8/13/2019 19501

1/59

4. Poisson Processes

4.1 Definition

4.2 Derivation of exponential distribution

4.3 Properties of exponential distribution

a. Normalized spacings

b. Campbells Theorem

c. Minimum of several exponential random variables

d. Relation to Erlang and Gamma Distribution

e. Guarantee Time

f. Random Sums of Exponential Random Variables

4.4 Counting processes and the Poisson distribution

4.5 Superposition of Counting Processes4.6 Splitting of Poisson Processes

4.7 Non-homogeneous Poisson Processes

4.8 Compound Poisson Processes

155

8/13/2019 19501

2/59

4 Poisson Processes

4.1 Definition

Consider a series of events occurring over time, i.e.

> Time0 X X X X

DefineTi as the time between the(i 1)st andith event. Then

Sn =T1+ T2+ . . . + Tn = Time tonth event.

DefineN(t) =no. of events in(0, t].

Then

P{Sn > t}=P{N(t)< n}

If the time for thenth event exceedst, then the number of events in(0, t]

must be less thann

.

156

8/13/2019 19501

3/59

P{Sn > t}=P{N(t)< n}

pt(n) =P{N(t) =n} =P{N(t)< n + 1} P{N(t)< n}

=P{Sn+1> t} P{Sn > t}

where Sn =T1+ T2+ . . . + Tn.

DefineQn+1(t) =P{Sn+1> t}, Qn(t) =P{Sn > t}

Then we can write

pt(n) =Qn+1(t) Qn(t)

and taking LaPlace transforms

ps(n) =Qn+1(s) Q

n(s)

157

8/13/2019 19501

4/59

If qn+1(t)andqn(t)are respective pdfs.

Qn+1(s) = 1 qn+1(s)

s , Qn(s) =

1 qn(s)

s

and

ps(n) = 1 qn+1(s)s

1 qn(s)s

= qn(s) q

n+1(s)

s

RecallT1 is time between0 and first event, T2 is time between first and

second event, etc.

158

8/13/2019 19501

5/59

Assume {Ti} i= 1, 2, . . . are independent and with the exception of

i= 1, are identically distributed with pdf q(t). Also assumeT1 has pdf

q1(t). Then

qn+1(s) =q1(s)[q

(s)]n, qn(s) =q1(s)[q

(s)]n1

and

ps(n) =qn(s) q

n+1(s)

s =q1(s)q

(s)n1

1 q(s)

s

Note thatq1(t)is a forward recurrence time. Hence

q1(t) = Q(t)

m and q1(s) =

1 q(s)

sm

ps(n) = q(s)n1

m

1 q(s)

s

2

159

8/13/2019 19501

6/59

Assume q(t) =et

fort >0 (m= 1/)= 0 otherwise.

Then q(s) = / + s, 1 q(s)

s = 1/( + s)

and ps(n) =

+ s

n1 1

+ s

2 =

1

( / + s)

n+1

.

ps(n) = 1(/ + s)n+1

However(/ + s)n+1 is the LaPlace transform of a gamma distribution

with parameters(, n + 1)

i.e.

f(t) = et(t)n+11

(n + 1) fort >0

160

8/13/2019 19501

7/59

.. pt(n) = L1{ps(n)} =

et(t)n

n!

which is the Poisson Distribution. HenceN(t)follows a Poissondistribution and

P{N(t)< n}=n1

r=0

pt(r) =P{Sn > t}

P{Sn > t}=n1

r=0et(t)r

r! .

We have shown that if the times between events are iid following an

exponential distribution theN(t)is Poisson withE[N(t)] =t.

161

8/13/2019 19501

8/59

Alternatively ifN(t) follows a Poisson distribution, thenSn has a

gamma distribution with pdff(t) = et(t)n1

(n) fort >0.

This implies time between events are exponential.

SinceP{Sn > t}=P{N(t)< n} we have proved the identity

P{Sn > t}= t

et

(t)n1

(n)

dx=

n1r=0

et

(t)r

r! .

This identity is usually proved by using integration by parts.

WhenN(t)follows a Poisson distribution withE[N(t)] =t, the

set {N(t), t >0} is called a Poisson Process.

