SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \psi_ {\mathrm {B A I}} ^ {*} = \inf _ {\lambda \in P _ {\mathcal {X}}} \max _ {z \in \mathcal {Z} \backslash \{z ^ {*} \}} \frac {| | z ^ {*} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}}{\left\{(z ^ {*} - z) ^ {\top} \theta^ {*} \right\} ^ {2}} \\ = \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {\ell \leq \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \max  _ {z \in \mathcal {S} _ {\ell}} \frac {| | z ^ {*} - z | | ^ {2}}{\left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}} \\ \geq \inf_{\lambda \in P_{\mathcal{X}}}\max_{\ell \leq \lceil \log_{2}(4\Delta^{-1})\rceil}\max_{z\in \mathcal{S}_{\ell}}\frac{\left|\left|z^{*} - z\right|\right|^{2}\left(\sum_{x\in\mathcal{X}}\lambda_{x}\frac{x x^{\top}}{\sigma_{x}^{2}}\right)^{-1}}{(4\times 2^{-\ell})^{2}} \\ \stackrel {(a)} {\geq} \frac {1}{\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \inf  _ {\lambda \in P _ {\mathcal {X}}} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \max  _ {z \in S _ {\ell}} \frac {| | z ^ {*} - z | | ^ {2}}{\left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}} (4 \times 2 ^ {- \ell}) ^ {2} \\ \stackrel {(b)} {\geq} \frac {1}{1 6 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z \in S _ {\ell}} | | z ^ {*} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1} \\ \stackrel {(c)} {\geq} \frac {1}{6 4 \left\lceil \log_ {2} \left(4 \Delta^ {- 1}\right)\right\rceil} \sum_ {\ell = 1} ^ {\left\lceil \log_ {2} \left(4 \Delta^ {- 1}\right)\right\rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z, z ^ {\prime} \in S _ {\ell}} | | z ^ {\prime} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x}{\sigma_ {x} ^ {2}}\right) ^ {- 1} \\ \stackrel {(d)} {\geq} \frac {1}{6 4 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z, z ^ {\prime} \in S _ {\ell}} \frac {1}{2} | | z ^ {\prime} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\hat {\sigma} _ {x} ^ {2}}\right) ^ {- 1} \\ \geq \frac {1}{1 2 8 \left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} \sum_ {\ell = 1} ^ {\left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z, z ^ {\prime} \in S _ {\ell}} | | z ^ {\prime} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x}{\hat {\sigma} _ {x} ^ {2}}\right) ^ {- 1} \\ = \frac {1}{1 2 8 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} q (\mathcal {S} _ {\ell}) \\ \end{array}
$$
here  $(a)$  follows from the fact that the maximum of a set of numbers is always greater than the average and  $(b)$  by the fact that the minimum of a sum is greater than the sum of the minimums. We used the fact that
$$
\max  _ {z, z ^ {\prime} \in \mathcal {S} _ {\ell}} | | z ^ {\prime} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1} \leq 4 \max  _ {z \in \mathcal {S} _ {\ell}} | | z ^ {*} - z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}
$$
by the triangle inequality in  $(c)$ . Finally,  $(d)$  follows from Lemma E.9.
Consequently, putting the previous pieces together we find that the sample complexity can be upper bounded as
$$
\begin{array}{l} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {x \in \mathcal {X}} \lceil \tau_ {\ell} \widehat {\lambda} _ {\ell , x} \rceil \leq \frac {d (d + 1)}{2} \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil \\ { + } { 3 \log \left\{ \frac { 8 \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil ^ { 2 } | \mathcal { Z } | } { \delta } \right\} \sum _ { \ell = 1 } ^ { \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil } 2 ^ { 2 \ell } q ( \mathcal { S } _ { \ell } ) } \\ \leq \frac {d (d + 1)}{2} \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil \\ { + } { 3 8 4 \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil \log \left\{ \frac { 8 \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil ^ { 2 } | \mathcal { Z } | } { \delta } \right\} \psi _ { \mathrm { B A I } } ^ { * } , } \\ \end{array}
$$
where the dependency on  $d(d + 1) / 2$  can reduced to  $d$  by Proposition A.1. This proves the result.
