Chapter 2. Poisson point processes

Transcription

1 Chapter 2. Poisson point processes Jean-François Coeurjolly Laboratoire Jean Kuntzmann (LJK), Grenoble University

2 Setting for this chapter To ease the presentation, assume S R d. The intensity measure is absolutely continuous with respect to the Lebesgue measure and denote ρ its density. ρ is locally integrable on S, i.e. B s.t. B <, µ(b) = B ρ(ξ)dξ < X is a simple point process, i.e. µ(ξ) = 0, ξ S.

3 Towards the Poisson point process Simplest point process = point process with one point one uniform point P(ξ B) = B S. one point with a density f on S P(ξ B) = B f (ξ)dξ S f (ξ)dξ.

4 Binomial point process Definition Let f a density on B S and let n 1. The point process corresponding to n i.i.d. points with density f is called the binomial point process on B and is denoted by binomial(b, n, f ). Why binomial? particular case : B <, ξ i U(B). For any A B ( ) n A k B \ A n k P(N (A) = k) = k B n. f (ξ) = ρ(ξ) µ(b). For any A B P(N (A) = k) = ( ) n µ(a) k µ(b \ A) n k k µ(b) n.

5 Poisson point process Definition A point process X is called a Poisson point process on S R d with intensity ρ( ) if 1 B S, µ(b) <, N (B) P(µ(B)). 2 n 1, B S s.t. 0 < µ(b) < then given N (B) = n, X B binomial(b, n, f ) with f (ξ) = ρ(ξ) µ(b).(if µ(b) = 0 then X B = ). We denote X Poisson(S, ρ). Remark : when ρ( ) is constant, X is said to be the homogeneous Poisson point process. The homogeneous Poisson point process X is stationary and isotropic.

6 Simulation on a bounded domain The objective is to generate on a domain S a Poisson point process defined on S with S S. We assume that S is a rectangle. Homogeneous case : 1 Generate N (S) P(ρ S ). Let n be this realization. 2 Generate n uniform points in S. Inhomogeneous case : assume ρ( ) c. The simulation is done by thinning (see Møller and Waagepetersen 2004) a homogeneous Poisson point process. 1 Generate Poisson(S, c). Let y = {y 1,..., y n } be this realization. 2 For i = 1,..., n, delete independently the point y i with the probability 1 ρ(y i ) c.

7 Homogeneous Poisson point process S = [0, 1] 2, ρ = 100. (recall En(X S ) = EN (S) = ρ S ) 110 points

8 Homogeneous Poisson point process S = [0, 1] 2, ρ = 200. (recall En(X S ) = EN (S) = ρ S ) 184 points

9 Homogeneous Poisson point process S = [0, 2] [0, 1], ρ = 100. (recall En(X S ) = EN (S) = ρ S ) 210 points

10 Inhomogeneous Poisson point process S = [ 1, 1] 2, ρ(ξ) = β exp ( ξ 1 ξ ξ3 1). The parameter β is adjusted such that En(X S ) = EN (S) = S ρ(ξ)dξ =

11 Inhomogeneous Poisson point process S = [ 1, 1] 2, ρ(ξ) = β exp ( 2ξ 2 1 ξ2 2). The parameter β is adjusted such that En(X S ) = EN (S) = S ρ(ξ)dξ =

12 Inhomogeneous Poisson point process S = [ 1, 1] 2, ρ(ξ) = β exp (2 sin(4πξ 1 ξ 2 )). The parameter β is adjusted such that En(X S ) = EN (S) = S ρ(ξ)dξ =

13 Probability distribution Proposition (i) X Poisson(S, ρ) iif B S s.t. µ(b) < and F N lf P(X B F ) = n 0 exp( µ(b)) B... 1({x 1,..., x n } F ) B n ρ(x i )dx i where the integral for n = 0 corresponds to 1( F ). (ii) If X Poisson(S, ρ) then for any function h : N lf R + and any B S s.t. µ(b) < Eh(X B ) = n 0 exp( µ(b)) n! B... h({x 1,..., x n }) B where the integral for n = 0 should be read as h( ). i=1 n ρ(x i )dx i i=1 Proof : (i) follows from P(X B F ) = n 0 P(X B F N (B) = n)p(n (B) = n and the definition of the Poisson point process. (ii) follows from standard measure theory.

14 Existence of a Poisson point process If S < the existence is given by the previous proposition. If S = (S = R d for example), the existence follows from the next result. Theorem X Poisson(S, ρ) exists and is uniquely determined by its void probabilities v(b) = P(N (B) = 0) = exp( µ(b)) B S Proof : see Møller and Waagepetersen (2004).

