I
UNIVERSITY OF NORTH CAROLINA
Department of Statistics
Chapel Hill, N. C.
PRESENT VAlliE OF A RENEWAL PROCESS
by
Giorgio Dall'Ag1io
May 1963
This research was supported by a grant from the NATO Research
Grants Programme under a joint arrangement between the Institute
of Probability of the University of Rome, the Institute of
Statistics of the University of Paris and the Department of
Statistics of the University of North Carolina.
tatistics
3~
.~
PBESENT VAIlJE OF A RENEWAL PROCESS
l
by
Giorgio Dall'Aglio
University of North Carolina
= == = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = - - =
==
=
O.
S'UIIl11lary•
This paper studies the present cost C of a renewal process, defined as the
sum of the values of the costs of the replacements considered at the starting time
of the renewal process, With a compound interest •.
The characteristic function of C is found when the inter-arrival times are
negatively exponentially distributed; the asymptotic properties of C as the rate
of interest tends to zero are studied in the general case.
e
1.
Introduction.
Let us consider a renewal process With inter-arrival times
Xl' X2, •••
identically and independently distributed With distribution function F(X)
(F(O) = 0).
Starting at time To = 0 we Will have a replacement at each of
the instants Tl = Xl' T2 = Xl + X2, ••• •
We suppose that each renewal has a constant cost, which we assume equal to 1 ..
So one Will have to pay one (dollar, say) at time T , one at time T2 , and so on.
l
It is of interest to study the present value, at time 0, of these payments, assuming
compound interest.
The present value A 1 at time 0, of a sum A at time T, is given by:
o
(1)
~is research was supported by a grant from the NATO Research Grants Programme under a joint arrangement between the Institute of Probability
of the University of Rome, the Institute of Statistics of the University of Paris
and the Department of Statistics of the University of North Carolina ..
2
where
p is the force of interest (see, for instance,
ai' at time
L"d.7).
SO the value
0, of a replacement which will take place at time T , is given by
i
(2)
0i
=
-p T
e
i.
The present value, at time
0, of the total cost of the renewal process,
which Will be denoted by 0, is
a =
a
E
i;;l
i
i
-p E
:
=
e
.LJ
J.;:1
X
j=l
Let us consider the random variables
(4)
Y
i
=
j
Y
i
defined by
-p X.
e
].
•
They are independent and identically distributed; we will denote by 0:
r
moments;,
¢X(t )
(5)
If X is a random variable with distribution function
= E~
-prX
=
E e
=
¢X(i r p)
With the exception of the trivial case p(X
(6)
F(X), and
is the characteristic function of X, we have
O:r
0<0: <1
r
•
= 0) = 1,
their
we shall have:
:5
e
With the introduction of the Y , we can write:
i
i
ex>
C =
IT
!:
i=l
Yj
j=l
,
and the moments of C are easily expressed in terms of the
putting
~;
(8)
we have
1
co
Y1 = E
!:
T-I
i=l
j=l
co
=
1
~
!:
i=l
(9)
e
Yj
(Xl
- 1 - C¥1
2
i
co
11
Y2 = E( !:
1=1
YJ
J
j=l
co
n
!:
= 2 E E
1=1 k=i+1
co
j=l
ex>
ex>
E
!:
1=1 k=1+1
= 2
C¥2
1 - C¥2
(1-a2)(l-~)
ex>
n
Yj
k
co
1
E
n-l.n Y+E 1=1
n
j=l j j=1+1 j
j=l ~
1
C¥2 ~-1 +
~
1- ~
C¥2(1 + C¥1)
=
k
1
ex>
!:
= 2E E
1=1 k=1+1
= 2
1
2
Y6 + E !: (n- Yj)
6=1
i=l j=l
1
ex>
•
+
co
E
1=1
~
1 • C¥2
t
C¥2
4
(10)
•
The first two moments are finite" because of (6).