F Proofs of Level Set Identification
This appendix contains the proofs for the results on transductive linear bandit level set identification. In this problem, we are interested in identifying the set  $G_{\alpha}(\theta^{*}) = \{z \in \mathcal{Z} : z^{\top}\theta^{*} > \alpha\}$ , i.e., the items with mean rewards that exceed threshold  $\alpha$ . We prove the lower bound on the sample complexity from Theorem 4.1 in App. F.1. The proof of correctness and the sample complexity upper bound for Alg. 2 are presented in App. F.2. These proofs resemble the transductive linear bandit identification proofs with slight modifications accounting for the problem structure. Recalling  $B_{\alpha}(\theta^{*}) = \{z \in \mathcal{Z} : z^{\top}\theta^{*} < \alpha\}$ , we assume that  $\mathcal{Z} = G_{\alpha}(\theta^{*}) \cup B_{\alpha}(\theta^{*})$ , which is equivalent to the condition  $\{z \in \mathcal{Z} : z^{\top}\theta^{*} = \alpha\} = \emptyset$ . Let  $\Delta = \min_{z \in \mathcal{Z}} |z^{\top}\theta^{*} - \alpha|$  denote the minimum absolute gap for the threshold  $\alpha$ .
F.1 Lower Bound
Proof of Theorem 4.1 for transductive level set identification. Label  $\mathcal{X} \subset \mathbb{R}^d$  as the measurement vectors such that  $K = |\mathcal{X}|$  and  $\mathcal{Z}$  as the decision item set. This proof is similar to the lower bound proofs of [35] and [12]. Let  $\mathcal{C}$  be the set of alternatives such that for  $\widetilde{\theta} \in \mathcal{C}$ ,  $G_{\alpha}(\theta^{*}) \neq G_{\alpha}(\widetilde{\theta})$ . This set of alternatives is decomposed by [35] as,
$$
\mathcal {C} = \left(\cup_ {z \in G _ {\alpha} (\theta^ {*})} \{\widetilde {\theta}: z ^ {\top} \widetilde {\theta} <   \alpha \}\right) \cup \left(\cup_ {z \notin G _ {\alpha} (\theta^ {*})} \{\widetilde {\theta}: z ^ {\top} \widetilde {\theta} > \alpha \}\right).
$$
We now recall Lemma 1 of [27]. Under a  $\delta$ -PAC strategy for bandit instance  $(\mathcal{X},\mathcal{Z},\theta^{*})$ , let  $T_{i}$  denote the random variable that is the number of times arm  $i$  is pulled. Defining  $\sigma_x^2 = x^\top \Sigma^* x$  for all  $x\in \mathcal{X}$  we denote  $\nu_{\theta ,i}$  as the reward distribution of the  $i$ -th arm of  $\mathcal{X}$ , i.e.,  $\nu_{\theta ,i} = \mathcal{N}\left(x_i^\top \theta ,\sigma_{x_i}^2\right)$ . For any  $\widetilde{\theta}\in \mathcal{C}$ , we know from [27] that
$$
\sum_ {i = 1} ^ {K} \mathbb {E} \left(T _ {i}\right) \geq \log (1 / 2. 4 \delta) \min  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {\tilde {\theta} \in \mathcal {C}} \frac {1}{\sum_ {i = 1} ^ {K} \lambda_ {i} K L \left(\nu_ {\theta^ {*} , i} \mid \mid \nu_ {\tilde {\theta} , i}\right)}. \tag {18}
$$
For  $\lambda \in P_{\mathcal{X}}$ , define  $A(\lambda) := \sum_{i=1}^{K} \lambda_i \frac{x_i x_i^\top}{\sigma_{x_i}^2}$ . For  $\epsilon > 0$  and  $z \in G_\alpha(\theta^*)$ , define
$$
\theta_ {z} (\epsilon , \lambda) := \theta^ {*} - \frac {\left(\theta^ {* ^ {\top}} z - \alpha + \epsilon\right) A (\lambda) ^ {- 1} z}{\| z \| _ {A (\lambda) ^ {- 1}} ^ {2}},
