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I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem found here (1) : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

Ergo in this case $cX$ should be $W(n,c\Sigma)$ distributed and then, for the covariance sample purpose should be $\frac{1}{\sqrt{n}}$. In this way $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.

I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem found here (1) : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

Ergo in this case $cX$ should be $W(n,c\Sigma)$ distributed. Relating to my example for the sample covariance, $C = \text{diag}(\frac{1}{\sqrt{n}})$. Hence, $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.

I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

Ergo in this case $cX$ should be $W(n,c\Sigma)$ distributed and then, for the covariance sample purpose should be $\frac{1}{\sqrt{n}}$. In this way $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.

I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem found here (1) : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

Ergo in this case $cX$ should be $W(n,c\Sigma)$ distributed and then, for the covariance sample purpose should be $\frac{1}{\sqrt{n}}$. In this way $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.

I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

So the diagonal ele. for the covariance sample purpose should be $\frac{1}{\sqrt{n}}$. In this way $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.

I think I just found an answer but I leave open the question for the sake of the community:

Actually using the theorem : $$CXC^t \sim W(n,C\Sigma C^t) $$ and taking $C$ as the diagonal matrix : $$\left( \begin{matrix} \sqrt{c} & \ldots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \sqrt{c} \\ \end{matrix} \right) $$

Ergo in this case $cX$ should be $W(n,c\Sigma)$ distributed and then, for the covariance sample purpose should be $\frac{1}{\sqrt{n}}$. In this way $\frac{1}{n}X=\hat{\Sigma}$ should be simply $W(n, \frac{1}{n}\Sigma) $ distributed.