This vignette
provides a quick demo of the truh
package. The example that we consider here is taken from Figure 3 of the paper: Trambak Banerjee, Bhaswar B. Bhattacharya, Gourab Mukherjee Ann. Appl. Stat. 14(4): 1777-1805 (December 2020)
We will consider a nonparametric two sample testing problem where the $d$ dimensional baseline (or uninfected) sample $\boldsymbol{U}=(U_1,\ldots,U_n)$ are i.i.d with cdf $F_0$ and the $d$ dimensional treated (infected) sample $\boldsymbol{V}=V_1,\ldots,V_m$ are i.i.d with cdf $G$. Here, we assume that the heterogeneity in the baseline population is reflected by $K$ different subgroups, each having unimodal distributions with distinct modes and cdfs $F_1,\ldots,F_K$, and mixing proportions $w_1,\ldots,w_K$ such that $$F_0=\sum_{a=1}^{K}w_aF_a~\text{where}~w_a\in(0,1)~\text{and}~\sum_{a=1}^{K}w_a=1. $$
The goal is to test the following composite hypothesis: $$H_0:G\in\mathcal{F}(F_0)~\text{versus}~H_1:G\notin\mathcal{F}(F_0), $$ where $\mathcal{F}(F_0)$ is the convex hull of $F_1,\ldots,F_K$. We take $d=2,n=2000,m=500$ and sample $U_1,\ldots,U_n$ from $F_0$ where $$F_0=0.3N(\boldsymbol{0},\boldsymbol{I}_2)+0.3N(\boldsymbol{\mu}_1,\boldsymbol{I}_2)+0.4N(\boldsymbol{\mu}_2,\boldsymbol{I}_2), $$ with $\boldsymbol{\mu}_1=(0,-4)$ and $\boldsymbol{\mu}_2=(4,-2)$.
n = 2000 d = 2 #Sampling the baseline (uninfected) set.seed(1) p<-runif(n,0,1) set.seed(10) U<- (p<=0.3)*matrix(rnorm(d*n),n,d)+ (p>0.3 & p<=0.6)*cbind(matrix(rnorm(n),n,1), matrix(rnorm(n,-4),n,1))+ (p>0.6)*cbind(matrix(rnorm(n,4),n,1), matrix(rnorm(n,-2),n,1))
To sample $V_1,\ldots,V_m$ we consider three settings for $G$.
# Sampling the treated (infected) m = 500 set.seed(50) V1<-cbind(matrix(rnorm(m,4),m,1), matrix(rnorm(m,-2),m,1)) #Scatter plot of the data grp = c(rep('Baseline',n), rep('Treated',m)) plot(c(U[,1],V1[,1]), c(U[,2],V1[,2]), pch = 19, col = factor(grp), xlab = 'X_1', ylab = 'X_2') # Legend legend("topright", legend = levels(factor(grp)), pch = 19, col = factor(levels(factor(grp))))
# Sampling the treated (infected) m = 500 set.seed(20) q<-runif(m,0,1) set.seed(50) V2<-(q<=0.5)*cbind(matrix(rnorm(m,2),m,1), matrix(rnorm(m,-2),m,1))+ (q>0.5)*cbind(matrix(rnorm(m,3),m,1), matrix(rnorm(m,3),m,1)) #Scatter plot of the data plot(c(U[,1],V2[,1]), c(U[,2],V2[,2]), pch = 19, col = factor(grp), xlab = 'X_1', ylab = 'X_2') # Legend legend("topright", legend = levels(factor(grp)), pch = 19, col = factor(levels(factor(grp))))
# Sampling the treated (infected) m = 500 set.seed(20) q<-runif(m,0,1) set.seed(50) V3<-(q<=0.8)*matrix(rnorm(d*m),m,d)+ (q>0.8 & q<=0.9)*cbind(matrix(rnorm(m),m,1), matrix(rnorm(m,-4),m,1))+ (q>0.9)*cbind(matrix(rnorm(m,4),m,1), matrix(rnorm(m,-2),m,1)) #Scatter plot of the data plot(c(U[,1],V3[,1]), c(U[,2],V3[,2]), pch = 19, col = factor(grp), xlab = 'X_1', ylab = 'X_2') # Legend legend("topright", legend = levels(factor(grp)), pch = 19, col = factor(levels(factor(grp))))
Let us now execute the truh
testing procedure for these scenarios. Recall that the goal is to test the following composite hypothesis:
$$H_0:G\in\mathcal{F}(F_0)~\text{versus}~H_1:G\notin\mathcal{F}(F_0).
$$
- Setting 1: Here we know that $G=F_0$ and so $H_0$ is true.
library(truh) truh.1 = truh(V1,U,B=200) truh.1$pval
So, truh
fails to reject the null hypothesis.
library(truh) truh.2 = truh(V2,U,B=200) truh.2$pval
We see that truh
rejects the null hypothesis.
library(truh) truh.3 = truh(V3,U,B=200) truh.3$pval
In this case, truh
makes the correct decision and fails to reject $H_0$.
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