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Renewal Process
Hu Jin
Department of Electronics and Communication Engineering
Hanyang University ERICA Campus
Renewal Process
Definition of a renewal process
Let {Xn, n ≥ 1} be a sequence of nonnegative i.i.d random variables
with a common distribution F(t) with F(0) < 1.
Let S0 = 0 and Sn = 𝑖=1𝑛 𝑋𝑛, n ≥ 1.
Then the process N(t) = sup{n | Sn ≤ t}, the number of renewals (or
arrivals) by time t, is called a renewal process.
2
time
Renewals (arrivals, events)
1X2X
3X4X 5X
1S 2S 3S4S 5S
Distribution of N(t)
Note that
Then,
Let Fn(t) be the distribution of Sn, then
Renewal function m(t) = E[N(t)]
3
nN t n S t
1
1
n n
P N t n P N t n P N t n
P S t P S t
1n nP N t n F t F t
1 1 1 1
n n
n nS t S tn n n n
m t E N t E I E I P S t F t
Contents
Renewal process N(t) < ∞ for all 0 ≤ t < ∞
N(t) =∞ for t = ∞ with probability 1
m(t) < ∞ for all 0 ≤ t < ∞
Alternating renewal process
Renewal reward process
4
1lim with probability 1t
N t
t
1lim t
m t
t
lim
n
tn n
E ZP t
E Z E Y
If and , then
i With probability 1, as
ii as
E R E X
E RR tt
t E X
E R t E Rt
t E X
Properties of a Renewal Process
Only finite number of renewals can occur in a finite time
N(t) < ∞ for all 0 ≤ t < ∞
Therefore,
5
By strong law of large numbers, with probability 1
= as
Therefore, goes to as goes to .
In orther words, can be less than or equal to
for at most a finite number of va
nn
n
n
SE X n
n
S n
S t
lues of . n
max : nN t n S t
Properties of a Renewal Process
N(t) =∞ for t = ∞ with probability 1
6
1
1
for some
0
n
nn
n
n
P N P X n
P X
P X
Properties of a Renewal Process
m(t) < ∞ for all 0 ≤ t < ∞ N(t) has finite expectation.
7
1 2
Sketch of the proof:
0 if
if
sup :
The average number of renewals at time is .
11 .
nn
n
n
n
n
XX
X
N t n X X X t
nt n
P X
tE N t
P X
Properties of a Renewal Process
8
1lim with probability 1t
N t
t
1
1 1
Sketch of the proof:
Strong law of large numbers: as
1
1
N t N t
N t
N t N t
S St
N t N t N t
St
N t
S S N t
N t N t N t
The Elementary Renewal Theorem
9
1lim t
m t
t
1
1
1
Sketch of the proof:
obtained by applying
1 1 liminf
if
if
1
1 Wald's
+ limsu
Equ on
p
ati
N tt
n nn
t Mt
N
n
t
N
m tS t m t t
t
X X MX
M X M
m tS t M m t t M
t
E S m t
1
M
The Elementary Renewal Theorem
10
1lim t
m t
t
10 if
1
if
where is a random variable that is uniformly distributed on 0,1
1 With probaiblity 1, 0, as
12 1
n
n
n
Un
Y
n Un
U
Y n
E Y nP Un
1lim with probability 1t
N t
t vs.
Wald’s Equation
Stopping time An integer-valued random variable N is said to be a stopping time of a
sequence X1, X2, … if the event {N=n} is independent of Xn+1, Xn+2,
… for all n=1,2,….
Examples
11
1 2
Let , 1,2,...., be independent and such that
1P 0 1 , 1, 2,....
2
Then min : .... 10 is a stopping time.
Let , 1,2,...., be independent and such that
1P 1 1 ,
2
n
n n
n
n
n n
X n
X P X n
N n X X X
X n
X P X
1 2
1, 2,....
Then min : .... 1 is a stopping time. n
n
N n X X X
Wald’s Equation
12
1 2
1 2
1
If , .... are indepdent and identically distributed random variables having < ,
and if is a stopping time for , .... such that , then
.
n
N
n
n
X X E X
N X X E N
E X E N E X
1 1 1 1 1
Sketch of the proof:
1 if
0 if
n
N
n n n n n n n n
n n n n n
n NI
n N
E X E X I E X I E X E I E X E I
E X P N n
1
1 1
1 1
1 if and only if we have not sopped after successively observing ,..., .
Therefore, is determined by ,..., and is indepdent of .
n
n n
n n n
E X E N
I X X
I X X X
Wald’s Equation
Examples
13
1 2
1
Let , 1,2,...., be independent and such that
1P 0 1 , 1, 2,....
2
Then min : .... 10 is a stopping time.
Let , 1,2,...., be independe
10
nt a
20
n
n
n
N
n
i
i
n
X
E N
n
X P X n
N n X
E X E X E N
X X
X n
1 2
1
nd such that
1P 1 1 , 1, 2,....
