Applied Probability - Jonathan Jordan · Applied Probability School of Mathematics and Statistics, University of She eld 2019{20 (University of She eld) Applied Probability 2019{201/59

Applied Probability

School of Mathematics and Statistics, University of Sheffield

2019–20

(University of Sheffield) Applied Probability 2019–20 1 / 59

Modelling sets of points

Aspects of many phenomena can be represented by sets of points, sopoint process models are widely useful.

These sets of points may refer to moments in time, in which casethey are thought of as points in one dimensional space, such asevents in a Poisson process.

More generally they may refer to locations in space, often two orthree dimensional.












Example: Floods in Burbage Brook


Example: Insurance claims

Major fire insurance claims in Denmark, 1980–1990


Example: Japanese pine saplings

Locations of saplings of Japanese black pines


Example: East Yorkshire leukaemia cases


Fitting a Poisson process model to data

In this section we consider how to fit a Poisson process model tosome data.

Given observation of a point process over an interval [0, t], how canwe fit a Poisson process model?

That is, how do we estimate the rate λ?

Having fitted a Poisson process, how can we assess the adequacy ofthe model?




















Two possibilities

Two initial possibilities for the estimation:

1 from the observed no. of events N(t) = n;

2 by the methods developed for continuous time Markov chains


Two possibilities

Two initial possibilities for the estimation:

1 from the observed no. of events N(t) = n;

2 by the methods developed for continuous time Markov chains


Number of events method

Fitting using N(t) ∼ Po(λt):

Log-likelihood having observedN(t) = n:

l = −λt + n log(λt) + constant

= −λt + n log(λ) + n log(t) + constant

= −λt + n log(λ) + constant

so the derivative ∂l∂λ

= −t + nλ

.

So the MLE λ̂ = n/t.

Also ese(λ̂) = λ̂/√n.



Fitting using N(t) ∼ Po(λt): Log-likelihood having observedN(t) = n:





= −t + nλ

.










= −t + nλ

.










= −t + nλ

.










= −t + nλ

.










= −t + nλ

.










= −t + nλ

.




Markov chain method

Fitting using the Markov chain method:

The durations in states i = 0, . . . , n are

a0 = T1, . . . , an−1 = Tn, an = t −n∑1

Ti .

Also, the transition rates in the chain are simply

gi i+1 = λ = −gii = gi


Markov chain method



a0 = T1, . . . , an−1 = Tn, an = t −n∑1

Ti .




Markov chain method



a0 = T1, . . . , an−1 = Tn, an = t −n∑1

Ti .




Log likelihood

So the log-likelihood is

l = −∑

giai +∑i 6=j

nij log gij

= −λ∑i

ai + log λ∑i

ni i+1

= −λ t + n log λ

since∑

i ai =∑n

1 Ti + (t −∑n

1 Ti) = t and∑

i ni i+1 = totalnumber of points in [0, t].

Since this is the same as the log-likelihood from approach 1,inferences from the two approaches are the same.


Log likelihood


l = −∑

giai +∑i 6=j

nij log gij

= −λ∑i

ai + log λ∑i

ni i+1


since∑

i ai =∑n

1 Ti + (t −∑n

1 Ti) = t and∑




Log likelihood


l = −∑

giai +∑i 6=j

nij log gij

= −λ∑i

ai + log λ∑i

ni i+1


since∑

i ai =∑n

1 Ti + (t −∑n

1 Ti) = t and∑




Log likelihood


l = −∑

giai +∑i 6=j

nij log gij

= −λ∑i

ai + log λ∑i

ni i+1


since∑

i ai =∑n

1 Ti + (t −∑n

1 Ti) = t and∑




Log likelihood


l = −∑

giai +∑i 6=j

nij log gij

= −λ∑i

ai + log λ∑i

ni i+1


since∑

i ai =∑n

1 Ti + (t −∑n

1 Ti) = t and∑




Conclusion

Thus, slightly surprisingly at first sight, what appears to be extrainformation here, the values of individual Ti , doesn’t actually makeany difference.

Checks on model adequacy: in notes


Conclusion

Thus, slightly surprisingly at first sight, what appears to be extrainformation here, the values of individual Ti , doesn’t actually makeany difference.

