CHAPTER 6:Modelling sets of points Flashcards

Question

Example 42: Freezes of Lake Constance 1300–1974 | Extra calculations

Answer 1

Adding in the years to 2018 (with no more freezes) changes the estimates to α^ = 7.437 (1.74) β^ = −0.95 (0.34) Slightly smaller standard errors; estimates give a slightly steeper line.

Answer 2

For the Lake Constance data the maximum likelihood estimators α^ and β^ above give the values of α and β that maximize l under H1, so we only need to find the maximizing values under H0, the restricted maximum likelihood estimators α˜ and β˜. When H0 is true, λ(t) = α so the Poisson process is time-homogeneous. From section 6.2 therefore the maximum likelihood estimator of α is α˜ =n/t_o=29/6.75 = 4.3 events/century (and necessarily β˜ = 0). Substitution into l (71) and use of (74) now gives w = −2(13.275 − 15.249) = 3.948. By comparison with χ^2_1 the p-value is slightly less than 0.05. Thus there is some evidence of a change in the rate of occurrence of freezing events – which from the sign of β^ must be a reduction – but the strength of the evidence is not overwhelming. Thus there is some evidence of a change in the rate of occurrence of freezing events, which from the sign of β^ must be a reduction. But the strength of the evidence is not overwhelming.

Answer 3

1. Other forms for λ(t) may be preferable. For example λ(t) = exp(α + βt) specifies a rate that could be increasing or decreasing (according to the sign of β) but can never give a negative value, unlike the linear λ(t) used in the example. If there is a possibility of periodicity in the rate of occurrence of points (as for floods, for example, which may be more likely in the winter), then a λ(t) incorporating sines and cosines might be useful. 2. To check adequacy of a non-homogeneous model we could transform the time-scale from t to s by the s = Λ(t) transformation. On the new scale the non-homogeneous process becomes homogeneous, and therefore all the checks for that case (section 6.2) become available.

Answer 4

Given an intensity function λ(·) ≥ 0, a general Poisson process on the line has the following properties: *for any interval I, the number of points N(I) of the process in I has a Poisson distribution with mean integral_I λ(u) du * for disjoint intervals I1, I2, . . . , Ik , the random variables N(I1),N(I2), . . . ,N(Ik ) are independent. * N is a counting process: the number of points it counts in an interval is the sum of the numbers in subintervals.

Answer 5

A Poisson process in the plane or in space (or indeed in d dimensions for any positive integer d) is defined as a counting process with these same properties, where the properties are merely re-phrased to make sense in higher dimensions.

Answer 6

Suppose that λ(·) is a real-valued non-negative function on R2 and that for each set B in the plane, N(B) is a random variable taking non-negative integer values (interpreted as the number of points of the process in B). If * N(B) has a Poisson distribution with mean ∫ _B λ(u) du; * when B1, B2, . . . , Bk are disjoint, the random variables N(B1), N(B2), . . . , N(Bk) are independent; • N has the additive property N(∪_{i=1,k} Bi) = ∑_{i=1,k} N(B_i) for disjoint Bi; then N is called a spatial Poisson process (on the plane), or a planar Poisson process, with intensity λ(·).

Answer 7

A homogeneous spatial Poisson process is the special case of Definition 17 when λ(·) is constant.

Answer 8

Notation: To save writing let us define, for sets B in the space of the points (R^d, d =2, 3, . . .), Λ(B) = ∫ _B λ(u) du

Answer 9

Let R denote the distance from the origin to the nearest point in a homogeneous planar Poisson process with intensity λ. Then the probability density function of R is h_R(r) = 2λπre^{−λπr^2}, r > 0, (75) (the density of a Rayleigh distribution). Reason: The number of points in a circle of radius r centred at the origin has a Poisson distribution with mean λπr^2 . If there are no points in this circle then R > r, and conversely. Thus P(R > r) = exp(−λπr2 ) and (75) follows by differentiation.