162

8/13/2019 19501

9/59

4.2 Derivation of Exponential Distribution

Define Pn(h) =Prob. ofn events in a time intervalh

Assume

P0(h) = 1 h + o(h); P1(h) =h + o(h); Pn(h) =o(h) forn >1

whereo(h)means a term(h)so that limh0

(h)

h = 0. Consider a finite

time interval(0, t). Divide the interval inton sub-intervals of lengthh.

Thent =nh.

0 t

h h h h

t=nh

The probability of no events in(0, t)is equivalent to no events in each

sub-interval; i.e.

Pn{T > t}=P{no events in(0, t)}

T =Time for

1

st event

163

8/13/2019 19501

10/59

Suppose the probability of events in any sub interval are independent of

each other. (Assumption of independent increments.) Then

Pn{T > t}= [1 h + o(h)]n = [1

t

n + o(h)]n

= (1 t

n)n + n o(h)(1

tn

)n1 + . . .

Since

limn(1

t

n)n

=et

and

limn

n o(h) = limh0

t

ho(h) = 0

We have P{T > t}= limh0

Pn{T > t}=et.

.. The pdf ofT is d

dtP{T > t}=et. (Exponential Distribution)

164

8/13/2019 19501

11/59

4.3 Properties of Exponential Distribution

q(t) =et t >0

E(T) = 1/=m, V(t) = 1/2 =m2

q(s) =/ + s

Considerr < t.

Then

P{T > r+ t|T > r}=Conditional distribution

= Q(r+ t)

Q(r) =

e(r+t)

er =et

i.e. P{T > r+ t|T > r}=P{T > t} for allr andt.

Also P{T > r+ t}=e(r+t) =Q(r)Q(t) =Q(r+ t)

Exponential distribution is only function satisfying Q(r + t) =Q(r)Q(t)

165

8/13/2019 19501

12/59

Proof:

Q

2

n

=Q

1

n

2and in generalQ

mn

=Q

1

n

m

Q(1) =Q 1

n + 1

n + . . . +1

n

=

Q 1

nn

, m=n

.. Q

m

n =

Q

1

n

nm/n=Q(1)m/n.

IfQ()is continuous or left or right continuous we can write

Q(t) =Q(1)t.

SinceQ(1)t =et logQ(1) we havelog Q(1)is the negative of the rateparameter. Hence

Q(t) =et where = log Q(1).

166

8/13/2019 19501

13/59

a. Normalized Spacings

Let {Ti} i= 1, 2, . . . , n be iid following an exponential distribution with

E(Ti) = 1/.

Define T(1)T(2). . . T(n) Order statistics

Then the joint distribution of the order statistics is

f(t(1), t(2), . . . , t(n))dt(1), t(2), . . . t(n) =P{t(1) < T(1), t(1)+dt(1), . . . }

= n!

1! 1! . . . 1!

= n!et(1) et(2) . . . et(n)dt(1) . . . d t(n)

f(t(1), . . . , t(n)) =n!nen

1 t(i) =n!neS

whereS=n

1t(i) =

n

1ti, 0t(1) . . . t(n)

167

8/13/2019 19501

14/59

f(t(1), . . . , t(n)) =n!n

eS

, 0 t(1) . . . t(n)

S =n1

t(i)

Consider

Z1=nT(1), Z2 = (n 1)(T(2) T(1)), ,

Z(i)

= (n i + 1)(T(i)

T(i1)

), , Z(n)

=T(n)

T(n1)

.

We shall show that {Zi} are iid exponential.

f(Z1, Z2, . . . , Z n) =f(t(1), . . . , t(n))(t(1), . . . , t(n))(Z1, . . . , Z n)

where (t(1), . . . , t(n))

(Z1, . . . , Z n) is the determinant of the Jacobian.168

8/13/2019 19501

15/59

We shall find the Jacobian by making use of the relation

(t(1), . . . , t(n))(Z1, . . . , Z n)

= (Z1, Z2, . . . Z n)t(1), . . . , t(n))

1

Zi= (n i + 1)(T(i) T(i1)), T(0) = 0

ZiT

(j)

=

n i + 1 j =i

(n i + 1) j =i 1

0 Otherwise

169

8/13/2019 19501

16/59

(Z1, . . . , Z n)

(t(1), . . . , t(n)=

n 0 0 0 . . . 0

(n 1) (n 1) 0 0 . . . 0

0 (n 2) (n 2) 0 . . . 0

...

