{k+l}
5*.
(n+m)( +b
m
a
)*a*m * b*n
5* .
a*
This proves the lemma..
note,
vIe
however, that the finaJ. algebraic steps
could be considerably simplified if a
and b
*
B
were manipulated as
though they 1-rere real nUI:lbers" instead of functions forming convolutions.
It is necessary, nevertheless, to
d~onstrate
that the desired result can
be achieved by legitimate operations on functions, and not by purely
t1
symbolic lf operations.
Lemma 5.
Proof.
g2;
~1
- f
t (Q)} is bounded in a neighborhood of o.
It is easy to see that
(1 - cos g x)
f(x)
(i
ax"
and, since the integrand on the right of this inequality is non-negative,
r,
we have that for any large positive
+r
J
-r
But, given any small
( 1 - cos y)
l
G
> "21
(1 - e) y
E
2
f(x)
>
we can find a
0
for all
I y I < B.
~
I
dx
5(e)
•
such that
Hence, if we choose
I < 8/r ,
\ 1 - / (9)
By choosing
I
>
+r
(1 - e)
2
x r(x) ax •
r large enough 'We can always make
+r
J
2
x f(x)
ax >
0,
-r
and so the lemma is proved.
As a matter of fact, we only strictly need in the sequel the weaker
result that
Q2/
t1
- ft" (Q)
r
is integrable in a neighborhood of the
origin; this weaker result can be inferred from the genual Lemma 8
of Smith (1960 b).
A desire for completeness, and the simplicity of the
above proof, suggest 'tve give it here.
Lernr.Ja 6.
1
-
1
-
sin (Qj2)
(g/2)
is a bounded function of Q •
Proof.
By a well-know property of the characteristic :f'unction of an
absolutely continuous distribution, given any
T}(e:) > 0
I
such that
Hence, for all
IQ
I
T}
IQf? e: •
for all
{Q 2
f
1
t (Q)
<
2
T}
lQ I < e, it is easily shown that
But, for
1
-
(sin }Qi2»
-
1
I~1
(Q)
> 0 there is an
I? e:,
-
1
ft
E
-
(Sin (Q/2»
-(Q!2)
=
o (g2) .
Thus the lennna follows from an appeal to Lemna. 5.
Lemma 7.
If the :f'unction k(x)
belongs to
L:t
then
*h
k
(x)
->
.Jl..!..l.
I-l1
as
x ->00
P and its transform
belongs to
( I-l
-1
l
'
is defined to mean zero when
III
= + 00)
•
k ..... (Q)
14
~
Thus
Ik (x)
I
J
1
k (x)
= 2n
II
is bounded, by ( 2nf 1
lemma shows that k(x) -> 0
k
t II,
as jxl ->
00
and the Riemann-Lebessue
•
Let us next write
~(x)
N
I:
=
* f*n
k
•
n=l
Then, because all terms involved are non-negative, we can introduce the
limit
Uoo (x)
=
~
(0)
= k * hex)
~ (x)
lim
N
•
_>00
Furthermore,.
and so it is plain that
~(x)
~ k t (ft)n
n=l
belongs to L and that we may write
1
=
uj
~
=
2;!
e..
iOx
k
t (0)
f
t
(0)
,
1
•
l
1 -
(ft (Q))N + 1
1
~
ft (0)
dO.
Next define
X+1/2
vN(x)
=
J
~(z)
d z
x-l/2
and note that, by the monotone convergence theorem, we can write
X+1/2
voo(x)
=
lim
N ->
VN(x)
00
=
J
u00 (z)
d z.
x-l/2
Routine calculation shows that
~t
so that, since
b
(sin (0/2»/(0/2)
(0)
,
is a bounded function of
Q,
v f (0)
N
15
6 and use the fact that Ift(Q) l:s 1
If we now appeal to LemDJa
it is clearly seen that the integrand on the right is dominated by some
Llultiple of the integrable function Ikt(Q)[.
