CondTest {IndTestPP}

Conditional test of independence between two Poisson process

Description

It calculates a test of independence between two Poisson process, based on the analysis of the occurrences in the second process, given that there is an occurrence in the first one. Two different approaches to calculate the p-value are implemented.

It calls the auxiliary function calcNmu, not intended for the users.

Usage

CondTest(posx, posy, lambday, r, changer = TRUE, type = "All", 
	plotRes = FALSE, ...)

Arguments

posx

Numeric vector. Occurrence points in the first process, N_x.

posy

Numeric vector. Occurrence points in the second process, N_y.

lambday

Numeric vector. Intensity at each time in N_y.

r

Numeric value. The radius of the intervals centered on the occurrence times in N_x.

changer

Optional. Logical flag. If it is TRUE, when the defined intervals overlap, their lengths are changed to obtain disjoint intervals. The two overlapping intervals are shortened by half of the overlapped period. In general, the resulting intervals are not centered.

type

Optional. Label "Poisson", "Normal" or "All". Approach to be used to calculate test p-values.

plotRes

Logical flag. If it is TRUE, the residual differences (y_i-\mu_i)/\mu_i^{(1/2)} are plotted.

...

Further arguments to be passed to the function plot.

Details

The underlying idea of the tests is to analyze the behaviour of the second process N_y, given that a point has occured in the first one, N_x. Under independence between N_x and N_y, N_y should be a Poisson process with intensity lambday.

Intervals of length 2r centered on each point in N_x are defined. To analyze the behaviour of N_y, two approaces are implemented, both based on the idea that the number of points in each interval should be a Poisson of mean \mu_i equal to the integral of lambday in the interval.

"Poisson" option: under the null, and if the intervals are independent (that is if they do not overlap) the number of points in all them should be a Poisson of mean \mu, equal to the sum of all the \mu_i. The p-values is calculated as 2*min ( (P(Y<yo)+P(Y=yo)/2), (P(X>yo)+P(Y=yo)/2)), where Y is a r.v. with distibution Poisson(\mu) and yo is the sum of the observed number of points in all the intervals. Since the p-values are based on a discrete distribution, they are valid but not exact p-values.

"Normal" option: under the null, the variables (N_i-\mu_i)/(\mu_i^{1/2}) must be zero mean and variance one variables but they are not identically distributed. Under general conditions, the mean of the variables (N_i-\mu_i)/(\mu_i^{1/2}) can be approximated by a Normal distribution using the Central limit theorem under the Lindeberg condition for r.v which are independent but not identically distributed. The conditions to have a valid Normal aprroximation are quite weak, even with a complex intensity, mean values of \mu_i around 0.6 are valid with n_x=50, and around 0.3 with n_x=100.

Value

A list with elements

pvP

P-value obtained with the 'Poisson' approach.

PvN

P-value obtained with the 'Normal' approach.

Ni

Number of occurrences in each interval.

mui

Theoretical mean of the number of occurrences in each interval under the independence assumption.

Res

Vector of th residual differences.

linf

Lower bound of each interval.

lsup

Upper bound of each interval.

mmu

Mean of the mui vector. It is used to check the conditions of the approximation of the Normal test.

References

Cebrian, A.C., Abaurrea, J. and Asin, J. (2020). Testing independence between two point processes in time. Journal of Simulation and Computational Statistics.

See Also

TestIndNH, NHK, NHJ, DutilleulPlot

Examples

#Two dependent Poisson processes from  a NHCPSP
set.seed(30)
lambdao1<-runif(3000)/20
set.seed(31)
lambdao2<-runif(3000)/10
set.seed(32)
lambda12<-runif(3000)/20
lambdaiM<-cbind(lambdao1,lambdao2,lambda12)
aux<-DepNHCPSP(lambdaiM=lambdaiM, d=2,fixed.seed=123, dplot=FALSE)

zz<-CondTest(posx=aux$posNH[[1]],posy=aux$posNH[[2]],lambday=aux$lambdaM[,2], r=2)
zz$pvP
zz$pvN

# Two independent non homogeneous  Poisson processes  
lambdao1<-runif(6000)/20
set.seed(124)
lambdao2<-runif(6000)/10
aux1<-simNHPc(lambda=lambdao1, fixed.seed=123)
aux2<-simNHPc(lambda=lambdao2, fixed.seed=124)

zz<-CondTest(posx=aux1$posNH, posy=aux2$posNH, lambday= aux2$lambda, r=3)
zz$pvP
zz$pvN