$$
where  $z^{\top}\theta_{z}(\epsilon ,\lambda) = \alpha -\epsilon$  implying that  $\theta_z(\epsilon ,\lambda)\in \mathcal{C}$ . For  $z\in G_{\alpha}^{c}(\theta^{*})$ , define
$$
\theta_ {z} (\epsilon , \lambda) := \theta^ {*} - \frac {\left(\theta^ {* ^ {\top}} z - \alpha - \epsilon\right) A (\lambda) ^ {- 1} z}{\| z \| _ {A (\lambda) ^ {- 1}} ^ {2}},
$$
where  $z^{\top}\theta_{z}(\epsilon ,\lambda) = \alpha +\epsilon$  implying that  $\theta_z(\epsilon ,\lambda)\in \mathcal{C}$
Defining  $\epsilon_z = \epsilon$  if  $z\in G_{\alpha}(\theta^{*})$  and  $\epsilon_z = -\epsilon$  if  $z\in G_{\alpha}^{c}(\theta^{*})$ , we find that the KL divergence between  $\nu_{\theta ,i}$  and  $\nu_{\theta_{z}(\epsilon ,\lambda),i}$  is given by:
$$
\begin{array}{l} K L (\nu_ {\theta , i} | | \nu_ {\theta_ {z} (\epsilon , \lambda), i}) = \frac {\left[ x _ {i} ^ {\top} \{\theta^ {*} - \theta_ {z} (\epsilon , \lambda) \} \right] ^ {2}}{2 \sigma_ {x _ {i}} ^ {2}} \\ = z ^ {\top} A (\lambda) ^ {- 1} \frac {(z ^ {\top} \theta^ {*} - \alpha + \epsilon_ {z}) ^ {2} \frac {x _ {i} x _ {i} ^ {\top}}{2 \sigma_ {x _ {i}} ^ {2}}}{\left\{z ^ {\top} A (\lambda) ^ {- 1} z \right\} ^ {2}} A (\lambda) ^ {- 1} z. \\ \end{array}
$$
Returning to Eq. 18,
$$
\begin{array}{l} \sum_ {i = 1} ^ {K} \mathbb {E} (T _ {i}) \geq \log (1 / 2. 4 \delta) \min  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {\widetilde {\theta} \in \mathcal {C}} \frac {1}{\sum_ {i = 1} ^ {K} \lambda_ {i} K L (\nu_ {\theta^ {*} , i} | | \nu_ {\widetilde {\theta} , i})} \\ \geq \log (1 / 2.4\delta)\min_{\lambda \in P_{\mathcal{X}}}\max_{z\in \mathcal{Z}}\frac{1}{\sum_{i = 1}^{K}\lambda_{i}KL(\nu_{\theta^{*},i}||\nu_{\theta_{z}(\epsilon_{z},\lambda),i})} \\ \geq \log (1 / 2. 4 \delta) \min _ {\lambda \in P _ {\mathcal {X}}} \max _ {z \in \mathcal {Z}} \frac {\left\{z ^ {\top} A (\lambda) ^ {- 1} z \right\} ^ {2}}{\sum_ {i = 1} ^ {K} (z ^ {\top} \theta^ {*} - \alpha + \epsilon_ {z}) ^ {2} z ^ {\top} A (\lambda) ^ {- 1} \lambda_ {i} \frac {x _ {i} x _ {i} ^ {\top}}{2 \sigma_ {x _ {i}} ^ {2}} A (\lambda) ^ {- 1} z ^ {\top} A (\lambda) ^ {- 1} z}. \\ \stackrel {(a)} {=} 2 \log (1 / 2. 4 \delta) \min _ {\lambda \in P _ {\mathcal {X}}} \max _ {z \in \mathcal {Z}} \frac {\left\{z ^ {\top} A (\lambda) ^ {- 1} z \right\} ^ {2}}{(z ^ {\top} \theta^ {*} - \alpha + \epsilon_ {z}) ^ {2} z ^ {\top} A (\lambda) ^ {- 1} \left(\sum_ {i = 1} ^ {K} \lambda_ {i} \frac {x _ {i} x _ {i} ^ {\top}}{\sigma_ {x _ {i}} ^ {2}}\right) A (\lambda) ^ {- 1} z ^ {\top} A (\lambda) ^ {- 1}} \\ = 2 \log (1 / 2. 4 \delta) \min _ {\lambda \in P _ {\chi}} \max _ {z \in \mathcal {Z}} \frac {z ^ {\top} A (\lambda) ^ {- 1} z}{(z ^ {\top} \theta^ {*} - \alpha + \epsilon_ {z}) ^ {2}}, \\ \end{array}
$$
where  $(a)$  uses the fact that  $\sum_{i=1}^{K} \lambda_i \frac{x_i x_i^\top}{\sigma_{x_i}^2} = A(\lambda)$ . Letting  $\epsilon_z \to 0$  establishes the result.
