The Poisson process

Report
CIS 2033 based on
Dekking et al. A Modern Introduction to Probability and Statistics. 2007
Instructor Longin Jan Latecki
C12: The Poisson process
From Baron book:
The number of rare events occurring within a fixed
period of time has Poisson distribution.
Essentially it means that two such events are
extremely unlikely to occur simultaneously or within a
very short period of time.
Arrivals of jobs, telephone calls, e-mail messages,
traffic accidents, network blackouts, virus attacks,
errors in software, floods, and earthquakes are
examples of rare events.
This distribution bears the name of a famous French
mathematician Siméon-Denis Poisson (1781–1840).
12.2 – Poisson Distribution
Definition: A discrete RV X has a Poisson distribution with parameter µ,
where µ > 0 if its probability mass function is given by
P(k) P(X  k) 
k
k!
e
for k = 0,1,2…,
where µ is the expected number of rare events, or number of successes,
occurring in time interval [0, t], which is fixed for X. We can express µ = t λ,
where t is the length of the interval, e.g., number of minutes. Hence
λ = µ / t = number of events per time unite = probability of success.
λ is also called the intensity or frequency of the Poisson process.
We denote this distribution: Pois(µ) = Pois(tλ).
Expectation E[X] = µ = tλ and variance Var(X) = µ = tλ
Dekking 12.6
A certain brand of copper wire has flaws about every 40 centimeters.
Model the locations of the flaws as a Poisson process. What is the
probability of two flaws in 1 meter of wire?
The expected numbers of flaws in 1 meter is 100/40 = 2.5,
and hence the number of flaws X has a Pois(2.5) distribution. The
answer is P(X = 2):
(2.5)2 2.5
P(X  2) 
e  0.256
2!
since
P(X  k) 
k
k!
e
Let X1, X2, … be arrival times such that the probability of k arrivals in a given
time interval [0, t] has a Poisson distribution Pois(tλ):
k
(t ) t
P(X  k) 
e
k!
The differences Ti = Xi – Xi-1 are called inter-arrival times or wait times.
The inter-arrival times T1=X1, T2=X2 – X1, T3=X3 – X2 … are independent RVs, each
with an Exp(λ) distribution.
Hence expected inter-arrival time is E(Ti) =1/λ. Since for Poisson
λ = µ / t = (number of events) / (time unite) = intensity
= probability of success,
we have for the exponential distribution
E(Ti) =1/λ = t / µ = (time unite) / (number of events) = wait time
Quick exercise 12.2
We model the arrivals of email messages at a server as a Poisson
process. Suppose that on average 330 messages arrive per minute.
What would you choose for the intensity λ in messages per second?
What is the expectation of the interarrival time?
Because there are 60 seconds in a minute, we have
λ = µ / t = (number of events) / (time unite) = 330 / 60 = 5.5
Since the interarrival times have an Exp(λ) distribution, the
expected time between messages is 1/λ = 0.18 second, i.e.,
E(T) =1/λ = t / µ = (time unite) / (number of events) = 60/330=0.18
Let X1, X2, … be arrival times such that the probability of k arrivals in a given
time interval [0, t] has a Poisson distribution Pois(λt):
(  t ) k  t
P(X  k) 
e
k!
Each arrival time Xi, is a random variable with Gam(i, λ) distribution for α=i :
We also observe that Gam(1, λ) = Exp(λ):
a) It is reasonable to estimate λ with (nr. of cars)/(total time in sec.) = 0.192.
b) 19/120 = 0.1583, and if λ = 0.192 then μ = 10 λ =1.92. Hence
P(K = 0) = e-0.192*10 = 0.147
c) Again μ = 10 λ =1.92 and we have P(K = 10) = ((1.92 )10/ 10!) * e-1.92 = 2.71 * 10-5.
12.2 –Random arrivals


Example: Telephone calls arrival times
Calls arrive at random times, X1, X2, X3…

Homegeneity aka weak stationarity: is the rate lambda at which arrivals
occur in constant over time: in a subinterval of length u the expectation of
the number of telephone calls is λu.

Independence: The number of arrivals in disjoint time intervals are
independent random variables.


N(I) = total number of calls in an interval I
Nt=N([0,t])
E[Nt] = t λ

Divide Interval [0,t] into n intervals, each of size t/n

12.2 –Random arrivals




When n is large enough, every interval Ij,n = ((j-1)t/n , jt/n] contains either
0 or 1 arrivals.
Arrival: For such a large n ( n > λ t),
Rj = number of arrivals in the time interval Ij,n, Rj = 0 or 1
Rj has a Ber(p) distribution for some p.
Recall: (For a Bernoulli random variable)
E[Rj] = 0 • (1 – p) + 1 • p = p
By Homogeneity assumption for each j
p = λ • length of Ij,n = λ ( t / n)

Total number of calls:
Nt = R1 + R2 + … + Rn.
By Independence assumption
Rj are independent random variables, so
Nt has a Bin(n,p) distribution, with p = λ t/n, hence λ = np/t

When n goes to infinity, Bin(n,p) converges to a Poisson distribution
Example form Baron Book:
Example 3.23 from Baron Book
Baron uses λ for μ, hence
and λ=np,
where we have Bin(n, p).
P(X  k) 
k
k!
e 

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