Checks on model adequacy: in notes


Example: Burbage Brook floods

Between 1925 and 1983 (inclusive) there were 48 flood events (flows≥ 4 cumecs) in Burbage Brook.

Since the observation period is t = 59 years, the maximum likelihoodestimator of the rate of occurrence is

λ̂ = 48/59 = 0.81, with ese = 0.12

events/year.


Example: Burbage Brook floods

Between 1925 and 1983 (inclusive) there were 48 flood events (flows≥ 4 cumecs) in Burbage Brook.

Since the observation period is t = 59 years, the maximum likelihoodestimator of the rate of occurrence is

λ̂ = 48/59 = 0.81, with ese = 0.12

events/year.


Inhomogeneous one-dimensional Poisson processes

The rate λ governs the probability that a point occurs in anarbitrarily short interval.

So far we have assumed that it is constant.

However in many applications it is plausible that events could occurrandomly and independently, but that the probability of occurrencecould change over time.

A flood, for example, might be more likely in winter, arrivals at acasualty department more likely in the rush hour, etc.




















Definition and properties

We now allow λ = λ(t) to depend on time. Assume that thecounting process N(t) satisfies

P(N(t + h) = i + 1 |N(t) = i) = λ(t)h + o(h)

andP(N(t + h) = i |N(t) = i) = 1− λ(t)h + o(h),

and that the probabilities of all other changes are o(h).

The resulting process is called an inhomogeneous Poisson process ofrate λ(t).




P(N(t + h) = i + 1 |N(t) = i) = λ(t)h + o(h)







P(N(t + h) = i + 1 |N(t) = i) = λ(t)h + o(h)







P(N(t + h) = i + 1 |N(t) = i) = λ(t)h + o(h)





Time change

Let N(t) be an inhomogeneous Poisson process of intensity λ(t), anddefine

Λ(t) =

∫ t

0

λ(u) du.

Suppose we change the time-scale and define a new counting process

M(s) = N(t) where s = Λ(t).

Let t(s) = Λ−1(s) be the inverse transformation of the time scale,taking the new time s back to t.

Thus M(s) is a homogeneous Poisson process with intensity 1.(Proof in notes.)


Time change


Λ(t) =

∫ t

0

λ(u) du.






Time change


Λ(t) =

∫ t

0

λ(u) du.






Time change


Λ(t) =

∫ t

0

λ(u) du.






Transferring properties

We can therefore transfer properties of the homogeneous processes tothe inhomogeneous case.

1 N(t) has a Poisson distribution with mean Λ(t) =∫ t

0λ(u)du.

Reason: N(t) = M(s) ∼ Po(s) = Po(Λ(t)).

2 The numbers of points in disjoint intervals I1, . . . , Ik areindependent and Poisson distributed with means∫Iiλ(u)du, i = 1, . . . , k .

Reason: independence follows from translation of thecorresponding property of the basic Poisson process, and thedistributions follow as above.





0λ(u)du.








0λ(u)du.








0λ(u)du.








0λ(u)du.





Distribution of points

Given that the total number of points in [0, t] is N(t) = n, thepositions of the points are independently distributed with pdfλ(v)/Λ(t), 0 ≤ v ≤ t.

Reason: conditional independence and identical distribution ofpositions in the N process follows from the same properties of thosein the M process.

Let V denote the position of a point in the N process. Then thecorresponding position for the M process is s(V ) and in the Mprocess positions are uniformly distributed over [0, s(t)].












Distribution of points cont.

Thus the distribution function of V is

P(V ≤ v) = P(s(V ) ≤ s(v)) =s(v)

s(t)=

Λ(v)

Λ(t)

and so the pdf isdP(V ≤ v)

dv=λ(v)

Λ(t).


Distribution of points cont.

Thus the distribution function of V is

P(V ≤ v) = P(s(V ) ≤ s(v)) =s(v)

s(t)=

Λ(v)

Λ(t)

and so the pdf isdP(V ≤ v)

dv=λ(v)

Λ(t).


Fitting to data

The last property enables us to write down the likelihood for λ(·)based on observations over [0, t].