F.2 Upper Bound
We now prove the upper bound on the sample complexity of Alg. 2 for level set estimation. This proof follows similarly to the upper bound on the sample complexity of Alg. 2 for best-arm identification shown in App. E.2. Note that in each round  $\ell \in \mathbb{N}$ , Alg. 2 maintains a set  $\mathcal{Z}_{\ell}$  of undecided items, a set  $\mathcal{G}_{\ell}$  of items estimated to have value exceeding the threshold  $\alpha$ , and a set  $\mathcal{B}_{\ell}$  of items estimated to have value falling below the threshold  $\alpha$ .
F.2.1 Proof of Correctness
To begin, we show that that for all rounds  $\ell \in \mathbb{N}$  of Alg. 2, the error in the estimates of  $z^{\top}\theta^{*}$  for all  $z\in \mathcal{Z}$  are bounded by  $\epsilon_{\ell}$  with probability at least  $1 - \delta$  for  $\delta \in (0,1)$ .
Lemma F.1. For all  $\ell \in \mathbb{N}$  and  $z\in \mathcal{Z}$ , with probability at least  $1 - \delta$  for  $\delta \in (0,1)$ ,
$$
| \langle z, \widehat {\theta} _ {\ell} - \theta \rangle | \leq \epsilon_ {\ell}.
$$
Proof. This proof follows identically to Lemma E.1 by replacing  $z^{*}$  with the zero vector.
We now apply the preceding result to show that with probability  $1 - \delta$  for  $\delta \in (0,1)$ , Alg. 2 does not place items in the wrong category.
Lemma F.2. With probability at least  $1 - \delta$  for  $\delta \in (0,1)$ , for any  $z\in \mathcal{Z}_{\ell}$ ,  $\ell \in \mathbb{N}$ , such that  $z^{\top}\theta^{*} > \alpha$ ,
$$
\langle z, \widehat {\theta} _ {\ell} \rangle + \epsilon_ {\ell} > \alpha ,
$$
and for any  $z \in \mathcal{Z}_{\ell}$  such that  $z^{\top} \theta^{*} < \alpha$
$$
\langle z, \widehat {\theta} _ {\ell} \rangle - \epsilon_ {\ell} <   \alpha .
$$
Proof. Consider  $z \in G_{\alpha}$  so that  $z^{\top} \theta^{*} > \alpha$ . Using Lemma F.1, with probability at least  $1 - \delta$  for  $\delta \in (0, 1)$  we have
$$
\begin{array}{l} \langle z, \widehat {\theta} _ {\ell} \rangle + \epsilon_ {\ell} = \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle + \epsilon_ {\ell} + \langle z, \theta^ {*} \rangle \\ > \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle + \epsilon_ {\ell} + \alpha \\ \geq \alpha . \\ \end{array}
$$
By the procedure of Alg. 2, this guarantees  $\mathcal{G}_{\ell +1}\subseteq G_{\alpha}$
Now, consider  $z \in B_{\alpha}$  so that  $z^{\top} \theta^{*} < \alpha$ . Using Lemma F.1, with probability at least  $1 - \delta$  for  $\delta \in (0, 1)$  we have
$$
\begin{array}{l} \langle z, \widehat {\theta} _ {\ell} \rangle - \epsilon_ {\ell} = \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle - \epsilon_ {\ell} + \langle z, \theta^ {*} \rangle \\ <   \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle - \epsilon_ {\ell} + \alpha \\ \leq \alpha . \\ \end{array}
$$
By the procedure of Alg. 2, this guarantees  $\mathcal{B}_{\ell +1}\subseteq B_{\alpha}$
We now show that any  $z \in \mathcal{Z}$  such that  $|z^{\top} \theta^{*} - \alpha| > 2\epsilon_{\ell}$  is removed from the active set,  $\mathcal{Z}_{\ell}$ , in round  $\ell \in \mathbb{N}$  of Alg. 2 with probability at least  $1 - \delta$  for  $\delta \in (0,1)$ . Since  $\epsilon_{\ell} \to 0$ , this proves that we will eventually identify  $G_{\alpha}$  correctly.