0 0 . . . . . . 1 1

170

8/13/2019 19501

17/59

Note: The determinant of a trangular matrix is the product of the maindiagonal terms

.. (Z1, . . . , Z n)

(t(1), . . . , t(n))

=n(n 1)(n 2) . . . 2 1 =n!

and

f(z1, z2, . . . , zn) =n!n eS

1

n!=ne

n1

zi=neS

asS=n

i=1

t(i) =z1+ . . . + zn.

The spacings Zi = (n i + 1)(T(i) T(i1))are sometimes called

normalized spacings.

171

8/13/2019 19501

18/59

Homework:

1. Suppose there aren observations which are iid

exponential(Ti = 1/). However there arer non-censored observations

and(n r) censored observations all censored att(r).

ShowZi = (n i + 1)(T(i) T(i1)) fori = 1, 2, . . . , r are iidexponential.

2. Show that

T(i) = Z1

n +

Z2n 1

+ . . . + Zi

n i + 1

and prove

E(T(i)) = 1

ij=1

1

n j+ 1

Find variances and covariances of{T(i)}.

172

8/13/2019 19501

19/59

b. Campbells Theorem

Let {N(t), t >0} be a Poisson Process. Assumen events occur in the

interval(0, t]. Note thatN(t) =nis the realization of a random variable

and has probabilityP{N(t) =n}=et(t)n

n!DefineWn =Waiting time forn

th event.

If{Ti} i= 1, 2, . . . , n are the random variables representing the time

between events

f(t1, . . . , tn) = n1e

ti =ne

n

1 ti

But

n1

ti =Wn, hence

f(t1, . . . , tn) =neWn

173

8/13/2019 19501

20/59

f(t1, . . . , tn) =neWn

Now consider the transformation

W1 =t1, W2=t1+ t2, . . . , W n =t1+ t2+ . . . + tn

The distribution ofW= (W1, W2, . . . , W n)is

f(W) =f(t)

(t)

W

where(t)W is the determinant of the Jacobian.

174

8/13/2019 19501

21/59

Note: (W)

t=

1 0 0 0 . . . 0

1 1 0 0 . . . 0

1 1 1 0 . . . 0

1 1 1 1 . . . 0...

1 1 1 1 . . . 1

and

(t)

W

=

(W)

t

1

= 1

.. f(w1, . . . , wn) =newn 0< w1. . . wn < t.

But there are no events in the interval(wn, t]. This carries probability

e(twn). Hence the joint distribution of theWis

f(W) =newn e(twn) =net

175

8/13/2019 19501

22/59

f(W) =net 0w1 w2. . .wn < t

Consider

f(W|N(t) =n) = net

et(t)n/n! =n!/tn.

This is the joint distribution of the order statistics from a uniform(0, t)distribution; i.e., f(x) =

1

t 0< x t.

Hence E(Wi|N(t) =n) = it

n + 1

i= 1, 2, . . . , n

We can consider the unordered waiting times, conditional onN(t) =n,

as following a uniform (0, t)distribution.

176

8/13/2019 19501

23/59

Sincew1 =t1, w2=t1+ t2, . . . , wn =t1+ t2+ . . . + tn

ti =wi wi1 (w0= 0)

The difference between the waiting times are the original timesti. Thesetimes follow the distribution conditional onN(t) =n; i.e.

f(t1, . . . , tn|N(t) =n) =n!/tn

Note that iff(ti) = 1/t 0< ti < t, the joint distribution

fori = 1, 2, . . . , n ofn independent uniform(0, t) random variables

isf(t) = 1/tn. If0 < t(1)t(2) . . . t(n) < tthe distribution of the

order statistics isf(t(1), . . . , t(n)) =n!/t

n

which is the same asf(t1, . . . , tn|N(t)).

177

8/13/2019 19501

24/59

c. Minimum of Several Exponential Random Variables

LetTi(i= 1, . . . , n)be ind. exponential r.v. with parameteri and let

T =min(T1, . . . , T n)

P{T > t}=P{T1> t, T2> t, . . . , T n > t}=n

i=1P{Ti> t}

=ni=1eit =et, =

n1

i

T is exponential with parameter

P{T > t}=et =n

i=1i

T= min(T1, . . . , T n)

If alli =0, =n0, P{T > t}=en0t

178

8/13/2019 19501

25/59

DefineNas the index of the random variable which is the smallest failure

time.

