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)