15 Independence scattering Theorem If X Poisson(S, ρ) then X B1, X B2,... are independent random variables B 1, B 2,... disjoint sets of S. Proof : Let B 1, B 2 disjoint bounded domains of S. Let F 1, F 2 N lf and h j = 1(. F j ) (j = 1, 2) and B = B 1 B 2 (the general case is done by induction and since S is a complete space) P(X B1 F 1, X B2 F 2, ) = E(h 1 (X B1 )h 2 (X B2 )) = Eh(X B ) 1 n = exp( µ(b 1 ) µ(b 2 )) h({x 1,..., x n }) ρ(x i )dx i n! n 0 (B 1 B 2 ) n i=1 n ( ) n 1 n = exp( µ(b 1 ) µ(b 2 )) h 1 ({x 1,..., x k })h 2 ({x k+1,..., x n }) ρ(x i )dx i k n! n 0 k=0 B 1 k B 2 n k i=1 1 1 = exp( µ(b 1 ) µ(b 2 )) h 1 ({x 1,..., x k 1! k 2! B k 1 1 B k k1 })h 2 ({y 1,..., y k2 }) 2 2 = Eh 1 (X B1 )Eh 2 (X B2 ). k 1 0 k 2 0 k 1 k 2 ρ(x i )dx i ρ(y i )dy i i=1 i=1

16 Slivnyak-Mecke Theorem Theorem Let X Poisson(S, ρ) and let h be a non-negative function h : S N lf R +, then E h(ξ, X \ ξ) = Eh(ξ, X )ρ(ξ)dξ. ξ X S

17 Slivnyak-Mecke Theorem Theorem Let X Poisson(S, ρ) and let h be a non-negative function h : S N lf R +, then E h(ξ, X \ ξ) = Eh(ξ, X )ρ(ξ)dξ. ξ X Exercise : a point ξ of a configuration of points x is said to be a R-closed neighbour if d(ξ, x \ ξ) R. Let X be a homogeneous Poisson point process on R d with intensity ρ. Compute the mean number of R-closed neighbours of X in B = [0, 1] 2. S

18 Slivnyak-Mecke Theorem Theorem Let X Poisson(S, ρ) and let h be a non-negative function h : S N lf R +, then E h(ξ, X \ ξ) = Eh(ξ, X )ρ(ξ)dξ. ξ X Exercise : a point ξ of a configuration of points x is said to be a R-closed neighbour if d(ξ, x \ ξ) R. Let X be a homogeneous Poisson point process on R d with intensity ρ. Compute the mean number of R-closed neighbours of X in B = [0, 1] 2. Answer : E 1(d(ξ, X \ ξ) R) = ρ(1 exp( ρπr 2 )). ξ X B S

19 Simulation ρ = 20, R = 0.1 ρ = 100, R = 0.05 ξ x 1(d(ξ, x \ ξ) R) = 9 ξ x 1(d(ξ, x \ ξ) R) = 60 ρ(1 exp( ρπr 2 )) 9.33 ρ(1 exp( ρπr 2 )) 54.41

20 Proof of Slivnyak-Mecke Theorem Proof : we only consider the case where X depend only on X B for some B S s.t. µ(b) < (which of course covers the case where µ(s) < ). The general case is more awkward (see Møller and Waagepetersen 2004). Idea : apply the formula for the expectation with g(x B ) = ξ X h(ξ, X \ ξ). Since g( ) = 0, we have exp( µ(b)) n n Eg(X B ) =... h(x i, {x 1,..., x n } \ x i ) ρ(x i )dx i n! n 1 B B i=1 i=1 exp( µ(b)) n =... h(x n, {x 1,..., x n 1 }) ρ(x i )dx i (n 1)! n 1 B B i=1 exp( µ(b)) m = h(y, {x 1,..., x m }) ρ(x i )dx i B m! ρ(y)dy m 0 B m i=1 = Eh(ξ, X )ρ(ξ)dξ. B

21 Generalized Slivnyak-Mecke Theorem Theorem Let h : S m N lf R + then E ξ 1,...,ξ m X h({ξ 1,..., ξ m },X \ {ξ 1,..., ξ m } = S... Eh(x 1,..., x m, X ) S Proof : left as an exercise. (proof by induction) m ρ(x i )dx i. i=1

22 Second-order intensity of a Poisson point process We recalll that the second order intensity ρ (2) (which is the density of the second order reduced moment measure α (2) ) of a Poisson(S, ρ)) ρ (2) (u, v)dudv can be interpreted as the probability to observe a distinct pair of points around u and v. Proposition For any u, v S by ρ (2) (u, v) = ρ(u)ρ(v). Proof : for any B 1, B 2 S, we have from the GSM Theorem α (2) (B 1 B 2 ) = E 1({u, v} B 1 B 2 ) u,v X = ρ(u)ρ(v)dudv B 1 B 2 def = ρ (2) (u, v)dudv. B 1 B 2