For example ifTr Ti for alli, thenN=r.

ConsiderP{T > t, Tr Ti alli}=P{N=r, T > t}

P{N=r, T > t}=P{T > t, Ti Tr, i=r}

= t P{T > t

r, i=r| tr}f(tr)dtr

=

t

e(r)trrertrdtr

=r t

etrdtr

=r

et

179

8/13/2019 19501

26/59

P{N=r, T > t}= r

et

P{N=r, T >0}=P{N=r}= r

, =n1

i

P{N=r, T > t}=P{N=r}P{T > t}

N (index of smallest) andTare independent

If i =0

P{N=r}= 0

n0=

1

n(All populations have the same prob. of being the smallest.)

180

8/13/2019 19501

27/59

D. Relation to Erlang and Gamma Distribution

ConsiderT =T1+ . . . + Tn

Sinceqi(s) =

ii+ s , q

T(s) =

ni=1

ii+ s

which is L.T. of Erlang distribution. Ifi are all distinct

q(t) =

ni=1

Aieit ,Ai =j=i

ij i

Ifi =, qT(s) =

+ sn

q(t) = (t)n1et

(n) Gamma Distribution

181

8/13/2019 19501

28/59

E. Guarantee Time

Consider the r.v. following the distribution having pdf

q(t) =e

(tG)

fort > G= 0 fort G

The parameterG is called a guarantee time

If the transformation Y =T Gis made then f(y) =ey fory >0.

.. E(Y) = 1/, V (Y) = 1/2, . . .

Since T =Y + G, E(T) = 1 + G

and central moments ifT andYare the same.

182

8/13/2019 19501

29/59

F. Random Sums of Exponential Random Variables

Let {Ti} i= 1, 2, . . . , N be iid withf(t) =et and consider

SN =T1+ T2+ . . . + TN

with P{N=n}=pn.

The Laplace Transform ofSN is (/ + s)n

for fixedN=n. Hence

f(s) =E

+ s

Nresulting in a pdf which is a mixture of gamma

distributions.

f(t) =n=1

(t)n1et(n)

pn

183

8/13/2019 19501

30/59

Supposepn =pn1q n= 1, 2, . . . (negative exponential distribution)

f(s) =

n=1

+ sn

pn1q=

n=1

q

p p

+ sn

= q

pp/ + s

1 p+s

= q

p

p

s + (1 p) =

q

s + q

f(s) = q

s + q

SN =T1+ T2+ . . . + TN, P{N=n}=pn1q

has exponential distribution with parameter(q).

184

8/13/2019 19501

31/59

4.4 Counting Processes and the Poisson Distribution

Definition: A stochastic process {N(t), T >0} is said to be a counting

process whereN(t)denotes the number of events that have occurred in

the interval(0, t]. It has the properties.

(i.) N(t)is integer value

(ii.) N(t) 0

(iii.) Ifs < t, N (s)N(t) and N(t) N(s) =number of events

occurring is(s, t].

185

8/13/2019 19501

32/59

A counting process has independent increments if the events in disjoint

intervals are independent; i.e.N(s) are N(t) N(s)are independent

events.

A counting process has stationary increments if the probability of the

number of events in any interval depends only on the length of the

interval; i.e.N(t)andN(s + t) N(s)

have the same probability distribution for alls. A Poisson process

is a counting process having independent and stationary

increments.

186

8/13/2019 19501

33/59

TH. Assume {N(t), t0} is a Poisson Process. Then the dsitribution of

Ns(t) =N(s + t) N(s) is independent ofs and only depends on the

length of the interval, i.e.

P{N(t + s) N(s)|N(s)}=P{N(t + s) N(s)}

for alls. This implies that knowledge ofN(u)for0 < usis also

irrelevant.

P{N(t + s) N(s)|N(u), 0< us}

=P{N(t + s) N(s)}.

This feature defines a stationary process.

187

8/13/2019 19501

34/59

TH. A Poisson Process has independent increments.

Consider0 t1< t2< t3 < t4

> Time

0

X X X X

t1 t2 t3 t4

Consider events in(t3, t4]; i.e.