~
for all Q
0
so that {f+(Q)} N
<1
Moreover, Ift(Q)(
=> Oas N ->
00
for all Q
~
0 •
Thus we can appeal to Lebesgt.le l s theorem on dominated convergence to
deduce that
=--!..
u (x) - v (x)
00
211:
00
J
+00.
e- iQx kt(Q)ft(g)
i
l..
sin (g/2)
. (g/2J:
ft(Q)
1-
1
d g
•
-00
However, the integrand on the right of this last equation is also
dominated by some multiple of Ikt (Q) I and so belongs to
~.
Thus vTe
can effect a further appeal to the Riemann-Lebesgue lemma to infer that
u00 (x)
as
Ixf ->
txl
->
->
.. v00 (x)
0
=
Moreover, since u00
00.
k *h , it follows that, as
00,
x+l/2
*h
k
(x)
J
...
x-l/2
*h
k
(z)
->
d z
O.
The len:n:na 'Will therefore be proved if we can show that, as
X+l/2
J
k
*h
(z)
d
z ->
J-l -
l
x-l/2
To this end, let us put
x+l/2
g(x)
=
J
x-l/2
h
(z)
d
z
•
l
II
k
II
•
x ->
00,
e
16
Then
g( x)
can be interpreted as the expected nULlber of ren~ls to
(x - 1/2 ~ x + 1/2), and we can deduce from
occur in the interval
rene'Wal theory that g(x)
-> Ill- l as
g(x)
is bounded and that, by Blackwell r s theorem,
->
x
(Blackwell
00
(195;)~ Smith (1960 b». We
can then appeal to a very easily proved lemma. of smith (1954~ Lemma. 1) to
infer that k
*g
-> Ill - l
(x)
II
k"
as
->
x
But routine
00.
com;putation 'Will show that
x+l/2
k
*g
=
(x)
J
k
*h
(z)
d Z
,
x-l/2
and this completes the proof.
Lemma
8.
If u
and v
P
are both members of
and if
'We
wite
then, for any integer N ,
6.
-r u
Proof.
-
+ v7
=(
+
K*k +
l*k*2 + •••
Like part of the proof of LenmJa 4 this lemma is easily seen if
'We treat u, v,
!,
k
as reals; but it is again, of course, desirable to
establish the result by justifiable operations.
We note first that
00
v + !:
u
*n
n=l
+
00
+ !: v
n=l
ClO
ClO
!:
m=l
!:
n=l
00
00
u*n + !:
=!:
m=o
n=l
00
=!: u
n=l
*n
00
00
+ !: !:
m=o n=o
*u
v* u
*n
00
+!: v* (u+v)
n=l
'*m
* (u+v)
00
!: v * u
n=o
'*m
ClO
!:
r=o
(n)
r
u
*n
*n
* (u+v)
*(mf-r)
*v
*n
*(n+l-r)
17
For any integers 0:" f3"
a? 0"
f3
~ 1" the term u*ex*v'*f3 can arise in
the triple summation on the right in
r = 0, 1, 2, ••• , 0:.
(1 + 0:) 'Ways" corresponding to
The total coefficient of' u*ex*v'*f3
turns out to be
and a simple inductive argument proves this to equal
t + k * I::. .Lu+v_7
= ; u *n + ~
~ (0: + f3) u*ex
n=l
a=o f3=1
0:
* v'*f3
00
=
!:
n=l
00
=
!:
n=l
We have thus proved that the lemma holds for N = 1, and the general
result follows by repeated applications of the special result.
For
example,
and so on.
Lemma 9.
in
fi
If h(x)
then
->
lim· inf k
x
_>00
-1
~l
*h
as
x
~
(x)
~
~l
00
-1
and if k(x)
II
k "
is any function
Proof.
Choose
Co
k (x)
r
=
k (x)
=
Since
1I1~;)1
k{x)
>
belongs to
IIkll -
E
,
such that
L
if
l
we can find
h ( x) -> ~l
e
k *h(x)
y
Ih(x) - ~l-lt< E
-> ~l-l
IlkI'll
E
-1
~
r,
r
large enough so tl1at
>
0 •
as
x ->
for all
vie then observe that for all x >
Thus
x
otherwise.
0
for any prescribed
Furtherw.ore, since
~ >r
r and define
large positive nUDbe;r
as
00,
we can also find
x >~ .
2~,
x ->
co.
But
k{x) > k (x)
-
I'
so that
liD
inf
x -> 00
The lemrla is proved by letting
4.
