SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\sum_ {i = 1} ^ {n} \sum_ {x \in X ^ {(i)}} \left| \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \right| \leq 2.
$$
Proof. First, we have
$$
\begin{array}{l} \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \\ = \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} \left(1 - \frac {\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right) - \sum_ {x \in X} \sigma_ {X} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)}\right) \\ = \sigma_ {X ^ {(i)}} (x) \frac {\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right) - \sum_ {x \in X} \sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x) \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)} - \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \left(\sigma_ {X ^ {(i)}} (x) - \sigma_ {X} (x)\right). \\ \end{array}
$$
We can bound this quantity from below and above by Lemma B.1,
$$
\begin{array}{l} \sigma_ {X ^ {(i)}} (x) \frac {\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime}) - \sum_ {x \in X} \sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x) \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} - \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \frac {\ell (\theta^ {(i)} , x _ {i}) \ell (\theta^ {(i)} , x)}{\sum_ {x ^ {\prime} \in X} \ell (\theta^ {(i)} , x ^ {\prime}) \cdot \sum_ {x ^ {\prime} \in X ^ {(i)}} \ell (\theta^ {(i)} , x ^ {\prime})} \\ \leq \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \\ \leq \sigma_ {X ^ {(i)}} (x) \frac {\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right) - \sum_ {x \in X} \sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x) \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)}. \\ \end{array}
$$
$$
\begin{array}{l} \left| \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)} \right| \\ \leq \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x \in X} \sigma_ {X} (x) \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \left| \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime}) - \sum_ {x \in X} \sigma_ {X} (x) \right| \\ + \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \frac {\ell (\theta^ {(i)} , x _ {i}) \ell (\theta^ {(i)} , x)}{\sum_ {x ^ {\prime} \in X} \ell (\theta^ {(i)} , x ^ {\prime}) \cdot \sum_ {x ^ {\prime} \in X ^ {(i)}} \ell (\theta^ {(i)} , x ^ {\prime})}. \\ \end{array}
$$
It then follows,
$$
\begin{array}{l} \sum_ {i = 1} ^ {n} \sum_ {x \in X ^ {(i)}} \left| \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \right| \\ \leq \sum_ {i = 1} ^ {n} \sum_ {x \in X ^ {(i)}} \left(\frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x \in X} \sigma_ {X} (x) \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)} \Bigg | \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right) - \sum_ {x \in X} \sigma_ {X} (x) \right| \\ \left. + \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \frac {\ell (\theta^ {(i)} , x _ {i}) \ell (\theta^ {(i)} , x)}{\sum_ {x ^ {\prime} \in X} \ell (\theta^ {(i)} , x ^ {\prime}) \cdot \sum_ {x ^ {\prime} \in X ^ {(i)}} \ell (\theta^ {(i)} , x ^ {\prime})}\right) \\ = \sum_ {i = 1} ^ {n} \left(\frac {\left| \sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right) - \sum_ {x \in X} \sigma_ {X} (x) \right|}{\sum_ {x \in X} \sigma_ {X} (x)} + \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \frac {\ell \left(\theta^ {(i)} , x _ {i}\right)}{\sum_ {x ^ {\prime} \in X} \ell \left(\theta^ {(i)} , x ^ {\prime}\right)}\right) \\ \leq 1 + \frac {1}{\sum_ {x \in X} \sigma_ {X} (x)} \\ \leq 2, \\ \end{array}
$$
where the second to last inequality is from Lemma B.2, and the last inequality is by  $\sum_{x\in X}\sigma_X(x)\geq 1$
Lemma B.4. For  $\epsilon > 0$ , let  $X$  and  $X'$  be sampled from the uniform distribution over  $[B, (1 + \epsilon)B]$  and  $[B', (1 + \epsilon)B']$ , respectively. Then, we have
$$
\operatorname {T V} \left(X, X ^ {\prime}\right) \leq \frac {1 + \epsilon}{\epsilon} \left| 1 - \frac {B ^ {\prime}}{B} \right|.
$$
Proof. The proof is implicit in Lemma 2.3 of Kumabe and Yoshida [2022].