N(t4) N(t3)

P{N(t4) N(t3) | N(u), o < u t3}

=P{N(t4) N(t3)}.

Distribution is independent of what happened prior tot3. Hence if the

intervals(t1, t2]and(t2, t4)are non-overlapping

N(t2) N(t1)andN(t4) N(t3) are independent.

188

8/13/2019 19501

35/59

TH. Cov(N(t), N(s + t)) =t (Poisson Process)

Proof N(s + t) N(t)is independent ofN(t)Cov(N(s + t) N(t), N(t)) = 0

= Cov(N(s + t), N(t)) V(N(t)) = 0

.. Cov(N(s + t), N(t)) =V(N(t)) =t

as variance ofN(t)ist.

An alternative statement of theorem is

Cov(N(s), N(t)) = min(s, t)

189

8/13/2019 19501

36/59

TH. A counting process {N(t), t 0} is a Poisson Process if and only if

(i) It has stationary and independent increments

(ii) N(0) = 0 and

P{N(h) = 0} = 1 h + 0(h)

P{N(h) = 1} =h + 0(h)

P{N(h) =j} = 0(h), j >1

Notes: The notation0(h)little o of h refers to some function(h)for

which

limh0

(h)

h = 0

Divide interval(0, t]inton sub-intervals of lengthh; i.e.nh =t

P{N(kh) N(k 1)h)}=P{N(h)}

190

8/13/2019 19501

37/59

T =Time to event beginning att = 0.

P{T > t} =P{N(t) = 0}=P{No events in each sub-interval}

P{N(t) = 0} =P{T > t}= [1 h + o(h)]n

= (1 h)n + n(1 h)n1o(h) + o(h2)

= (1 h)n{1 + n o(h)

1 h+ . . . }

=

1 tnn

1 + t1 tn

o(h)h + . . .

et as n , h 0

P{T > t}=et

HenceT is exponential; i.e. Time between events is exponential.

{N(t), t 0} is Poisson Process

191

8/13/2019 19501

38/59

4.5 Superposition of Counting Processes

Suppose there arek counting processes which merge into a single

counting process; e.g.k = 3.

X X

X

X

X XX X X

X

Process 1:

Process 2:

Process 3:

Merged Process:

The merged process is called the superposition of the individual counting

processes

N(t) =N1(t) + N2(t) + . . . + Nk(t)

192

8/13/2019 19501

39/59

A. Superposition of Poisson Processes

N(t) =N1(t) + . . . + Nk(t)

Suppose {Ni(t), t0} i= 1, 2, . . . , k are Poisson Processes with

E[Ni(t)] =it.

Note that each of the counting processes has stationary and independent

increments.

AlsoN(t)is Poisson with parameter

E(N(t)) =k

i=1

(it) =t, =k

i=1

i

N(t)is a Poisson Process

Hence {N(t), t 0} has stationary and independent increments.

193

8/13/2019 19501

40/59

B. General Case of Merged Process

Consider the merged process fromk individual processes

XX

XX XX X X

X Vk> v}=Gk(v) =P{Tf(1)> v, Tf(2)> v, . . . , T f(k)> v}

=Qf(v)k

195

8/13/2019 19501

42/59

Gk(v) =P{Vk > v}=Qf(v)k where

Qf(v) =

v

qf(x)dx, q f(x) =Q(x)/m

Let gk(x) =pdf of merged process

Gk(v) =

v

gk(x)dx=Qf(v)k

d

dvGk(v) =gk(v) =kQf(v)

k1qf(v)

gk(v) =kQf(v)k1Q(v)

m

196

8/13/2019 19501

43/59

Consider transformation z= Vkm/k

= kVk

m ,

dz

dv =

k

m

gk(z) =gk(v) Vz = k

mQfmz

kk1 Qmz

k m

k

gk(z) =Qmz

k

1 mz

k

o

Q(x)

m dx

k1

as Qf

mzk

=

mz

k

Q(x)

m dx= 1

mzk

0

Q(x)

m dx

For fixedz,ask ,

zm

k 0 and Q

mzk

1

197

8/13/2019 19501

44/59

Also mzk

0

Q(x)

m dx

Qmzk

m

mz

k =Qmz

k

zk

z

k

.. ask

gk(z)

1 z

k

k1ez

Thus ask , the forward recurrence time (multiplied bymk)z =

mkVk is distributed as a unit exponential distribution. Hence for

largek,Vk = kmz has an asymptotic exponential distribution with

parameter =k/m. Since the asymptotic forward recurrence time isexponential, the time between events (of the merged process), is

asymptotically exponential.