Proof of Theorem 1.
E
---> 0 .
Vie dea.l with the necessity part first.
The proof that condition (i) must necessarily hold bas been attributed to Feller (1941), but we ba.ve recently noticed that Feller I s proof
needs to assume the boundedness of hex); most of our difficulties in the
present investigation arise precisely because we may not assume
to be bounded.
hex)
However, the necessity of (i) is easy to prove by an
appeal to Lemma 9.
We observe that
hex)
satisfies the farJiliar integral
equation
hex)
and so, if hex) -> ~l
-1
= f(x)
as
+
I' * hex)
x --->
00,
it follows that
19
lim
sup
f'(x)
x ->00
An appeal to Let1li1a. 9
=
III
-1
lim
inf'
x ->00
then shows that we must have
* hex)
f'
f'(x) -> O.
Next, to prove the necessity of' (ii), choose any
thatfl
be II <
N = 1, u
= bo'
Where
=v
k
The functions
class
in
~
r>.
=a
=
c::. [""£7
(4.1
e>0
such
We appeal to Lennna 8, making in it the substitutions
1.
v
,
+ c e'
This yields
v + v* c::. ["y] + c::. ["~7 + k * c::. [""£7 ,
+ v * c::. ["~7
u(x)
,
and v(x), a.s presently def'ined, are clearly in the
Thus, by an a.pPeal to Lemma 1 it follows that k(x)
is also
and that
,
=
1,
We are at the moment ass'Ul'ling that
h ( x)
->
III
-1
as
and so it follows from the properties just deduced for k(x)
x
->
00
and from
Le:mma 9 that
lim-" inf
x
->00
Therefore, from ( 4.1) we infer that
lim
sup
x ->00
vex)
+ v* c::. ["u(xt7 + c::.
L u(xt7
<
0 .
,
20
The non-negativity of all the terms involved in the last fornula
6["'}3;7'rJ£ A £bo-7
allows the deduction that
-> 0
as
x - > eo, and
thereby proves the necessity of (ii).
To conclude the necessity part of the proof we deal with (iii).
begin by observing that if hex) - > ~l
must be some 'Y
such that hex)
is trivial that
h(x):? c
*n
a
(x)
<
1
-1
as
x ->
for all
x:? 'Y.
for all
a > 0
m is the next integer greater than 'Y /0.
all x? 'Y.
of ID,
c
However, it
n = m,
c;m(x)
say,
In partiwbere
is bounded for
But it is not difficult to see that, because of our choice
'*m
a
Thus
then there
co
and all n.
cular, this last inequality will hold if we take
We
(x)
vanishes identically for
(x) is bounded
< r. Thus c '*m
a
x
everywhere and, being already known to be a member of
be a member of
L .
2
then follows that
L ; this shows that
2
By
the reciprocity of
~
~,
Fourier transforms it
(Cat (Q»m , the Fourier transform of
c: (Q)
is a member of
L
p
it must also
with p
c;m(x), is in
= 2m.
The
necessity of (iii) is thus established.
Let us now assume that, for some
a > 0, the conditions (i), (ii),
and (iii) hold, and prove them sufficient to :make hex) - > ~l-l as
x
->
co •
Suppose first that
that
II
....1
--
= 1.
hex) = A £b (xL7 - > 0
a
this result that if k(x)
as
IIbali
x ->
,
00'
co.
as
Then b a
x ->
=f
00.
and we see from (ii)
It follows easily from
is any bounded function in
j>
then k
* h(x)-> 0
Comparison with Lemma 7 then shows that '\re r.ro.st have
but this is impossible, since
f
vanishes outside the interval
(o, 0). Thus it must be that IIboll < 1.
Next we shall show that, with no loss of generality, it may be
supposed that
(a + ba)
e 1) and that
IIcoll > O.
For, since it is
21
-> 0 as x ->
given that t::. .Lbf~}xt7
00,
< 1, it is an easy matter to show that t::. fb~7
IIboll
by Lemma 4, we can choose
this
0*
0* can be chosen small enough to make
IIc o* II
f '£.7
t::..Lb,&-7, since we are assuming t::. .Lbo(xt7
-> 00
class
and they allow us to infer that
,
E
'J) . Then,
> 0 such that t::. .La + boL7
necessity arguments we applied above to t::.
x
and since we r,lUst have
>
O.
