Poisson ProcessesPoisson Processes

1

Summary

The Poisson Process The Poisson Process Properties of the Poisson Process

I t i l ti Interarrival times Memoryless property and the residual lifetime

paradoxp Superposition of Poisson processes Random selection of Poisson Points Bulk Arrivals and Compound Poisson Processes

2

The Poisson Counting Process

Let the process {N(t)} which counts the number of events p { ( )}that have occurred in the interval [0,t). For any 0 ≤ t1 ≤ …≤ tk ≤ … 1 20 ... ...0 kN N N N tt t

6 Process with independent increments: The random

4

2

variables N(t1), N(t1,t2),…, N(tk-1,tk),… are mutually independent.

0t1 t2 t3 tk-1… tk

Process with stationary independent increments: The random variable

1 1,k k k kN t t N t N t

N(tk-1, tk), does not depend on tk-1, tk but only on tk -tk-1

3

The Poisson Counting Process

Assumptions:p At most one event can occur at any time instant (no two

or more events can occur at the same time) A process with stationary independent increments

1 1Pr , Prk k k kN t t n N t t n 1 1k k k k

Given that a process satisfies the above assumptions Given that a process satisfies the above assumptions, find

Pr , 0,1, 2,...nP N n nt t

4

The Poisson Process

Step 1: Determine Step 1: Determine 0 Pr 0P Nt t

Starting from g Pr 0N t s Pr 0 and 0,N N t t st

Pr 0 Pr 0N Nt s Stationary independent i t

increments

0 0 0P P Pt s t s Lemma: Let g(t) be a differentiable function for all t≥0

such that g(0)=1 and g(t) ≤ 1 for all t >0. Then for any t, s≥0s≥0

( ) tg t s g g g et s t for some λ>05

The Poisson Process

Therefore P 0 tP N Therefore 0 Pr 0 tP N et t

Step 2: Determine P0(Δt) for a small Δt. 2 3t t Pr 0 tN et 1 ...

2! 3!t tt

1 .t o t 1 .t o t Step 3: Determine Pn(Δt) for a small Δt. For n=2,3,… since by assumption no two events can

PrnP N n ot t t occur at the same time

A lt f 1 1 Pr 1P N t ot t t

As a result, for n=1

6

The Poisson Process

Step 4: Determine Pn(t+Δt) for any np n( ) y

PrnP N nt t t t 0

n

n k kk

P Pt t

P P P P ot t t t t 0 1 1 .n nP P P P ot t t t t

Mo ing terms bet een sides 1 .1 n nP P ot o t ot t tt t

Moving terms between sides,

1 .n nn n

P P ot t t tP Pt tt t

Taking the limit as Δt 0t t

dP t 1n

n ndP t P Pt t

dt

7

The Poisson Process

Step 4: Determine Pn(t+Δt) for any np n( ) y

PrnP N nt t t t 0

n

n k kk

P Pt t

P P P P ot t t t t 0 1 1 .n nP P P P ot t t t t

Mo ing terms bet een sides 1 .1 n nP P ot o t ot t tt t

Moving terms between sides,

1 .n nn n

P P ot t t tP Pt tt t

Taking the limit as Δt 0t t

dP t 1n

n ndP t P Pt t

dt

8

The Poisson Process

Step 5: Solve the differential equation to obtain p q

Pr , 0, 0,1, 2,...!

nt

ntP N n e t nt t

n

This expression is known as the Poisson distribution and it full characterizes the stochastic process {N(t)} in [0,t)

!n

p { ( )} [ , )under the assumptions that No two events can occur at exactly the same time, and Independent stationary increments Independent stationary increments

You should verify that E tN t and var ( ) tN t E tN t and var ( ) tN t

Parameter λ has the interpretation of the “rate” that events arrive. 9

Properties of the Poisson Process: Interevent TimesInterevent Times

Let tk 1 be the time when the k-1 event has occurred and let k-1Vk denote the (random variable) interevent time between the kth and k-1 events. Wh t i th df f V G ( )? What is the cdf of Vk, Gk(t)? Pr 1 Prk k kG V t V tt