198

8/13/2019 19501

45/59

Note: A forward recurrence time is exponential if and only if the time

between events is exponential; ie.

qf(x) = Q(x)

m =ex ifQ(x) =ex

and ifqf(x) =ex

Q(x) =ex

Additional Note: The merged process isN(t) =k

i=1

Ni(t). Suppose

E(Ni(t)) =t. Units ofare no. of events per unit time

The units ofm are time per event

Thus E(N(t)) = (k)tand(k)is mean events per unit time. The units

of 1k or 1

is mean time per event. Hencem = 1/for an

individual process and the mean of the merged process is1/k.

Ex. =6 events per year m= 126 = 2months (mean time between

events).

199

8/13/2019 19501

46/59

5. Splitting of Poisson Processes

Example: Times between births (in a family) follow an exponential

distribution. The births are categorized by gender.

Example: Times between back pain follow an exponential distribution.

However the degree of pain may be categorized as the required

medication depends on the degree of pain.

Consider a Poisson Process {N(t), t 0} where in addition to observingan event, the event can be classified as belonging to one ofr possible

categories.

DefineNi(t) = no. of events of type i during(0, t]fori = 1, 2, . . . , r

N(t) =N1(t) + N2(t) + . . . + Nr(t)

200

8/13/2019 19501

47/59

This process is referred to as splitting the process.

Bernoulli Splitting Mechanism

Suppose an event takes place in the interval(t, t + dt]. Define the

indicator random variableZ(t) =i(i= 1, 2, . . . , r)such that

P{Z(T) =i|event at(t, t + dt]}=pi.

Notepi is independent of time.

Then ifN(t) =r

i=1

Ni(t) the counting processes {Ni(t), t0} are

Poisson process with parameter(pi)for i= 1, 2, . . . , r.

201

8/13/2019 19501

48/59

Proof: Suppose over time(0, t],n events are observed of whichsi are

classified as of timei withr

i=1 si =n.

P{N1(t) =s1, N2(t) =s2, . . . , N r(t) =sr|N(t) =n}

= n!

s1!s2! . . . sr!ps1

1 ps2

2 . . . psr

r

Hence P{Ni(t) =si, i= 1, . . . , r andN(t) =n}

=

n!ri=1 si!p

s11 p

s22 . . . p

srr

et(t)n

n!

=r

i=1(pit)

siepit

si! =

r

i=1P{Ni(t) =si}

which shows that the {Ni(t)} are independent and follow Poisson

distributions with parameters {pi}.

{Ni(t), t0} are Poisson Processes.

202

8/13/2019 19501

49/59

Example of Nonhomogenous Splitting

Suppose a person is subject to serious migraine headaches. Some of theseare so serious that medical attention is required. Define

N(t) = no. of migraine headaches in(0, t]

Nm(t) = no. of migraine headaches requiring medical attention

p() = prob. requiring medical attention if

headache occurs at(, + d).

Suppose an event occurs at(, + d); then Prob.of requiring

attention= p()d.

Note that conditional on a single event taking place in(0, t], is uniform

over(0, t]; i.e.

f(|N(t) = 1) = 1/t 0< t and = 1

t

t0

p()d

203

8/13/2019 19501

50/59

= 1

t

t0

p()d

0x

t

No event(, t)Time to event

.. P{Nm(t) =k|N(t) =n}=

n

k

k(1 )(nk)

P{Nm(t) =k, N(t) =n}=n

k

k

(1 )nk e

t(t)n

n!

204

8/13/2019 19501

51/59

P{Nm(t) =k}=n=k

n

k

k(1 )nk

et(t)n

n!

= k

k!et

n=k

(t)n

(n k)!(1 )nk

= k

k!e

t(t

)k

n=k

(t)nk(1 )nk

(n k)!

= k

k!(t)ket et(1)

P{Nm(t) =k}=et (t)

k

k!

205

8/13/2019 19501

52/59

4.7 Non-homogeneous Poisson Processes

Preliminaries

LetN(t) follow a Poisson distribution; i.e.