E
J'J , and
Now the
work equally well upon
tends to a limit as
(be - b *)+ belongs to some
o
ct
But we are given in condition (iii) that
belongs to
~
L . In view of the fact that both (be - b *)
and c: are necessarily
e
l'
bounded functions of Q it is evident that they both belong to Lr , where
L
.
q
r
= Max
(1', q).
ct
Thus, by a familiar application of Minkowski t s inequality,
is also in Lr (we may assume r > 1).
Thus the conditions (ii) and (iii) hold with 0 replaced by 0*, and,
ct. =
+
(be
-
be*)t
(a + be*) € 1) and that IIce*' > O. This
completes out justification of the claim made at the start of this :Paragraph.
in addition, we have that
vIe shall tnke use in our argument of the functional identity proved
8.
in Lernna
In this lemma we take N as any integer greater than p
= a + bo
and let
u
(4.2) k
= ce
and v
= ce'
Then, in the notation of lemma
+ c e * t::. fa + b,&-7
and it is easy to deduce from Lemma 1 that k
as hI
8,
is in
.P .
Moreover,
-> 00, the function
M
E
n=l
increases to the
L:J.
Co * (a.
+ be)*n
limit-function k;
thus it is a simple matter to
justify the following term-by-ter111 calculation of k t
=
c et
00
+
1::
n=l
:
=
1 -
iai" + bet'
Therefore, if we use the fact that
1/
II =
be
II
1 -
C
to Lp
I ~ II cell
so that it is obvious now that
( x) ->
h
Ikl' IN
Thus
so that
(4 .4)
=
k 1/
1/
c
e
1/
1
(
1
assumed in
'JJ
1 -
lIa
1
=
+ ball
.
->
1J
-1
1
+ bell
proves
t =k
8 we have
,
We have shawn that !::J. .La + be-7 may be
k E
c
e
is in
j'.) .
Thus an
1j . Employing further notation from
+ !::J. fa + be-7
t+t*k+l
Thus) by Lemma 2)
1
.
and condition (i) shows that
appeal to Lemma. 2
to ~ .
I/a
(
Let us return to (4.2).
Lemma
+
1
1, that
*N
x)
, and we
l
-1
1J
can actually be written more s1mply as
*h
L
is also a number of
* k*2 + . •• +
t * k *(N-l)
1'3 .
belongs
But Lemma 8 shows that
hex) = lex)
+
t * k(x)
belongs
00,
(4.3)
k.
Jk -t I
belongs to
-->
x
However, it can be shown by using LeImll8.
II
+
ICe+1/ II Ce II .
can appeal to LeImna 7 to infer that, as
*N *
II
a
~
and is bounded by unity.
k
II
1
<
ICet
<
< 1, we can deduce that
e 1/
<
But
I
+ .•• +
l * k*(N-l)
so the theorem is proved by an appeal to (4.4).
(x)
+
k*N * hex) ,
,.
p
Proof of Theorem 2.
?
2.
Choose
!wi
Since
23
(Q) I ~ IIwl(x)1 < 1
'We
can suppose
> 0 so that
8*
8*
J
o
w (x)
2
1
<
d x
and then define
W
(x) =
2
=
Wl(x)
'W
2
if x < 8*
(x)
otherwise
0
=
b 8 (X)
-
w2(x)
=
wl(X)
+
(w2 (x) -
,
;
w2 (x»
It is apparent from the hypothesis that
w (x) - W (X)
2
2
integrable :f'unction, and therefore it belongs to
+
w (Q) - w2f(Q),
2
+ (Q) - W2+(Q)
Iw
2
belongs to
two
also belongs to
r ~ IIw2
p
Lp - :f'unctions is another
Wl (x) = wl (x) + (w2 (X) transform is in
L.
p
w2(x»
Thus its transform,
L and, since
2
- w211 < 1,
L (recall that P
L .
