For now, we will implement a function that will get the observations $Z = (Z_1, Z_2, ..., Z_n)$ or matrix $U = {\sum_{i=1}}^n Z_i^T \cdot Z_i$ and will find the best permutational group $\Gamma$ to project the matrix $U$ on $\mathcal{P}_{\Gamma}$.
However, if $n < p $, then $\frac{1}{n}\cdot U$ is not the maximum likelihood Covariance estimator because such the likelihood function does not exist.
The projection $U$ on $\mathcal{P}_{\Gamma}$ will be the maximum likelihood estimator, iff $p\ge n_0$, where $n_0$ depends on $\Gamma$, according to the paper.
It also is, that $\forall_{\Gamma} \text{ }n_0 \le p$
It may be that the found group $\Gamma$ will still have $n_0 > n$. In such a case, the function should return the best Gamma among those with $n_0\le n$. Such a group always exists because for $\Gamma = \text{ }< (1,2,...,p)>$, we have $n_0 = 1$, which is an acceptable Gamma for every number $n$.
For now, we will implement a function that will get the observations$Z = (Z_1, Z_2, ..., Z_n)$ or matrix $U = {\sum_{i=1}}^n Z_i^T \cdot Z_i$ and will find the best permutational group $\Gamma$ to project the matrix $U$ on $\mathcal{P}_{\Gamma}$ .
However, if$n < p $ , then $\frac{1}{n}\cdot U$ is not the maximum likelihood Covariance estimator because such the likelihood function does not exist.
The projection$U$ on $\mathcal{P}_{\Gamma}$ will be the maximum likelihood estimator, iff $p\ge n_0$ , where $n_0$ depends on $\Gamma$ , according to the paper.
It also is, that$\forall_{\Gamma} \text{ }n_0 \le p$
It may be that the found group$\Gamma$ will still have $n_0 > n$ . In such a case, the function should return the best Gamma among those with $n_0\le n$ . Such a group always exists because for $\Gamma = \text{ }< (1,2,...,p)>$ , we have $n_0 = 1$ , which is an acceptable Gamma for every number $n$ .