23 Density of a Poisson point process Concept useful when dealing with parametric estimation. Proposition If µ(s) < then X Poisson(S, ρ) is absolutley continuous with respect to Poisson(S, 1) and the density is given by f (x) = exp ( S µ(s)) ρ(ξ). Proof : Let F N lf ξ x and let Y denote a Poisson(S, 1). Then exp( S ) E (1(Y F )f (Y )) =... n! n 0 S S n 1({x 1,..., x n } F ) exp ( S µ(s)) ρ(x i ) 1dx 1... dx n i=1 = P(X F ).

24 Homogeneous case We consider here the problem of estimating the parameter ρ of a homogeneous Poisson point process defined on S and observed on a window W S. Since N (W ) P(ρ W ), the natural estimator of ρ is ρ = N (W )/ W Properties

25 Homogeneous case We consider here the problem of estimating the parameter ρ of a homogeneous Poisson point process defined on S and observed on a window W S. Since N (W ) P(ρ W ), the natural estimator of ρ is ρ = N (W )/ W Properties (i) ρ corresponds to the maximum likelihood estimate.

26 Homogeneous case We consider here the problem of estimating the parameter ρ of a homogeneous Poisson point process defined on S and observed on a window W S. Since N (W ) P(ρ W ), the natural estimator of ρ is ρ = N (W )/ W Properties (i) ρ corresponds to the maximum likelihood estimate. (ii) ρ is unbiased.

27 Homogeneous case We consider here the problem of estimating the parameter ρ of a homogeneous Poisson point process defined on S and observed on a window W S. Since N (W ) P(ρ W ), the natural estimator of ρ is ρ = N (W )/ W Properties (i) ρ corresponds to the maximum likelihood estimate. (ii) ρ is unbiased. (iii)var ρ = ρ W. Proof : (i) follows from the definition of the density (ii-iii) can be checked using the Campbell formulae.

28 Homogeneous case (2) Asymptotic results

29 Homogeneous case (2) Asymptotic results For large N (W ), ρ W N(ρ W, ρ W ) and so W 1/2 ( ρ ρ) N(0, ρ). (the approximation is actually a convergence as W R d )

30 Homogeneous case (2) Asymptotic results For large N (W ), ρ W N(ρ W, ρ W ) and so W 1/2 ( ρ ρ) N(0, ρ). (the approximation is actually a convergence as W R d ) Variance stabilizing transform : 2 W 1/2 ( ρ ρ) N(0, 1)

31 Homogeneous case (2) Asymptotic results For large N (W ), ρ W N(ρ W, ρ W ) and so W 1/2 ( ρ ρ) N(0, ρ). (the approximation is actually a convergence as W R d ) Variance stabilizing transform : 2 W 1/2 ( ρ ρ) N(0, 1) We deduce a 1 α (α (0, 1)) confidence interval for ρ IC 1 α (ρ) = ( ρ ± z α/2 ) 2. 2 W 1/2

32 A simulation example We generated m = replications of homogeneous Poisson point processes with intensity ρ = 100 on [0, 1] 2 (blcak plots) and on [0, 2] 2 (red plots). Histograms of ρ Histograms of 2 W 1/2 ( ρ ρ) Density [0,1]^2 [0,2]^2 N(100,100) N(100,100/4) Density [0,1]^2 [0,2]^2 N(0,1)

33 A simulation example We generated m = replications of homogeneous Poisson point processes with intensity ρ = 100 on [0, 1] 2 (black plots) and on [0, 2] 2 (red plots). W = [0, 1] 2 W = [0, 2] 2 Emp. Mean of ρ Emp. Var. of ρ Emp. Coverage rate of 95% confidence intervals 95.31% 94.78%

34 Inhomogeneous case : parametric estimation Assume that ρ is parametrized by a vector θ R p (p 1). The most well-known model is the log-linear one : ρ(ξ) = ρ(ξ; θ) = exp(θ z(ξ)) where z(ξ) = (z 1 (ξ), z 2 (ξ),..., z p (ξ)) correspond to known spatial functions or spatial covariates. θ can be estimated by maximizing the log-likelihood on W l W (X, θ) = log ρ(ξ; θ) (1 ρ(ξ; θ))dξ ξ X W W = W + θ z(ξ) exp(θ z(ξ))dξ. ξ X W W } {{ } :=l W (X,θ) In other words θ = Argmax θ l W (X, θ).

