HHG (version 2.3.7)

hhg.univariate.ks.pvalue: The p-value computation for the K-sample problem using a fixed partition size

Description

The p-value computation for the K-sample test of Heller et al. (2016) using a fixed partition size m.

Usage

hhg.univariate.ks.pvalue(statistic, NullTable,m)

Value

The p-value.

Arguments

statistic

The value of the computed statistic by the function hhg.univariate.ks.stat. The statistic object includes the score type (one of "LikelihoodRatio" or "Pearson"), and the aggregation type (one of "sum" or "max").

NullTable

The null table of the statistic, which can be downloaded from the software website (http://www.math.tau.ac.il/~ruheller/Software.html) or computed by the function

hhg.univariate.ind.nulltable. See vignette('HHG') for a method of computing null tables on multiple cores.

m

The partition size.

Author

Barak Brill Shachar Kaufman.

Details

For the test statistic, the function extracts the fraction of observations in the null table that are at least as large as the test statistic, i.e. the p-value.

References

Heller, R., Heller, Y., Kaufman S., Brill B, & Gorfine, M. (2016). Consistent Distribution-Free K-Sample and Independence Tests for Univariate Random Variables, JMLR 17(29):1-54 https://www.jmlr.org/papers/volume17/14-441/14-441.pdf

Brill B. (2016) Scalable Non-Parametric Tests of Independence (master's thesis) https://tau.userservices.exlibrisgroup.com/discovery/delivery/972TAU_INST:TAU/12397000130004146?lang=he&viewerServiceCode=AlmaViewer

Examples

if (FALSE) {

#Two groups, each from a different normal mixture:
N0=30
N1=30
X = c(c(rnorm(N0/2,-2,0.7),rnorm(N0/2,2,0.7)),c(rnorm(N1/2,-1.5,0.5),rnorm(N1/2,1.5,0.5)))
Y = (c(rep(0,N0),rep(1,N1)))
plot(Y,X)

#I)p-value for fixed partition size using the sum aggregation type
hhg.univariate.Sm.Likelihood.result = hhg.univariate.ks.stat(X,Y)
hhg.univariate.Sm.Likelihood.result


sum.nulltable = hhg.univariate.ks.nulltable(c(N0,N1), nr.replicates=100)
#default nr. of replicates is 1000, but may take several seconds.
#For illustration only, we use 100 replicates, but it is highly recommended
#to use at least 1000 in practice. 

#p-value for m=4 (the default):
hhg.univariate.ks.pvalue(hhg.univariate.Sm.Likelihood.result, sum.nulltable, m=4)


#p-value for m=2:
hhg.univariate.ks.pvalue(hhg.univariate.Sm.Likelihood.result, sum.nulltable, m=2)

#II) p-value for fixed partition size using the max aggregation type

hhg.univariate.Mm.likelihood.result = hhg.univariate.ks.stat(X,Y,aggregation.type = 'max')

hhg.univariate.Mm.likelihood.result

max.nulltable = hhg.univariate.ks.nulltable(c(N0,N1), aggregation.type = 'max',
  score.type='LikelihoodRatio', mmin = 3, mmax = 5, nr.replicates = 100)
#default nr. of replicates is 1000, but may take several seconds.
#For illustration only, we use 100 replicates,
#but it is highly recommended to use at least 1000 in practice.

#p-value for m=3:
hhg.univariate.ks.pvalue(hhg.univariate.Mm.likelihood.result, max.nulltable ,m = 3) 

#p-value for m=5:
hhg.univariate.ks.pvalue(hhg.univariate.Mm.likelihood.result, max.nulltable,m = 5)


#III) p-value for sum and max aggregation type, using large data variants:

#Two groups, each from a different normal mixture, total sample size is 10^4:
X_Large = c(c(rnorm(2500,-2,0.7),rnorm(2500,2,0.7)),
c(rnorm(2500,-1.5,0.5),rnorm(2500,1.5,0.5)))
Y_Large = (c(rep(0,5000),rep(1,5000)))
plot(Y_Large,X_Large)

# for these variants, make sure to change mmax so that mmax<= nr.atoms

hhg.univariate.Sm.EQP.Likelihood.result = hhg.univariate.ks.stat(X_Large,Y_Large,
variant = 'KSample-Equipartition',mmax=30)
hhg.univariate.Mm.EQP.likelihood.result = hhg.univariate.ks.stat(X_Large,Y_Large,
aggregation.type = 'max',variant = 'KSample-Equipartition',mmax=30)

N0_large = 5000
N1_large = 5000

Sm.EQP.null.table = hhg.univariate.ks.nulltable(c(N0_large,N1_large), nr.replicates=200,
variant = 'KSample-Equipartition', mmax = 30)
Mm.EQP.null.table = hhg.univariate.ks.nulltable(c(N0_large,N1_large), nr.replicates=200,
aggregation.type='max', variant = 'KSample-Equipartition', mmax = 30)

hhg.univariate.ks.pvalue(hhg.univariate.Sm.EQP.Likelihood.result, Sm.EQP.null.table, m=5)
hhg.univariate.ks.pvalue(hhg.univariate.Mm.EQP.likelihood.result, Mm.EQP.null.table, m=5)

}