Lemma F.3. With probability at least  $1 - \delta$  for  $\delta \in (0,1)$ , for any  $\ell \in \mathbb{N}$  and  $z\in \mathcal{Z}_{\ell}$  such that  $|\langle z,\theta^{*}\rangle -\alpha | > 2\epsilon_{\ell}$ ,
$$
\left| \langle z, \widehat {\theta} _ {\ell} \rangle - \alpha \right| > \epsilon_ {\ell}.
$$
Proof. Set  $\delta \in (0,1)$ . Let us begin by considering any  $z \in \mathcal{Z}_{\ell}$  such that  $\langle z,\theta^{*}\rangle -\alpha >2\epsilon_{\ell}$ . Using Lemma F.1, with probability at least  $1 - \delta$
$$
\begin{array}{l} \langle z, \widehat {\theta} _ {\ell} \rangle - \epsilon_ {\ell} = \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle + \langle z, \theta^ {*} \rangle - \epsilon_ {\ell} \\ \geq - \epsilon_ {\ell} + \langle z, \theta^ {*} \rangle - \epsilon_ {\ell} \\ > \alpha . \\ \end{array}
$$
Now, consider any  $z \in \mathcal{Z}_{\ell}$  such that  $\langle z, \theta^{*} \rangle - \alpha < -2\epsilon_{\ell}$ . Using Lemma F.1, with probability at least  $1 - \delta$ ,
$$
\begin{array}{l} \langle z, \widehat {\theta} _ {\ell} \rangle + \epsilon_ {\ell} = \langle z, \widehat {\theta} _ {\ell} - \theta^ {*} \rangle + \langle z, \theta^ {*} \rangle + \epsilon_ {\ell} \\ \leq \epsilon_ {\ell} + \langle z, \theta^ {*} \rangle + \epsilon_ {\ell} \\ <   \alpha . \\ \end{array}
$$
Hence, for  $z \in \mathcal{Z}_{\ell}$  such that  $|\langle z, \theta^{*} \rangle - \alpha| > 2\epsilon_{\ell}$ , we have that  $|\langle z, \widehat{\theta}_{\ell} \rangle - \epsilon_{\ell}| > \alpha$  with probability at least  $1 - \delta$ .
By the elimination conditions of Alg. 2, this guarantees that the  $z \in \mathcal{Z}_{\ell}$  such that  $|\langle z, \theta^{*} \rangle - \alpha| > 2\epsilon_{\ell}$  are not included in  $\mathcal{Z}_{\ell + 1}$ .
F.2.2 Proof of Upper Bound on Sample Complexity
Now we prove the upper bound on the sample complexity of Alg. 2 for the transductive level set identification problem.
Proof of Theorem 4.2 for transductive level set identification. Set  $\delta \in (0,1)$ . Assume that  $\max_{z\in \mathcal{Z}_{\ell}}|\langle z,\theta^{*}\rangle -\alpha |\leq 2$ . Define  $\mathcal{S}_{\ell} = \{z\in \mathcal{Z}_{\ell}:|\langle z,\theta^{*}\rangle -\alpha |\leq 4\epsilon_{\ell}\}$  for  $\ell \in \mathbb{N}$ . Note that by assumption  $\mathcal{Z} = \mathcal{S}_1 = \mathcal{Z}_1$ . Lemma F.3 implies that with probability at least  $1 - \delta$ , we know  $\mathcal{Z}_{\ell}\subset \mathcal{S}_{\ell}$  for all  $\ell \in \mathbb{N}$ . This means
$$
\begin{array}{l} q(\mathcal{Z}_{\ell}) = \inf_{\lambda \in P_{\mathcal{X}}}  \max_{z\in \mathcal{Z}_{\ell}}||z||^{2}\left(\sum_{x\in \mathcal{X}}\lambda_{x}\frac{x x^{\top}}{\hat{\sigma}_{x}^{2}}\right)^{-1} \\ \leq \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z \in \mathcal {S} _ {\ell}} | | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x}{\sigma_ {x} ^ {2}} ^ {\top}\right) ^ {- 1} \\ = q \left(\mathcal {S} _ {\ell}\right). \\ \end{array}
$$
For  $\ell \geq \lceil \log_2(4\Delta^{-1})\rceil$  , we know that  $S_{\ell} = \emptyset$  , thus the sample complexity to identify the set  $G_{\alpha}$  is upper bounded as follows.