Suppose that we have observed N(t) = n points and that theirpositions are v1, . . . , vn.

Then

L = P(N(t) = n)× P(positions of the points |N(t) = n)

= e−Λ(t) Λ(t)n

n!×

n∏1

λ(vi)

Λ(t)

= e−Λ(t)

∏n1 λ(vi)

n!,


Fitting to data



Then


= e−Λ(t) Λ(t)n

n!×

n∏1

λ(vi)

Λ(t)

= e−Λ(t)

∏n1 λ(vi)

n!,


Fitting to data



Then


= e−Λ(t) Λ(t)n

n!×

n∏1

λ(vi)

Λ(t)

= e−Λ(t)

∏n1 λ(vi)

n!,


Fitting to data



Then


= e−Λ(t) Λ(t)n

n!×

n∏1

λ(vi)

Λ(t)

= e−Λ(t)

∏n1 λ(vi)

n!,


Fitting to data



Then


= e−Λ(t) Λ(t)n

n!×

n∏1

λ(vi)

Λ(t)

= e−Λ(t)

∏n1 λ(vi)

n!,


Log likelihood

The log-likelihood is

l = −Λ(t) +n∑1

log λ(vi) + const.

If λ(t) is a constant λ, then Λ(t) reduces to Λ(t) = λt and thisbecomes becomes

l = −λt + n log λ + const

as before.


Log likelihood

The log-likelihood is

l = −Λ(t) +n∑1

log λ(vi) + const.

If λ(t) is a constant λ, then Λ(t) reduces to Λ(t) = λt and thisbecomes becomes

l = −λt + n log λ + const

as before.


Notes

In most applications of inhomogeneous Poisson models the functionλ(·) is specified in terms of a finite number of parameters.

Their maximum likelihood estimates are then the values maximizing l .

The asymptotic properties of likelihood inference carry over to suchinhomogeneous Poisson processes under reasonable conditions.

Approximate standard errors may then be found from the informationmatrix and tests may be based on twice the difference oflog-likelihoods.


Notes






Notes






Example: Freezes of Lake Constance 1300–1974

Years between 1300AD and 1974AD when major freezes of LakeConstance occurred

Year

1300 1400 1500 1600 1700 1800 1900


Histogram

1300 1400 1500 1600 1700 1800 1900 2000

01

23

45

6

Freeze date

Fre

quen

cy

Figure: Histogram of Lake Constance Freeze Dates


Uniform QQ plot

++

+++

+++

+++

++

+++++++

++

++

++

++

+

1300 1400 1500 1600 1700 1800 1900

1300

1400

1500

1600

1700

1800

1900

Freeze Dates

Uni

form

qua

ntile

s

Figure: QQ plot of Lake Constance Freeze Dates


Model

The plots do not appear to support a Poisson process model with aconstant intensity.

Consider therefore a non-homogeneous Poisson process model.

To represent a changing rate in a simple form, we assume

λ(t) = α + βt

where α and β are constants.

t, for convenience in this problem, measures years post 1300 in unitsof 100 years.


Model




λ(t) = α + βt




Model




λ(t) = α + βt




Model




λ(t) = α + βt




Model




λ(t) = α + βt




Log likelihood

The cumulative intensity function Λ(t) is

Λ(t) =

∫ t

0

λ(u)du = αt +1

2βt2,

so that the log-likelihood is

l = −αto −1

2βt2

o +n∑1

log(α + βvi),

where to denotes the length of the observation period, 6.75 centuries,n denotes the number of events, and the vi denote the times ofoccurrence of the events.


Log likelihood


Λ(t) =

∫ t

0

λ(u)du = αt +1

2βt2,


l = −αto −1

2βt2

o +n∑1

log(α + βvi),



Log likelihood


Λ(t) =

∫ t

0

λ(u)du = αt +1

2βt2,


l = −αto −1

2βt2

o +n∑1

log(α + βvi),



Likelihood equations

To find the maximum likelihood estimators α̂ and β̂, consider thelikelihood equations:

−to +n∑1

1

α̂ + β̂vi= 0

− 1

2t2o +

n∑1

vi

α̂ + β̂vi= 0.


