1. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
AMA528
PROBABILITY AND STOCHASTIC MODELS
DEPARTMENT OF APPLIED MATHEMATICS
Lecturer & Tutor: Dr. Catherine LIU
Contact: 2766 6931 (O); Office Venue: HJ616
Consultation Hours: 7:45pm-8:45pm, Mon. & 4:00pm-5:00pm, Tues.
16/11/2011
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 1 / 15

2. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Chapter 7
The Poisson Process
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 2 / 15

3. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Outline
1
Review of exponential Distribution
2
Counting Process
3
Poisson Process
4
Nonhomogeneous Poisson Process
5
Compound Poisson Process
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 3 / 15

4. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Review on exponential distribution
pdf: 0-rate parameter, Y exp ; 0.
y 1 x
e if y 0 e if x 0
f y g x
0 elsewhere 0 elsewhere
2 2
E Y 1 Var Y 1 ;E X Var X .
cdf:
y x
1 e if y 0 1 e if x 0
F y G x
0 elsewhere 0 elsewhere
t 1 1
MGF: MY t 1 ; MX s 1 s .
Memoryless: For all s t 0,
P X s tX t P X s or P X s t P X s P X t
.
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 4 / 15

5. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
A note
Suppose X1 Xn i.i. and Xi exp i for i 1 n.
n
P min X1 Xn x P X1 x exp i x
i 1
i
P Xi Xj
i j
Eg1 (example 5.5, pp.2): Suppose one has a stereo system consisting of two main
parts, a radio and a speaker. If the lifetime of the radio is exponential with mean 1000
hours and the lifetime of the speaker is exponential with mean 500 hours independent
of the radio’s lifetime, then what is the probability that the system’s failure (when it
occurs) will be caused by the radio failing?
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 5 / 15

6. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Counting Process
A stochastic process N t t 0 is said to be a counting process if N t represents
the total number of ”events” that have occurred up to time t if
1 N t 0;
2 N t is integer valued;
3 If s t, then N s N t ;
4 For s t, N t N s equals the number of events that have occurred in the
interval s t .
Eg: Let N t equal # of persons who enter 7-11 shop at or prior to (or by) time t, then
N t t 0 is a counting process;
But if N t equal # of persons in the store at time t, then N t t 0 would not be a
counting process.
Independent increments: if # of events that occur in disjoint time intervals are
independent.
Eg: N 10 N 3 is independent of N 15 N 10 .
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 6 / 15

7. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Poisson Process
A counting process N t t 0 is said to be a Poisson Process having rate 0,
if
1 N 0 0;
2 The process has independent increments;
3 The # of events in any interval of length t is Poisson distributed with mean t.
That is, for all s t 0,
n
t t
P N t s N s n e n 0 1
n
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 7 / 15

8. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Alternative definition
f x f x
Infinite small o 1 : f x o x or x
o 1 or lim x
0.
x 0
A counting process N t t 0 is said to be a Poisson Process having rate 0,
if
1 N 0 0;
2 The process has independent and stationary increments;
3 P N h 1 h o h ;
4 P N h 2 o h .
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 8 / 15

9. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Interarrival and waiting times
Consider a Poisson process: Let T1 be the time of the first event; Let Tn denote the
elapsed time between the n 1 st and the nth event for n 1.
The sequence Tn n 1 2 is called the sequence of interarrival times.
Distribution of Tn : Tn n 1 2 i.i.d. exp .
n
Let Sn Ti , n 1, the arrival time of the n-th event, then Sn is called
i 1
the waiting time until the n-th event.
Distribution of Sn : Sn Gammar n . The pdf of Sn is
n 1
t t
fsn t e I t 0
n 1
A useful result:
Sn t N t n
Remark: Based on Tn n 1 with rate , we can set up a Poisson process with rate .
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 9 / 15

10. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Property 1
Suppose two Poisson processes
1
N1 t t 0 with rate 1 ; Sn : the time of the n-th event of the 1st process;
2
N2 t t 0 with rate 2; Sm : the time of the m-th event of the 2nd process.
N1 t t 0 and N2 t t 0 are independent.
n m 1 k n m 1 k
1 2 n m 1 1 2
P Sn Sm
k 1 2 1 2
k n
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 10 / 15

11. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Property 2
Let X1 Xn i.i.d. U 0 t and the corresponding order statistics X 1 Xn.
Then S1 Sn N t n X1 Xn .
That is, the conditional joint pdf of S1 Sn given that N t n is
n
f s1 sn n 0 s1 sn t
tn
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 11 / 15

12. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Nonhomogeneous Poisson Process
Nonstationary Poisson proess: allow the arrival rate at time t to be a function of t.
A counting process N t t 0 is said to be a nonhomogeneous Poisson Process
with intensity function t t 0, if
1 N 0 0;
2 The process has independent increments;
3 P N t h N t 1 t h o h ;
4 P N t h N t 2 o h .
t
Let m t 0
y dy. Then m t is called the mean value function of the
nonhomogeneous Poisson process. And
n
m s t m s m s t m s
P N s t N s n e n 0
n
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 12 / 15

13. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Example 2
Eg 2 (example 5.20, pp.15): Siegbert runs a hot dog stand that opens at 8 am. From 8
am until 11am customers seem to arrive, on the average, at a steadily increasing rate
that starts with an initial rate of 5 customers per hour at 8 am and reaches a maximum
of 20 customers per hour at 11 am. From 11 am until 1 pm the (average) rate seems
to remain constant at 20 customers per hour. However, the (average) rate seems to
remain constant at 20 customers per hour. However, the (average) arrival rate then
drops steadily from 1pm until closing time at 5pm at which time it has the value of 12
customers per hour. If we assume that the numbers of customers arriving at
Siegbert’s stand during disjoint time periods are independent, then what is a good
probability model for the above? What is the probability that no customers arrive
between 8:30 am and 9:30 am on Monday morning? What is the expected number of
arrivals in this period?
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 13 / 15

14. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Compound Poisson Process
Let Yi i 1 be a family of i.i.d. r.v.s which are independent of a Poisson process
N t t 0 .
N t
Let X t Yi t 0. Then the r.v. X t is said to be a compound Poisson r.v. and
i 1
the stochastic process X t t 0 is said to be a compound Poisson process.
Eg: Let N(t) be # of customers leave a supermarket by time t distributed with
Poisson t . Let Yi i 1 2 , the amount spent by the i-th customer, i.i.d. Let X t
N t
be the total amount of money spent by time t. Then X t Yi t 0. And
i 1
X t t 0 is a compound Poisson process.
Remark: Let Yi 1, the X t N t , a usual Poisson process.
X t X t N t N t Y1 t Y1
2
Var X t Var X t N t Var X t N t t Y1
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 14 / 15

15. Outline Exponential distribution Counting Process Poisson Process Nonhomogeneous PP Compound PP
Example 3
Eg 3 (exampel 5.22, pp. 19) Suppose that families migrate to an area at a Poisson
rate 2 per week. if the number of people in each family is independent and takes
on values 1, 2, 3, 4 with respective probabilities 1 , 3 , 1 , 6 , then what is the expected
6
1
3
1
value and variance of the number of individuals migrating to this area during a fixd
five-week period?
AMA528 (By Catherine Liu) Lecture 10 The Poisson Process 16/11/2011 15 / 15