P{N(t) =k}=et(t)k/k!

Holdingt fixed, the generating function of the distribution is

N(t)(s) =E[esN(t)] =

k=0

et(t)k

k!

esk

=et

k=0(est)k

k! =etee

st

N(t)(s) =et[es1] =et(z1) if z=es

The mean isE[N(t)] =t

206

{

8/13/2019 19501

53/59

Consider the Counting Process {N(t), t0} having the Laplace

Transform

(*) N(t)(s) =e(t)[es1] =e(t)[z1]

E[N(t)] = (t), P{N(t) =k}=e(t) [(t)]k/k!

For the Poisson Process(t) =tand the mean is proportional tot.

However whenE[N(t)]=twe call the process {N(t), t 0} a

non-homogenized Poisson Process and E(N(t)] = (t)

(t)can be assumed to be continuous and differentiable

d

dt(t) = (t) =(t).

The quantity(t)is called intensity function.(t)can be represented by

(t) =

t0

(x)dx

207

8/13/2019 19501

54/59

IfN(t)has the Transform given by()then

P{N(t) =k}=e(t)(t)k/k!

SinceP{Sn > t}=P{N(t)< n}

We have P{S1> t}=P{N(t) t) =e(t)

Thus pdf of time between events is

f(t) =(t)e t

0(x)dx, (t) =

t0

(x)dx

Note that ifH= (t), thenHis a random variable following a unit

exponential distribution.

208

A i d d i i N (t ) N ( ) d N ( )

8/13/2019 19501

55/59

Assume independent increments; i.e.N(t + ) N()andN()are

independent

L.T. Transform (z, t) =e(t)[z1] z=es

Generating function =E[esN(t)] =E[zN(t)]

e(t+u)(z1) =E[zN(t+u)] =E[zN(t+u)N(u)+N(u)]

=E[zN(t+u)N(u)] E[zN(u)]

=e

(u)[z1]

.. =E[zN(t+u)N(u)]] = e(t+u)(z1)

e(u)(z1) =e[(t+u)(u)][z1]

where(t + u) (u) = t+uu

(x)dx

.. P{N(t + u) N(u) =k}= e[(t+u)(u)][(t + u) (u)]k

k!

209

8/13/2019 19501

56/59

Axiomatic Derivation ofNon-Homogenized Poisson Distribution

Assume counting process {N(t), t 0}(i)N(0) = 0

(ii) {N(t), t0} has independent increments; i.e.N(t + s) N(s)

andN(s)are independent

(iii)P{N(t + h) =k|N(t) =k}= 1 (t)h + 0(h)

P{N(t + h) =k+ 1|N(t) =k}=(t) + 0(h)

P{N(t + h) =k+j|N(t) =k}=o(h) j 2

p{N(t + s) N(s) =k}=e[(t+s)(s)]2[(t + s) (s)]k

k!

210

8/13/2019 19501

57/59

4.8 Compound Poisson Process

Example. Consider a single hypodermic needle which is shared. The

times between use follow a Poisson Process. However at each use, severalpeople use it. What is the distribution of total use?

Let {N(t), t 0} be a Poisson process and {Zn, n 1} be iid random

variables which are independent ofN(t). Define

Z(t) =

N(t)n=1

Zn

The processZ(t)is called a Compound Poisson Process. It will be

assumed that {Zn} takes on integer values.

211

Define A(s) E[eszn ] Then

8/13/2019 19501

58/59

Define A(s) =E[e szn ]. Then

(s|N(t) =r) =E[esZ(t)] =A(s)r

(s|N(t) =r) =A(s)r

(s) =r=0

(s|N(t) =r)P(N(t) =r)

=

r=0A

(s)

r et(t)r

r! =e

tr=0

(A(s)t)r

r!

(s) =eteA(s)t = et(1A

(s))

A(s) =E(eszN) = 1 sm1+ s2

2m2+ . . .

t(1 A(s)) =t[sm1 s2

2m2+ . . . ], mi =E(zin)

212

8/13/2019 19501

59/59

Cumulant function =K(s) = log (s)

K(s) =sm +s2

22 + . . .

where(m, 2)refer toZ(t).

K(s) =t[sm1s2

2

m2+ . . . ]

E[Z(t)] =tm1

V[Z(t)] =tm2

mi =E(zin)

213