SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Furthermore, as  $(\lambda_n^* - \lambda_n) / \lambda_n \to 0$ , we obtain  $|\Delta v_n^*| / |\Delta v_n - \Delta v_n^*| \to \infty$  and  $|\Delta v_n^*| / |\Delta \mu_n - \Delta \mu_n^*|^2 \to \infty$ , i.e.,  $\Delta v_n / \Delta v_n^* \to 1$ . With all of these results, we can check that  $\overline{\mathcal{Q}}(G_n, G_{*,n}) \lesssim (\lambda_n^* - \lambda_n)|\Delta v_n^*|^2$ . Denote  $(\Delta v_n - \Delta v_n^*) = k_1^n |\Delta v_n^*$ ,  $(\Delta \mu_n - \Delta \mu_n^*) = k_2^n |\Delta v_n^*|^{1/2}$ , and  $\Delta \mu_n^* = k_3^n |\Delta v_n^*|^{1/2}$  for all  $n$ . From all the assumptions we have thus far, we get  $k_1^n \to 0$ ,  $k_2^n \to 0$ , and  $|k_3^n| \not\to \infty$ . Additionally, as  $B \not\to 0$ , we further have  $\lambda_n |k_1^n| / (\lambda_n^* - \lambda_n) \not\to \infty$  and  $\lambda_n (k_2^n)^2 / (\lambda_n^* - \lambda_n) \not\to \infty$ . By dividing both the numerator and the denominator of  $F_3$  and  $F_4$  respectively by  $(\lambda_n^* - \lambda_n) |\Delta v_n^*|^{3/2}$  and  $(\lambda_n^* - \lambda_n)|\Delta v_n^*|^2$ , as the new denominators of  $F_3, F_4$  do not go to infinity, we obtain the new numerators of these terms go to 0, i.e., the following holds
$$
\left\{- (k _ {3} ^ {n}) ^ {3} + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (k _ {2} ^ {n}) ^ {3} \right\} / 6 + \left\{k _ {3} ^ {n} + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} k _ {1} ^ {n} k _ {2} ^ {n} \right\} / 2 \rightarrow 0,
$$
$$
\left\{(k _ {3} ^ {n}) ^ {4} + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (k _ {2} ^ {n}) ^ {4} \right\} / 2 4 + \Bigg \{- (k _ {3} ^ {n}) ^ {2} + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} k _ {1} ^ {n} (k _ {2} ^ {n}) ^ {2} \Bigg \} / 4 + \Bigg \{1 + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (k _ {1} ^ {n}) ^ {2} \Bigg \} / 8 \rightarrow 0.
$$
With the assumptions with  $k_1^n, k_2^n$ , and  $k_3^n$ , we would have
$$
\frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (k _ {2} ^ {n}) ^ {i} \rightarrow 0, \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} k _ {1} ^ {n} (k _ {2} ^ {n}) ^ {j} \rightarrow 0, \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (k _ {1} ^ {n}) ^ {2} \rightarrow 0
$$
for any  $3 \leq i \leq 4$  and  $1 \leq j \leq 2$ . If we denote  $k_3^n \to k_3$ , by combining all the above results we achieve the following system of equations
$$
- k _ {3} ^ {3} / 6 + k _ {3} / 2 = 0, k _ {3} ^ {4} / 2 4 - k _ {3} ^ {2} / 4 + 1 / 8 = 0,
$$
which does not admit a solution, a contradiction. Hence, Case 2.1.1.2 cannot hold.
