SlowGuess/NeurIPS_2023_OCR huggingface dataset:
If  $q_{|T|-m_1} - q_{m_1} \leq R$  and  $\mathbf{E}_{X \sim T}[f(x)] > C'R^2/\alpha^3$  for  $C' \geq 720/\epsilon_0$ , that is, in the case where the check in Line 12 fails, then, the function  $f(\cdot)$  satisfies
$$
\sum_ {\widetilde {x} \in \widetilde {T}} f (\widetilde {x}) > \frac {4 0}{\epsilon_ {0}} \frac {1}{\alpha^ {3}} \sum_ {\widetilde {x} \in \widetilde {S} \cap \widetilde {T}} f (\widetilde {x}).
$$
Proof. The core argument for the proof is that  $\sum_{\widetilde{x} \in \widetilde{S} \cap \widetilde{T}} f(\widetilde{x})$  can be upper bounded under the assumption that  $q_{|T| - m_1} - q_{m_1} \leq R$ , as shown in the following claim.
Claim E.5. Assuming the conditions in Lemma E.4, we have  $\sum_{\widetilde{x}\in \widetilde{S}\cap \widetilde{T}}f(\widetilde{x})\leq 9|S|R^2$
Proof. Let  $\mu' = \mathbf{E}_{\widetilde{X} \sim \widetilde{S}}[\widetilde{X}^\top A \widetilde{X}]$ . We have the following series of inequalities:
$$
\begin{array}{l} \sum_ {\widetilde {x} \in \widetilde {S} \cap \widetilde {T}} f (\widetilde {x}) \leq \sum_ {\widetilde {x} \in \widetilde {S}} f (\widetilde {x}) \quad \left(\text {s i n c e} f (x) \geq 0\right) \\ = \sum_ {\widetilde {x} \in \widetilde {S}} \left(\widetilde {x} ^ {\top} A \widetilde {x} - \widetilde {y} _ {\text {m e d i a n}}\right) ^ {2} \\ \end{array}
$$
$$
\begin{array}{l} \leq 2 \sum_ {\tilde {x} \in \tilde {S}} \left(\left(x ^ {\top} A x - \mu^ {\prime}\right) ^ {2} + \left(\mu^ {\prime} - \tilde {y} _ {\text {m e d i a n}}\right) ^ {2}\right) \quad (\text {b y} (a + b) ^ {2} \leq 2 a ^ {2} + 2 b ^ {2}) \\ = 2 | S | \underset {\widetilde {X} \sim \widetilde {S}} {\mathbf {V a r}} [ \widetilde {X} ^ {\top} A \widetilde {X} ] + 2 \sum_ {\widetilde {x} \in \widetilde {S}} (\mu^ {\prime} - \widetilde {y} _ {\text {m e d i a n}}) ^ {2} \\ \leq 2 | S | \underset {\widetilde {X} \sim \widetilde {S}} {\operatorname {V a r}} \left[ \widetilde {X} ^ {\top} A \widetilde {X} \right] + 8 R ^ {2} | S | \quad (\text {b y}) \\ \leq 2 | S | \left(\frac {7 2 C _ {1}}{\alpha^ {2}}\right) + 8 R ^ {2} | S | \quad (\text {b y}) \\ \leq 9 | S | R ^ {2} , \quad (\text {b y}) \\ \end{array}
$$
where the last line uses that  $R \geq 12\sqrt{C_1} / \alpha$ , which is satisfied for  $C \geq 12\sqrt{C_1}$ . This completes the proof of Claim E.5.