Answer 10

Thinning of a Poisson process refers to the random deletion of some of the points. A simple form of thinning is to remove or retain each point independently with fixed probabilities 1 − p and p, say. If the original process has intensity function λ(·), then the point process resulting from such independent thinning is a Poisson process with intensity pλ(·). Reason: Let N denote the original process and N* the thinned process. Independence and additivity of N* follow immediately from independence and additivity of N and the fact that thinning is carried out independently. Thus the only thing to show is that, for each set B, N*(B) has a Poisson distribution with mean pΛ(B). For each r ≥ 0, ``` P(N*(B) = r) = X∞ k=r P(N ∗ (B) = r | N(B) = k)P(N(B) = k) = X∞ k=r ``` k r ``` p r (1 − p) k−r e −Λ(B)Λ k (B) k! since, given k points, the number of points retained has a Binomial Bi(k, p) distribution = e −Λ(B) (pΛ(B))r r! X∞ k=r {(1 − p) Λ(B)} k−r (k − r)! = e −Λ(B) (pΛ(B))r r! e (1−p) Λ(B) = e −p Λ(B) (pΛ(B))r r! . ```

Answer 11

Conditional property (cf section 2.1.3 and section 6.3.1). Given the total number of points of a spatial Poisson process in a region B, the positions V of the points are independently distributed over B with probability density function f_V (v) = λ(v)/Λ(B) v ∈ B.

Answer 12

if we know the intensity function, the conditional distribution property 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 Poisson process using the comparison of distributions method suggested by the first contact distribution. The approach is feasible even for processes on sets B with irregular shapes.

Answer 13

The conditional distribution property also gives the likelihood for λ based on an observed pattern of points. If n points are observed in a region B and their positions are vi , i = 1, . . . , n, then the likelihood function is L = e^{−Λ(B})Λ^n(B)/n! × ∏_{1,n} λ(v_i)/Λ(B) = (1/n!) e^{−Λ(B)} ∏_{1,n} λ(vi), and the log-likelihood l = −Λ(B) + sum from i=1 to n of log λ(vi) + constant.

Answer 14

The λ function is often specified in terms of a small number of parameters. In that case fitting of the model by maximization of l and subsequent inference go ahead along the same lines as before.

Answer 15

Several examples can be thought of as a sequence of time points at each 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 higher dimensions – with intensity λ(·), associate with each point Xi of the process a random variable Yi, called the mark at Xi Then the new process {N, Y1, . . . } is called a marked Poisson process. consists of the mark values and also the poisson process points

Answer 16

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 marks at different points, and in possible changes in the distribution of marks with position of the point.

Answer 17

The arrival of claims at an insurance company and the sizes of the claims might be modelled as a marked Poisson process. An initial assumption, to be checked, might be that marks (claim amounts) are independent and identically distributed. The difference between premium income and claim payouts is the key to financial viability of the company. If the ith claim is made at time Xi and is of size Yi, and premiums bring income at a steady rate ρ net of running costs, then the assets A(t) of the insurance company at time t are A(t) = A(0) + ρt −∑_{1,N(t)} Y_i where N(t) is the number of claims up to time t. The probability that A(t) remains positive for a long time, and of how large the reserves A(0) need to be to make this probability large, are of great interest.

Answer 18

Sums of the form ∑_{1,N(t)} Y_i for a Poisson process N with marks Yi arise in many contexts. They are called compound Poisson processes.

Answer 19

A sequence of earthquakes could be modelled by attaching marks representing earthquake magnitude to the times of occurrence. Questions about dependence between magnitudes close in time are highly relevant to predictability and the possibility of warning systems. The same question arises too about the times themselves and motivates further development of the Poisson models we have considered in this course.

Answer 20

Floods could be modelled by a marked Poisson process, the mark for a flood occurrence being the magnitude of the flood. Marks could include more information too, becoming multi-dimensional. If further data were available, for example about weather conditions at the times of floods, or environmental conditions such as dryness/wetness of the ground in the period before the flood, then it too could be modelled as part of a multi-dimensional mark

Answer 21

A point process model for rainfall attempts to mimic the occurrence and heaviness of rain at a place in terms of the passage of rain cells over the place. The arrivals of rain cells are modelled by a Poisson process and the time a rain cell takes to pass over the place and the intensity of the rain it brings are attached as random marks. Marks in this case are two-dimensional.

Answer 22

An initial model for the times and severities of the Burbage Brook flood events is based on a marked point process. Dates of floods are assumed to come from a Poisson process, and the excess flood flows over 4 cumecs are modelled as conditionally independent marks with exponential distributions whose means 1/µ(t) may depend on time.

Answer 23

if indep and indep of locations of the points then the marks give a second dimension to the poisson process If the original Poisson process has intensity λ(x) and the mark probability density is k(y) then the intensity of the two-dimensional Poisson process (Xi , Yi) is µ(x, y) = λ(x) k(y).