do not depend on $m$ nor $n$). We assume that the parameters of the Dirichlet prior distribution $ \mathcal{D}ir(\alpha_1,\dots,\alpha_d) $ satisfy

-   $\forall i = 1,\ldots,d : \alpha_i \leq 1$,

-   $  \sum_{i=1}^d\alpha_i =D_1$,

-   $ \prod_{i=1}^d \alpha_i \geq e^{-D_2d \log(d)} $.

Note that this assumption is satisfied for $ \alpha_1 = \ldots = \alpha_d = 1/d$ with $ D_1 = D_2 = 1$.

The first theorem provides the concentration bound on the square error of the first estimator $\tilde{\rho}^{prob}_{\lambda}$. The proof of this theorem is left to the .

\[thrm: rate 1\] Fix a small $\epsilon \in (0,1) $. Under , for $ \lambda = \lambda^*: =m/2 $, with probability at least $ 1-\epsilon$, one has $$\begin{aligned}
 \| \tilde{\rho}^{prob}_{\lambda^*} - \rho^0\|_F^2
 \leq
  C^{prob}_{D_1,D_2}\frac{   3^n {\rm rank}(\rho^0)
  \log\left(
\frac{{\rm rank}(\rho^0) N}{2^n}
\right)
 + (1.5)^n\log(2/\epsilon)}{ N},\end{aligned}$$ where $ C^{prob}_{D_1,D_2}$ is a constant that depends only on $D_1,D_2$.

As said in the introduction, the best known rate up-to-date in this problem is $\frac{   3^n {\rm rank}(\rho^0)}{N}$, so our estimator $ \tilde{\rho}^{prob}_{\lambda^*}$ reaches this rate (up to log terms). This rate is actually $\left(\frac{3}{2}\right)^n
\frac{   rd}{N}$ and the best lower bound known in this case is $\frac{rd}{N}$ [@butucea2015spectral] (we remind that $d=2^n$).

The 