$$
\begin{array}{l} \sum_ {\tilde {x} \in \tilde {T}} f (\tilde {x}) > C ^ {\prime} R ^ {2} | T | / \alpha^ {3} \quad \text {(b y a s s u m p t i o n)} \\ \geq 0. 5 C ^ {\prime} R ^ {2} | S | / \alpha^ {3} \quad \left(\text {s i n c e} | T | \geq | S \cap T | \geq | S | \left(1 - 2 \epsilon_ {0}\right) \geq | S | / 2\right) \\ > \left(3 6 0 / \epsilon_ {0}\right) R ^ {2} | S | / \alpha^ {3} \quad \left(\text {s i n c e} C ^ {\prime} \geq 7 2 0 / \epsilon_ {0}\right) \\ \geq 4 0 / \left(\epsilon_ {0} \alpha^ {3}\right) \sum_ {\widetilde {x} \in \widetilde {S} \cap \widetilde {T}} f (\widetilde {x}). \quad (u s i n g C l a i m E. 5) \\ \end{array}
$$
This completes the proof of Lemma E.4.
F Proof of Theorem 1.2
In this section, we present the main algorithmic result of the paper. As previously stated, the theorem holds for distributions beyond Gaussians, as long as the input set of points satisfies the deterministic conditions from Section 2.1 and the distribution meets a mild requirement. We now restate and prove the theorem in this more general form.
Theorem F.1 (Weak List Decodable Covariance Estimation). Let the ambient dimension be  $d \in \mathbb{Z}_+$ , let  $C$  be a sufficiently large absolute constant, and let the parameters  $\alpha \in (0, 1/2)$ ,  $\epsilon \in (0, \epsilon_0)$  for a sufficiently small positive constant  $\epsilon_0$ , and failure probability  $\delta \in (0, 1/2)$  be known to the algorithm. Let  $D$  be a distribution with mean  $\mu \in \mathbb{R}^d$ , covariance  $\Sigma \in \mathbb{R}^{d \times d}$ , and assume that  $\operatorname{Var}_{X \sim D}[X^\top AX] = O(\| \Sigma^{1/2}A\Sigma^{1/2}\|_F^2 + \| \Sigma^{1/2}A\mu \|_2^2)$ . There is a polynomial-time algorithm such that, on input  $\alpha, \delta$  as well as an  $(\alpha, \epsilon)$ -corruption of a set  $S$  (Definition 1.1) with  $|S| > C / (\alpha^6\epsilon_0^2)\log(1 / (\alpha\delta))$  which is  $((\epsilon_0 / 40)\alpha^3, 2\epsilon_0)$ -stable with respect to  $D$  (Definition 2.1), with probability at least  $1 - \delta$  over the randomness of the algorithm, the algorithm returns a list of at most  $O(1 / \alpha)$  many sets  $T_i$  which are disjoint subsets of samples in the input, each  $T_i$  has size at least  $0.5\alpha m$ , and there exists a  $T_i$  in the output list such that:
- Recall the notation in the corruption model (Definition 1.1) where  $n$  is the size of the original inlier set  $S$  and  $\ell$  is the number of points in  $S$  that the adversary replaced— $n$  and  $\ell$  are unknown to the algorithm except that  $n \geq \alpha m$  and  $\ell \leq \epsilon n$ . The set  $T_i$  satisfies that  $|T_i \cap S| \geq (1 - 0.01\alpha)(n - \ell)$ .
- Denote  $H_{i} \coloneqq \mathbf{E}_{X \sim T_{i}}[XX^{\top}]$ . The matrix  $H_{i}$  satisfies
$$
\max \left(\| (H _ {i} ^ {\dagger}) ^ {1 / 2} \Sigma (H _ {i} ^ {\dagger}) ^ {1 / 2} - \mathbb {P} _ {H} \| _ {\mathrm {F}}, \| (H _ {i} ^ {\dagger}) ^ {1 / 2} \Sigma (H _ {i} ^ {\dagger}) ^ {1 / 2} - \mathbb {P} _ {\Sigma} \| _ {\mathrm {F}}\right) \leq O ((1 / \alpha^ {4}) \log (1 / \alpha)).
$$
The Gaussian distribution  $\mathcal{N}(\mu, \Sigma)$  satisfies the distributional assumption of Theorem F.1 (c.f. Fact B.1), and a set of size  $m = C' \frac{d^2 \log^5(d / \alpha \delta)}{\alpha^6}$  from  $\mathcal{N}(\mu, \Sigma)$  is  $((\epsilon_0 / 40) \alpha^3, 2 \epsilon_0)$ -stable with probability  $1 - \delta$  for a constant  $\epsilon_0$  (c.f. Lemma 2.2). Thus, Theorem F.1 theorem covers the Gaussian
case  $D = \mathcal{N}(\mu, \Sigma)$  with polynomial sample complexity, and it directly implies Theorem 1.2, the main result of this work.
