Lecture 6. Poisson processes
Mathematical Statistics and Discrete Mathematics
November 18th, 2015
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Motivation
We want to have a probabilistic model describing the number of calls arriving at a call
center within a given time range, say, 15 minutes between 1pm and 1.15pm. We can
represent the incoming calls as marks on the timeline:
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5
10
15
So that it is a realistic model, it should posses the following properties:
• The numbers of calls arriving in disjoint time intervals should be independent.
0
5
10
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• The distribution of the number of calls arriving in a certain time interval depends
only on the length of the interval and not its placement in time.
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5
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We will define a stochastic process which satisfies these properties, namely the
Poisson process.
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Poisson distribution
A discrete random variable X has Poisson distribution with parameter λ > 0 if
fX (k) = P(X = k) =
λk −λ
e ,
k!
for k = 0, 1, 2, . . .
We denote this by writing X ∼ Pois(λ).
Properties:
• E[X] = Var[X] = λ.
• If X ∼ Pois(λ1 ), Y ∼ Pois(λ2 ), and X and Y are independent, then
X + Y ∼ Pois(λ1 + λ2 ).
Pois(λ) can be well approximated by Binom(n, p) with very large n and small p such
that np = λ. Hence, it models well the number of successes in a large number of trials
where the success probability is small. This is sometimes called the law of rare events.
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Poisson distribution
Poisson distribution can be used to model the number of
X people winning the lottery each week,
X people older than 100 years in some community,
X deaths in a given age group within a year,
X mutations in a piece of DNA after exposure to radiation,
X times a server is accessed per minute,
X telephone calls arriving at a call center within 15 minutes (the population
covered by one call center is large and each person has a small probability of
making a call within a particular 15 minutes period).
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Exponential distribution
A continous random variable X has exponential distribution with parameter λ > 0 if
fX (x) = λe−λx
for x > 0.
We denote this by writing X ∼ Exp(λ).
Properties:
• E[X] = λ−1 ,
• Var[X] = λ−2 .
• FX (x) = P(X ≤ x) = 1 − e−λx for x > 0.
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Exponential distribution
A continuous random variable X has exponential distribution, if and only if it is
memoryless, that is
P(X > s + t | X > t) = P(X > s),
for any s, t > 0.
Proof. Suppose X has exponential distribution with parameter λ. Then, we have for
any x > 0,
P(X > x) = 1 − P(X ≤ x) = e−λx ,
and hence
P(X > s + t) = P(X > s) · P(X > t)
by the property of the exponential function.
The opposite implication follows since the exponential function is the only one
satisfying this property, and hence the CDF function is uniquely identified.
The only discrete memoryless distribution is the geometric distribution.
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Exponential distribution
X Let X be the time until the arrival of the first call at the call center. Since we
assume that the arrivals are independent in disjoint intervals, the knowledge that
there was no call until time s (this is the event {X > t}, has no influence on the
arrivals after time s). Moreover, since the arrivals after time s in an interval of
length t should be distributed as the arrivals in the interval [0, t], we get
P(X > s + t | X > t) = P(X > s),
for any s, t > 0,
and hence X has exponential distribution.
X Time until a radioactive particle decays can be modelled with exponential
distribution.
X Time between server requests can be modelled with exponential distribution.
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Counting process
A counting process is stochastic process {Ct }t≥0 that keeps count of events that have
occurred up to time t. In particular,
• Ct it takes values in {0, 1, 2, . . .},
• Ct it is non-decreasing in t,
• Ct − Cs is the number of events in the interval (s, t] for s < t.
Here, the word event has the traditional, and not probabilistic, meaning. Counting
processes are also called jump processes.
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Counting process
X The graph below shows an instance of a counting process together with the times
when the events occurred:
30
25
20
15
10
5
2
0
4
6
5
8
10
10
12
14
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Poisson process
A Poisson process {Nt }t≥0 with intensity parameter λ > 0 is a counting process with
the following additional properties:
• (starting at 0) N0 = 0.
• (independent increments) The random variables (Nt2 − Nt1 ) and (Nt4 − Nt3 ) are
independent for all t1 < t2 < t3 < t4 .
• (stationarity) Fix r > 0. The random variable (Nt+r − Nt ) has the same
distribution for all t.
• (no double jumps/events)
P(Nh = 1)
→ λ,
h
and
P(Nh ≥ 2)
→ 0,
h
as h & 0.
The last property says that the probability that two events counted by the Poisson
process happen at the same time is 0.
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Poisson process
X Calls arriving at a call center can be modelled by a Poisson process.
X Request arriving at a server can be modelled by a Poisson process.
X Arrivals of students for this lecture cannot be modelled by a Poisson process.
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Properties of Poisson processes
If {Nt }t≥0 is a Poisson process with intensity λ, then:
• Nt ∼ Pois(λt) for all t. In particular, E[Nt ] = λt.
• Nt − Ns ∼ Pois(λ(t − s)) for all s < t. In particular, E[Nt − Ns ] = λ(t − s).
• Let T0 = 0, T1 , T2 , T3 , . . . be the (random) times when the events counted by the
process occur. Let
τi = Ti − Ti−1 ,
for i = 1, 2, 3, . . .
Then τi ∼ Exp(λ) for all i, and τi , τj are independent for i 6= j. In particular,
E[τi ] = λ−1 for all i.
The larger the intensity λ, the more events counted by the process.
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X It is a slow day at a call center of a taxi company. There is 1 call every 10
minutes on average. Bob is the only person present and he needs to use the
restroom every hour for 5 minutes. How many calls on average will he miss if he
works 8 hours? What is the probability that he will not miss any call? What is
the probability that he will miss more than 2 calls? Use a Poisson process to
answer the questions.
Let X1 , X2 , . . . , X8 be the numbers of calls that arrive when Bob is in the
restroom each hour. By the properties of the Poisson process, they are all
independent, and all have the same distribution Pois(1/2). Hence, by the
property of the Poisson distribution
X = X1 + X2 + . . . + X8 ∼ Pois(8 · 1/2) = Pois(4).
Therefore, Bob will miss E[X] = 4 calls on average, and
40 −4
e = e−4 ∼ 0.018, and P(X > 2) = 1 − P(X ≤ 2) =
0!
40
41
42
= 1 − e−4 − e−4 − e−4 = 1 − (1 + 4 + 4)e−4 ∼ 0.835.
0!
1!
2!
P(X = 0) =
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X Bob is waiting for the first call already 10 minutes. What is the probability that
he will still wait more than 20 minutes? Let τ1 be the time that Bob waits for the
first call. By the property of the Poisson process τ1 has exponential distribution
with parameter 1 (here one unit is 10 minutes). We have
P(τ1 > 1 + 2 | τ1 > 1) = P(τ1 > 2) = 1 − P(τ1 ≤ 2) = e−2 ∼ 0.135.
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