Likelihood equations

To find the maximum likelihood estimators α̂ and β̂, consider thelikelihood equations:

−to +n∑1

1

α̂ + β̂vi= 0

− 1

2t2o +

n∑1

vi

α̂ + β̂vi= 0.


MLEsThese equations do not have a simple explicit solution, but numericalmaximization of l gives

α̂ = 7.015 (1.76)

β̂ = −0.81 (0.38)

where the values in brackets are estimated standard errors obtainedby inversion of the observed information matrix J :(

Var(α̂) Cov(α̂, β̂)

Cov(α̂, β̂) Var(β̂)

)= J−1,

where

J = −

(∂2l∂α2

∂2l∂α∂β

∂2l∂β∂α

∂2l∂β2

).



α̂ = 7.015 (1.76)

β̂ = −0.81 (0.38)

where the values in brackets are estimated standard errors obtainedby inversion of the observed information matrix J :

(Var(α̂) Cov(α̂, β̂)


)= J−1,

where

J = −

(∂2l∂α2

∂2l∂α∂β

∂2l∂β∂α

∂2l∂β2

).



α̂ = 7.015 (1.76)

β̂ = −0.81 (0.38)




)= J−1,

where

J = −

(∂2l∂α2

∂2l∂α∂β

∂2l∂β∂α

∂2l∂β2

).



α̂ = 7.015 (1.76)

β̂ = −0.81 (0.38)




)= J−1,

where

J = −

(∂2l∂α2

∂2l∂α∂β

∂2l∂β∂α

∂2l∂β2

).


Information matrix

We could calculate J by differentiating the log likelihood again andsubstituting α̂ and β̂.

However numerical differentiation is available as a by-product of thenumerical maximization of l in R.


Information matrix

We could calculate J by differentiating the log likelihood again andsubstituting α̂ and β̂.

However numerical differentiation is available as a by-product of thenumerical maximization of l in R.


Extra calculations

Adding in the years to 2020 (with no more freezes) changes theestimates to

α̂ = 7.459 (1.74)

β̂ = −0.952 (0.34)

Slightly smaller standard errors; estimates give a slightly steeper line.


Extra calculations


α̂ = 7.459 (1.74)

β̂ = −0.952 (0.34)



Extra calculations


α̂ = 7.459 (1.74)

β̂ = −0.952 (0.34)

Slightly smaller standard errors

; estimates give a slightly steeper line.


Extra calculations


α̂ = 7.459 (1.74)

β̂ = −0.952 (0.34)



Changing rate?

The estimated rate λ̂(t) = 7.0− 0.81t is decreasing, in agreementwith the indications from the plots.

To assess the objective strength of evidence for a decreasing rate wecan carry out a test of the hypothesis H0 : β = 0, against thealternative H1 : that there is no restriction on the value of β.

Use a generalized likelihood ratio test.


Changing rate?





Changing rate?





Generalized Likelihood Ratio Test

The test statistic is

W = −2(l(α̃, β̃)− l(α̂, β̂)),

Here l(α̂, β̂) is the maximum log-likelihood under H1, and l(α̃, β̃) themaximum under the restriction imposed by H0.

Under H0, the distribution of W is approximately χ2 with the degreesof freedom being the no. parameters under H1 minus the no.parameters under H0, and under H1 it tends to be larger.

Thus large values of W discredit H0 and the p-value corresponding toan observed value of W may be found from the χ2 distribution.




W = −2(l(α̃, β̃)− l(α̂, β̂)),







W = −2(l(α̃, β̃)− l(α̂, β̂)),







W = −2(l(α̃, β̃)− l(α̂, β̂)),





Example: Lake Constance revisited

For the Lake Constance data the maximum likelihood estimators α̂and β̂ above give the values of α and β that maximize l under H1.

We only need to find the maximizing values under H0, the restrictedmaximum likelihood estimators α̃ and β̃.

When H0 is true, λ(t) = α so the Poisson process istime-homogeneous.

Therefore the maximum likelihood estimator of α is

α̃ =n

to=

29

6.75= 4.3 events/century

(and necessarily β̃ = 0).