It can be easily seen
that all the moments of C are finite; in fact" as for V2" each moment can be
expressed as the sum of a finite number of terms 'Which are all finite. The same
could
result .f. be obtained by an argument similar to the one used to prove that NT
(i. e.
(see
the number of renewals between 0 and T) has finite moments of all orders
15.7"
p. 245).
From C~) we obtain moreover
C =
(11)
Where Ct
2.
e-~(l+ct)
has the same distribution as C, and is independent of Xl-
The Oase of the negative exponential distribution.
A particular case of great importance in applications is given by the negative
exponential distribution" 'Which 'Will be studied in this section.
In this case"
we have
(x > 0)
with A. > O.
We have also
O/r
= ¢x (i
=
A.
A. +
EC
=
A.
P
0-2(C)
=
A.
2p
p
r)
,
rp
,
•
5
In order to study the distribution of C, we can use (II).
For the charac-
teristic function of C we have
therefore
~
(12)
eite-PX ¢C(t e-px) d F(x) •
o
Introducing the density function of the negative exponential distribution, we
obtain, by a change of variables:
-+eo
[
o
e
ite-PX
¢c(t e- px.)
A-
e-AX
d
x
t
=
AP
dY •
S
o
In order to solve this integral equation, we t:r:ansform it intR a differential
equation, by differentiating on both sides after mUltiplying by t p • From
A- - 1
yP
d y
we have
A-
A-
A-
- - 1
A- t P
¢~(t) + tP~' (t)
pvC
=
t
- - 1
P
e
6
and hence
with the condition
We obtain finally
t
¢c (t)
(13)
=
exp
£
~
p
f.
o
eix - 1
d
x
%_7 •
From this expression we can derive the moments ofC; expanding the exponentials in series I we have
- 1
1 t3 .
- ~ '3T
=
=
1 +
P~ (i
t -
+... L7:'
2
t
i t3
'4 - ., y.- + •••• )
•••
•
Thus the first moments are
A.
2p
•
The same results could have been obtained from . (9) I (10) •
An interesting question concerning the random variable C is its asymptotic
behaviour as
stu~
p tends to zero.
For the negative exponential distribution the
of this problem is made very easy by formula (13) •
Let us consider the new
e
7
variable Z defined as
- EO
z = oer(O}
=
•
er
Using (13) we have
t
e
ix
-
1
x
Thus
-VI
-
-
t
it + '"p ("~-er
er
- 4er2
1
'2
=
_ ~ (.?e)
p
=
t2
-2
-
1
"3
t3
.
~~+
31
1
t i +
'2
!: (~)
p
'"
- i
2
t
3
2
2
-3
1
",- '2
'"
1
)
3
2
t3
i
- - " ? : ~(~)
•••
'3T+
4 :;
2
2p t
ti - -p ~
'"
...
p
'"
t3
- + •••
'2
P
31
All the terms in the last expression, except the first one, contain
a positive exponent;
of the form
0
~
p
~
the series is uniformly convergent for
p with
p in any interval
t:,; hence we have:
2
- -t2 .
lim
P
->
0
Tbus we have proved:
Theorem 1.
If X has a negative exponential distribution,
normally distributed as
0
is asymptotically
p tends to zero.
3. The general case.
It can be easily seen that the negative exponential distribution is the only
one which permits a s:1Itl.ple solution of the integral equation (12) by a transforms.tion into a differential equation.
In the general case, since the integral equation
e
(12) is a homogeneous Volterra equation of second kind, the solution must be
searched for among the singular solutions.
Thus it appears rather difficult to
find by this method the distribution of C for distributions of X other than
the negative exponential one.
On the other hand, if we want to study the asymptotic properties of the
distribution of C, other means are available; for instance, the investigation of
the behaviour of the moments of C_
Let us first establish some lemmas.
~k =
If, for an integer k ,
Lemma 1.