Case 2.1.2:  $M_n' = |\Delta v_n|$  for all  $n$  (by the subsequence argument). From (27), we would have  $|\Delta v_n - \Delta v_n^*| / |\Delta v_n| \to 0$ , i.e.,  $\Delta v_n / \Delta v_n^* \to 1$ , and  $|\Delta \mu_n - \Delta \mu_n^*|^2 / |\Delta v_n| \to 0$ . The argument under this case is rather similar to that of Case 2.1; therefore, we only sketch the key steps. By using the result that  $A_1 \to 0$  and  $A_2 \to 0$ , we would obtain that
$$
\begin{array}{l} \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n} ^ {*}) ^ {4} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {4}}{2 4 \overline {{Q}} (G _ {n} , G _ {\ast , n})} \to \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \Big [ (\Delta \mu_ {n}) ^ {2} - 3 \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} + 3 (\Delta \mu_ {n} ^ {*}) ^ {2} \Big ]}{2 4 \overline {{Q}} (G _ {n} , G _ {\ast , n})}, \\ \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta v _ {n} ^ {*}) ^ {2} + \lambda_ {n} (\Delta v _ {n} - \Delta v _ {n} ^ {*}) ^ {2}}{8 \overline {{Q}} (G _ {n} , G _ {\ast , n})} \to \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} ^ {*} [ \Delta \mu_ {n} \Delta v _ {n} - \Delta \mu_ {n} ^ {*} \Delta v _ {n} - \Delta u _ {n} \Delta v _ {n} ^ {*} ]}{8 \overline {{Q}} (G _ {n} , G _ {\ast , n})}, \\ \frac {\lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {2} (\Delta v _ {n} - \Delta v _ {n} ^ {*})}{4 \overline {{Q}} (G _ {n} , G _ {*, n})} \rightarrow \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) [ \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \Delta v _ {n} ^ {*} - \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \Delta v _ {n} + \Delta v _ {n} \Delta v _ {n} ^ {*} ]}{4 \overline {{Q}} (G _ {n} , G _ {*, n})}. \\ \end{array}
$$
As  $F_{4}\to 0$  , we equivalently have
$$
\begin{array}{l} {A _ {4}} {:=} {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Bigg \{\frac {\Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \bigg [ (\Delta \mu_ {n}) ^ {2} - 3 \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} + 3 (\Delta \mu_ {n} ^ {*}) ^ {2} \bigg ]}{2 4 \overline {{\mathcal {Q}}} (G _ {n} , G _ {*, n})}} \\ + \left. \frac {\Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \Delta v _ {n} - 2 (\Delta \mu_ {n} ^ {*}) ^ {2} \Delta v _ {n} - \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \Delta v _ {n} ^ {*} + \Delta v _ {n} \Delta v _ {n} ^ {*}}{8 \overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \right\}\rightarrow 0. \\ \end{array}
$$
Under Case 2.1.2, we only consider the setting when  $(\Delta \mu_n)^2 /\Delta v_n\to 0$  as other settings of this term can be argued in the similar fashion as that of Case 2.1.2. Since  $|\Delta \mu_{n} - \Delta \mu_{n}^{*}|^{2} / |\Delta v_{n}|\to 0$
we have  $(\Delta \mu_n^*) / \Delta v_n\to 0$  . As  $\Delta v_{n} / \Delta v_{n}^{*}\rightarrow 1$  , we also further have that  $(\Delta \mu_n^*)^2 /\Delta v_n^*\to 0$  and  $(\Delta \mu_n)^2 /\Delta v_n\to 0$  . Therefore, we have  $\Delta \mu_n\Delta \mu_n^* /\Delta v_n\to 0$  and  $\Delta \mu_n\Delta \mu_n^* /\Delta v_n^*\to 0$  . Now, from the formulation of  $\overline{Q} (G_n,G_{*,n})$  , we achieve
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} \right| ^ {2} / \bar {\mathcal {Q}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta \mu_ {n} \right| ^ {2} / \left| \Delta v _ {n} ^ {*} \right| ^ {2} \rightarrow 0,
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta v _ {n} \left(\Delta \mu_ {n} ^ {*}\right) ^ {2} \right| / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta \mu_ {n} ^ {*} \right| ^ {2} / \left| \Delta v _ {n} ^ {*} \right|\rightarrow 0,
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta v _ {n} ^ {*} \left(\Delta \mu_ {n}\right) ^ {2} / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta \mu_ {n} \right| ^ {2} / \left| \Delta v _ {n} \right|\rightarrow 0, \right.
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta \mu_ {n} \right| ^ {3} \left| \Delta \mu_ {n} ^ {*} \right| / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta \mu_ {n} \right|\left| \Delta \mu_ {n} ^ {*} \right| / \left| \Delta v _ {n} ^ {*} \right|\rightarrow 0,
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta \mu_ {n} \right|\left| \Delta \mu_ {n} ^ {*} \right| ^ {3} / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta \mu_ {n} \right|\left| \Delta \mu_ {n} ^ {*} \right| / \left| \Delta v _ {n} \right|\rightarrow 0.
$$
Combining these results with  $A_4 \to 0$ , we achieve  $(\lambda_n^* - \lambda_n)\Delta v_n\Delta v_n^* / \overline{\mathcal{Q}}(G_n, G_{*,n}) \to 0$ . From here, we can easily verify that all the results in (31) hold. Thus, by using the same argument as that of Case 2.1.1, we would get  $\overline{\mathcal{Q}}(G_n, G_{*,n}) / \overline{\mathcal{Q}}(G_n, G_{*,n}) \to 0$ , a contradiction. As a consequence, Case 2.1.2 cannot happen.