α̃ =n

to=

29









α̃ =n

to=

29









α̃ =n

to=

29




Results

Substitution into l now gives

w = −2(13.275− 15.249) = 3.948.

By comparison with χ21 the p-value is slightly less than 0.05.

Thus there is some evidence of a change in the rate of occurrence offreezing events, which from the sign of β̂ must be a reduction.

But the strength of the evidence is not overwhelming.

Extending the data to 2020 makes the p-value smaller: just under0.01. So the evidence seems to be getting stronger.


Results


w = −2(13.275− 15.249) = 3.948.






Results


w = −2(13.275− 15.249) = 3.948.






Results


w = −2(13.275− 15.249) = 3.948.






Results


w = −2(13.275− 15.249) = 3.948.






Spatial Poisson processes

Given an intensity function λ(·) ≥ 0, a general Poisson process onthe line has the following properties:

for any interval I , the number of points N(I ) of the process in Ihas a Poisson distribution with mean

∫Iλ(u) du

for disjoint intervals I1, I2, . . . , Ik , the random variablesN(I1),N(I2), . . . ,N(Ik) are independent.

N is a counting process: the number of points it counts in aninterval is the sum of the numbers in subintervals.





∫Iλ(u) du







∫Iλ(u) du







∫Iλ(u) du




Higher dimensions

A Poisson process in the plane or in space is defined as a countingprocess with these same properties, where the properties are merelyre-phrased to make sense in higher dimensions.

Suppose that λ(·) is a real-valued non-negative function on R2.

For each set B in the plane, let N(B) be a random variable takingnon-negative integer values (interpreted as the number of points ofthe process in B).


Higher dimensions





Higher dimensions





Definition

If

N(B) has a Poisson distribution with mean∫Bλ(u) du;

when B1,B2, . . . ,Bk are disjoint, the random variablesN(B1),N(B2), . . . ,N(Bk) are independent;

N has the additive property N(∪ki=1Bi) =

∑ki=1 N(Bi) for

disjoint Bi ;

then N is called a spatial Poisson process, with intensity λ(·).


Definition

If



N has the additive property N(∪ki=1Bi) =

∑ki=1 N(Bi) for

disjoint Bi ;



Definition

If



N has the additive property N(∪ki=1Bi) =∑k

i=1 N(Bi) fordisjoint Bi ;



Definition

If



N has the additive property N(∪ki=1Bi) =∑k

i=1 N(Bi) fordisjoint Bi ;



Homogeneous case

A homogeneous spatial Poisson process is the special case of whenλ(·) is constant.


More dimensions

A Poisson process in a three- or more- dimensional Euclidean spaceS = Rd , d = 3, . . . is defined in the same way.

λ is a function of the spatial coordinates, and the B are sets in S .


More dimensions

A Poisson process in a three- or more- dimensional Euclidean spaceS = Rd , d = 3, . . . is defined in the same way.

λ is a function of the spatial coordinates, and the B are sets in S .


Notation

To save writing let us define, for sets B in the space of the points(Rd , d = 2, 3, . . . ),

Λ(B) =

∫B

λ(u) du.

A Poisson process may be defined on an arbitrary subset of the plane,for example the region of East Yorkshire in the leukaemia example, bysimply restricting the definition above to that subset.


Notation


Λ(B) =

∫B

λ(u) du.



Notation


Λ(B) =

∫B

λ(u) du.



Nearest point

Let R denote the distance from the origin to the nearest point in ahomogeneous planar Poisson process with intensity λ.

Then the probability density function of R is

hR(r) = 2λπre−λπr2

, r > 0,

(the density of a Rayleigh distribution).

Reason: The number of points in a circle of radius r centred at theorigin has a Poisson distribution with mean λπr 2.

If there are no points in this circle then R > r , and conversely. ThusP(R > r) = exp(−λπr 2) and the result follows by differentiation.


Nearest point




, r > 0,





Nearest point




, r > 0,





Nearest point




, r > 0,





First contact

The distribution of R is called the first contact distribution of thePoisson process.