1 Pr 0 arrivals in the interval [ )t t t 1 11 Pr 0 arrivals in the interval [ , )k kt t t

1 Pr 0N t

1 t

Stationary independent increments

1 te

tk tk + t 1 tG et

Vk

tk-1 tk-1 + t 1G et N(tk-1,tk-1+t)=0

Exponential Distribution 10

Properties of the Poisson Process: Exponential Interevent TimesExponential Interevent Times

The process {Vk} k=1,2,…, that corresponds to the p { k} , , , pinterevent times of a Poisson process is an iid stochastic sequence with cdf

P 1 tG V Therefore, the Poisson is also a renewal process

Pr 1 tkG V t et

The corresponding pdf is p

, 0tg e tt

O il h th t One can easily show that

1kE V

2

1var kV

and

11

Properties of the Poisson Process: Memoryless PropertyMemoryless Property

Let tk be the time when previous event has occurred and let V denote the time until the next event.

Assuming that we have been at the current state for z time units, let Ybe the remaining time until the next event. V

What is the cdf of Y?

Pr Pr |YF Y t V z t V zt tk tk+z

V

Y=V-z |Y t

Pr and Pr

V z V z tV z

Pr

1 Prz V z t

V z

1

G Gt z zG z

Pr V z 1 1

1 1

zt z

z

e ee

Memoryless! It does not matter that we have already spent z time units at the current state.

1 tYF e Gt t

12

Memoryless Property

This is a unique property of the exponential distribution. If q p p y pa process has the memoryless property, then it must be exponential, i.e.,

Pr | Pr Pr 1 tV z t V z V t V t e Pr | Pr Pr 1V z t V z V t V t e

Poisson Exponential M lPoissonProcess

λ

ExponentialInterevent Times

G(t)=1-e-λt

MemorylessProperty

13

Superposition of Poisson Processes

Consider a DES with m events each modeled as a Poisson Process with rate λi, i=1,…,m. What is the resulting process?

Suppose at time tk we observe event 1. Let Y1 be the time until the next event 1. Its cdf is G1(t)=1-exp{-λ1t}.

Let Y1,…,Ym denote the residual time until the next occurrence of the corresponding event.

Their cdfs are: e1ej

Vj

1 i tiG et

tk

V1= Y1

e1ej

Memoryless Property Yj=Vj-zj

p y Yj Vj zj

Let Y* be the time until the next event (any type). * min iY Y

Therefore, we need to find *

*PrY

G Y tt 14

Superposition of Poisson Processes

*PrG Y tt Pr min Y t * PrY

G Y tt Pr min iY t

1 Pr min iY t

11 Pr ,..., mY t Y t

1 Prm

Y t 1 i

mte

1

1 Pr ii

Y t

1

1i

e

t hm

Independence

The superposition of m Poisson processes is also a

* 1 tY

G et 1

where = ii

The superposition of m Poisson processes is also a

Poisson process with rate equal to the sum of the rates of the individual processes 15

Superposition of Poisson Processes

Suppose that at time tk an event has occurred. What is Suppose that at time tk an event has occurred. What is the probability that the next event to occur is event j?

Without loss of generality, let j=1 and define Y’=min{Y : i=2 m}

m

Pr next event is 1j

Y =min{Yi: i=2,…,m}.

1Pr Y Y

2

~1-exp 1 expii

t t

Pr next event is 1j 1Pr Y Y

1 11 1

0 0

yy ye e dy dy

0 0

1 1

01 yy e dye

1 1

where =m

ii

16

Residual Lifetime Paradox

Suppose that buses pass by the V

pp p ybus station according to a Poisson process with rate λ. A passenger arrives at the bus station at some random point

Vbk bk+1p

random point. How long does the passenger has

to wait?YZ

S l ti 1 Solution 1: E[V]= 1/λ. Therefore, since the passenger will (on average) arrive

in the middle of the interval, he has to wait for E[Y]=E[V]/2= 1/(2λ). But using the memoryless property the time until the next bus is But using the memoryless property, the time until the next bus is

exponentially distributed with rate λ, therefore E[Y]=1/λ not 1/(2λ)!