$$
\begin{array}{l} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {x \in \mathcal {X}} \lceil \tau_ {\ell} \widehat {\lambda} _ {\ell , x} \rceil \stackrel {(a)} {=} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \left\{\frac {d (d + 1)}{2} + \tau_ {\ell} \right\} \\ = \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \left\{\frac {d (d + 1)}{2} + 2 \left(\frac {3}{2}\right) \epsilon_ {\ell} ^ {- 2} q (\mathcal {Z} _ {\ell}) \log (8 \ell^ {2} | \mathcal {Z} | / \delta) \right\} \\ \leq \frac {d (d + 1)}{2} \left[ \log_ {2} \left(4 \Delta^ {- 1}\right)\right] + \sum_ {\ell = 1} ^ {\left\lceil \log_ {2} \left(4 \Delta^ {- 1}\right)\right\rceil} 3 \epsilon_ {\ell} ^ {- 2} q \left(\mathcal {S} _ {\ell}\right) \log \left(8 \ell^ {2} | \mathcal {Z} | / \delta\right) \\ \leq \frac {d (d + 1)}{2} \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil + \\ 3 \log \left\{\frac {8 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil^ {2} | \mathcal {Z} |}{\delta} \right\} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} q (\mathcal {S} _ {\ell}), \\ \end{array}
$$
where  $(a)$  follows from Lemma E.10. We now relate this quantity to the lower bound and explain below. With probability  $1 - \delta$ ,
$$
\begin{array}{l} \psi_ {\mathrm {L S}} ^ {*} = \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z \in \mathcal {Z}} \frac {| | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}}{\left(z ^ {\top} \theta^ {*} - \alpha\right) ^ {2}} \\ = \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {\ell \leq \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \max  _ {z \in S _ {\ell}} \frac {| | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1}}{\left(z ^ {\top} \theta^ {*} - \alpha\right) ^ {2}} \\ \geq \inf_{\lambda \in P_{\mathcal{X}}}\max_{\ell \leq \lceil \log_{2}(4\Delta^{-1})\rceil}\max_{z\in S_{\ell}}\frac{\left|\left|z\right|\right|^{2}\left(\sum_{x\in\mathcal{X}}\lambda_{x}\frac{x x^{\top}}{\sigma_{x}^{2}}\right)^{-1}}{(4\times 2^{-\ell})^{2}} \\ \stackrel {(a)} {\geq} \frac {1}{1 6 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \inf  _ {\lambda \in P _ {\mathcal {X}}} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} \max  _ {z \in S _ {\ell}} | | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1} \\ \stackrel {(b)} {\geq} \frac {1}{1 6 \left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} \sum_ {\ell = 1} ^ {\left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z \in \mathcal {S} _ {\ell}} | | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\sigma_ {x} ^ {2}}\right) ^ {- 1} \\ \stackrel {(c)} {\geq} \frac {1}{3 2 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} 2 ^ {2 \ell} \inf  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {z \in S _ {\ell}} | | z | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x} \frac {x x ^ {\top}}{\hat {\sigma} _ {x} ^ {2}}\right) ^ {- 1} \\ = \frac {1}{3 2 \left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} \sum_ {\ell = 1} ^ {\left\lceil \log_ {2} (4 \Delta^ {- 1}) \right\rceil} 2 ^ {2 \ell} q (\mathcal {S} _ {\ell}). \\ \end{array}
$$
where  $(a)$  follows from the fact that the maximum of a set of numbers is always greater than the average,  $(b)$  by the fact that the minimum of a sum is greater than the sum of the minimums, and  $(c)$  from Lemma E.9. Consequently, putting the previous pieces together we find that the sample
complexity can be upper bounded as
$$
\begin{array}{l} \sum_ {\ell = 1} ^ {\lceil \log_ {2} (4 \Delta^ {- 1}) \rceil} \sum_ {x \in \mathcal {X}} \lceil \tau_ {\ell} \widehat {\lambda} _ {\ell , x} \rceil \leq \frac {d (d + 1)}{2} \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil \\ { + } { 3 \log \left\{ \frac { 8 \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil ^ { 2 } | \mathcal { Z } | } { \delta } \right\} \sum _ { \ell = 1 } ^ { \lceil \log _ { 2 } ( 4 \Delta ^ { - 1 } ) \rceil } 2 ^ { 2 \ell } q ( \mathcal { S } _ { \ell } ) } \\ \leq \frac {d (d + 1)}{2} \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil \\ + \quad 9 6 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil \log \left\{\frac {8 \lceil \log_ {2} (4 \Delta^ {- 1}) \rceil^ {2} | \mathcal {Z} |}{\delta} \right\} \psi_ {\mathrm {L S}} ^ {*}, \\ \end{array}
$$
where the dependency on  $d(d + 1) / 2$  can be reduced to  $d$  by Proposition A.1. This proves the result.
