# Clustering of unknown subpopulations in admixture models

The clustering of populations following admixture models is, for now, based on the K-sample test theory. Consider $$K$$ samples. For $$i=1,...,K$$, sample $$X^{(i)} = (X_1^{(i)}, ..., X_{n_i}^{(i)})$$ follows $L_i(x) = p_i F_i(x) + (1-p_i) G_i, \qquad x \in \mathbb{R}.$

We still use IBM approach to perform pairwise hypothesis testing. The idea is to adapt the K-sample test procedure to obtain a data-driven method that cluster the $$K$$ populations into $$N$$ subgroups, characterized by a common unknown mixture component. The advantages of such an approach is twofold:

• the number $$N$$ of clusters is automatically chosen by the procedure,
• Each subgroup is validated by the K-sample testing method, which has theoretical guarantees.

This clustering technique thus allows to cluster unobserved subpopulations instead of individuals. We call this algorithm the K-sample 2-component mixture clustering (K2MC).

# Algorithm

We now detail the steps of the algorithm.

1. Initialization: create the first cluster to be filled, i.e. $$c = 1$$. By convention, $$S_0=\emptyset$$.
2. Select $$\{x,y\}={\rm argmin}\{d_n(i,j); i \neq j \in S \setminus \bigcup_{k=1}^c S_{k-1}\}$$.
3. Test $$H_0$$ between $$x$$ and $$y$$.
If $H_0$ is not rejected then $S_1 = \{x,y\}$,\\
Else $S_1 = \{x\}$, $S_{c+1} = \{y\}$ and then $c=c+1$.
1. While $$S\setminus \bigcup_{k=1}^c S_k = \emptyset$$ do
Select $u={\rm argmin}\{d(i,j); i\in S_c, j\in S\setminus \bigcup_{k=1}^c S_k\}$;
Test $H_0$ the simultaneous equality of all the $f_j$, $j\in S_c$ :\\
If $H_0$ not rejected, then put $S_c=S_c\bigcup \{u\}$;\\
Else $S_{c+1} = \{u\}$ and $c = c+1$.

# Applications

## On $$\mathbb{R}^+$$

We present a case study with 5 populations to cluster on $$\mathbb{R}^+$$, with Gamma-Exponential and Gamma-Gamma mixtures.

## Simulate data (chosen parameters indicate 3 clusters (populations (1,3), (2,5) and 4)!):
list.comp <- list(f1 = "gamma", g1 = "exp",
f2 = "gamma", g2 = "exp",
f3 = "gamma", g3 = "gamma",
f4 = "exp", g4 = "exp",
f5 = "gamma", g5 = "exp")
list.param <- list(f1 = list(shape = 16, rate = 4), g1 = list(rate = 1/3.5),
f2 = list(shape = 14, rate = 2), g2 = list(rate = 1/5),
f3 = list(shape = 16, rate = 4), g3 = list(shape = 12, rate = 2),
f4 = list(rate = 1/2), g4 = list(rate = 1/7),
f5 = list(shape = 14, rate = 2), g5 = list(rate = 1/6))
A.sim <- rsimmix(n=6000, unknownComp_weight=0.8, comp.dist = list(list.comp$f1,list.comp$g1),
comp.param = list(list.param$f1, list.param$g1))$mixt.data B.sim <- rsimmix(n=6000, unknownComp_weight=0.5, comp.dist = list(list.comp$f2,list.comp$g2), comp.param = list(list.param$f2, list.param$g2))$mixt.data
C.sim <- rsimmix(n=6000, unknownComp_weight=0.65, comp.dist = list(list.comp$f3,list.comp$g3),
comp.param = list(list.param$f3, list.param$g3))$mixt.data D.sim <- rsimmix(n=6000, unknownComp_weight=0.75, comp.dist = list(list.comp$f4,list.comp$g4), comp.param = list(list.param$f4, list.param$g4))$mixt.data
E.sim <- rsimmix(n=6000, unknownComp_weight=0.55, comp.dist = list(list.comp$f5,list.comp$g5),
comp.param = list(list.param$f5, list.param$g5))$mixt.data ## Look for the clusters: list.comp <- list(f1 = NULL, g1 = "exp", f2 = NULL, g2 = "exp", f3 = NULL, g3 = "gamma", f4 = NULL, g4 = "exp", f5 = NULL, g5 = "exp") list.param <- list(f1 = NULL, g1 = list(rate = 1/3.5), f2 = NULL, g2 = list(rate = 1/5), f3 = NULL, g3 = list(shape = 12, rate = 2), f4 = NULL, g4 = list(rate = 1/7), f5 = NULL, g5 = list(rate = 1/6)) clusters <- admix_clustering(samples = list(A.sim,B.sim,C.sim,D.sim,E.sim), n_sim_tab = 10, comp.dist=list.comp, comp.param=list.param, parallel=FALSE, n_cpu=2) clusters #>$n_clust
#> [1] 3
#>
#> $clusters #> [,1] [,2] [,3] [,4] [,5] #> Id_sample 1 2 3 4 5 #> Id_cluster 3 1 3 2 1 #> #>$discrepancy_matrix
#>             [,1]        [,2]        [,3]       [,4]        [,5]
#> [1,]   0.0000000 106.2503840   0.4559752  17.198006 212.5783544
#> [2,] 106.2503840   0.0000000  68.5687703   5.998583   0.2265777
#> [3,]   0.4559752  68.5687703   0.0000000 227.404200  74.4947230
#> [4,]  17.1980060   5.9985831 227.4041998   0.000000   9.5940432
#> [5,] 212.5783544   0.2265777  74.4947230   9.594043   0.0000000