Transcript Q-Layout

The Poisson Process
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Example 2: N(t) is the number of individuals born at or prior
to time t. An event occurs whenever a child is born.
Counting Processes
A stochastic process {N(t), t ≥ 0} is said to be a counting
process if N(t) represents the total number of “events” that
occur by time t (i.e., in the time interval [0, t]).
Example 1: N(t) is the number of customers that enter a store
at or prior to time t. An event corresponds to a person
entering the store.
Example 2: N(t) is the number of individuals born at or prior
to time t. An event occurs whenever a child is born.
Example 3: N(t) is the number of calls made to a technical
help line at or prior to time t. An event occurs whenever a
call is placed.
Properties of counting processes
A counting process satisfies the following properties.
(i) N(t) ≥ 0.
(ii) N(t) is integer valued.
(iii) If s < t, then N(s) ≤ N(t).
(iv) For s < t, N(t) – N(s) equals the number of events that
occurs in the time interval (s, t].
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
Example 1: The number of customers N(10) that enter the
store in the interval [0, 10] is independent from the number
of customers N(15) – N(10) that enter the store in the interval
(10, 15] if customers arrive “randomly.”
A counting process is said to possess independent increments
if the number of events that occur in disjoint intervals are
independent.
Example 1: The number of customers N(10) that enter the
store in the interval [0, 10] is independent from the number
of customers N(15) – N(10) that enter the store in the interval
(10, 15] if customers arrive “randomly”.
Example 2: The number of individuals N(2000)-N(1996) born
in the interval [1996, 2000] is not independent from the
number of individuals N(2004) – N(2000) born in the interval
(2000, 2004].
A counting process is said to possess stationary increments
if the distribution of the number of events that occur in an
interval depend only on the length of the interval and not the
starting time of the interval.
A counting process is said to possess stationary increments
if the distribution of the number of events that occur in an
interval depend only on the length of the interval and not the
starting time of the interval.
Example 1: The number of customers N(t) – N(s) that enter
the store in the interval (s, t] does not depend on s (this is
true if there is not a particular time of day where more
customers enter the store).
The Poisson processes
The counting process {N(t) t ≥ 0} is said to be a Poisson process
having rate l, l > 0, if
(i) N(0) = 0.
(ii) The process has independent increments.
(iii) The number of events in any interval of length t is Poisson
distributed with mean lt. That is for all s, t ≥ 0
n
(
l
t
)
 lt
P{N (t  s )  N (t )  n}  e
n!
,
for n  0,1,...
The Poisson processes
The number of customers that call a help line forms a Poisson
process with rate 4 customers per hour. What is the probability
that more 10 customers arrive in the first 5 hours? What is the
probability that less than 4 customers arrive in the interval (6, 8].
The Poisson processes
The number of customers that call a help line forms a Poisson
process with rate 4 customers per hour. What is the probability
that more 10 customers arrive in the first 5 hours? What is the
probability that less than 4 customers arrive in the interval (6, 8].
N (5)  N (0)  N (5);
N (5)
Poisson(20)
 P{N (5)  10}  1  P{N (5)  10}  1   n  0 e 20
9
(20) n
,
n!
P{N (6)  N (8)  n}  P{N (2)  n}
N (2)
.
n
(8)
Poisson(8)  P{N (2)  4}   n  0 e 8
,
n!
2
The distribution of interarrival times
• Let Tn describe the time that elapses between (n-1)th event and
the nth event for n > 1 and let T1 be the time of the first event.
• The sequence {Tn , n = 1, 2, ...} is called the sequence of
interarrival times.
The distribution of interarrival times
• Let Tn describe the time that elapses between (n-1)th event and
the nth event for n > 1 and let T1 be the time of the first event.
• The sequence {Tn , n = 1, 2, ...} is called the sequence of
interarrival times.
Example: if T1 = 5 and T2 = 10  the first event arrives at time t
= 5 and event 2 occurs at time t = 15.
The distribution of interarrival times
for a Poisson process
• P(T1 > t) = P(N(t) = 0) = e-lt  T1 has the exponential
distribution.
The distribution of interarrival times
for a Poisson process
• P(T1 > t) = P(N(t) = 0) = e-lt  T1 has the exponential
distribution.
• P(T2 > t) = E[P(T2 > t|T1) ]
Since P(T2 > t|T1=s) = P(0 events in (s, s+t]|T1=s)
= P(0 events in (s, s+t])
= e-lt
Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-lt  T2 has the exponential
The distribution of interarrival times
for a Poisson process
• P(T1 > t) = P(N(t) = 0) = e-lt  T1 has the exponential
distribution.
• P(T2 > t) = E[P(T2 > t|T1) ]
Since P(T2 > t|T1=s) = P(0 events in (s, s+t]|T1=s)
= P(0 events in (s, s+t])
= e-lt
Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-lt  T2 has the exponential
•The same applies to other values of n  Tn has the exponential
distribution.
The distribution of interarrival times
for a Poisson process
Let Tn denote the inter-arrival time between the (n-1)th event
and the nth event of a Poisson process, then the Tn (n=1, 2, ...)
are independent, identically distributed exponential random
variables having mean 1/l.
Arrival times
• Let Sn describe the time at which the nth arrival takes place.
• The sequence {Sn , n = 1, 2, ...} is called the sequence of arrival
times.
Example: If T1 = 5 and T2 = 10  S1 = 5 and S2 = 15.
The distribution of arrival times
 In general, Sn  i 1Tn .
n
The distribution of arrival times
 In general, Sn   i 1Tn .
n
 Since the Tn 's are i.i.d. exponential random variables,
Sn has the gamma distribution with parameters n and l.
The distribution of arrival times for a
Poisson process
 In general, Sn   i 1Tn .
n
 Since the Tn 's are i.i.d. exponential random variables,
Sn has the gamma distribution with parameters n and l.
 f Sn (t )  l e  lt
(lt ) n 1
,
(n  1)!
t  0.
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P ( N (t )  0)  P (T1  t )  e  lt
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P ( N (t )  0)  P (T1  t )  e  lt
 P ( N (t )  1)  P (T1  t , T1  T2  t )
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P ( N (t )  0)  P (T1  t )  e  lt
 P ( N (t )  1)  P(T1  t , T1  T2  t )
 E[ P (T1  t , T1  T2  t | T1  y )]
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P( N (t )  0)  P(T1  t )  e  lt
 P( N (t )  1)  P(T1  t , T1  T2  t )
 E[ P (T1  t , T1  T2  t | T1  y )]
 E[ P (T2  t  y )]
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P( N (t )  0)  P(T1  t )  e  lt
 P( N (t )  1)  P(T1  t , T1  T2  t )
 E[ P(T1  t , T1  T2  t | T1  y )]
 E[ P(T2  t  y )]

