next theorem presents the square error bound of the second estimator $\tilde{\rho}^{dens}_{\lambda}$. Here again, see the appendix for the proof.

\[rate 2\] Fix a small $\epsilon \in (0,1) $. Under , for $ \lambda = \lambda^*: = \frac{N}{5^{n}4}$, with probability at least $ 1-\epsilon$, $$\|\tilde{\rho}^{dens}_{\lambda^*} -\rho^0\|_F^2
\leq
C^{dens}_{D_1,D_2} \frac{10^n {\rm rank}(\rho^0) \log\left(
\frac{{\rm rank}(\rho^0) N}{2^n}
\right)
+ 5^n \log(2/\varepsilon)}{N}$$ where $ C^{dens}_{D_1,D_2}$ is a constant that depends only on $D_1,D_2$.

The guarantee for $\tilde{\rho}^{dens}_{\lambda^*}$ is far less satisfactory. However, as this estimator is easier to compute, we think it is interesting to provide a convergence rate, even if it is far from optimal: note that for a fixed $d$, the bound goes to $0$ when $m\rightarrow \infty$.

Experiments show that $ \lambda = \lambda^*: = \frac{N}{5^{n}4}$ is actually not the best choice for dens-estimator. The choice $ \lambda = \frac{N}{4}$ (heuristically motivated by [@dalalyan2008aggregation]) leads to results comparable to the prob-estimator in Section \[num\]. This leads to the conjecture that the rate of $ \tilde{\rho}^{dens}_{N/4} $ is much better than $  \frac{10^n {\rm rank}(\rho^0) }{N} $ but this is still an open question.

Numerical Experiments {#num}
=====================

Metropolis-Hastings Implementation
----------------------------------

We implement the two proposed estimators via the Metropolis-Hasting (MH) 