Case 2.2:  $\lambda_{n}^{*} / \lambda_{n}\to \infty$ . Remind that  $M_n^\prime = \max \left\{\left|\Delta \mu_n\right|^2,\left|\Delta v_n\right|,\left|\Delta \mu_n^*\right|^2,\left|\Delta v_n^*\right|\right\}$ . We can verify that  $\overline{\mathcal{Q}} (G_n,G_{*,n})\lesssim \lambda_n^* (M_n')^4$ . By dividing both the numerator and the denominator of  $A_{1}$  and  $A_{2}$  respectively by  $\lambda_{n}^{*}(M_{n}^{\prime})^{1 / 2}$  and  $\lambda_{n}^{*}M_{n}^{\prime}$ , given that the new denominators go to 0 we would obtain the new numerators also go to 0, i.e., we have the following results
$$
\lambda_ {n} \Delta \mu_ {n} ^ {n} / \left\{\lambda_ {n} ^ {*} (M _ {n} ^ {\prime}) ^ {1 / 2} \right\} - \Delta \mu_ {n} ^ {*} / (M _ {n} ^ {\prime}) ^ {1 / 2} \rightarrow 0,
$$
$$
\left[ (\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} + \lambda_ {n} \Delta v _ {n} - \lambda_ {n} ^ {*} \Delta v _ {n} ^ {*} \right] / \{\lambda_ {n} ^ {*} M _ {n} ^ {\prime} \} \rightarrow 0.
$$
Since  $\lambda_n / \lambda_n^* \to 0$ , the first limit implies that  $\Delta \mu_n^* / M_n' \to 0$ . Combining this result with the second limit, we obtain  $\Delta v_n^* / M_n' \to 0$ . Therefore, we would have  $M_n' = \max \left\{|\Delta \mu_n|^2, |\Delta v_n|\right\}$ . Without loss of generality, we assume that  $M_n' = |\Delta \mu_n|^2$  as the argument for other possibility of  $M_n'$  can be argued in the similar fashion. With these assumptions,  $|\Delta v_n - \Delta v_n^*| / |\Delta \mu_n|^2 \nrightarrow \infty$ , i.e., as  $n$  is large enough we get  $|\Delta v_n - \Delta v_n^*| \lesssim |\Delta \mu_n|^2$ . Now, we have two distinct cases
Case 2.2.1:  $\lambda_{n}^{*}\max \left\{|\Delta \mu_{n}^{*}|^{2},|\Delta v_{n}^{*}|\right\} /\left(\lambda_{n}|\Delta \mu_{n}|^{2}\right)\to \infty$  . Due to this assumption, we can check that as  $n$  is large enough,  $\overline{\mathcal{Q}} (G_n,G_{*,n})\asymp \lambda_n^* |\Delta \mu_n|^2\max \left\{|\Delta \mu_n^* |^2,|\Delta v_n^* |\right\}$  . If  $\max \left\{|\Delta \mu_n^* |^2,|\Delta v_n^* |\right\} = |\Delta \mu_n^* |^2$  for all  $n$  , then by dividing both the numerator and denominator of  $A_{1}$  by  $\lambda_{n}^{*}\Delta \mu_{n}^{*}$  , given that the new denominator of  $A_{1}$  goes to O, its new numerator must go to 0,i.e.,we have
$$
\lambda_ {n} \Delta \mu_ {n} / (\lambda_ {n} ^ {*} \Delta \mu_ {n} ^ {*}) \rightarrow 1,
$$
which cannot hold since  $\lambda |\Delta \mu_n|^2 /(\lambda_n^* |\Delta \mu_n^* |^2)\to 0$  (assumption of Case 2.2.1) and  $|\Delta \mu_n| / |\Delta \mu_n^* |\to$ $\infty$ . Therefore, we must have  $\max \left\{|\Delta \mu_n^* |^2,|\Delta v_n^* |\right\} = |\Delta v_n^* |$  for all  $n$ . By dividing both the numerator and denominator of  $A_{2}$  by  $\lambda_n^*\Delta v_n^*$ , as the new denominator of  $A_{2}$  goes to 0, we would have
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*}}{\lambda_ {n} ^ {*} \Delta v _ {n} ^ {*}} + \frac {\lambda_ {n} \Delta v _ {n}}{\lambda_ {n} ^ {*} \Delta v _ {n} ^ {*}} - 1 \rightarrow 0.