E
-r- <
(l)
then
,
k
(14)
••• + ( -1)
where 0 ~ Q(p) ~ 1 , and 11.m
P -> 0
Proof:
e-
rpx
Q
(1')
r
k
k
~k P Q(p)
k~
= 1.
We have
i
=k-1
I: (-1)
i=o
ri
IT
i i
k r~ k k
x I' Ql ( x, 1')
x I' + (-1) kl
,
where
If we integrate this equation, and use the mean value theorem, then we find
+co
~
...
e-
rpX
d p(x)
=
o
k-l
I: (-1)
i=l
k-1
=
I:
i=l
where 0 ~ Q(p) ~ 1. Moreover
iii
k rk k
!...~. I' + (-1) f I'
i~ 1
k.
(_l)i
ri
i1
-
k
J~
k
X- Ql(x,P) d F(x)
o
k
~i pi + (_l)k!:... P (3k Q(p) ,
k~
9
= ~k
lI{p)
fCIO
k
x "l (x,p) d F{X)
)
•
o
Hence it follows from the hypotheses and the properties of
=
lim Q (p)
p->
The lemma
if
<
(k
00
->
Proof:
{E(X - 13 1 )k
e -rpx
= ..
1
0
0
1
if h
=k
if h
>k
We can write, as in lemma 1 ,
<
Q
1
( X,p)
5 1;
and
~ = 1 - P 13 1 Q (p)
where
1.
~ 1) then, for every t ~ 0
p-k E e-ptX(e-pX - a )
lim
where
(XIP):
0
h
P
1
is thus proved.
If E
Lemma 2.
Q
0 ~ Q (p) ~ 1
and
.-lim
p-> 0
Q
,
(p) = 1 •
Then
-lL -ptX( e -pX-a )h = E e-ptX( e -pX
p-""Ee
l
_ tx _PX
= E e P
(e
-a )h-k (
l
. - a1 )
-pX
e
- ex
k
1 )
p
h-k
f
1-pX Q1(X'P) - l+p 131Q(p)
..
_7
p
k
10
Moreover, for
h-k? 0,
~ ttlf)~Qk(p)
i
~ 2k(f)~
2k(f)~
Since the expectation of
+ it)
+
+
itQ~(X,p) I
it) •
is finite, we can interchange the
expectation and lim signs, and the lemma is proved.
yr
Let us consider now the central moments
of C.
We have
From (ll) we obtain a recurrence formula for the moments y :
r
=
-r p X
E e l (1 + C,)r
By the well-knovm relations between moments and central moments
yr
=
r
~
j=o
.
(_l)r- J (r)
j
yr-
j
1
Yj
,
,
we may derive a recurrence formula for the central moments of C:
11
Substituting Y1 by (9)" we have:
r
Z
=
i=o
r (r
1
-Vi; (r-i)
= Z
i) r-i
LI
j i
i=o
(1-0: )
j=i"
1
= O:r -V
r
We obtain finally
(15)
1
1-0:r
r-1
+ Z
i=o
(r)
i .
r
1
(1-01)
O:J"
r-i
V-I
...
( ) r -j
-0:1
i
(r-)
j i
j=i""
E
O:j (_o:..)rU.L
j
•
12
Since
=
(15)
Ee
-piX { -pX
)r-i
e
- al
'
can also be written as
-
(15 1 )
r~l (r)
1
= 1 - ex
Yr
r
J.'
i=o
1
r i
(l~l) -
E e
YJ.'
-piX ( -pX
e. - al
)r-i
•
We can now proceed to study the convergence of the central moments Yr.
Theorem 2.
If, for
r > 0,
E
=
Kr
r! + l <
(Xl
,
then
r
(16)
'2
p
lim
p->
0
Yr
r
22
where
,
Kr
=
J
\
t
'"
Proof:
~
if
r
is even
if
r
is odd
13i
0
We will prove this theorem by induction.