We achieve the guarantee of Theorem F.1 by combining the procedure of the previous section in a recursive algorithm (Algorithm 1). We maintain the notation of Assumption 3.1 with the addition of subscripts that indicate the number of filtering steps performed.
We restate the algorithm below, before proving Theorem F.1:
Algorithm 3 List-Decodable Covariance Estimation in Relative Frobenius Norm
1: Constants:  $m, \alpha, \eta, R \coloneqq C(1 / \alpha^2) \log (1 / (\epsilon_0 \alpha))$  for  $C > 6000\sqrt{C_1}$ ,  $C' > 720 / \epsilon_0$  (where  $C_1, \epsilon_0$  are defined in Assumption 3.1).  
2: function COVLISTDECODING( $T_0$ )  
3:  $t \gets 0$ .  
4: loop  
5: Compute  $H_t = \mathbf{E}_{X \sim T_t}[XX^\top]$ .  
6: Let  $\widetilde{T}_t = \{H_t^{\dagger/2}x : x \in T_t\}$  be the transformed set of samples.  
7: Let  $A$  be the symmetric matrix corresponding to the top eigenvector of  $\mathrm{Cov}_{\widetilde{X} \sim \widetilde{T}_t}[\widetilde{X}^{\otimes 2}]$ .  
8:Normalize  $A$  so that  $\|A\|_{\mathrm{F}} = 1$ .  
9:Compute the set  $\widetilde{Y}_t = \{\widetilde{x}^\top A\widetilde{x} : \widetilde{x} \in \widetilde{T}_t\}$   
10:Compute the  $\alpha m/9$ -th smallest element  $q_{\text{left}}$  as well as the  $\alpha m/9$ -th largest element  $q_{\text{right}}$ , as well as the median  $\widetilde{y}_{\text{median of }}\widetilde{Y}_t$ .  
11:Define the function  $f(\widetilde{x}) = (\widetilde{x}^\top A\widetilde{x} - \widetilde{y}_{\text{median}})^2$ .  
12:if  $\mathbf{E}_{\widetilde{X} \sim \widetilde{T}_t}[f(\widetilde{X})] \leq C'R^2/\alpha^3$  then ▷c.f. Lemma 3.4  
13:If  $|T_t| \geq 0.5\alpha m$  then return  $\{T_t\}$  else return the empty list.  
14:else if  $q_{\text{right}} - q_{\text{left}} \leq R$  then ▷c.f. Lemma 3.5  
15:Let the probability mass function  $p(\widetilde{x}) := f(\widetilde{x}) / \sum_{\widetilde{x} \in \widetilde{T}_t} f(\widetilde{x})$ .  
16:Pick  $x_{\text{removed}} \in T_t$  according to  $p(H_t^{\dagger/2}x)$ .  
17:  $T_{t+1} \gets T_t \setminus \{x_{\text{removed}}\}$ .  
18:else  
19:  $\tau \gets \text{FindDivider}(\widetilde{Y}_t, \alpha m/9)$ . ▷c.f. Lemma 3.6  
20: $T' \gets \{H_t^{1/2}\widetilde{x} : \widetilde{x} \in \widetilde{T}_t, \widetilde{x}^\top A\widetilde{x} \leq \tau\}, T'' \gets \{H_t^{1/2}\widetilde{x} : \widetilde{x} \in T_t, \widetilde{x}^\top A\widetilde{x} > \tau\}$ .  
21:L1← COVLISTDECODING( $T'$ ),  $L_2 \gets$  COVLISTDECODING( $T''$ ).  