$$
Since  $\frac{\lambda_n|\Delta v_n|}{\lambda_n^*|\Delta v_n^*|} \leq \frac{\lambda_n|\Delta\mu_n|^2}{\lambda_n^*|\Delta v_n^*|} \to 0$  and  $(\lambda_{n}^{*} - \lambda_{n}) / \lambda_{n}^{*} \to 1$ , the above limit shows that  $\Delta \mu_{n}\Delta \mu_{n}^{*} / \Delta v_{n}^{*} \to 1$ . Since  $(\Delta \mu_{n})^{2} / |\Delta v_{n}^{*}| \to \infty$ , it implies that  $(\Delta \mu_{n}^{*})^{2} / \Delta v_{n}^{*} \to 0$ . Now, by combing the result that  $A_{1} \to 0$  and  $A_{2} \to 0$ , since  $F_{3} \to 0$ , we can verify that it is equivalent to
$$
A _ {3} := \frac {\left[ (\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} (\Delta \mu_ {n} - 2 \Delta \mu_ {n} ^ {*}) \right] / 3 + (\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} ^ {*} \Delta v _ {n}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \to 0.
$$
By dividing both the numerator and the denominator of  $A_{3}$  by  $\lambda_n^*\Delta \mu_n\Delta v_n^*$ , we obtain
$$
\frac {\left[ (\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} (\Delta \mu_ {n} - 2 \Delta \mu_ {n} ^ {*}) \right] / 3 + (\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} ^ {*} \Delta v _ {n}}{\lambda_ {n} ^ {*} \Delta \mu_ {n} \Delta v _ {n} ^ {*}} \rightarrow 0.
$$
As  $(\Delta \mu_n^*)^2 /\Delta v_n^*\to 0$  and  $\Delta \mu_{n}\Delta \mu_{n}^{*} / \Delta v_{n}^{*}\rightarrow 1$ , the above limit leads to  $\Delta \mu_n^*\Delta v_n / (\Delta \mu_n\Delta v_n^*)\to -1 / 3$ . Now, by studying  $A_4\to 0$  with the assumption that  $\overline{\mathcal{Q}} (G_n,G_{*,n})\asymp \lambda_n^* |\Delta \mu_n|^2 |\Delta v_n^*|$ , we eventually get the equation  $1 / 24 - 1 / 12 = 0$ , which is a contradiction. Therefore, Case 2.2.1 cannot hold.
Case 2.2.2:  $\lambda_{n}^{*}\max \left\{\left|\Delta \mu_{n}^{*}\right|^{2},\left|\Delta v_{n}^{*}\right|\right\} /\lambda_{n}\left|\Delta \mu_{n}\right|^{2}\not\to\infty$ . Therefore, as  $n$  is large enough, we would have  $\lambda_{n}^{*}\max \left\{\left|\Delta \mu_{n}^{*}\right|^{2},\left|\Delta v_{n}^{*}\right|\right\} \lesssim (\lambda_{n}|\Delta \mu_{n}|^{2})$ . Hence, we achieve under this case that  $\overline{\mathcal{Q}} (G_n,G_{*,n})\asymp \lambda_n|\Delta \mu_n|^4$ . Denote  $\Delta \mu_n^* = l_1^n\Delta \mu_n$ ,  $\Delta v_{n} = l_{2}^{n}(\Delta \mu_{n})^{2}$ , and  $\Delta v_{n}^{*} = l_{3}^{n}(\Delta \mu_{n})^{2}$ . From the assumptions of Case 2.2.2, we would have  $l_1^n\rightarrow 0$  and  $l_3^n\rightarrow 0$  while  $l_2^n\neq \infty$ . Additionally,  $\lambda_{n}^{*}\max \left\{(l_{1}^{n})^{2},|l_{3}^{n}|\right\} /\lambda_{n}\neq 0$ . By dividing the numerators and denominators of  $A_{i}$  by  $\lambda_{n}(\Delta \mu_{n})^{i}$  for  $1\leq i\leq 3$ , we achieve the following system of limits
$$
