View source: R/mean2_2014CLX.R
mean2.2014CLX | R Documentation |
Given two multivariate data X and Y of same dimension, it tests
H_0 : μ_x = μ_y\quad vs\quad H_1 : μ_x \neq μ_y
using the procedure by Cai, Liu, and Xia (2014) which is equivalent to test
H_0 : Ω(μ_x - μ_y)=0
for an inverse covariance (or precision) Ω. When Ω is not given and known to be sparse, it is first estimated with CLIME estimator. Otherwise, adaptive thresholding estimator is used. Also, if two samples are assumed to have different covariance structure, it uses weighting scheme for adjustment.
mean2.2014CLX( X, Y, precision = c("sparse", "unknown"), delta = 2, Omega = NULL, cov.equal = TRUE )
X |
an (n_x \times p) data matrix of 1st sample. |
Y |
an (n_y \times p) data matrix of 2nd sample. |
precision |
type of assumption for a precision matrix (default: |
delta |
an algorithmic parameter for adaptive thresholding estimation (default: 2). |
Omega |
precision matrix; if |
cov.equal |
a logical to determine homogeneous covariance assumption. |
a (list) object of S3
class htest
containing:
a test statistic.
p-value under H_0.
alternative hypothesis.
name of the test.
name(s) of provided sample data.
cai_twosample_2014SHT
## CRAN-purpose small example smallX = matrix(rnorm(10*3),ncol=3) smallY = matrix(rnorm(10*3),ncol=3) mean2.2014CLX(smallX, smallY, precision="unknown") mean2.2014CLX(smallX, smallY, precision="sparse") ## Not run: ## empirical Type 1 error niter = 100 counter = rep(0,niter) # record p-values for (i in 1:niter){ X = matrix(rnorm(50*5), ncol=10) Y = matrix(rnorm(50*5), ncol=10) counter[i] = ifelse(mean2.2014CLX(X, Y)$p.value < 0.05, 1, 0) } ## print the result cat(paste("\n* Example for 'mean2.2014CLX'\n","*\n", "* number of rejections : ", sum(counter),"\n", "* total number of trials : ", niter,"\n", "* empirical Type 1 error : ",round(sum(counter/niter),5),"\n",sep="")) ## End(Not run)
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