that the error in estimating the correct covariance matrix $\Sigma_g$ scales as $1/\sqrt{m}$ where $m$ is the number of samples used. In order to obtain an error of the same order of magnitude as the one due to Berry-Esseen theorem, one should use a constant fraction of the data for the parameter estimation. In that case, the covariance matrix would be estimated with a precision $1/\sqrt{n}$.

We now prove that this error for the covariance matrices translates into an error (computed with respect to the variational distance) of the same magnitude for the probability distributions.

Let us first consider the case of univariate normal distributions, that is two normal distributions $\mathcal{N}(0,\sigma_1^2)$ and $\mathcal{N}(0,\sigma_2^2)$ with $\sigma_2 > \sigma_1$. Let us note $g_1$ and $g_2$ their respective density function. One has: $$\begin{aligned}
&& \int_{-\infty}^{\infty} |g_1(x) - g_2(x)| \mathrm{d}x \nonumber \\
&&= 2 \mathrm{erf} \left(\sigma_2 \sqrt{\frac{\ln (\sigma_2/\sigma_1)}{\sigma_2^2-\sigma_1^2}} \right)-2 \mathrm{erf} \left(\sigma_1 \sqrt{\frac{\ln (\sigma_2/\sigma_1)}{\sigma_2^2-\sigma_1^2}} \right)  \nonumber\end{aligned}$$ where $\mathrm{erf}(x) = 2/\sqrt{\pi} \int_0^x e^{-t^2}\mathrm{d}t$ is the error function. In particular, if one has $\sigma_2 = \sigma_1 + \delta$ where $\delta=O(1/\sqrt{n})$ is a small error, then a first order expansion of the expression above gives $$\int_{-\infty}^{\infty} |g_1(x) - g_2(x)| \mathrm{d}x = \delta \sqrt{\frac{8}{e \pi \sigma_1^2}} + o\left(\frac{1}{\sqrt{n}}\right).$$

One 