\frac {\lambda_ {n} ^ {*} l _ {1} ^ {n}}{\lambda_ {n}} - 1 \rightarrow 0, \frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) l _ {1} ^ {n}}{\lambda_ {n}} + l _ {2} ^ {n} - \frac {\lambda_ {n} ^ {*} l _ {3} ^ {n}}{\lambda_ {n}} \rightarrow 0, \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} \left\{\frac {l _ {1} ^ {n} - \left(l _ {1} ^ {n}\right) ^ {2}}{3} + l _ {1} ^ {n} l _ {2} ^ {n} \right\}\rightarrow 0. \tag {33}
$$
As  $l_1^n \to 0$ , the first limit in the above system implies that  $\lambda_n^*(l_1^n)^2 / \lambda_n \to 0$ . If we have  $\max \left\{ (l_1^n)^2, |l_2^n| \right\} = |l_1^n|^2$  for all  $n$ , the previous result would mean that  $\lambda_n^* l_3^n / \lambda_n \to 0$ . Therefore, the second limit in (33) demonstrates that  $l_2^n \to -1$ . However, plugging these results to the third limit in this system would yield  $1/3 - 1 = 0$ , which is a contradiction. Hence, we must have  $\max \left\{ (l_1^n)^2, |l_2^n| \right\} = |l_3^n|$  for all  $n$ . Under this setting, by denoting  $\frac{\lambda_n^* l_3^n}{\lambda_n} \to a$  as  $n \to \infty$ , the first and second limit in (33) leads to  $l_2^n \to a - 1$ . With this result, the third limit in this system shows that  $a = 2/3$ . With these results, by dividing both the numerator and denominator of  $A_4$  by  $\lambda_n (\Delta \mu_n)^4$ , we quickly achieve the equation  $1/24 - 5/72 = 0$ , which is a contradiction. Therefore, Case 2.2.2 cannot hold.
In sum, not all the coefficients of  $\frac{\partial^{|\beta|}f}{\partial\mu^{\beta}} (x|\mu_n^*,v_n^*)$  as  $1\leq |\beta |\leq 8$  go to 0. From here, by using the same argument as that of Proposition C.1 and Proposition C.3, we achieve the result of part (b) of the proposition. As a consequence, we reach the conclusion of the theorem.
D Proofs for Convergence Rates of Parameter Estimation and Minimax Lower Bounds
In this appendix, we provide the proofs for the convergence rates of the MLE as well as the corresponding minimax lower bounds introduced in Section D.
(a) For any  $G_{1} = G_{1}(\lambda_{1},\mu_{1},\Sigma_{1})$  and  $G_{2} = G_{2}(\lambda_{2},\mu_{2},\Sigma_{2})$ , we denote the following distance
$$
d _ {1} \left(G _ {1}, G _ {2}\right) = \lambda_ {1} \left\| \left(\mu_ {1}, \Sigma_ {1}\right) - \left(\mu_ {2}, \Sigma_ {2}\right) \right\|,
$$
$$
d _ {2} \left(G _ {1}, G _ {2}\right) = \left| \lambda_ {1} - \lambda_ {2} \right| ^ {2}.
$$
Even though  $d_{2}(G_{1}, G_{2})$  is a proper distance, it is clear that  $d(G_{1}, G_{2})$  is not symmetric and only satisfies a weak triangle inequality, i.e. we have
$$
d _ {1} \left(G _ {1}, G _ {3}\right) + d _ {1} \left(G _ {2}, G _ {3}\right) \geq \min  \left\{d _ {1} \left(G _ {1}, G _ {2}\right), d _ {1} \left(G _ {2}, G _ {1}\right) \right\}.