2
is a bounded
it follows that
w: (Q) -
wt
(Q)
> 2). By Minkowski' s inequality the sum of
Ip
:f'unction (for
p
> 1). Thus
is a non-negative functi. on 'Whose Fourier
In other words, with no
loss of generality, we may
suppose it possible to write b (X) = w (x) + w (x) 'Where in addition to
2
8
l
the properties required of w (x) and 'W (X) in the enunciation of Theorem
l
2
2 we also have IIw 11 ~ 2-(M+2).
2
We shall next prove the intermediate result that, for some constant
'Y ,
(5.1)
for all
*n
w
(x)
2
n and all
x >0 •
<
'Y
24
We may, by hypothesis, choose a
all x > O.
for
n
= 1,
so that
y
< Y x -M for
w ( x)
2
Let us make the inductive hypothesis that
2, ... , m.
(5.1)
is true
Then
J2 xl
x
W*(m+l) ( x )
2
=
*m
W
2
+
0
W
(z)
w (z)
2
d z
+
J2
y
<
(x - z)
2m- l (X/2)M
2
d z
y
(X/2)M
w~
(z) d z
0
0
Evidently, by Lemma 1,
J2
=
IIw;m1l
IIw2l1m;
and we have shown it
possible to suppose with no loss of generality that
II w 11
2
< 2-(M+2) .
Thus
y
+
and, since mM + 2m - M> m + 1
for all
m~
y
,
1, it follovm that
so that the inductive hypothesis is continued.
true for m = 1, the validity of
Since this hypothesis is
(5.1) is thereby established for all m.
Hence,
co
2 {xL7
b. Lw
<
-
<
-
y
E
n-l M
x
2
n=l
2y
M
x
,
I-
and so
b,.
L 'W,}xL7
-> 0 as
->
x
00
•
Let us next choose an integer N > P and appeal to Lemma
*
'W
l
b,.
L 'We-7
Thence it follows that
f=
k + b,.
Le:mma
2,
(5.2)
r (x) + f * k{x) + "•.
as
->
x
00.
u = 'W
and v = 'W . Then k = 'Wl
l
and so, in vie'W of the result just proved, k € f)
this time 'With the substitutions
+
8 again,
L'We-7
+
Furthermore,
2
also belongs to ~~
f * k*(N-l)
k*N*
b,.
L wl
(x)
.
Thus, by
---> 0,
+ 'Wr;:-7 is an Ll
function
whose Fourier transform can be shown to be dominated by
This Fourier transform must therefore belong to
tion about
wi.
We saw at the beginning of the proof of Lemma
a function whose Fourier transform is in
tend to
as
x
zero as
-> 00
j
L , because of our assumpl
x
L
l
7 that
must be bounded and must
-> 0 0 . Thus kN * b,. L wl (x) + w2 (xL7 -> 0
(5" 2) and Lennna 8,
this fact, coupled 'With
is
enough to show that
The lemma is therefore proven.
Corollary
f(x)
2.1.
a >0
If, tor some
for which
"ball < 1, the function
is monotone decreasing in (0,0) then A Lb'O-7
Proof.
If f(x)
00
E
n=l
l'
(~)
n
2
is monotone decreasing in
0
2
n
E
Z2
n=l
n
00
<
(0,0) then we observe that
n 1
- f{x)
a 2
e 1.)
dx
,
26
a
J
=
f (x)
,
x
d
o
<
since
00
,
n
L . Thus, for some 'Y > 0 and all n ~ 1, f(a/2 ) <
l
Therefore if x is any number in (0, a), and we choose the in-
f(x)
n
2 'Y /a.
€
a/,zrrtl < x < a/2rJ.,
teger m so that
2m+l 'Y/a < 2'Y/x .
Hence
f(x)
=
then
< f(a/2m+l):s
f(x)
0 (x- l )
(o,a)
and the
in (o,a)
then
in
corollary follows from Theorem 2.
Notice that if
must be bounded
in
once from Theorem 2
f(x)
is monotone increasing
( 0, a*)
for any
f(x)
a* < a, and it then follows at
that 6. I"baixL7
->
0
as
x
->
co
6. Some examples. It is of interest to consider briefly some cases in
which certain of the conditions of Theorem 1 fail to hold.
