chapter3(7)

Chapter 3 - Poisson process

Chapter 3 - Poisson process

Nial Friel Stochastic Models - STAT30090 - STAT40680

Chapter 3 - Poisson processDefinition of a Poisson processMain propertiesSome extensions

Counting processes

Counting processes are increasing processes {Xt , t ∈ R+} withXt ∈ N.

NotationThe idea is to count something over time (eg, the arrivals ofcustomers, ...) : Xt is the number of arrivals between time 0and time t so naturally we set X0 = 0.Let N(s,t] be the number of arrivals during the interval (s, t] :

N(s,t] = Xt − Xs .

Note that Xt = N(0,t].



Counting processes


NotationThe idea is to count something over time (eg, the arrivals ofcustomers, ...) : Xt is the number of arrivals between time 0and time t so naturally we set X0 = 0.

Let N(s,t] be the number of arrivals during the interval (s, t] :

N(s,t] = Xt − Xs .




Counting processes


NotationThe idea is to count something over time (eg, the arrivals ofcustomers, ...) : Xt is the number of arrivals between time 0and time t so naturally we set X0 = 0.Let N(s,t] be the number of arrivals during the interval (s, t] :

N(s,t] = Xt − Xs .




Definition of a Poisson process

Poisson processA Poisson process with intensity λ > 0 is a counting process{Xt} with

1 independent increments ;2 P(Xt+h − Xt = 1) = λh + o(h) when h→ 0 ;3 P(Xt+h − Xt > 1) = o(h) when h→ 0.



Some remarks about (1)

Recall that by “independent increments” we mean that, for anyt1 < t1 ≤ t3 < t4,

Xt2 − Xt1 indep. of Xt4 − Xt3 .

In other words :

N(t1,t2] is indep. of N(t3,t4].

This is sometimes referred as the loss of memory property.



Remarks on (2) and (3)Recall (2) and (3) :

P(Xt+h − Xt = 1) = λh + o(h),

P(Xt+h − Xt > 1) = o(h).

We can write (3) in a different way :

P(Xt+h − Xt > 1)

h−−→h→0

0.

Two customers will never arrive exactly at the same time, andthe probability that two customers arrive at a very small periodof time is negligible. Remark that (2) and (3) imply that :

P(Xt+h − Xt = 0) = 1− λh + o(h).




P(Xt+h − Xt = 1) = λh + o(h),

P(Xt+h − Xt > 1) = o(h).


P(Xt+h − Xt > 1)

h−−→h→0

0.

Two customers will never arrive exactly at the same time, andthe probability that two customers arrive at a very small periodof time is negligible.

Remark that (2) and (3) imply that :

P(Xt+h − Xt = 0) = 1− λh + o(h).




P(Xt+h − Xt = 1) = λh + o(h),

P(Xt+h − Xt > 1) = o(h).


P(Xt+h − Xt > 1)

h−−→h→0

0.

Two customers will never arrive exactly at the same time, andthe probability that two customers arrive at a very small periodof time is negligible. Remark that (2) and (3) imply that :

P(Xt+h − Xt = 0) = 1− λh + o(h).



About Poisson

Siméon Denis Poisson (1781-1840), the French mathematicianand physicist, who received the Copley medal from the RoyalSociety of London in 1832.



Fundamental theorem

A reminder : the Poisson distribution

U ∼ P(`)⇔ for any k ∈ N, P(U = k) =`k

k!e−`.

Then :E(U) = Var(U) = `.U ∼ P(`)⇔ for any s > 0, E

(esU)

= eλ(es−1).

TheoremLet {Xt} be a Poisson process with intensity λ andN(s,t] = Xt − Xs for s < t. Then

N(s,t] ∼ P(λ(t − s)).



An immediate consequence

The arrival rateThe arrival rate between times s and t is :

Xt − Xs

t − s.

Consequence of the theoremFor a Poisson process, the expected arrival rate is constant,equal to λ :

E[Xt − Xs

t − s

]= λ.



Arrival times

DefinitionLet S1, S2, ... be the arrival times :

Sk = inf{t ≥ 0 : Xt = k}

(and by convention S0 = 0).

Note that :

Xt =∞∑i=1

1Si≤t .