$$
Therefore, we will utilize the modification of Le Cam method for nonsymmetric loss in Lemma 6.1 of [16] to deal with such distance. We start with the following proposition
Proposition D.1. Given that  $f$  satisfies assumption (S.1) in Theorem 4.1, we achieve for any  $r < 1$  that
(i)  $\lim_{\epsilon \to 0}\inf_{G_1 = (\lambda ,\mu_1,\Sigma_1),G_2 = (\lambda ,\mu_2,\Sigma_2)}\{h(p_{G_1},p_{G_2}) / d_1^r (G_1,G_2):d_1(G_1,G_2)\leq \epsilon \} = 0.$
(ii)  $\lim_{\epsilon \to 0}\inf_{G_1 = (\lambda_1,\mu ,\Sigma),G_2 = (\lambda_2,\mu ,\Sigma)}\{h(p_{G_1},p_{G_2}) / d_2^r (G_1,G_2):d_2(G_1,G_2)\leq \epsilon \} = 0.$
Proof. (i) For any sequences  $G_{1,n} = (\lambda_n, \mu_{1,n}, \Sigma_{1,n})$  and  $G_{2,n} = (\lambda_n, \mu_{2,n}, \Sigma_{2,n})$ , we have
$$
\begin{array}{l} h ^ {2} \left(p _ {G _ {1, n}}, p _ {G _ {2, n}}\right) \leq \frac {1}{\lambda_ {n}} \int \frac {\left(p _ {G _ {1 , n}} (x) - p _ {G _ {2 , n}} (x)\right) ^ {2}}{f \left(x \mid \mu_ {2 , n} , \Sigma_ {2 , n}\right)} d x \\ = \lambda_ {n} \int \frac {\left(f (x | \mu_ {1 , n} , \Sigma_ {1 , n}) - f (x | \mu_ {2 , n} , \Sigma_ {2 , n})\right) ^ {2}}{f (x | \mu_ {2 , n} , \Sigma_ {2 , n})} d x \\ \end{array}
$$
where the first inequality is due to  $\sqrt{p_{G_{1,n}}(x)} + \sqrt{p_{G_{2,n}}(x)} > \sqrt{\lambda_n f(x|\mu_{2,n},\Sigma_{2,n})}$ . By Taylor expansion up to the first order, we have
$$
\begin{array}{l} f (x | \mu_ {1, n}, \Sigma_ {1, n}) - f (x | \mu_ {2, n}, \Sigma_ {2, n}) = \sum_ {| \alpha | = 1} \frac {\left(\mu_ {1 , n} - \mu_ {2 , n}\right) ^ {\alpha_ {1}} \left(\Sigma_ {1 , n} - \Sigma_ {2 , n}\right) ^ {\alpha_ {2}}}{\alpha_ {1} ! \alpha_ {2} !} \frac {\partial f}{\partial \mu^ {\alpha_ {1}} \partial \Sigma^ {\alpha_ {2}}} (x | \mu_ {2, n}, \Sigma_ {2, n}) \\ + \sum_ {| \alpha | = 1} \frac {(\mu_ {1 , n} - \mu_ {2 , n}) ^ {\alpha_ {1}} (\Sigma_ {1 , n} - \Sigma_ {2 , n}) ^ {\alpha_ {2}}}{\alpha_ {1} ! \alpha_ {2} !} \int_ {0} ^ {1} \frac {\partial f}{\partial \mu^ {\alpha_ {1}} \partial \Sigma^ {\alpha_ {2}}} (x | \mu_ {2, n} + t (\mu_ {1, n} - \mu_ {2, n}), \Sigma_ {2, n} + t (\Sigma_ {1, n} - \Sigma_ {2, n})) d t \\ \end{array}
$$
Now, by choosing  $\lambda_{n}^{1 - 2r}\| (\mu_{1,n},\Sigma_{1,n}) - (\mu_{2,n},\Sigma_{2,n})\|^{2 - 2r}\to 0$  and  $\| (\mu_{1,n},\Sigma_{1,n})-$ $(\mu_{2,n},\Sigma_{2,n})\| \rightarrow 0$  and using condition (S.1), we can easily verify that  $h(p_{G_{1,n}},p_{G_{2,n}}) / d_1^r (G_{1,n},G_{2,n})\to 0.$  Therefore, we achieve the conclusion of part (i).
