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Returning acceptable permutational group #2

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PrzeChoj opened this issue Apr 13, 2022 · 1 comment
Open

Returning acceptable permutational group #2

PrzeChoj opened this issue Apr 13, 2022 · 1 comment
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enhancement New feature or request

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@PrzeChoj
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PrzeChoj commented Apr 13, 2022

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 $.

@PrzeChoj PrzeChoj added the enhancement New feature or request label Jun 2, 2022
@PrzeChoj
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PrzeChoj commented Aug 5, 2022

One can see the $n_0$ in summary(g).

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