22:return  $L_1 \cup L_2$ .  
23:t← t+1.
Proof of Theorem F.1. Since the final error guarantees do not depend on  $\epsilon$ , without loss of generality we use the maximum level of corruptions  $\epsilon = \epsilon_0 = 0.01$  in the following proof. We will also use the letter  $\eta$  to denote the expression  $(\epsilon_0 / 40)\alpha^3$  as in the stability assumption in the theorem statement on the inlier set. We will proceed to prove Theorem F.1 by induction, and crucially use the following fact that "union bounds also work under conditioning".
Fact F.2. If event  $A$  happens with probability  $1 - \tau_{1}$  and event  $B$  happens with probability  $1 - \tau_{2}$  conditioned on event  $A$ , then the probability of both  $A$  and  $B$  happening is at least  $1 - \tau_{1} - \tau_{2}$ .
The technical bulk for proving Theorem F.1 is Lemma F.4, an inductive claim over the levels of recursion, that (in addition to some other basic properties) with high probability over any level, either (i) there was a recursive call at a prior level whose input contains most of the inliers, and by definition that call has terminated, or (ii) there exists some recursive call at this current level whose input contains most of the inliers. We will use the following notation (Definition F.3) to denote various objects in the recursive execution of our algorithm.
Definition F.3 (Notation for Recursion Tree). We define a rooted binary tree  $\mathcal{T}$  that corresponds to the execution of our recursive algorithm (Algorithm 1). The root node at level  $O$  corresponds to the top-level call of COVLISTDECODING (Algorithm 1) and for every non-leaf node, its left child corresponds to a call of COVLISTDECODING from Line 21 and its right child corresponds to a call
from Line 21. Thus, every node is uniquely specified by the level in the tree and its position within that level. We denote by  $\mathcal{T}_{i,j}$  the  $i^{th}$  node of level  $j$ , where the node numbers are also 0-indexed (e.g., the root node is  $\mathcal{T}_{0,0}$ ). For the node  $\mathcal{T}_{i,j}$ , we denote by  $T_{(0)}^{(i,j)}$  the input data set to the corresponding recursive call. In order to refer to the working data set at the  $t^{th}$  iteration of the main loop in the execution of node  $\mathcal{T}_{i,j}$ , we use the notation  $T_{(t)}^{(i,j)}$  and  $\widetilde{T}_{(t)}^{(i,j)}$  exactly in the same way as  $T_{(t)}$  and  $\widetilde{T}_{(t)}$  are used in Algorithm 1. Finally,  $\mathcal{T}_j$  (i.e., using a single subscript) refers to the subtree growing from the root and including all the nodes up to and including the  $j^{th}$  level.
Lemma F.4. In the context of Theorem F.1, consider the recursion tree of COVLISTDECODING (Definition F.3). Then, for any  $j \in \{0, \dots, 9 / \alpha\}$ , we have that with probability at least  $1 - 0.01\alpha j\delta$  the following holds:
1. (Size decrease under recursion) The input of every node  $\mathcal{T}_{i,j}$  of level  $j$  satisfies  $|T_{(0)}^{(i,j)}|\leq m - j\alpha m / 9$
2. (Disjoint inputs) Consider the subtree  $\mathcal{T}_j$  growing from the root and truncated at (but including) level  $j$ . All the leaves in  $\mathcal{T}_j$  have disjoint input sets. Note that the leaves may not all be at level  $j$ , and some might be at earlier levels.