(ii) The argument for this part is essentially similar to that in part (i). In fact, for any two sequences  $G_{1,n}^{\prime} = (\lambda_{1,n},\mu_n,\Sigma_n)$  and  $G_{2,n}^{\prime} = (\lambda_{2,n},\mu_n,\Sigma_n)$ , we also obtain
$$
\begin{array}{l} \frac {h ^ {2} \left(p _ {G _ {1 , n} ^ {\prime}} , p _ {G _ {2 , n} ^ {\prime}}\right)}{d _ {2} ^ {2 r} \left(G _ {1 , n} ^ {\prime} , G _ {2 , n} ^ {\prime}\right)} \leq \frac {\left(\lambda_ {1 , n} - \lambda_ {2 , n}\right) ^ {2 - 2 r}}{\left(1 - \lambda_ {1 , n}\right) \wedge \lambda_ {1 , n}} \int \frac {\left(h _ {0} \left(x | \mu_ {0} , \Sigma_ {0}\right) - f \left(x | \mu_ {n} , \Sigma_ {n}\right)\right) ^ {2}}{h _ {0} \left(x | \mu_ {0} , \Sigma_ {0}\right) + f \left(x | \mu_ {n} , \Sigma_ {n}\right)} d x \\ \leq \frac {2 (\lambda_ {1 , n} - \lambda_ {2 , n}) ^ {2 - 2 r}}{(1 - \lambda_ {1 , n}) \wedge \lambda_ {1 , n}} \\ \end{array}
$$
By choosing  $(\lambda_{1,n} - \lambda_{2,n})^{2 - 2r} / \{(1 - \lambda_{1,n})\wedge \lambda_{1,n}\} \to 0$  , we also achieve the conclusion of part (ii).
Now, given  $G_{*} = (\lambda^{*},\mu^{*},\Sigma^{*})$  and  $r < 1$ . Let  $C_0$  be any fixed constant. According to part (i) of Proposition D.1, for any sufficiently small  $\epsilon > 0$ , there exists  $G_{*}' = (\lambda^{*},\mu_{1}^{*},\Sigma_{1}^{*})$  such that  $d_{1}(G_{*},G_{*}') = d_{1}(G_{*}',G_{*}) = \epsilon$  and  $h(p_{G_*},p_{G_*'}) \leq C_0\epsilon^r$ . By means of Lemma 6.1 of [16], we achieve
$$
\inf  _ {\widehat {G} _ {n} \in \Xi} \sup  _ {G \in \Xi} \mathbb {E} _ {p _ {G}} \left(\lambda^ {2} \| (\widehat {\mu} _ {n}, \widehat {\Sigma} _ {n}) - (\mu , \Sigma) \| ^ {2}\right) \geq \frac {\epsilon^ {2}}{2} \left(1 - V (p _ {G _ {*}} ^ {n}, p _ {G _ {*} ^ {\prime}} ^ {n})\right).
$$
where  $p_{G_*}^n$  denotes the density of the  $n$ -iid sample  $X_1, \ldots, X_n$ . From there,
$$
\begin{array}{l} V \left(p _ {G _ {*}} ^ {n}, p _ {G _ {*} ^ {\prime}} ^ {n}\right) \leq h \left(p _ {G _ {*}} ^ {n}, p _ {G _ {*} ^ {\prime}} ^ {n}\right) \\ = \sqrt {1 - \left(1 - h ^ {2} \left(p _ {G _ {*}}, p _ {G _ {*} ^ {\prime}}\right)\right) ^ {n}} \\ \leq \sqrt {1 - (1 - C _ {0} ^ {2} \epsilon^ {2 r}) ^ {n}}. \\ \end{array}
$$
$$
\inf  _ {\widehat {G} _ {n} \in \Xi} \sup  _ {G \in \Xi} \mathbb {E} _ {p _ {G}} \left(\lambda^ {2} \| (\widehat {\mu} _ {n}, \widehat {\Sigma} _ {n}) - (\mu , \Sigma) \| ^ {2}\right) \geq \frac {\epsilon^ {2}}{2} \sqrt {1 - (1 - C _ {0} ^ {2} \epsilon^ {2 r}) ^ {n}}.
$$
By choosing  $\epsilon^{2r} = \frac{1}{C_0^2n}$ , we achieve
$$
\inf _ {\widehat {G} _ {n} \in \Xi} \sup _ {G \in \Xi} \mathbb {E} _ {p _ {G}} \left(\lambda^ {2} \| (\widehat {\mu} _ {n}, \widehat {\Sigma} _ {n}) - (\mu , \Sigma) \| ^ {2}\right) \geq c _ {1} n ^ {- 1 / r}.
$$
for any  $r < 1$  where  $c_{1}$  is some positive constant. Using the similar argument, with the result of (ii) in Proposition D.1 we also immediately obtain the result  $\inf_{\widehat{G}_n\in \Xi}\sup_{G\in \Xi}\mathbb{E}_{p_G}\left(|\widehat{\lambda}_n - \lambda |^2\right)\geq c_2n^{-1 / r}.$  As a consequence, we reach the conclusion of part (a) of the theorem.