Proof. The average sensitivity of importance sampling can be bounded by the sum of the average total variation distance between selected elements and that between assigned weights, conditioned on that the selected elements are the same. The former is bounded by
$$
\begin{array}{l} \frac {1}{n} \sum_ {i = 1} ^ {n} | C | \cdot \left(\frac {\sigma_ {X} (x _ {i})}{\sum_ {x \in X} \sigma_ {X} (x)} + \sum_ {x \in X ^ {(i)}} \left| \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} (x ^ {\prime})} \right|\right) \\ = O \left(\frac {| C |}{n}\right) + O \left(\frac {| C |}{n}\right) \\ = O \left(\frac {| C |}{n}\right), \\ \end{array}
$$
where the first equality is Lemma B.3. With Lemma B.4, we have the latter as
$$
\frac {1}{n} \sum_ {i = 1} ^ {n} | C | \cdot \left(\sum_ {x \in X ^ {(i)}} \min  \left\{\frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)}, \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)} \right\} \cdot \frac {1 + \epsilon}{\epsilon} \left| 1 - \frac {\sigma_ {X ^ {(i)}} (x) / \left(\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)\right)}{\sigma_ {X} (x) / \sum_ {x \in X} \sigma_ {X} (x)} \right|\right)
$$
$$
\begin{array}{l} \leq \frac {1}{n} \sum_ {i = 1} ^ {n} | C | \cdot \left(\sum_ {x \in X ^ {(i)}} \frac {1 + \epsilon}{\epsilon} \left| \frac {\sigma_ {X} (x)}{\sum_ {x \in X} \sigma_ {X} (x)} - \frac {\sigma_ {X ^ {(i)}} (x)}{\sum_ {x ^ {\prime} \in X ^ {(i)}} \sigma_ {X ^ {(i)}} \left(x ^ {\prime}\right)} \right|\right) \\ = O \left(\frac {| C |}{\epsilon n}\right). \\ \end{array}
$$
Combining the two terms, we have the average sensitivity of importance sampling be bounded as
$$
O \left(\frac {| C |}{n}\right) + O \left(\frac {| C |}{\epsilon n}\right) = O \left(\frac {| C |}{\epsilon n}\right).
$$
C Proofs for Section 6
C.1 Algorithm
We now introduced the detailed version of Algorithm 3. In Algorithm 7, we provide a detailed description of the coreset construction method for clustering, which is based on Huang and Vishnoi [2020]. To make the coreset construction enjoys small average sensitivity, we perturb the weights assigned (Line 6 and Line 11). We show that this preserves the approximation ratio while makes the overall algorithm insensitive. With this, we obtain the online consistent clustering Algorithm 6, which clusters data from a coreset at each step.
Algorithm 6: Online consistent  $(k,z)$ -clustering
Input: PTAS algorithm  $\mathcal{D}$  for  $(k,z)$ -clustering, approximation ratio  $\epsilon, \delta \in (0,1)$ .  
for  $t = 1,\dots,n$  do  
    Construct an  $\epsilon$ -coreset  $C_{t-1} = (S_{t-1},\omega_{t-1})$  by running Algorithm 7 on  $X_{t-1}$ .  
    Obtain cluster set  $Z_t$  by running a PTAS  $\mathcal{D}$  with approximation ratio of  $(1+\epsilon)$  on  $C_{t-1}$ .  
    Receive  $x_t$  and  $\ell(Z_t,x_t)$ .
Algorithm 7: Coreset construction for clustering Huang and Vishnoi [2020]
Input: A set of point  $X$ , approximation parameter  $\epsilon, \delta \in (0,1)$ , integer  $k, z$ .  
1 Set  $\epsilon = \epsilon / c$  for some large constant  $c > 0$ .  
2 Compute a  $k$ -center set  $C_t^* \subseteq \mathbb{R}^d$  as an  $\epsilon$ -approximation of the  $(k,z)$ -clustering problem over  $X$  with the  $D^z$ -sampling algorithm with an approximation ratio of  $O(2^z \log k)$ .  
3 For each  $x \in X$ , compute the closest point to  $x$  in  $C^*$ ,  $c^*(x)$ , with ties broken arbitrarily. For each  $c \in C^*$ , denote  $X^c$  to be the set of points  $x \in X$  with  $c^*(x) = c$ .  
4 For each  $x \in X$ , let  $\sigma_{1,X}(x) = 2^{2z + 2}\epsilon^2\left(\frac{\|x - c^*(x)\|_2^2}{\sum_{x \in X} \ell(C^*, x)} + \frac{1}{|X_{c^*(x)}|}\right)$ .  