It is in fact the distribution of the distance from any fixed point inthe plane to the nearest point in the process, as can be seen bysimply re-defining the origin to be at the fixed point.

The distribution of distance from an arbitrary point of the processitself to its nearest neighbour may be found too, and for thehomogeneous Poisson process this distribution turns out to be exactlythe same as the first contact distribution.

The two distributions will not necessarily be equal generally, so a testfor a Poisson process could be based on seeing whether estimates ofthe distributions based on observed distances are similar or not.


First contact






First contact






First contact






Thinning

Thinning of a Poisson process refers to the random deletion of someof the points.

A simple form of thinning is to remove or retain each pointindependently with fixed probabilities 1− p and p, say.

If the original process has intensity function λ(·), then the pointprocess resulting from such independent thinning is a Poisson processwith intensity pλ(·).

(Proof in notes)


Thinning




(Proof in notes)


Thinning




(Proof in notes)


Conditional distribution of points

Given the total number of points of a spatial Poisson process in aregion B , the positions V of the points are independently distributedover B with probability density function

fV (v) =λ(v)

Λ(B)v ∈ B .


Conditional distribution of points

Given the total number of points of a spatial Poisson process in aregion B , the positions V of the points are independently distributedover B with probability density function

fV (v) =λ(v)

Λ(B)v ∈ B .


Simulation

If we know the intensity function, the conditional distributionproperty gives a way to simulate a Poisson process over any region.

First generate a number n from the Poisson distribution Po(Λ(B)),then simulate n independent values from the density fV .

With this ability we could implement a simulation test for a Poissonprocess using the comparison of distributions method suggested bythe first contact distribution.

The approach is feasible even for processes on sets B with irregularshapes.


Simulation






Simulation






Simulation






LikelihoodThe conditional distribution property also gives the likelihood for λbased on an observed pattern of points.

If n points are observed in a region B and their positions arev i , i = 1, . . . , n, then the likelihood function is

L =e−Λ(B)Λn(B)

n!×

n∏1

λ(v i)

Λ(B)

=1

n!e−Λ(B)

n∏1

λ(v i),

and the log-likelihood

l = −Λ(B) +n∑1

log λ(v i) + constant.




L =e−Λ(B)Λn(B)

n!×

n∏1

λ(v i)

Λ(B)

=1

n!e−Λ(B)

n∏1

λ(v i),


l = −Λ(B) +n∑1





L =e−Λ(B)Λn(B)

n!×

n∏1

λ(v i)

Λ(B)

=1

n!e−Λ(B)

n∏1

λ(v i),


l = −Λ(B) +n∑1





L =e−Λ(B)Λn(B)

n!×

n∏1

λ(v i)

Λ(B)

=1

n!e−Λ(B)

n∏1

λ(v i),


l = −Λ(B) +n∑1



Parametrization

The λ function is often specified in terms of a small number ofparameters.

In that case fitting of the model by maximization of l and subsequentinference go ahead along the same lines as before.


Parametrization

The λ function is often specified in terms of a small number ofparameters.

In that case fitting of the model by maximization of l and subsequentinference go ahead along the same lines as before.


Marked Poisson processes

Several examples can be thought of as a sequence of time points ateach of which another variable is observed.

A marked Poisson process is a simple model for this.

Given a Poisson process N – on the line, plane or in higherdimensions – with intensity λ(·), associate with each point X i of theprocess a random variable Yi , called the mark at X i .

Then the new process {N ,Y1, . . . } is called a marked Poissonprocess.




















Dependence

For some modelling problems it might be appropriate to take the Yi

to be independent and identically distributed.

In others there may be interest in possible dependence between marksat different points, and in possible changes in the distribution ofmarks with position of the point.


Dependence

For some modelling problems it might be appropriate to take the Yi

to be independent and identically distributed.

In others there may be interest in possible dependence between marksat different points, and in possible changes in the distribution ofmarks with position of the point.


Example: Insurance risk

The arrival of claims at an insurance company and the sizes of theclaims might be modelled as a marked Poisson process.

An initial assumption, to be checked, might be that marks (claimamounts) are independent and identically distributed.


Example: Insurance risk

The arrival of claims at an insurance company and the sizes of theclaims might be modelled as a marked Poisson process.