(b) The proof of this part is a direct consequence of Theorem 3.3 and Theorem 2.3. Indeed, for  $\widehat{G}_n = (\widehat{\lambda}_n,\widehat{\mu}_n,\widehat{\Sigma}_n)$  being the MLE as in equation (3), we have
$$
\begin{array}{l} \mathbb {E} _ {p _ {G _ {*}}} \left(\left| \hat {\lambda} _ {n} - \lambda^ {*} \right| + \lambda^ {*} \| \left(\widehat {\mu} _ {n}, \widehat {\Sigma} _ {n}\right) - \left(\mu^ {*}, \Sigma^ {*}\right) \|\right) \stackrel {{\text {T h m 3 . 3}}} {{\lesssim}} \mathbb {E} _ {p _ {G _ {*}}} V \left(p _ {\widehat {G} _ {n}}, p _ {G _ {*}}\right) \leq \mathbb {E} _ {p _ {G _ {*}}} h \left(p _ {\widehat {G} _ {n}}, p _ {G _ {*}}\right) \\ \stackrel {\text {T h m 2 . 3}} {\lesssim} \frac {\log n}{\sqrt {n}} \\ \end{array}
$$
Because all inequalities are uniform in  $G_{*}$ , we achieve the conclusion of part (b) of the theorem.
D.2 Proof of Theorem A.1
(a) Similar to the proof argument of part (a) of Theorem 4.1, we define
$$
d _ {3} \left(G _ {1}, G _ {2}\right) = \lambda_ {1} \| \left(\Delta \mu_ {1}, \Delta \Sigma_ {1}\right) \| \| \left(\mu_ {1}, \Sigma_ {1}\right) - \left(\mu_ {2}, \Sigma_ {2}\right) \|,
$$
$$
d _ {4} \left(G _ {1}, G _ {2}\right) = | \lambda_ {1} - \lambda_ {2} | \| (\Delta \mu_ {1}, \Delta \Sigma_ {1}) \| ^ {2}.
$$
for any  $G_{1} = G_{1}(\lambda_{1},\mu_{1},\Sigma_{1})$  and  $G_{2} = G_{2}(\lambda_{2},\mu_{2},\Sigma_{2})$ . It is clear that both  $d_{3}(G_{1},G_{2})$  and  $d_{4}(G_{1},G_{2})$  still satisfy weak triangle inequality. To achieve the conclusion of this part, it suffices to demonstrate the following results
(i) There exists two sequences  $G_{1,n} = (\lambda_n, \mu_{1,n}, \Sigma_{1,n}) \in \Xi_1(l_n)$  and  $G_{2,n} = (\lambda_n, \mu_{2,n}, \Sigma_{2,n}) \in \Xi_1(l_n)$  such that  $d_3(G_{1,n}, G_{2,n}) \to 0$  and  $h(p_{G_{1,n}}, p_{G_{2,n}}) / d_3^r(G_{1,n}, G_{2,n})$  as  $n \to \infty$ .
(ii) There exists two sequences  $G_{1,n}^{\prime} = (\lambda_{1,n}, \mu_n, \Sigma_n) \in \Xi_1(l_n)$  and  $G_{2,n}^{\prime} = (\lambda_{2,n}, \mu_n, \Sigma_n) \in \Xi_1(l_n)$  such that  $d_4(G_{1,n}, G_{2,n}) \to 0$  and  $h(p_{G_{1,n}^{\prime}}, p_{G_{2,n}^{\prime}}) / d_4^r(G_{1,n}, G_{2,n})$  as  $n \to \infty$ .
for any  $r < 1$ . The proof argument for the above results can proceed in a similar fashion as that of Proposition D.1; therefore, it is omitted. We achieve the conclusion of part (a) of the theorem.
(b) Combining the result of Theorem 3.5 and the fact that  $D(G, G_{*}) \asymp \overline{D}(G, G_{*})$  for any  $G$  and  $G_{*}$ , we immediately achieve the following convergence rates