About learning a single Gaussian in total-variation distance I am looking for the proof of this following result which I saw as being claimed as a "folklore" in a paper. It would be helpful if someone can share a reference where this has been shown! 


  Let $G$ be an $n-$dimensional Gaussian and let $\delta &gt;0$. Then there
  exists a polynomial time algorithm that given $O(\frac {n^2}{\delta^2})$ independent samples from $G$ returns a probability
  distribution ${\bf P}$ so that with probability at least $\frac 2 3$
  we have, $d_{\text{TV}}(G,{\bf P}) &lt; \delta$




To clarify the notation of "TV" used above : 

Given two distributions $\bf P$ and $\bf Q$ with p.d.fs (which we denote by the same symbols) we define the ``Total-Variation" distance between them as, 

$ d_{\text{TV}}({\bf P},{\bf Q}) := \frac {1}{2} \int_{\bf x \in \mathbb{R}^n} \vert {\bf P}(\bf x) - {\bf Q}(\bf x) \vert d{\bf x} =: \frac {1}{2} ||{\bf P} - {\bf Q}||_1 $
Essentially, this follows from three facts:


learning a Gaussian in total variation distance $\delta$ is equivalent to learning its two parameters, $\mu,\Sigma$, to (respectively) $\ell_2$ and relative Frobenius norms $O(\delta)$. (Since then the "empirical Gaussian" with the mean and covariance you estimated will be $\delta$-close to the true Gaussian).
learning the mean $\mu$ to $\ell_2$ distance $\delta$ can be done with $O(\frac{n}{\delta^2})$ samples. (This is tight)
learning the covariance to relative Frobenius distance $\delta$ can be done with $O(\frac{n^2}{\delta^2})$ samples. (This is tight)


I suggest you try to prove these yourself. The first one follows from relatively "standard" facts about Gaussians [1], the second two are good exercises.

(If you really want a specific reference, I can try to dig some up.)



[1] Theorem 1.3 in this recent paper is overkill, but will do the job. https://arxiv.org/abs/1810.08693