  P (T2  t  y ) fT1 ( y )dy
0
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P( N (t )  0)  P(T1  t )  e  lt
 P( N (t )  1)  P(T1  t , T1  T2  t )
 E[ P(T1  t , T1  T2  t | T1  y )]
 E[ P(T2  t  y )]

  P (T2  t  y ) fT1 ( y )dy
0

  P (T2  t  y )l e  l y dy
0
Equivalence
A counting process with N(t) = 0 and exponentially i.i.d. interarrival times is a Poisson process.
 P( N (t )  0)  P(T1  t )  e  lt
 P( N (t )  1)  P(T1  t , T1  T2  t )
 E[ P (T1  t , T1  T2  t | T1  y )]
 E[ P (T2  t  y )]

  P (T2  t  y ) fT1 ( y )dy
0

  P (T2  t  y )l e  l y dy
0

  e  l ( t  y ) l e  l y dy  l te  lt
0
Equivalence
 P( N (t )  n)  P( S n  t , S n  Tn 1  t )
 E[ P( S n  t , S n  Tn 1  t | S n  y)]
 E[ P(Tn 1  t  y)]

  P(Tn 1  t  y ) f Sn ( y ) dy
0

  P(Tn 1  t  y )l e  l y
0

 e
 l (t  y )
0
 e  lt
(l t ) n
n!
le
l y
(l y ) n 1
dy
(n  1)!
(l y ) n 1
dy
(n  1)!
Composition of independent
Poisson processes
Suppose N1(t) and N2(t) are independent Poisson process
processes with rates l1 and l2. Then the composite process
N1(t)+N2(t) is also a Poisson process with rates l1+l2.
This follows from the fact that interval arrival times are i.i.d.
and exponentially distributed with rate l1+l2.
Random partition of a Poisson
distribution
Suppose N is a random variable with the Poisson distribution
with parameter l. Suppose each item is of type i (i= 1, ..., N)
with probability pi. Then Ni the number of items of type i has
the Poisson distribution with parameter pil, and Ni and Nj are
independent for i≠j.
Random partition of a Poisson process
Suppose N(t) is a Poisson process processes with rate l.
Suppose events can be of different types. The probability that
an event is of type i is pi. Then, the number of events of type i
Ni(t) is a Poisson process with rate pil, and and Ni(t) and Nj(t)
are independent for i≠j.