5 Pick a non-uniform random sample  $D^1$  of  $N_1 = O\left((168z)^{10z}\epsilon^{-5z - 15}k^5 \log \frac{k}{\delta}\right)$  points from  $X$ , where each  $x \in X$  is selected with probability  $\frac{\sigma_{1,X}(x)}{\sum_{y \in X} \sigma_{1,X}(y)}$ .  
6 For each  $x \in D^1$ , sample  $\tilde{u}_X(x)$  from  $\left[\frac{\sum_{y \in X} \sigma_{1,X}(y)}{|D_t^1| \cdot \sigma_{1,X}(x)}, (1 + \epsilon)\frac{\sum_{y \in X} \sigma_{1,X}(y)}{|D_t^1| \cdot \sigma_{1,X}(x)}\right]$ .  
7 Set  $u_X(x) = \tilde{u}_X(x)$   
8 For each  $c \in C^*$ , compute  $D_c$  to be the set of points in  $D^1$  whose closest point in  $C^*$  is  $c$  with ties broken arbitrarily.  
9 For each  $x \in D^1$ , let  $\sigma_{2,X}(x) = \frac{u_X(x) \cdot \ell(C^*, x)}{\sum_{y \in D^1} u_X(y) \cdot \ell(C^*, y)}$   
10 Pick a non-uniform random sample  $D^2$  of  $N_2 = O\left(\epsilon^{-2z - 2}k \log k \log \frac{k}{\epsilon\delta}\right)$  points from  $X_t$ , where each  $x \in X_t$  is selected with probability  $\frac{\sigma_{2,X}(x)}{\sum_{y \in D^1} \sigma_{2,X}(y)}$   
11 For each  $x \in D^2$ , sample  $\tilde{w}_X(x)$  from  $\left[\frac{\sum_{y \in D^1} \sigma_{2,X}(y)}{|D^2| \cdot \sigma_{2,X}(x)}, (1 + \epsilon)\frac{\sum_{y \in D^1} \sigma_{2,X}(y)}{|D^2| \cdot \sigma_{2,X}(x)}\right]$ .  
12 Set  $w_X(x) = \tilde{w}_X(x)$   
13 For each  $c \in C^*$ , let  $w_X(c) = (1 + 10\epsilon) \sum_{x \in D_c} u_X(x) - \sum_{x \in D^2 \cap D_c} w_X(x)$ .  
14  $S = D^2 \cup C^*$   
15  $w(x) = w_X(x)$   
16 Output ( $S, w$ )
C.2 Analysis
Theorem C.1. Algorithm 7 outputs an  $\epsilon$ -coreset with probability at least  $1 - \delta$  and has an average sensitivity of
$$
O \left(\frac {k + (1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {6} \log \frac {k}{\delta} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k}{\epsilon \delta} (1 + \epsilon^ {- 2 z - 2} k ^ {2} \log k \log \frac {k}{\epsilon \delta})}{n}\right).
$$
Proof. We first show that our coreset construction method gives an  $\epsilon$ -coreset.
We remark that the way we assigned the weights is a perturbed version of the method presented in Huang and Vishnoi [2020]. Then, to show that the perturbed version of the assignment of the weight still preserves the coreset approximation ratio, we show that the perturbed version still satisfies Corollary 15.3 of Feldman and Langberg [2011] and Theorem 14.5. Then using the same argument as Theorem 15.4 of Feldman and Langberg [2011], our coreset construction method gives an  $\epsilon$ -coreset.
In the first importance sampling stage, we only perturb  $\tilde{u}_X(x)$  by a ratio of  $1 + \epsilon$ , for each  $x \in D^1$ . The same is applied in the second stage with  $\tilde{w}_X(x)$ , for all  $x \in D^2$ . For  $c \in C^*$ , we have
$$
(1 + 10 \epsilon) \sum_ {x \in D _ {c}} u _ {X} (x) - \sum_ {x \in D ^ {2} \cap D _ {c}} w _ {X} (x) \leq (1 + \epsilon) \left((1 + 1 0 \epsilon) \sum_ {x \in D _ {c}} \tilde {u} _ {X} (x) - \sum_ {x \in D ^ {2} \cap D _ {c}} \tilde {w} _ {X} (x)\right).