2
Then min : .... 1 is a stopping time.
1
N
i
i
n n
n
X P
E N E X E X
X n
N n X X X
E N
Wald’s Equation
14
11
N tE S m t
1 1 1 1
1 1
Sketch of proof:
1 is a stopping time. ?
1 1
,
1 dependes only on ,..., and is independent of ,....
From Wald's equatio
n n n
n n
N t why
N t n N t n
X X t X X X t
N t X X X
1 1
1
n we can obtain
1
1
N t
N t
E X X E X E N t
E S m t
The Blackwell’s Theorem
Lattice distribution A nonnegative random variable is said to be lattice if there exists d ≥
0 such that 𝑖=1𝑛 𝑃 𝑋 = 𝑛𝑑 = 1.
The largest d is called period.
15
If is not lattice, then lim
If is lattice with span , then lim
t
t
aF x m t a m t
kdF x d m t kd m t
Intuitive illustration:
lim
lim lim
1 obtained from the elementary re
t
t t
g a m t a m t
g a b m t a b m t m t a b m t b m t b m t
g a g b
g a ca
c
newal theorem.
The Key Renewal Theorem
16
0 0
01
If is not lattice, and if is directly Riemann integrable, then
1lim
where and
t
n
n
t
F h t
h d
m x F
h t x d
x F t t
x
d
m
0
, 0.N t
s
F t y dmP S s F t t sy
1
0
, .....
N t
n nN tn
S
P S s P S s S t
dF y F t y dm y
Alternating Renewal Process
an i.i.d. sequence {Zn, n≥ 1} for ON periods (distribution H)
an i.i.d. sequence {Yn, n ≥ 1} for OFF periods (distribution G)
For each n, Zn and Yn may be dependent
{(Zn, Yn, n ≥ 1)} is called an alternating renewal process
an alternating renewal process is a good model for an ON and
OFF source.
17
0
(off)
1
(on)
Limiting Theorem
for Alternating Renewal Process
Limiting Theorem for alternating renewal process Assume that the system is on at time 0.
Let P(t) = P{The system is on at time t}.
Assume that F(x) is distribution of Zn+Yn, which is non-lattice with
finite mean. If E[Zn+Yn]<∞, then
18
lim
n
tn n
E ZP t
E Z E Y
S0
1 1 1
0
Sketch of the proof:
on at | 0 0 on at |
on at | 0 |
on at | |
0
N tN t N t N t
N t
N t
t
P t P t S P S P t S y dF y
H tP t S P Z t Z Y t
F t
H t yP t S y P Z t y Z Y t y
F t y
P t H t H t y dm y H t
0 0
as
1The key renewal theorem lim lim
tn
t tF n n
t
E ZP t H t y dm y H d
E Z E Y
Exponential ON and OFF source
The limiting distribution
For N homogeneous ON and OFF sources,
19
1/
The source is in off state .1/ 1/
P
sources are in OFF state .
n N nN
P nn
0
(off)
1
(on)
Excess Life and Age
Age A(t) is the time from t since the last renewal
A(t) = t- SN(t)
Excess life Y(t) is the time from t until the next renewal
Y(t) = SN(t)+1- t
20
0
Model: The system is "on" at time if the age at is less than or equal to .
min ,lim
x
t
t t x
F y dyE X xP A t x
E X
0
Model: The system is "off" the last units of a renewal cycle
min ,lim lim off at
x
t t
x
F y dyE X xP Y t x P t
E X
XN(t)+1
21
present time t
A t Y t
1
N tX
1
0
1
Model: "on" for the total cycle if that time is greater than and is zero otherwise.
on time in cycle |lim
lim
x
N tt
x
N tt
x
ydF yE E X X xP X x
E X
ydF yP X x
Renewal Reward Process
Each time a renewal occurs we receive a reward.
Let Rn be the reward earned with the n-th renewal.
Rn, n≥1 are i.i.d
Rn may depend on Xn.
22
1
N t
n
n
R t R
Properties of a Renewal Reward Process
23
If and , then
i with probability 1, as
ii as
E R E X
E RR tt
t E X
E R t E Rt
t E X
Sketch of the proof:
i Strong law of large numbers
ii Stopping time Walds' equation Elementary renewal theorem
Example
Alternating renewal process Suppose that we earn at a rate of one per unit time when the system is
on.
Then, the total reward earned by t is just the total on time [0, t].
24
By the property of a renewal reward process
average amount of on time in 0,
where is an "on" time and is an "off" time in a cycle.
Limiting probability of the system being on
= to the lo
t E Z
t E Z E Y
Z Y
ng-run proportion of time it is on
Exponential ON and OFF source
In the steady state
For N homogeneous ON and OFF sources,
25
1/
The source is in off state .1/ 1/
P
sources are in OFF state .
n N nN
P nn
0
(off)
1
(on)