Distribution of the arrival times

TheoremFor any k ≥ 1, Sk ∼ Γ(k , λ) where we recall that the Γ(k , λ)distribution has density :

f (x) =λk

Γ(k)e−λxxk−1.



Gap between arrival times

DefinitionFor any k ≥ 1, Tk = Sk − Sk−1.

TheoremThe Tk are iid with distribution E(λ), where we recall that theexponential distribution E(λ) has density

f (x) = λe−λx .

In particular, E(Tk) = 1/λ and Var(Tk) = 1/λ2.



Subdividing an interval with a given count

Question : suppose we know that Xt = n. How are the arrivaltimes S1, S2, ..., Sn distributed within the interval [0, t] ?

Example : Xt = 1. Then for 0 < s < t,

P (S1 < s|Xt = 1) =P (S1 < s,Xt = 1)

P (Xt = 1)

=P(N(0,s] = 1,N(s,t] = 0

)P(N(0,t] = 1

)=λse−λse−λ(t−s)

λte−λt =st

in other words, L(S1|Xt = 1) = U [0, t] the uniformdistribution on [0, t].




Question : suppose we know that Xt = n. How are the arrivaltimes S1, S2, ..., Sn distributed within the interval [0, t] ?Example : Xt = 1. Then for 0 < s < t,

P (S1 < s|Xt = 1) =P (S1 < s,Xt = 1)

P (Xt = 1)

=P(N(0,s] = 1,N(s,t] = 0

)P(N(0,t] = 1

)=λse−λse−λ(t−s)

λte−λt =st

in other words, L(S1|Xt = 1) = U [0, t] the uniformdistribution on [0, t].




TheoremLet U1, U2, ..., Un be an iid sample with a uniform distributionU [0, t]. Let U(1), ..., U(n) be the corresponding order statistic,ie :{U1, . . . ,Un} = {U(1), . . . ,U(n)} ;U(1) ≤ · · · ≤ U(n).

Then L(S1, . . . , Sn|Xt = n) = L(U(1), . . . ,U(n)).

In other words, conditional on Xt = n, the n arrivals areuniformly distributed on [0, t].



Marked Poisson process

Defintion - marked Poisson processA marked Poisson process is defined as :

a Poisson process {Xt , t ≥ 0} with arrival times S1, S2, ...a collection of iid random variables M1, M2, ... indep. of{Xt}.

For example :Sk is the arrival time of the k-th customer.Mk is the money spent by the k-th customer.

Or :Sk is the time of occurence of the k-th earthquake ;Mk is its magnitude.

...



Marked Poisson process

Defintion - marked Poisson processA marked Poisson process is defined as :

a Poisson process {Xt , t ≥ 0} with arrival times S1, S2, ...a collection of iid random variables M1, M2, ... indep. of{Xt}.

For example :Sk is the arrival time of the k-th customer.Mk is the money spent by the k-th customer.

Or :Sk is the time of occurence of the k-th earthquake ;Mk is its magnitude.

...Nial Friel Stochastic Models - STAT30090 - STAT40680


Thinned Poisson process

Definition - thinned Poisson processLet {Xt} be a marked Poisson process with arrival times S1,S2, ... and marks M1, M2, ... ∼ Be(p). We define the thinnedPoisson process {Yt} by

Yt =∞∑i=1

1Si≤t,Mi=1.

In other words, {Yt} is obtained from {Xt} by erasing thearrivals Si with Mi = 0.

TheoremThe process {Yt} is actually a Poisson process with parameterpλ.



Superposition of Poisson process

TheoremLet {Xt} and {Yt} be two Poisson processes independent ofeach other, with intensity given by λ and µ, respectively. Letus put Zt = Xt + Yt .

Then {Zt} is a Poisson process withintensity λ + µ.



Superposition of Poisson process

TheoremLet {Xt} and {Yt} be two Poisson processes independent ofeach other, with intensity given by λ and µ, respectively. Letus put Zt = Xt + Yt . Then {Zt} is a Poisson process withintensity λ + µ.



Campbell’s Theorem

Campbell’s TheoremFor a Poisson process {Xt} with arrival times S1, S2, ... andfor any function f integrable on [0, t], we have

E

[Xt∑i=1

f (Si)

]= λ

∫ t

0f (s)ds.