Suppose that
for some fixed
{x
Ix
f(x)
> 'Y
I
does not tend to zero as
> 0 and all large
€
€}
>
, f(x)
belongs to
of f(x)
f(x)
L
l
x
->
'Y, the set of
x-points
has always strictly positive measure.
which trails
+00
must contain an infinite sequence of
which are always of a height greater than
+
f(x)
does not satisfy
co ,
but
The density function
€
will also exhibit these occasional thin
If
Since
we must conclude that the graph of the 'tail'
spikes' which grow thinner and thinner as we go to'Wards
hex)
co
I
spikes I
as
X
->
00
(iii) (we skip (ii) for the moment)
then we see, from the proof of the necessity of (iii), that there is
no large
such a
'Y
'Y
such that
hex) < 'Y
for all
x
> 'Y. For, if there were
then a very slight variant of that necessity argument would
27
shO'tv that (iii) is sa-tiafied.
Thus" when (iii) is unsatisfied the renewal
density is unbounded for all large positive
Kolmogorov and Gnedenko (1951l." p. 22; )
x.
An example due to
and used by them in another
connection provides us with an example of a density which fails to
satisfy (iii).
They show that if we define a frequency function
g(x)
1
=
o
=
then there is" for fNery
g
for all
x
*n
(x)
>
for
n
lxl
<
1
e
otherwise ,
> 1, a constant cn > 0 such that
c
n
Ix logt n+1 ) Ixl
in a neighborhood of the origin.
Thus
gt (Q)
can be in
L, for otherwise g*n( x) would be bounded for all n > p,
P
which is not the case. Thus we can take f(x) = g(x - 1), say, for
no class
our example of a density which does not satisfy (iii).
Let us now assume, temporarily, that Theorem:3
has been proved, and
turn our attention to constructing a probability density function which
does not satisfy condition (ii)
can choose
'l)
of Theorem 1.
For any integer
such that
1 -
1
-2=--2n - 1
<
<
'l)
1
1
Then the function
for
=
0
0
< x < 1"
otherwise
,
n
we
28
is in
Y and has a Fourier transfoI'I";l
from Titchmarsh (1948, p. 172)
Ig t (Q)
IN
1
Y Igf 71 -
as
gt (Q) .
Furthermore it follows
that there is a constant
such that
-> 00. Thus it appears that gt (Q)
IQI
2
when r is an integer ~ 2n - 1.
r
2n
For n 1, 2, ... , let f (x) be a :f'unction of
whose integral
n
2- n , which vanishes outside the interval I :: (n- 1 - n-' , n-l+n-'),
is in L
2' . but in no class
L
'9
=
is
y
n
and whose Fourier transfonn is in L 2 but in no class L when r
2n
r
2
is an integer < 2n - 1. Such a :f'unction f (x) can obviously be
-
n
constructed by a suitable shift and scale changes on the example
g(x)
just discussed.
Define
f(x)
Z
=
f
n
n=l
Then f(x)
(x)
is evidently a probability density function, and it is not
difficult to see that
f(x)
satisfies conditions (i) and (iii) of
Theorem 1 ( the latter condition for every 6 > 0 ).
we consider any smil interval
integers
rand n
r
~2
a +
Then the function
0: < 13.
We can choose large
such that
< 1,
.!4 (13 -
~
(0:, (3), 0
However, suppose
a)
<
f *r (x)
n
r
n
<
;
<
13 -
(13 - 0:) ,
1
-4
(13 - 0:)
vanishes outside the interval
D:r)
+
~
n
which lies entirely within
transfonn would be in L
2
(a, (3).
If f*r (x)
n
and so f:
and so we may conclude that f*r(x)
n
is clear that h (x) ~ ~ £f(xL7
?
were bounded then its
would be in L
2r
is not bounded in
f:
r
(x).
2
But
r < n
(0:,13).
But it
.
Thus the rene'Wal density
29
hex)
(i)
is not bounded in any interval.
and
Since
f(x)
sat:t.sfies conditions
(iii) of Theorem 1 it appears that condition (ii)
rllust fail to
be satisfied.
7. Proof of Theorem 3. For
If we suppose
x~a + a
any
n >1
then, in order that the integrand in (7.1) be
non-zero, there must be an integer ra such that (if we define
z 1 < a and a < z
m-
-
b;n(X)
m
< a + a.
