The variance of variance-covariance matrix Looking at answer on the standard error of the variances and this answer on the standard error of the covariances, and knowing that both are part of the variance-covariance matrix, $\Sigma$, I am wondering if both formulas can be summarized in a single equation (probably in a matrix form).
The equation for the variance of the variance is
$\text{var}(s_{xx}^2) =  \frac{1}{n}\left(\mu_4 -\frac{n-3}{n-1} \sigma^4\right) $
where $\mu_4 = E(x-\mu_x)^4$ and $\sigma_x^4 = ((x-\mu_x)^2)^2)$
and the standard error of the covariance is
$\text{var}(s_{XY})=\frac{(n−1)^2}{n^3}(μ_{22}-μ_{11}^2)+ \frac{(n−1)}{n^3} ( μ_{20} μ_{02}-μ_{11}^2 )$
where $\mu_{rs}=E[(X-\mu_{_X})^r\,(Y-\mu_{_Y})^s]$.
Thank you,

EDIT:
I have found a good approximation as $(n^{-1}X^{2\prime}X^2)-(X^{\prime}Xn^{-1})^2$, where $X$ is centered, $X=X-\bar{X}$, and $^2$ is element-wise. In R:
round((t(X^2)%*%(X^2)*N^-1-((t(X)%*%X)*N^-1)^2),3)

 A: If you take this (I personally haven't checked it):

$\text{var}(s_{XY})=\frac{(n−1)^2}{n^3}(μ_{22}-μ_{11}^2)+
 \frac{(n−1)}{n^3} ( μ_{20} μ_{02}-μ_{11}^2 )$
where $\mu_{rs}=E[(X-\mu_{_X})^r\,(Y-\mu_{_Y})^s]$.

Then
$\text{var}(s_{XX})=\frac{(n−1)^2}{n^3}(μ_{22}-μ_{11}^{2})+\frac{(n−1)}{n^3} ( μ_{20} μ_{02}-μ_{11}^2 )$
should still hold, since it is already generalized.
\begin{cases}
\mu_{11}=E[(X-\mu_{_X})\,(X-\mu_{_X})] = E[(X-\mu_{_X})^2]=s_{XX}=\sigma^2\\
\mu_{22}=E[(X-\mu_{_X})^2\,(X-\mu_{_X})^2]=\kappa s_{XX}=\mu_4\\
\mu_{20}=\mu_{02}=E[(X-\mu_{_X})^0\,(X-\mu_{_X})^2] = E[(X-\mu_{_X})^2] = s_{XX} = \sigma^2
\end{cases}
So
$$\text{var}(s_{XX})=\frac{(n−1)^2}{n^3}(\mu_4-(\sigma^2)^{2})+\color{red}{\frac{(n−1)}{n^3} ( (\sigma^2)(\sigma^2)-(\sigma^2)^2 )}\\
=\frac{(n−1)^2}{n^3}(\mu_4-\sigma^4)
$$
Discrepancies are due to Bessel's correction (probably).

In their answer, glen_b says that

Note that correcting this for the $\frac{1}{n-1}$ version is a simple matter of multiplying the above result by $(\frac{n}{n-1})^2$.

This would lead to:
$$\text{var}(s_{XX})=\frac{n^2}{(n−1)^2}\frac{(n−1)^2}{n^3}(\mu_4-\sigma^4)\\
=\frac{1}{n}(\mu_4-\sigma^4)
$$
So still not quite your first result.