$$
In all cases, the result from [Feldman and Langberg, 2011] still holds, as the weights are scaled by  $(1 + \epsilon)$  at most. By the same argument that all perturbed weights are scaled by  $(1 + \epsilon)$  at most, we have Theorem 15.4 of Feldman and Langberg [2011] holds with a ratio of  $\epsilon (1 + \epsilon)$ . This results in an approximation of  $1 + \epsilon (1 + \epsilon)$  by applying the argument of Theorem 15.4 of Feldman and Langberg [2011]. Rescale  $\epsilon$  to  $\epsilon /c$  for some constant  $c > 0$  gives an approximation ratio of  $1 + \epsilon$  and completes our argument that Algorithm 7 produces an  $\epsilon$ -coreset with probability at least  $1 - \delta$ .
To show that our method enjoys low average sensitivity, we upper bound the average sensitivity of the two importance sampling stage of Algorithm 7 separately.
The average sensitivity of importance sampling in the first stage can be bounded by the average total variation distance between selected elements and that between assigned weights, conditioned on the selected elements are the same. By Lemma 5.2, we can upper bound the average sensitivity of the first importance sampling stage to be  $O(N_1 / n)$ , where  $N_{1} = O\left((168z)^{10z}\epsilon^{-5z - 15}k^{5}\log (\delta^{-1}k)\right)$ .
In the second stage, as points are selected with probability  $\frac{\sigma_{2,X}(x)}{\sum_{y\in X}\sigma_{2,X}(y)}$ , we bound the total variation distance between selected points in a similar way as to that for the first stage. This gives a bound of  $O(N_2 / n)$ , where  $N_{2} = O\left(\epsilon^{-2z - 2}k\log k\log \frac{k}{\epsilon\delta}\right)$ . To bound the average total variation distance between assigned weights, we again apply the same argument as above and obtain a bound of  $O(N_2 / n)$ .
Combining these and that the average sensitivity of Line 2 of Algorithm 7 is  $O(k / n)$  (by Lemma 2.2 of Yoshida and Ito [2022]), we have the average sensitivity as
$$
O \left(\frac {k + N _ {1} + N _ {2}}{n}\right) = O \left(\frac {k + (1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log \frac {k}{\delta} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k}{\epsilon \delta}}{n}\right).
$$
$$
\mathrm {R e g r e t} _ {\epsilon} (n) \leq O \left(\left((1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log \frac {k n}{\epsilon} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k n}{\epsilon}\right) \log n\right).
$$
Proof. First, we note that the approximation ratio of the algorithm with respect to the aggregated loss is at most
$$
(1 - \delta) \left(1 + \frac {\epsilon}{3}\right) \left(1 + \frac {\epsilon}{3}\right) + \delta n \leq 1 + \epsilon
$$
from the choice of  $\delta$ , i.e.,  $\delta = O(\epsilon / n)$ .
Also, we note that the overall average sensitivity is
$$
\begin{array}{l} \sum_ {t = 1} ^ {n} \beta (t) = \sum_ {t = 1} ^ {n} O \left(\frac {(1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log \frac {k}{\delta} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k}{\epsilon \delta}}{t}\right) \\ = O \left(\left((1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log {\frac {k}{\delta}} + \epsilon^ {- 2 z - 2} k \log {k} \log {\frac {k}{\epsilon \delta}}\right) \log {n}\right). \\ \end{array}
$$
Then substituting this into Theorem 4.1, for  $z \geq 1$ , we have a regret bound of
$$
\begin{array}{l} \operatorname {R e g r e t} _ {\epsilon} (n) \leq O \left(\left((1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log \frac {k}{\delta} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k}{\epsilon \delta}\right) \log n\right) \\ = O \left(\left((1 6 8 z) ^ {1 0 z} \epsilon^ {- 5 z - 1 5} k ^ {5} \log \frac {k n}{\epsilon} + \epsilon^ {- 2 z - 2} k \log k \log \frac {k n}{\epsilon}\right) \log n\right). \\ \end{array}
$$
For the second claim about inconsistency, we first convert the average sensitivity bound to an inconsistency bound of the same order. This can be done by Lemma 4.2 of [Yoshida and Ito, 2022] and by arguing in a similar way as Lemma 4.5 of [Yoshida and Ito, 2022] by showing that there exists a computable probability transportation for the output of Algorithm 3. This thus yields an inconsistency of  $O\left(\left((168z)^{10z}\epsilon^{-5z - 15}k^5\log (\epsilon^{-1}kn)\epsilon^{-2z - 2}k\log k\log (\epsilon^{-1}kn)\right)\log n\right)$ .