Campbell’s Theorem for marked processes

Campbell’s Theorem for marked processesFor a marked Poisson process {Xt} with arrival times S1, S2,..., marks M1, M2, ... ∈M and for any function f integrableon [0, t]×M, we have

E

[Xt∑i=1

f (Si ,Mi)

]= λ

∫ t

0E [f (s,M1)] ds.



Application of Campbell’s TheoremA squirrel drinks in a river at random times given by a Poissonprocess : S1, S2...

Unfortunately, the river is contaminated by adangerous substance, and at each time he absorbs a randomquantity Mi of poison. Each dose of poison is eliminated by hisbody at an exponential rate.In other words, the quantity of poison in his body at time t is :

Q =Xt∑i=1

Mie−α(t−Si ).

Thanks to Campbell’s Theorem one can compute

E(Q) = λ

∫ t

0E(Mie−α(t−s))ds =

λE(M1)

α

(1− e−αt) .



Application of Campbell’s TheoremA squirrel drinks in a river at random times given by a Poissonprocess : S1, S2... Unfortunately, the river is contaminated by adangerous substance, and at each time he absorbs a randomquantity Mi of poison.

Each dose of poison is eliminated by hisbody at an exponential rate.In other words, the quantity of poison in his body at time t is :

Q =Xt∑i=1

Mie−α(t−Si ).


E(Q) = λ

∫ t


λE(M1)

α

(1− e−αt) .



Application of Campbell’s TheoremA squirrel drinks in a river at random times given by a Poissonprocess : S1, S2... Unfortunately, the river is contaminated by adangerous substance, and at each time he absorbs a randomquantity Mi of poison. Each dose of poison is eliminated by hisbody at an exponential rate.

In other words, the quantity of poison in his body at time t is :

Q =Xt∑i=1

Mie−α(t−Si ).


E(Q) = λ

∫ t


λE(M1)

α

(1− e−αt) .



Application of Campbell’s TheoremA squirrel drinks in a river at random times given by a Poissonprocess : S1, S2... Unfortunately, the river is contaminated by adangerous substance, and at each time he absorbs a randomquantity Mi of poison. Each dose of poison is eliminated by hisbody at an exponential rate.In other words, the quantity of poison in his body at time t is :

Q =Xt∑i=1

Mie−α(t−Si ).


E(Q) = λ

∫ t


λE(M1)

α

(1− e−αt) .



Inhomogeneous Poisson process : definition

In some cases, the assumption that the expected arrival rate isconstant is not realistic. For example, customers are morelikely to come to a shop after 5pm than before.

DefinitionLet λ(t) > 0 be a function of the time t. An inhomogeneousPoisson process with intensity λ(t) is a counting process {Xt}with

1 independent increments ;2 P(Xt+h − Xt = 1) = λ(t)h + o(h) when h→ 0 ;3 P(Xt+h − Xt > 1) = o(h) when h→ 0.



Inhomogeneous Poisson process : properties

Let {Xt} be an inhomogeneous Poisson process with intensityλ(t). Then we can prove that

N(s,t] = Xt − Xs ∼ P(∫ t

sλ(x)dx

).

In particular, when λ(t) = λ is constant, {Xt} is a Poissonprocess and we have :

N(s,t] = Xt − Xs ∼ P ((t − s)λ) .



Planar Poisson point process

It is also possible to extend the Poisson process to the casewhere the index t does not belong to R but to R2. In this case,t does not refer to the time but to a position in the plane.

It is more convenient to extend the definition of the arrivaltimes S1, S2, ...A Poisson process on the plane is a random family of points Ssuch that

the number of each point in the region K is distributedaccording to P(λ.area(K )) ;inside a region K , the points are uniformly distributed ;the number of points in two disjoints regions K1 and K2 isindependent.




It is also possible to extend the Poisson process to the casewhere the index t does not belong to R but to R2. In this case,t does not refer to the time but to a position in the plane.It is more convenient to extend the definition of the arrivaltimes S1, S2, ...

A Poisson process on the plane is a random family of points Ssuch that





It is also possible to extend the Poisson process to the casewhere the index t does not belong to R but to R2. In this case,t does not refer to the time but to a position in the plane.It is more convenient to extend the definition of the arrivaltimes S1, S2, ...A Poisson process on the plane is a random family of points Ssuch that



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