J
Therefore, for
z
o
= 0),
x _> a + a ,
J
= n..
.E l
••••
m=l U<z <a+8
-
m
zm- 1
n-l
<
.E
m=l
< a
J ... J
ba(Zl)'" b 8 {X - Zn_l) dz l ... dzn_l
a<z<a+8
-
(7.2)
=
n-l
.E
m=l
t1
JOb;;" (z)
b:(n-m)
(x - z) dz
For this proof let us write A £b (xL7 more simply as
8
x > a + 8
Then we can conclude fram (7.2) that for
00
= .E
n=2
00
<.E
n=2
ha(x).
b~ (x)
n-l
.E
m=l
a+8
b;m (z)
J
a
b;<n..m) (x - z) dz
30
No'W suppose we are given that hex) < 'Y
a < ({3-a).
Then
in
(ex,{3).
Let us choose
( 7.3) proves that
ex +
:s
'Y
J
a
h a (x - z)
dz
ex
But it is 'Well-known in rene-wal theory ( and a fact
for all x > ex + a.
already utilized in the proof of Lemma 7)
that
a
h(z)
dz
x
is a bounded function of x.
Thus, since ha(x):S hex), it follows
that h (X) is bounded for all x > ex + a.
5
However, ha(X) satisfies the integral equation ha(x)
from (7.4)
+ ba
* ha(X),
and so, for
x
a
=
ha(X)
J
= ba(x)
> a 'We have
~
(x - z)
ba(~)
dz
o
Thus, if 'We assume
y > > a + a,
so that ha(X)
is bounded in the
domain of our discussion,
If we choose
a >0
smll enough to make
Ilball
< 1 then it is clear that
the last inequality implies that
.
sup
y + a ~ x
:s y + 2a
h (x) < IIbal!
a
-
and hence that ha(X)
-> 0 as x -> +
co ,
at an exponential rate.
This proves the theorem.
We can thus conclude that if there is no
D. .Lba(xL7
a >0
such the,t
-> 0 then there is no interval to the right of the origin
within which hex)
is bounded.
We have incidentally proved:
If t::. Lb e {xL7 -> 0
Corollary 3.1.
> 0 such that, as x ->
T)
x ->
as
then there is some
co
co,
t::. Lb (xL7 =
0 (e-1'\x) .
e
As a matter of fa.ct one can shmv that
1'\ = e-
l
log (l/Ilbell); we leave this
point a.s an exercise for the interested reader.
8.
4.
Proof of Theorem
rex)
and
a,
be'
Let us write
f(x) /1I¢(x) f(x) II
;:: ¢(x)
ce
(i)
for the mutilations of f
be' ce of f. Obviou.sly rex)
Further, if we write
a,
=
A
it is clear that
class
t::.
L
as
e.
L
p
rc:
b e(xL7:5 t::.
->
0
corresponding to the mutilations
as
I <
A
x
->
Icet 1 and therefore
LA be (xL7
if f(x) - > O.
co
,
sup _ ....
¢_(x-.)_
1I¢(x)f(x)II
c~ is a ·nleLlber.
of which
x ->
,
ct
belongs to any
Finally we notice that
and so, by Lemma. 3, t::.
L
b e (xL7
e, provided t::. Lb e (xL7 -> 0
This proves part (i) of Theorem 4.
(ii)
co,
for all small
Here we write rex)
= P:l\(x)
tation to that just adopted in (i).
->
0
for some
+ qf (X) and enploy similar no2
Plainly rex) -> 0
f l (x) -> 0
as
x ->
t
co
and f 2 {x) -> O. If clt € L and c
€ L
q
2
1'
then it is easy to show cet € L , where r = Max (1',q,2). Finally
r
we note that
if both
co
=
Z
n=l
z
r+s=n
( n ) r s b*r
r
l' q
le
*
00
<
L:
L:
n=l
r+s=n
*
*
But, if 6. ["b 8-7
l
*
6. ["b l 8-7
as
x
--->
00.
(iii)
belong to
and
1J
2
This proves (ii).
E
and hence 6.
f .
