---
title: "demo-truh"
output: rmarkdown::html_vignette
vignette: >
  %\VignetteIndexEntry{demo-asus}
  %\VignetteEngine{knitr::rmarkdown}
  %\VignetteEncoding{UTF-8}
---

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) <DOI: 10.1214/20-AOAS1362>. 

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 $\boldsymbol{\mu}_2=(4,-2)$. 

```{r class.source="bg-warning",fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}

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 $G$. 

- Setting 1: $G=N(\boldsymbol{\mu}_2,\boldsymbol{I}_2)$ which is the third component cdf of $F_0$. In this setting clearly $G\in\mathcal{F}(F_0)$ and the null hypothesis $H_0$ is true.

```{r class.source="bg-warning",fig.height = 4, fig.width = 6, fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
# 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))))
```


- Setting 2: $G=0.5N(\boldsymbol{\mu}_3,\boldsymbol{I}_2)+0.5N(\boldsymbol{\mu}_4,\boldsymbol{I}_2)$ where $\boldsymbol{\mu}_3=0.25\boldsymbol{\mu}_1+0.5\boldsymbol{\mu}_2$ and $\boldsymbol{\mu}_4=(3/4)\boldsymbol{\mu}_1+(9/8)\boldsymbol{\mu}_2$. Clearly in this case $G\notin\mathcal{F}(F_0)$.

```{r class.source="bg-warning",fig.height = 4, fig.width = 6,fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
# 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))))
```


- Setting 3: $G=0.8N(\boldsymbol{0},\boldsymbol{I}_2)+0.1N(\boldsymbol{\mu}_1,\boldsymbol{I}_2)+0.1N(\boldsymbol{\mu}_2,\boldsymbol{I}_2)$. This is the most interesting setting as here $G\in\mathcal{F}(F_0)$ but $G\neq F_0$ because the mixing weights differ.

```{r class.source="bg-warning",fig.height = 4, fig.width = 6,fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
# 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.
```{r class.source="bg-warning",fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
library(truh)
truh.1 = truh(V1,U,B=200)
truh.1$pval
```
So, `truh` fails to reject the null hypothesis.

- Setting 2: Here we know that $G\notin\mathcal{F}(F_0)$ and so $H_0$ is false.
```{r class.source="bg-warning",fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
library(truh)
truh.2 = truh(V2,U,B=200)
truh.2$pval
```
We see that `truh` rejects the null hypothesis.

- Setting 3: Here $G\in\mathcal{F}(F_0)$ but $G\neq F_0$. The null hypothesis $H_0$ is true in this setting.

```{r class.source="bg-warning",fig.align="center",eval=TRUE,echo=TRUE,message=FALSE,warning=FALSE}
library(truh)
truh.3 = truh(V3,U,B=200)
truh.3$pval
```
In this case, `truh` makes the correct decision and fails to reject $H_0$.