= 1,
Clearly (16) holds for
Yl = O. Now let us assume that it holds for
2, and prove it m"JSt consequently hold for r.
since Yo
r
«(32 - (3 1 ) r~
~r
r
2 (~) t
r
= 0,
1,
0, 1, ... r-l, with
13
From (15')
we have
r-l
(17)
E
1=1
where
(~)r-i (~)
1
l-ar
p
( I-a
, 1.
= (~)
)
Vi E e-piX (e-pX - "J.)1'-i
~
r-1
-(~ + 1 - ~)E
-
p
-p1X( -pX
pee
'Vi
- ~
)r-i
By lemma. 1 we obtain:
e
lim
p
_> 0 l-ar
,
1
=
so that
.(r + 1 _ '!)
1
K
Q.r-i
i
lim P
"'1
r
1
Since 2 + 1 - "2
~
and we can use lemma
2.
->
0
r
2 + 1, we have
r
E
P
2 E e- piX (e- pX _ a.. )r-1
. .L
"2:
15+ 1
r
i
-2
<
co
r
1
So, 1f' i < r-2, we have 2 + 1 - '2 < r-1, and then
r
11m
P
If' on the contrary i
r
lim
p
->
0
p2 Tr _2
=r
= (~)
t;;
->
0
'2
P Ti
=
r
O.
i
- 2, i.e.. "2 + 1 - "2
= r-1,
rl
•
we have:
•
14
Finally" for
=r
i
.. 1 , we can write
E e-(r-1)pX
(e-pX _ QL)
n.L
=
Or -
a:r _1
~
and, by lemma 1 , we have
:5
=
--
lim
P
=
->
2 £1 -
p
0
~lP
.. 1 +
~lP +
2
(r-1) f3 1 P + O(p
L7
0
Hence
r
lim P Tr _1
'2
P
->
=
0
•
0
We thus have:
r
1-
1111
p
->
~P
'V
0
r
=
.=
If r
have
is odd" so is
( r)
2
1:r
r-2" and then Kr
K
=
r-2
•
Kr- 2
= O.
If r
is even, we
15
(~)
;
r!
1
=
r
2
(~)!
= Kr
•
The theorem is thus proved.
Theorem 2 enables us to establish a sufficient condition for the asymptotic
normality of C•
Theorem ;.
If all the moments of X are finite, then the distribution of
(C .. EC)/cr(C)
Proof:
tends to the standard normal distribution as
Theorem 2 gives, for
2
l-im p cr (C)
P
->
r
p
decreases to zero.
= 2:
2
(32 .. (31
=
2 (3;
0
1
Thus we have:
p
lim
->
E.L
0
C _ EC
r
cr(c)_7
which is equal to the r ..th moment of the normal variate N(O,l).
of
Thus the moments
(C .. EC)/cr(C) converge to the moments of the normal distribution, and this is
sufficient to ensure that the distribution of the variate converge to the normal
16
£];7 p.
distribution (see, for instance,
110).
The theorem is thus proved.
The asymptotic normality is of course an important feature of C; if permits
us to approximate Cwith a normal variate when
p is small.
It would be in-
teresting to have some necessary condition in order that the asymptotic distribution
be normal; unfortunately a condition of this ki.nd appears rather difficult to
establish.
However, the proof given above requires as an essential condition the existence of all the moments of X; so that one should expect that, if this condition
is dropped, the moments of (C - EC)/O"(C)
normal variate.
Lemma
That this is actually true will be seen later.
3. If EX<
P
->
00
aId
k
-t
pX
p
E (e- ell)
lim
q<t
q
EX <
00
We will show f'irst that, under the hypotheses above, we bave:
(18)
lim pP
where
(0 < t ~ k)
= 0
0
then, for every
Proof':
do not converge to the moments of the
o<
->
A<
t
E
Ie-pX -
~
k
I
= A
0
00.
In fact, we can write:
1
- - log ex.