Solution 2: Using the memoryless property, the time until the next bus is

exponentially distributed with rate λ, therefore E[Y]=1/λ. But note that E[Z]= 1/λ therefore E[V]= E[Z]+E[Y]= 2/λ not 1/λ! 17

Random selection of Poisson Points

Let represent random arrival points of a , ,, , 21 ittt Let represent random arrival points of a Poisson process X(t) with parameter λt. Associated with each point, we define an independent Bernoulli R.V. Niwhere

,,,, 21 i

where

{ 1} , { 0} 1 .i iP N p P N p q 1

ti

tt

2t

Define )()()1()( ; )(

)(

1

)(

1tYtXNtZNtY

tX

ii

tX

ii

We claim that both Y(t) and Z(t) are independent Poisson processes with parameters λpt and λqt respectively.

18

Random selection of Poisson Points

Proof: ( ( ) ) { ( ) | ( ) } { ( ) )}P Y t k P Y t k X t n P X t n

Proof: ( ( ) ) { ( ) | ( ) } { ( ) )}.n k

P Y t k P Y t k X t n P X t n

( ){ ( ) } .n

t tP X t n e

{ ( ) | ( ) } , 0 ,k n knkP Y t k X t n p q k n

{ ( ) }!n

( ) ( )!( )! ! ! ( )!{ ( ) } ( )

!n n k

q t

k tt q tt k n k kn

n k k n n kn k n k

p eP Y t k e p q tk

(1 ) ( ) ( ) , 0, 1, 2, ! !

q teq t k

k pte ptpt e kk k

~ ( ).P pt19

Random selection of Poisson Points

Similarly: { ( ) } ~ ( ).P Z t m P qt

More generally,

{ ( ) , ( ) } { ( ) , ( ) ( ) }P Y t k Z t m P Y t k X t Y t m { ( ) , ( ) } { ( ) , ( ) ( ) } { ( ) , ( ) } { ( ) | ( ) } { ( ) }

P Y t k X t k mP Y t k X t k m P X t k m

( ) ( ) ( ) ( )! ! !

k m n nk m t pt qtk m

kt pt qtp q e e e

k m k m

( ( ) ) ( (P Y t k P Z

) )

{ ( ) } { ( ) },t m

P Y t k P Z t m

20

Bulk Arrivals and Compound Poisson ProcessesProcesses

Consider a random number of events Ci occur Consider a random number of events Ci occur simultaneously at any event instant time of a Poisson process.

31 C 22 C 4iC

t1

t2

tn

t

t

1t

2t

nt

(a) Poisson Process (b) Compound Poisson Process

Let , then is a compound Poisson process where N(t) is an ordinary

,2,1,0 },{ kkCPp ik

(a) Poisson Process ( ) p

( )

1( ) ,

N t

ii

X t C

compound Poisson process, where N(t) is an ordinary Poisson process.

{ ( ) } { ( ) | ( ) } { ( ) }P X t m P X t m N t n P N t n

0

{ ( ) } { ( ) | ( ) } { ( ) }n

P X t m P X t m N t n P N t n

21

Bulk Arrivals and Compound Poisson ProcessesProcesses

( ) { } { }iC kC iz E z P C k z

0

( )

( ) { } { }

( ) { } { ( ) }X

C ik

X t m

z E z P C k z

z E z P X t m z

0

0 0{ ( ) | ( ) } { ( ) }

m

m

m nP X t m N t n P N t n z

0 0

0 0 1( { } ) { ( ) }

m nn

mi

n m iP C m z P N t n

1

0 0( { }) { ( ) } ( { } ) { ( ) }

n

ii i

C mC n

n nE z P N t n E z P N t n

0( ( ))n

Cn

z

(1 ( )){ ( ) } Ct zP N t n e

22

Bulk Arrivals and Compound Poisson Processes

21 2 (1 )(1 ) (1 )( )

kk t zt z t z

Processes

We can write 1 2 ( )( ) ( )( ) kX

t tz e e e We can write

where which shows that the compound Poisson b d th f i t l d

, kk pprocess can be expressed as the sum of integer-scaled independent Poisson processes Thus.),(),( 21 tmtm

.)()(1

k

k tmktX

M ll li bi ti f i d d t More generally, every linear combination of independent Poisson processes represents a compound Poisson process.

23