3. (Bound on inliers removed) Consider the subtree  $\mathcal{T}_j$  growing from the root and truncated at (but including) level  $j$ . There exists a leaf  $\mathcal{T}_{i',j'}$  in  $\mathcal{T}_j$  (but the level of the leaf  $j'$  is not necessarily equal to  $j$ ) such that its input  $T_{(0)}^{(i',j')}$  satisfies
$$
\alpha^ {3} \left(\epsilon_ {0} / 4 0\right) \left| T _ {(0)} ^ {\left(i ^ {\prime}, j ^ {\prime}\right)} \right| + \left| S \setminus T _ {(0)} ^ {\left(i ^ {\prime}, j ^ {\prime}\right)} \right| \leq (j + 1) \alpha^ {3} \left(\epsilon_ {0} / 2 0\right) m + \ell \tag {12}
$$
where as in Definition 1.1,  $m$  is the size of the original input set of the root node  $\mathcal{T}_{0,0}$  and  $\ell \leq \epsilon_0 n$  is the number of samples that the adversary replaced in  $S$ . In particular, the number of inliers removed by the algorithm until  $T_{(0)}^{(i',j')}$  is at most  $(j + 1)\alpha^3 (\epsilon_0 / 20)m$ .
Remark F.5. We remark on why (12) uses a cubic power of  $\alpha$ , which is the same cubic power appearing in Lemma E.4 that states that the total score of all the outliers is a  $1 / \alpha^3$  factor more than the total score of the inliers. The three powers of  $\alpha$  are due to the following reasons: 1) the inliers can be as few as an  $\alpha$ -fraction of all the samples, 2) the depth of the recursion tree can be as much as  $O(1 / \alpha)$ , meaning that a set can be filtered  $O(1 / \alpha)$  times, and 3) Theorem F.1 guarantees that we remove at most an  $O(\alpha)$  fraction of inliers. We also remark that, by tweaking parameters in the algorithm and increasing the power of  $\mathrm{poly}(1 / \alpha)$  in our covariance estimation guarantees, we can make the score function ratio (and hence the  $\mathrm{poly}(\alpha)$  factor in (12)) as large a power of  $\alpha$  as we desire. This would guarantee that we remove at most a smaller  $\mathrm{poly}(\alpha)$  fraction of inliers.
Proof. We prove Lemma F.4 by induction on the level number  $j$ . For the base case of  $j = 0$ , Conditions 1 and 2 are trivial. As for Condition 3, at  $j = 0$  we have  $|T_{(0)}^{(0,0)}| = m$  and  $|S \setminus T_{(0)}^{(0,0)}| \leq \ell$  by the definition of the contamination model (Definition 1.1), which directly implies  $\alpha^3(\epsilon_0/40)|T_{(0)}^{(0,0)}| + |S \setminus T_{(0)}^{(0,0)}| \leq \alpha^3(\epsilon_0/40)m + \ell \leq \alpha^3(\epsilon_0/20)m + \ell$  as desired.
To show the inductive case, we assume the lemma statement is true for some  $j$  and we will show the statement for the case  $j + 1$ .
Conditions 1 and 2 These are trivially true, by virtue of the fact that the recursive algorithm, after it is done with removing points via filtering, it partitions the input set using FINDDIVIDER which guarantees that both output sets have size at least  $m_{1} = \alpha m / 9$  (Lemma 3.6).
Condition 3 Recall the notation  $T_{(t)}^{(i,j)}$  from Definition F.3 that is used to denote the set of points in the variable  $T_{t}$  after the  $t^{\text{th}}$  iteration in the  $i^{\text{th}}$  recursive call in level  $j$ .
By Fact F.2, we will condition on Condition 3 being true for  $\mathcal{T}_j$ , the subtree truncated at (but including) level  $j$  and show that Condition 3 holds also for  $\mathcal{T}_{j + 1}$ , the subtree truncated at level  $j + 1$ , except with probability  $0.01\alpha \delta$ . Concretely, the conditioning implies that there exists a leaf  $\mathcal{T}_{i',j'}$  of the subtree
$\mathcal{T}_j$  whose input set  $T_{(0)}^{(i',j')}$  satisfies the inequality in Condition 3 for  $j' \leq j$ . If  $j' < j$ , then we are done (since it continues to be a leaf and the desired bound on the right hand size in Equation (12) is only larger). Otherwise, in the case of  $j' = j$ , we have to analyze the execution of the recursive call associated with this leaf node in the recursion tree  $\mathcal{T}_j$ .