D Proofs for Section 7
Proof. Let  $\delta > 0$  be determined later. By Theorem 6 from Cohen et al. [2017], with probability  $1 - \delta$ , for any rank- $k$  orthogonal projection  $\mathbf{X}$ , by sampling  $m = O\left(\epsilon^{-2} k \log (\delta^{-1} k)\right)$  columns, in Line 4 of Algorithm 5, we have,
$$
\left(1 - \frac {\epsilon}{2}\right) \| \mathbf {A} _ {t} - \mathbf {X A} _ {t} \| _ {F} ^ {2} \leq \| \mathbf {C} _ {t} - \mathbf {X C} _ {t} \| _ {F} ^ {2} \leq \left(1 + \frac {\epsilon}{2}\right) \| \mathbf {A} _ {t} - \mathbf {X A} _ {t} \| _ {F} ^ {2}.
$$
With this, we note that the algorithm has an approximation of  $1 + \epsilon$  as  $1 + \epsilon /2 + \delta n\leq 1 + \epsilon$  by choosing the hidden constant in  $\delta$  small enough.
Note that by applying Lemma 5.2, this routine has an average sensitivity of  $O(m / t) = O\left(\epsilon^{-2}k\log (\delta^{-1}k) / t\right)$  at any step  $t$ . Then set  $\mathbf{Z}_t$  to the top  $k$  left singular vectors of  $\mathbf{C}_t$ , and we have
$$
\left\| \mathbf {A} _ {t} - \mathbf {Z} _ {t} \mathbf {Z} _ {t} ^ {\top} \mathbf {A} _ {t} \right\| _ {F} \leq (1 + \epsilon) \left\| \mathbf {A} _ {t} - \mathbf {U} _ {k} \mathbf {U} _ {k} ^ {\top} \mathbf {A} _ {t} \right\| _ {F}.
$$
To obtain the regret, we calculate the overall average sensitivity as  $\sum_{t=1}^{n} \beta(t) = \sum_{t=1}^{n} O\left(\epsilon^{-2} k \log (\delta^{-1} k) / t\right) = O\left(\epsilon^{-2} k \log n \log (\delta^{-1} k)\right)$ . Applying Theorem 4.1, and from the choice  $\delta = O(\epsilon / n)$ , we have  $\operatorname{Regret}_{\epsilon}(n) = O\left(\epsilon^{-2} k \log n \log (\epsilon^{-1} k n)\right)$ .
For the second claim about inconsistency, we prove this convert the average sensitivity bound to an inconsistency bound of the same order. This can be done by Lemma 4.2 of [Yoshida and Ito, 2022] and by arguing in a similar way as Lemma 4.5 of [Yoshida and Ito, 2022] by showing that there exists a computable probability transportation for the output of Algorithm 3.
E Proof for Section 8
Proof. By Theorem 8.4, the algorithm has an approximation of  $1 + \epsilon$  as  $1 + \epsilon /2 + \delta n\leq 1 + \epsilon$  by choosing the hidden constant in  $\delta$  small enough.
Similar to the low-rank case, the average sensitivity of leverage score sampling at step  $t$  is  $O\left(\frac{d\log(d / \delta)}{\epsilon^2t}\right)$ . Summing over the average sensitivity
$$
\sum_ {t = 1} ^ {n} O \left(\frac {d \log (d / \delta)}{\epsilon^ {2} t}\right) = O \left(\frac {d \log n \log (d / \delta)}{\epsilon^ {2}}\right).
$$
Take this into Theorem 1, and with  $\delta = O(\epsilon /n)$ , we have a  $\epsilon$ -regret of  $O\left(\epsilon^{-2}d\log n\log (dn / \epsilon)\right)$ .
Similar to that of Theorem 6.1 and Theorem 7.2, we prove the second claim by converting the average sensitivity bound to an inconsistency bound of the same order. This can be done by Lemma 4.2 of [Yoshida and Ito, 2022] and by arguing in a similar way as Lemma 4.5 of [Yoshida and Ito, 2022] by showing that there exists a computable probability transportation for the output of Algorithm 3.