Description Usage Arguments Details Value References See Also Examples

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.

1 2 |

`posx` |
Numeric vector. Occurrence points in the first process, |

`posy` |
Numeric vector. Occurrence points in the second process, |

`lambday` |
Numeric vector. Intensity at each time in |

`r` |
Numeric value. The radius of the intervals centered on the occurrence times in |

`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 |

`...` |
Further arguments to be passed to the function |

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 *μ_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 *μ*, equal to the sum of all the *μ_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(*μ*) 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-μ_i)/(μ_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-μ_i)/(μ_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 *μ_i* around 0.6 are valid with *n_x=50*, and around
0.3 with *n_x=100*.

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. |

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

`TestIndNH`

, `NHK`

, `NHJ`

, `DutilleulPlot`

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 | ```
#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
``` |

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