To begin with the analysis, we verify that the input set  $T_{(0)}^{(i',j)}$  satisfies Assumption 3.1. We first check that  $|S \cap T_{(0)}^{(i',j)}| \geq n(1 - 2\epsilon_0)$ :
$$
| S \cap T _ {(0)} ^ {(i ^ {\prime}, j)} | = | S | - | S \setminus T _ {(0)} ^ {(i ^ {\prime}, j)} |
$$
$\geq n - (j + 1)\alpha^3 (\epsilon_0 / 20)m - \ell$  (by the inductive hypothesis)
$\geq n - (9 / \alpha +1)\alpha^3 (\epsilon_0 / 20)m - \ell$ $(j\leq 9 / \alpha)$
$\geq n - \alpha^{2}\epsilon_{0}m - \epsilon_{0}n$ $(\ell \leq \epsilon_0n)$
$\geq n - \alpha \epsilon_0n - \epsilon_0n$  （20  $n\geq \alpha m)$
$\geq n - 1.5\epsilon_0n$  ， (13)
where the last line uses that  $\alpha \leq 1/2$ . Then it remains to check that  $|T_{(0)}^{(i',j)}| \leq (1/\alpha)|S|$ , but this is trivially satisfied since it holds for the input set  $T_{(0)}^{(0,0)}$  that we start with:  $|T_{(0)}^{(i',j)}| \leq |T_{(0)}^{(0,0)}| = m \leq (1/\alpha)n = (1/\alpha)|S|$ .
In a recursive call, the algorithm iteratively removes points (through Line 16) before either terminating (Line 12) or calling FINDDIVIDER and recursing (Line 21). Denote the iteration just before either terminating or running FINDDIVIDER by  $t_{i}^{*}$ . We need to argue that the iterative filtering of points prior to iteration  $t_{i}^{*}$  still roughly preserves Condition 3. Lemma F.6, that is stated below and proved in Lemma F.6, captures the standard martingale argument for filtering-based robust algorithms, and it informally states that if the input set  $T_{(0)}^{(i',j)}$  satisfies Assumption 3.1, then Condition 3 is roughly preserved.
Lemma F.6. Recall the notations and assumptions from Assumption 3.1. Consider an execution of COVLISTDECODING $(T_0)$  with  $T_0 = T$  (with  $T$  satisfying the assumptions from Assumption 3.1), and suppose that  $|T_0| \geq C' / (\alpha^6 \epsilon_0^2) \log (1 / (\alpha \delta))$ . Further assume that  $|S \setminus T_0| \leq 1.5 \epsilon_0 |S|$  (which is a strengthening over Assumption 3.1). Moreover, denote by  $T_t$  the dataset through the loop iterations of Algorithm 1. Then, with probability at least  $1 - 0.01 \alpha \delta$ , it holds that
$$
\alpha^ {3} \left(\epsilon_ {0} / 4 0\right) \left| T _ {t} \right| + \left| S \setminus T _ {t} \right| \leq \alpha^ {3} \left(\epsilon_ {0} / 4 0\right) \left| T _ {0} \right| + \left| S \setminus T _ {0} \right| + \alpha^ {3} \left(\epsilon_ {0} / 4 0\right) \left| T _ {0} \right|,
$$
simultaneously for all iterations  $t$  until the execution of COVLISTDECODING enters either Line 13 or Line 19.
