the poisson process. a stochastic process { n ( t ), t ≥ 0} is said to be a counting process if n...

The Poisson 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 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.

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].

Properties of counting processes

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 , > 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 t. That is for all s, t ≥ 0

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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].

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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-t T1 has the exponential distribution.

• P(T1 > t) = P(N(t) = 0) = e-t 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-t Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-t T2 has the exponential

The distribution of interarrival times for a Poisson process

• P(T1 > t) = P(N(t) = 0) = e-t 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-t Then, P(T2 > t) = E[P(T2 > t|T1) ] = e-t 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/.

The distribution of interarrival times for a Poisson process

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

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The distribution of arrival times

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The distribution of arrival times for a Poisson process

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Equivalence

A counting process with N(t) = 0 and exponentially i.i.d. inter-arrival times is a Poisson process.

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Composition of independent Poisson processes

Suppose N1(t) and N2(t) are independent Poisson process processes with rates 1 and 2. Then the composite process N1(t)+N2(t) is also a Poisson process with rates 1+2.

This follows from the fact that interval arrival times are i.i.d.and exponentially distributed with rate 1+2.

Random partition of a Poisson distribution

Suppose N is a random variable with the Poisson distribution with parameter . 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 pi, 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 . 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 pi, and and Ni(t) and Nj(t) are independent for i≠j.