35 Inhomogeneous case : parametric estimation (2) Why would θ be a good estimate?

36 Inhomogeneous case : parametric estimation (2) Why would θ be a good estimate? Compute the score function s W (X, θ) = l W (X, θ) = z(ξ) ξ X W W z(ξ) exp(θ z(ξ)) dξ. } {{ } :=ρ(ξ)

37 Inhomogeneous case : parametric estimation (2) Why would θ be a good estimate? Compute the score function s W (X, θ) = l W (X, θ) = z(ξ) ξ X W W z(ξ) exp(θ z(ξ)) dξ. } {{ } :=ρ(ξ) The true parameter θ 0 (i.e. X P θ0 ) minimizes the expectation of the score function. Indeed from Campbell formula

38 Inhomogeneous case : parametric estimation (2) Why would θ be a good estimate? Compute the score function s W (X, θ) = l W (X, θ) = z(ξ) ξ X W W z(ξ) exp(θ z(ξ)) dξ. } {{ } :=ρ(ξ) The true parameter θ 0 (i.e. X P θ0 ) minimizes the expectation of the score function. Indeed from Campbell formula Es W (X, θ) = z(ξ) ( exp(θ0 z(ξ)) exp(θ z(ξ)) ) dξ = 0 when θ = θ 0. W

39 Inhomogeneous case : parametric estimation (2) Why would θ be a good estimate? Compute the score function s W (X, θ) = l W (X, θ) = z(ξ) ξ X W W z(ξ) exp(θ z(ξ)) dξ. } {{ } :=ρ(ξ) The true parameter θ 0 (i.e. X P θ0 ) minimizes the expectation of the score function. Indeed from Campbell formula Es W (X, θ) = z(ξ) ( exp(θ0 z(ξ)) exp(θ z(ξ)) ) dξ = 0 when θ = θ 0. W Rathbun and Cressie (1994) showed the strong consistency and the asymptotic normality of θ as W R d.

40 A simulation example As an example, consider the model log ρ(ξ) = θz(ξ) with θ = 2 and z(ξ) = ξ 2 1 ξ We generated m = 1000 replications of this model and estimate θ W = [0, 1] 2 W = [0, 2] 2 EN (W ) = 200 EN (W ) = 800 Emp. Mean of θ Emp. Var. of θ

41 Data example : dataset bei A point pattern giving the locations of 3605 trees in a tropical rain forest. Accompanied by covariate data giving the elevation (altitude) (z 1 ) and slope of elevation (z 2 ) in the study region. elevation, z1 elevation gradient, z Assume an inhomogeneous Poisson point process (which is not true, see the next chapter) with intensity log ρ(ξ) = β + θ 1 z 1 (ξ) + θ 2 z 2 (ξ). Question : how can we prove that each covariate has a significant influence?

42 Data example : dataset bei (2) Using the MLE we get θ 1 = and θ 2 =

43 Data example : dataset bei (2) Using the MLE we get θ 1 = and θ 2 = Without any estimation of the standard error of estimates, one may use a parametric Bootstrap approach 1 generate B = 1000 inhomogeneous Poisson point processes with intensity ρ(ξ) = β + θ 1 z 1 (ξ) + θ 2 z 2 (ξ). 2 for each replication we estimate the parameters which allows us to estimate the standard errors of the initial estimates σ θ 1 = σ θ 2 = Using the asymptotic normality result, we can evaluate p-values of the tests H 1 : θ 1 0 and H 1 : θ 2 0 : we get p-value 1 0% p-value 2 35%. The elevation is clearly significant.

44 Inhomogeneous case : nonparametric estimation (Diggle 2003) Idea is to mimic the kernel density estimation to define a nonparametric estimator of the spatial function ρ. Let k : R d R + a symmetric kernel with intensity one. Examples of kernels Gaussian kernel : (2π) d/2 exp( y 2 /2). Cylindric kernel : 1 π 1( y 1). Epanecnikov kernel : 3 4 1( y < 1)(1 y 2 ). Let h be a positive real number (which will play the role of a bandwidth window), then the nonparametric estimate (with border correction) at the location η is defined as ( ) ρ h (η) = K h (η) 1 1 η ξ h k d h ξ X W

45 Intuitively, this works... Indeed, using the Campbell formula and a change of variables we can obtain ( ) E ρ h (η) = K h (η) 1 1 η ξ E h k d h ξ X W ( ) = K h (η) 1 1 η ξ W h k ρ(ξ)dξ d h = K h (η) 1 k ( ω ) ρ(ωh + η)dω h small ρ(η). W η h K h (η) 1 W η h k ( ω ) ρ(η)dω More theoretical justifications and properties and a discussion on the bandwidth parameter and edge corrections can be found in Diggle (2003).

46 Illustration on the bei dataset Using a Gaussian kernel and a bandwidth parameter chosen using a cross-likelihood