Concretely, applying Lemma F.6 for input set  $T_{(0)}^{(i',j)}$  and with failure probability  $0.01\alpha \delta$  in place of  $\delta$ . The lemma is applicable since its first requirement  $|S \setminus T_{(0)}^{(i',j)}| \leq 1.5\epsilon_0|S|$  has been already checked in (13), and for its second requirement we have that  $|T_{(0)}^{(i',j)}| \geq |T_{(0)}^{(i',j)} \cap S| \geq (1 - 1.5\epsilon_0)|S| \geq C' / (\alpha^6\epsilon_0^2)\log (1 / (\alpha \delta))$ , where the last one uses (13) and the next step uses our assumption  $|S| > C / (\alpha^6\epsilon_0^2)\log (1 / (\alpha \delta))$ . The lemma then yields that, with probability at least  $1 - 0.01\alpha \delta$ ,  $T_{(t_{i'}^*)}^{(i',j)}$  is such that
$$
\alpha^ {3} \left(\epsilon_ {0} / 4 0\right) \left| T _ {\left(t _ {i ^ {*}} ^ {*}\right)} ^ {\left(i ^ {\prime}, j\right)} \right| + \left| S \backslash T _ {\left(t _ {i ^ {*}} ^ {*}\right)} ^ {\left(i ^ {\prime}, j\right)} \right| \leq (j + 1) \alpha^ {3} \left(\epsilon_ {0} / 2 0\right) m + \ell + \left(\epsilon_ {0} / 4 0\right) \alpha^ {3} m. \tag {14}
$$
Now, either the recursive call terminates at iteration  $t_{i'}^*$  or it goes into Line 19. In the former case, we are done (since it is now a leaf node). Otherwise, the recursive call uses FINDDIVIDER to partition  $T_{(t_{i'}')}^{(i',j)}$  into  $T'$  and  $T''$ . We need to show that at least one of  $T'$  and  $T''$  satisfies Condition 3 (for case  $j + 1$ ), using Lemma 3.6. Let us now derive what parameters and set to invoke Lemma 3.6 with.
Using (i) Condition (L.3) of the stability property of the original inlier set  $S$ , and (ii) the fact that  $(H^{\dagger /2}X)^{\top}A(H^{\dagger /2}X) - \mathbf{E}_{X\sim \mathcal{N}(\mu ,\Sigma)}[(H^{\dagger /2}X)^{\top}A(H^{\dagger /2}X)]$  is an even quadratic in  $X$  for any  $H$
and  $A$ , it must be the case that a  $1 - \eta$  fraction (recall  $\eta$  is a shorthand for  $(\epsilon_0 / 40)\alpha^3$ ) of the points  $x \in S$  satisfies
$$
\begin{array}{l} \left. \left(H ^ {\dagger / 2} x\right) ^ {\top} A \left(H ^ {\dagger / 2} x\right) - \underset {X \sim \mathcal {N} (\mu , \Sigma)} {\mathbf {E}} \left[ \left(H ^ {\dagger / 2} X\right) ^ {\top} A \left(H ^ {\dagger / 2} X\right) \right] \right| (15) \\ \leq 10 \log (2 / \eta) \sqrt {\operatorname {V a r} _ {X \sim D} \left[ \left(H ^ {\dagger / 2} X\right) ^ {\top} A \left(H ^ {\dagger / 2} X\right) \right]} \\ \leq 5 0 \log (2 / \eta) \sqrt {\frac {C _ {1}}{\alpha^ {2}}} . (Lemma3.2) \\ \end{array}
$$
Denote the set of points  $x \in S$  with the above property by  $S_{H,A} \subseteq S$ , and let  $S_{\mathrm{core}}^{(i',j)} = T_{(t_{i'}^*)}^{(i',j)} \cap S_{H,A}$ . We will therefore apply Lemma 3.6 with  $S_{\mathrm{proj}} = \{(H^{\dagger/2}x)^{\top}A(H^{\dagger/2}x) : x \in S_{\mathrm{core}}^{(i',j)}\}$ ,  $T = \{(H^{\dagger/2}x)^{\top}A(H^{\dagger/2}x) : x \in T_{(t_{i'}^*)}^{(i',j)}\}$  and diameter  $r = 100\log(2/\eta)\sqrt{C_1}/\alpha$ . It remains to check that for  $q_{m_1} = q_{\mathrm{left}}$  and  $q_{|T_{(t_{i'}^*)}| - m_1} = q_{\mathrm{right}}$  of  $T_{(t_{i'}^*)}^{(i',j)}$ , we have  $q_{|T_{(t_{i'}^*)}| - m_1} - q_{m_1} \geq 10(|T_{(t_{i'}^*)}^{(i',j)}| / |S_{\mathrm{proj}}|)r$ , to make sure that the application of Lemma 3.6 is valid.