An initial assumption, to be checked, might be that marks (claimamounts) are independent and identically distributed.


Assets over timeThe difference between premium income and claim payouts is the keyto financial viability of the company.

If the ith claim is made at time Xi and is of size Yi , and premiumsbring income at a steady rate ρ net of running costs, then the assetsA(t) of the insurance company at time t are

A(t) = A(0) + ρt −N(t)∑

1

Yi ,

where N(t) is the number of claims up to time t.

The probability that A(t) remains positive for a long time, and ofhow large the reserves A(0) need to be to make this probability large,are of great interest.




A(t) = A(0) + ρt −N(t)∑

1

Yi ,






A(t) = A(0) + ρt −N(t)∑

1

Yi ,






A(t) = A(0) + ρt −N(t)∑

1

Yi ,




Compound Poisson processes

Sums of the formN(t)∑

1

Yi ,

for a Poisson process N with marks Yi arise in many contexts.

They are called compound Poisson processes.


Compound Poisson processes

Sums of the formN(t)∑

1

Yi ,

for a Poisson process N with marks Yi arise in many contexts.

They are called compound Poisson processes.


Example: Earthquakes

A sequence of earthquakes could be modelled by attaching marksrepresenting earthquake magnitude to the times of occurrence.

Questions about dependence between magnitudes close in time arehighly relevant to predictability and the possibility of warning systems.

The same question arises too about the times themselves andmotivates further development of the Poisson models we haveconsidered in this course.












Example: Floods

Floods could be modelled by a marked Poisson process, the mark fora flood occurrence being the magnitude of the flood.

Marks could include more information too, becomingmulti-dimensional.

If further data were available, for example about weather conditionsat the times of floods, or environmental conditions such asdryness/wetness of the ground in the period before the flood, then ittoo could be modelled as part of a multi-dimensional mark.


Example: Floods





Example: Floods





Example: Rainfall

A point process model for rainfall attempts to mimic the occurrenceand heaviness of rain at a place in terms of the passage of rain cellsover the place.

The arrivals of rain cells are modelled by a Poisson process and thetime a rain cell takes to pass over the place and the intensity of therain it brings are attached as random marks.

Marks in this case are two-dimensional.


Example: Rainfall





Example: Rainfall





Example: Burbage floods

An initial model for the times and severities of the Burbage Brookflood events is based on a marked point process.

Dates of floods are assumed to come from a Poisson process, and theexcess flood flows over 4 cumecs are modelled as conditionallyindependent marks with exponential distributions whose means1/µ(t) may depend on time.


Example: Burbage floods

An initial model for the times and severities of the Burbage Brookflood events is based on a marked point process.

Dates of floods are assumed to come from a Poisson process, and theexcess flood flows over 4 cumecs are modelled as conditionallyindependent marks with exponential distributions whose means1/µ(t) may depend on time.


Special case

Another way to view a marked Poisson process is as a point processin a higher dimensional space.

If the Poisson process is one-dimensional with points at Xi , i = 1, . . .and the marks Yi are also one-dimensional, then the points (Xi ,Yi)form a point process in two dimensions.

When the marks Yi are independent and identically distributed thistwo-dimensional process is itself a Poisson process.


Special case

Another way to view a marked Poisson process is as a point processin a higher dimensional space.

If the Poisson process is one-dimensional with points at Xi , i = 1, . . .and the marks Yi are also one-dimensional, then the points (Xi ,Yi)form a point process in two dimensions.

When the marks Yi are independent and identically distributed thistwo-dimensional process is itself a Poisson process.


2D Poisson process

If the original Poisson process has intensity λ(x) and the markprobability density is k(y) then the intensity of the two-dimensionalPoisson process (Xi ,Yi) is µ(x , y) = λ(x) k(y).

Proof in notes.

The result generalizes to higher dimensions.


2D Poisson process


Proof in notes.



2D Poisson process


Proof in notes.



Documents

Applied Probability - Jonathan Jordan · Applied Probability School of Mathematics and Statistics, University of She eld 2019{20 (University of She eld) Applied Probability 2019{201/59