(19)
-t (_pX
)k
-t
P Ee
-ell =p
J(P
o
.
-~
(e- px -~)
k
i~
dF{x) + p-t
1
k
(e-px-o:t.) dF(x)
- p log
~
17
Now, since 13
1
1
lim (- -
P
-> 0
=E X <
log ~)
,
~
= -
P
lim
P
->
0
r
1
'1.
-JQ:l
0
=
11m
x e-PX dF(x)
-4<0-----
P-> OJ o
e-PX dF(x)
=
Moreover
Hence
1
- - log ~
P
-t
fo
P -> °
1im
P
k
(e-px -~) d
rex) =
A
'2 '
where
=t
if k
A
'2
=
if k > t •
Since" by hypothesis, the left side of (19) tends to zero, we have also
dF(x) = - 2A
We have finally
•
and
(18) follows.
lim a = 1, given a number K with
1
p -> 0
small enough, we can make a > K. Then we have:
1
Now, since
Ete-pX - a11k
~
Jof,Q)
1
,
0
< K < 1,
(a - e-px)k d F(X)
1
1
P log ie
f '
(too
> (~~ l()k)
)1
d F(x)
1
P log ie
=
If we put
~
log
~
=
(a1 - K)
k1
1
F(p log f{L7
Ll -
•
x, we obtain
1 t
<
(log Ie)
(C1.- K)k
A•
This establishes the lemma.
by making
p
19
~e
Theorem. 4.
E:l <
If
Q)
~
then in order that all the moments of
(C .. EC)/<r(C)
converge to the moments of the normal distribution N(O,l), it is necessary that
all the moments of X be finite.
Proof:
The proof will be by induction.
Y?- <
Since E
Q)
,
the convergence of
to the k.. th central moment of the normal variate N(O,l}
Ef{C .. EC)/o-{Ctt-
is equivalent to
k
(20)
~
lim
P
We will
->
p
0
r
assum~ that
-?
E
<
for an integer
Q)
r > 4." and that
(20)
holds for every integer k, and we will prove that
r+l
e
E X
2"'
<
Q).
The theorem. will thus be proved.
since (20) holds for k < r,
With the notation of theorem. 2, (17) holds;
we have again:
r
2
11m P Ti
p->O
For
i
Hence, for
p
->
0
i
l
1
i
p
r
->
2, that is 2+l"2~ 2'
i
~
-> 0
r
p2 'V r
i)
2 E e..piX(e-px .. O1)r-i •
r
P ~
(
-~+l--
ltm
~
0
we have E X
r
1
2+l-~
2, the conclusions of theorem 2 hold, that is
"2r
p
{K
=
if i = . r ..2
r
Ti
0
if i
Moreover, for
lim
K
f'r-i
r
lim
P
(r)
=.
f'1
J.r
1
i = 1,
'V
r-l
=
I:
= 0; thus
1
r
r
lim
i=o p->O
I
p"'2 T1
= Kr
+ lim :- To
P -> 0
r-2
<
Q)
•
20
.e
Since (20) holds for k:= r, the last limit above must be zero" that is:
r
p2
lim
p
->
0
1
1 -
1
0;
r
q
By lemma 3, this implies that E X
.
cular for
4.
q
r+l
=~
•
Acknowledgements.
<.0) for every q < ~
+ 1, in parti..
The theorem is thus proved.
I would like to thaok Dr. Walter L. Smith for proposing
this problem and for his kind suggestions.
REF.ERENCES
£];7
Kendall, M. G. (1948).
The Advanced Theory of Statistics, 1 (4th ed.)
C. Griffin and Co., London.
f~7
Smith, Walter L. (1958).
Rene-wal theory and its ramifications.
J. Roy.
Statist. Soc. Sere B. 20, 243-302•
.L~7
Todhunter, R. (1931).
(;rd ed.).
Textbook on Canpound Interest and Annuities-eerta:i.n
The University Press, Cambridge.