To show this, recall that Line 14 fails whenever FINDDIVIDER is called, meaning that  $q_{|T_{(t_{i'}')}^{*(i',j)}| - m_1} = q_{m_1} \geq R$ . Thus, we need to verify that the definition of  $R$  in Algorithm 1 satisfies  $R \geq 10(|T_{(t_{i'}')}^{(i',j)}| / |S_{\mathrm{proj}}|)r$ . The main step is to lower bound  $|S_{\mathrm{proj}}| = |S_{\mathrm{core}}^{(i',j)}|$ . Since  $S_{\mathrm{core}}^{(i',j)} \subset S$ , we can lower bound the above by
$$
\begin{array}{l} | S _ {\text {c o r e}} ^ {(i ^ {\prime}, j)} | \geq | S | - | S \setminus S _ {H, A} | - | S \setminus T _ {(t _ {i ^ {\prime}} ^ {*})} ^ {(i ^ {\prime}, j)} | \hskip 2 8. 4 5 2 7 5 6 p t (S _ {\text {c o r e}} ^ {(i ^ {\prime}, j)} = T _ {(t _ {i ^ {\prime}} ^ {*})} ^ {(i ^ {\prime}, j)} \cap S _ {H, A}) \\ \geq (1 - \eta) | S | - | S \setminus T _ {\left(t _ {i ^ {\prime}} ^ {*}\right)} ^ {\left(i ^ {\prime}, j\right)} | \quad (\text {b y}) \\ \geq (1 - \eta) | S | - (j + 1) \alpha^ {3} \left(\epsilon_ {0} / 2 0\right) m - \ell - \left(\epsilon_ {0} / 4 0\right) \alpha^ {3} m \tag {by(14)} \\ \geq (1 - \eta) n - (j + 2) \alpha^ {3} \left(\epsilon_ {0} / 2 0\right) m - \ell \quad \left(| S | = n \text {a n d b a s i c i n e q u a l i t y}\right) \\ \geq (1 - \eta) n - (j + 2) \alpha^ {2} \left(\epsilon_ {0} / 2 0\right) n - \ell \quad (n \geq \alpha m) \\ \geq (1 - \eta) n - \alpha \epsilon_ {0} n - \ell \quad (j \leq 9 / \alpha \text {a n d} \alpha <   1 / 2) \\ \geq (1 - \eta) n - \alpha \epsilon_ {0} n - \epsilon_ {0} n \quad (\ell \leq \epsilon_ {0} n) \\ \geq (1 - \eta - 2 \epsilon_ {0}) n \geq n / 2 \quad (\eta \leq 0. 0 0 1 \text {a n d} \epsilon_ {0} = 0. 0 1) \\ \end{array}
$$
Combining with the fact that  $|T_{(t_{i'}^*)}^{(i',j)}| \leq m$ , we have that  $10(|T_{(t_{i'}^*)}^{(i',j)}| / |S_{\mathrm{proj}}|)r \leq 20(m/n)r \leq (20/\alpha)r$ . Recalling that  $r = 100\log(2/\eta)\sqrt{C_1}/\alpha$ ,  $\eta = (\epsilon_0/40)\alpha^3$  and  $\epsilon_0 = 0.001$ , the definition of  $R$  in Algorithm 1 satisfies  $R \geq 6000\sqrt{C_1}(1/\alpha^2)\log(1 / (\epsilon_0\alpha)) \geq (20/\alpha)r \geq 10(|T_{(t_{i'}^*)}^{(i',j)}| / |S_{\mathrm{proj}}|)r$ .
Knowing that the application of Lemma 3.6 is valid, the lemma then guarantees that either  $T'$  or  $T''$  contains all of  $S_{\mathrm{core}}^{(i',j)}$ . Without loss of generality, we assume this happens for  $T'$ . We now check Condition 3 for case  $j + 1$  on  $T'$ :