CHAPTER 4 POISION Processes
CHAPTER 4 POISION Processes
CHAPTER 4 POISION Processes
POISSON PROCESSES
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The Bernoulli Process
independent identically distributed (IID)
binary Random Variables.
We often visualize a Bernoulli process as evolving in discrete time with the event
representing an arriving customer at time i and {Zi = 0} representing no arrival.
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Counting Processes
For example, you might have a random process N(t) that shows the number of
customers who arrive at a supermarket by time t starting from time 0.
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Definition:
N(0)=0;
N(t)∈{0,1,2,⋯}, for all t∈[0,∞);
for 0≤s<t, N(t)−N(s) shows the number of events that occur in the
interval (s,t].
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A counting process has independent increments if the numbers of arrivals in non-
overlapping (disjoint) intervals are independent.
A counting process has stationary increments if, for all t2>t1≥0, N(t2)−N(t1) has
the same distribution as N(t2−t1).
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Basic Concepts of the Poisson Process
For example, suppose that from historical data, we know that earthquakes occur
in a certain area with a rate of 2 per month. Other than this information, the timings
of earthquakes seem to be completely random. Thus, we conclude that the Poisson
process might be a good model for earthquakes.
Example:
Each subinterval corresponds to a time slot of length δ. Thus, the intervals are
(0,δ], (δ,2δ], (2δ,3δ], ⋯. More generally, the kth interval is ((k−1)δ, kδ].
We assume that in each time slot, we toss a coin for which P(H)=p=λδ. If the
coin lands heads up, we say that we have an arrival in that subinterval.
Otherwise, we say that we have no arrival in that interval. Figure 11.3 shows
this process. Here, we have an arrival at time t=kδ, if the kth coin flip results in
a heads.
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Now, let N(t) be defined as the number of arrivals (number of heads) from time 0
to time t. There are n≈ t/δ time slots in the interval (0,t].
Thus, N(t) is the number of heads in n coin flips. We conclude that
N(t)∼Binomial(n,p). Note that here p=λδ, so
Example
The number of customers arriving at a grocery store can be modeled by a Poisson
process with intensity λ=10 customers per hour.
Find the probability that there are 2 customers between 10:00 and 10:20.
Find the probability that there are 3 customers between 10:00 and 10:20 and 7 customers
between 10:20 and 11.
Solution:
1. Here, λ=10 and the interval between 10:00 and 10:20 has length τ=1/3 hours. Thus, if X
is the number of arrivals in that interval, we can write X∼Poisson(10/3). Therefore,
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2, Here, we have two non-overlapping intervals I1=(10:00 a.m., 10:20 a.m.] and I2= (10:20
a.m., 11 a.m.]. Thus, we can write
Since the lengths of the intervals are τ1=1/3 and τ2=2/3 respectively, we obtain
λτ1=10/3 and λτ2=20/3. Thus, we have
inter arrival time = 1/arrival rate = 1/12= 0.083(hours)== 0.083 x 60 minutes= 5 (minutes)
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Therefore from the arrival rate of 12 per hour, the time between each arrival is 5 minutes.
Arrival and Interarrival Times:
Let N(t) be a Poisson process with rate λ. Let X1 be the time of the first interarrival.
Then,
Therefore, X1∼Exponential(λ). Let X2 be the time elapsed between the first and
the second arrival (Figure 11.4).
Figure 11.4 - The random variables X1, X2, ⋯ are called the interarrival times of the counting process
N(t).
Let s>0 and t>0. Note that the two intervals (0,s] and (s,s+t] are disjoint. We can
write
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We conclude that X2∼Exponential(λ), and that X1 and X2 are independent. The random
variables X1, X2, ⋯ are called the interarrival times of the counting process N(t). Similarly, we
can argue that all Xi's are independent and Xi ∼Exponential(λ) for i=1,2,3, ⋯.
If N(t) is a Poisson process with rate λ, then the interarrival times X1, X2, ⋯ are
independent and
Xi∼Exponential(λ), for i=1,2,3,⋯.
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Example:
Let N(t) be a Poisson process with intensity λ=2, and let X1, X2, ⋯ be the
corresponding interarrival times.
a. Find the probability that the first arrival occurs after t=0.5, i.e., P(X1>0.5).
b. Given that we have had no arrivals before t=1, find P(X1>3).
c. Given that the third arrival occurred at time t=2, find the probability that the
fourth arrival occurs after t=4.
d. I start watching the process at time t=10. Let T be the time of the first arrival that
I see. In other words, T is the first arrival after t=10. Find ET and Var(T).
(a)
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(c)
(d) When I start watching the process at time t=10, I will see a Poisson process.
Thus, the time of the first arrival from t=10 is Exponential (2). In other words, we
can write
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Now that we know the distribution of the interarrival times is exponential , we
can find the distribution of arrival times:
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Since Tn=X1+X2+⋯+Xn, we conclude that
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Merging and Splitting Poisson Processes
Merging Independent Poisson Processes:
Let N1(t) and N2(t) be two independent Poisson processes with rates λ1 and λ2
respectively. Let us define N(t)=N1(t)+N2(t). That is, the random process N(t) is
obtained by combining the arrivals in N1(t) and N2(t) (Figure 11.5). We claim
that N(t) is a Poisson process with rate λ=λ1+λ2.
To see this, first note that
N(0)=N1(0)+N2(0)=0+0=0.
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Figure 11.6 - Splitting a Poisson process to two independent Poisson processes.
Problem 1
Let {N(t),t∈[0,∞)} be a Poisson process with rate λ=0.5.
Find the probability of no arrivals in (3,5].
Find the probability that there is exactly one arrival in each of the following
intervals: (0,1], (1,2], (2,3], and (3,4].
2. Let Y1, Y2, Y3 and Y4 be the numbers of arrivals in the intervals (0,1], (1,2], (2,3], and
(3,4]. Then Yi∼Poisson(0.5) and Yi's are independent, so
P(Y1=1,Y2=1,Y3=1,Y4=1) = P(Y1=1)⋅P(Y2=1)⋅P(Y3=1)⋅P(Y4=1)
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Problem 2
Let N1(t) and N2(t) be two independent Poisson processes with rates λ1=1 and λ2=2,
respectively. Let N(t) be the merged process N(t)=N1(t)+N2(t).
Find the probability that N(1)=2 and N(2)=5.
Given that N(1)=2, find the probability that N1(1)=1.
b.
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Nonhomogeneous Poisson Processes
Let N(t) be the number of customers arriving at a fast food restaurant by time t. We
think that the customers arrive somewhat randomly, so we might want to model N(t)
as a Poisson process. However, we notice that this process does not have stationary
increments.
For example, we note that the arrival rate of customers is larger during lunch time
compared to, say, 4 p.m. In such scenarios, we might model N(t) as a
nonhomogeneous Poisson process. Such a process has all the properties of a Poisson
process, except for the fact that its rate is a function of time, i.e., λ=λ(t).
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For a nonhomogeneous Poisson process with rate λ(t), the number of arrivals in any
interval is a Poisson random variable; however, its parameter can depend on the
location of the interval. More specifically, we can write
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