G Comparing Identification Lower Bounds
In this section, we use the general notation for the level set and best-arm identification problems developed in Section 4. Define
$$
\rho_ {\mathrm {O B J}} ^ {*} := \sigma_ {\max } ^ {2} \min  _ {\lambda \in P _ {\mathcal {X}}} \max  _ {h \in \mathcal {H} _ {\mathrm {O B J}}, q \in \mathcal {Q} _ {\mathrm {O B J}}} \frac {| | h - q | | ^ {2} \left(\sum_ {x \in \mathcal {X}} \lambda_ {x x x} ^ {-}\right) ^ {- 1}}{\left\{(h - q) ^ {\top} \theta^ {*} - b _ {\mathrm {O B J}} \right\} ^ {2}},
$$
which is the lower bound for both BAI and LS by [12] and [35] in the homoskedastic case when upper bounding the variance.
Lemma G.1. In the same setting as Theorem 4.1,
$$
\sigma_ {\min } ^ {2} \cdot \rho_ {\mathrm {O B J}} ^ {*} \leq \psi_ {\mathrm {O B J}} ^ {*} \leq \sigma_ {\max } ^ {2} \cdot \rho_ {\mathrm {O B J}} ^ {*},
$$
and both the upper and lower bounds are tight.
Proof. Consider semidefinite matrices  $A$ ,  $B$ , and  $C$  in  $\mathbb{R}^{d \times d}$  such that  $A \preceq B \preceq C$ . Then  $C^{-1} \preceq B^{-1} \preceq A^{-1}$ . Hence, for  $x \in \mathbb{R}^d$ ,  $\|x\|_{C^{-1}} \leq \|x\|_{B^{-1}} \leq \|x\|_{A^{-1}}$ . Defining
$$
A (\lambda) := \sum_ {i = 1} ^ {K} \lambda_ {i} \frac {x _ {i} x _ {i} ^ {\top}}{\sigma_ {x _ {i}} ^ {2}},
$$
we note that
$$
\frac {1}{\sigma_ {\max } ^ {2}} \sum_ {i = 1} ^ {K} \lambda_ {i} x _ {i} x _ {i} ^ {\top} \preceq A (\lambda) \preceq \frac {1}{\sigma_ {\min } ^ {2}} \sum_ {i = 1} ^ {K} \lambda_ {i} x _ {i} x _ {i} ^ {\top}.
$$
$$
\sigma_ {\min } ^ {2} \| x \| _ {\left(\sum_ {i = 1} ^ {K} \lambda_ {i} x _ {i} x _ {i} ^ {\top}\right) ^ {- 1}} ^ {2} \leq \| x \| _ {A (\lambda) ^ {- 1}} ^ {2} \leq \sigma_ {\max } ^ {2} \| x \| _ {\left(\sum_ {i = 1} ^ {K} \lambda_ {i} x _ {i} x _ {i} ^ {\top}\right) ^ {- 1}} ^ {2}.
$$
Applying this to the result of Theorem 4.1 and invoking Lemma E.8, we see that