2
c
2
t
But it is clear that
-
Ce)f~
le
c le
* f2 +
* f2 +
c
r
c
L
*
*
2e
2e
fl
If
and there
r
-> 0
be(xL7
and
clt
f2
L
Il
r = r.aaX(Il"q,2).
L , say, where
f t also belongs to
l
integer m, say, such that (c
L
If both
'J.
belongs to
Lq then they both belong to
f t +
2
clt
oJ ,
1 = fl *
f
then, by Lemma 2,
1.) then, by LelJ:]J;JB. 2,
belong to
6. Lb 8-7 belongs to
For this Ilart we write
c2et
Thus
and 6. ["b 8-7
2
E
~ therefore an
fl)*m. is a bounded function.
/,F ~ .. ;: 2.S
~
f l " so we infer that
-*m
ce _
Thus -"*m
c *" belongs to L , which implies that
2
e
Finally we must show that, for some e > 0"
is also a bounded function.
-t
ce. belongs
6.
to
L- be (xL7
L ,
2m
->
0
as
x
--->
1\
and denote mutilations of f(x)
~
that
flex) +
Thus 6. ["
~
f 2 (X)
00,
1
(2
Denote
flex) +
in an obvious way.
belongs to
~
1
'2
f 2 (X»
*2
A
by f(x),
Then it is clear
, bY:Part (i) of this proof.
~ f l (x) + ~ f 2 (xL7 is bounded for all large x. But it is
1\
not difficult to see that 6. [" f(xL7 ~ 6.
1\
.
f
I
'2
f
I
l (x) + '2 f 2 (xL7
Thus 6. Lf(xt7 is bounded for all large x and so, by
A
-'
Theorem:5, 6. Lb (xL7 -> 0 as x -> 00. However, ~ = ~ f~2 +
e
I2 f
* f + '41 f *2 ' so that '2I -be
Cx) ~ be
I\()
x"
and we have
l
2
2
for all
6.
L
x.
~ be (xL7 -> 0
to infer that 6.
small
e.
L
as
x
->
be (xL7 -> 0
This proves the theorem.
00.
as
lie can now aIlIleal to LeJ.mU8. :5
x
-> 00 for all sufficiently
"
9.
Concluding observations.
of' hex)
as
x ~ -
smith (1955) discussed also the behavior
This is equivalent to supposing
00.
> +
instead of' III > 0" and still letting
x -
our arguments where we have used that
III > 0
00.
III < 0"
The only place in
is in the proof' of' Theorem
1" where we wanted to conclude that if'
x+1/2
g( x)
then
g(x)
->
=
J
h (z)
x-1/2
1l1-1. as
x
d
Z
-> too. It is an easy consequence of'
renewal theory that if' we were to have
g(x)
-> 0 as x -> +
00.
III
<0
then we would have
The rest of' the proof' of' Theorem 1 would
hold without change and allow us to conclude that if' III < 0
ditions (i)" (ii), and (iii)
to have
hex)
of' that theorem are necessary and suf'ficient
-> 0 as x -> +
00
•
Feller and Orey (1961) have recently shown that
bounded and
g(x)
-> 0 as x -> +
~ X cannot be def'ined.
00
g(x)
is uniformly
for certain situations in which
It should be clear that we can use our re-
sults here a1so,to conclude that for densities
distributions of the kind
then con-
f(x)
associated with
envisaged by Feller and Orey" the conditions
we give are still necessary and suff'icient f'or the convergence of' the
renewal density.
In conclusion we make the following observation.
If' F(x)
is the
distribution: function associated with the probability density f'(x) then
we have been assuming" of course, that
F(x)
is absolutely continuous.
However, it is possible to prove our theorems assutling that F(x) is
absolutely continuous only over the open interval (0, co).
F(x)
to be singular over (-co,
07.
vTe can allow
Admittedly the renewal density func-
tion is no longer defined for negative x, but this need not trouble us.
vTe
are only concerned with hex)
mains defined.
for positive x, and here hex)
Perusal of our arguments will show that the only
needed is to change the Ll-function a(x)
re~dcation
into a certain monotone non-
decreasing function A(x), say, and to change all convolutions v1ith a(x)
into Stieltjes convolutions with A(x).
35
REFERENCES