SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Also recall that for the standard unweighted maximum information gain  $\gamma_T$ , we have that [35]
$$
\gamma_ {T} = \frac {1}{2} \log \det  \left(I _ {T} + \frac {1}{\lambda} K _ {T}\right) \leq \frac {1}{2} D \log \left(1 + \frac {T}{D \lambda}\right) + \frac {1}{2} T \frac {\delta_ {D}}{\lambda}. \tag {24}
$$
Therefore, the upper bound on the weighted information gain  $\widetilde{\gamma}_m$  is obtained by replacing the  $T$  from the upper bound on standard maximum information gain  $\gamma_T$  by  $\sum_{s=1}^{m} \frac{1}{\epsilon_s^2}$ . This is formalized by the following Theorem.
Theorem 8. [More formal statement of Theorem 1 (a)] Given  $\mathcal{X}_m \triangleq \{x_1, \ldots, x_m\}$ , then we have that
$$
\widetilde {\gamma} _ {m} \leq \frac {1}{2} D \log (1 + \frac {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{D \lambda}) + \frac {1}{2} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \frac {\delta_ {D}}{\lambda},
$$
$$
\gamma_ {\sum_ {s = 1} ^ {m} 1 / \epsilon_ {s} ^ {2}} \leq \frac {1}{2} D \log \left(1 + \frac {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{D \lambda}\right) + \frac {1}{2} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \frac {\delta_ {D}}{\lambda}.
$$
Lastly, note that it has been shown that  $\gamma_{T} = \mathcal{O}(d\log T)$  for the linear kernel and  $\gamma_{T} = \mathcal{O}((\log T)^{d + 1})$  for the SE kernel. Therefore, plugging in Lemma 2, which allows us to upper-bound  $\sum_{s = 1}^{m}1 / \epsilon_s^2$  by  $T^2$ , leads to Corollary 1.
Our proof here follows the outline of the proof of Lemma 2 from the work of [38]. To begin with, we show that  $\operatorname{det}(V_{\tau + 1}) = 2 \operatorname{det}(V_{\tau}), \forall \tau = 0, \dots, m - 1$ , i.e., the amount of collected information is doubled after every stage.
$$
\begin{array}{l} \det  (V _ {\tau + 1}) = \det  \left(\lambda I + \sum_ {j = 1} ^ {\tau + 1} \frac {1}{\epsilon_ {j} ^ {2}} \phi (x _ {j}) \phi (x _ {j}) ^ {\top}\right) \\ = \det  \left(\lambda I + \sum_ {j = 1} ^ {\tau} \frac {1}{\epsilon_ {j} ^ {2}} \phi (x _ {j}) \phi (x _ {j}) ^ {\top} + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} \phi (x _ {\tau + 1}) \phi (x _ {\tau + 1}) ^ {\top}\right) \\ = \det  \left(V _ {\tau} + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} \phi (x _ {\tau + 1}) \phi (x _ {\tau + 1}) ^ {\top}\right) \\ = \det  \left(V _ {\tau} ^ {1 / 2} \left(I + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} V _ {\tau} ^ {- 1 / 2} \phi \left(x _ {\tau + 1}\right) \phi \left(x _ {\tau + 1}\right) ^ {\top} V _ {\tau} ^ {- 1 / 2}\right) V _ {\tau} ^ {1 / 2}\right) \tag {25} \\ = \det  \left(V _ {\tau}\right) \det  \left(I + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} V _ {\tau} ^ {- 1 / 2} \phi \left(x _ {\tau + 1}\right) \phi \left(x _ {\tau + 1}\right) ^ {\top} V _ {\tau} ^ {- 1 / 2}\right) \\ \stackrel {(a)} {=} \det  \left(V _ {\tau}\right) \left(1 + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} \phi \left(x _ {\tau + 1}\right) ^ {\top} V _ {\tau} ^ {- 1 / 2} V _ {\tau} ^ {- 1 / 2} \phi \left(x _ {\tau + 1}\right)\right) \\ = \det  \left(V _ {\tau}\right) \left(1 + \frac {1}{\epsilon_ {\tau + 1} ^ {2}} \left\| \phi \left(x _ {\tau + 1}\right) \right\| _ {V _ {\tau} ^ {- 1}} ^ {2}\right) \\ \stackrel {(b)} {=} \det  \left(V _ {\tau}\right) \left(1 + \frac {\widetilde {\sigma} _ {\tau} ^ {2} \left(x _ {\tau + 1}\right) / \lambda}{\widetilde {\sigma} _ {\tau} ^ {2} \left(x _ {\tau + 1}\right) / \lambda}\right) \\ = 2 \det  \left(V _ {\tau}\right). \\ \end{array}
$$
Step  $(a)$  has made use of the matrix determinant lemma:  $\operatorname*{det}(I + ab^{\top}) = \operatorname*{det}(I + a^{\top}b)$ , and  $(b)$  follows from the expression of the weighted GP posterior variance (6) and our choice of  $\epsilon_{\tau +1} =$
$\widetilde{\sigma}_{\tau}(x_{\tau +1}) / \sqrt{\lambda}$  (line 3 of Algo. 1). An immediate consequence of (25) is that  $\operatorname *{det}(V_m) = 2^m\operatorname *{det}(V_0)$  Recall we have defined earlier that  $V_{0} = \lambda I$  (App. A).
Recall that following the notations of App. A, we have that  $V_{m} = \lambda I + \Phi_{m}^{\top}W_{m}\Phi_{m}$ . Next, we can also show that
$$
\begin{array}{l} \log \frac {\det  (V _ {m})}{\det  (V _ {0})} = \log \frac {\det  (\lambda I + \Phi_ {m} ^ {\top} W _ {m} \Phi_ {m})}{\det  (\lambda I)} \\ = \log \det  \left(I + \frac {1}{\lambda} \left(\Phi_ {m} ^ {\top} W _ {m} ^ {1 / 2}\right) \left(W _ {m} ^ {1 / 2} \Phi_ {m}\right)\right) \\ \stackrel {(a)} {=} \log \det  \left(I + \frac {1}{\lambda} W _ {m} ^ {1 / 2} \Phi_ {m} \Phi_ {m} ^ {\top} W _ {m} ^ {1 / 2}\right) \tag {26} \\ = \log \det  \left(I + \frac {1}{\lambda} \widetilde {K} _ {m}\right) \\ = 2 \widetilde {\gamma} _ {m}, \\ \end{array}
$$
in which step  $(a)$  has again made use of the Weinstein-Aronszajn identity (similar to (13)), and the last two equalities have simply plugged in the expressions for  $\widetilde{K}_s = W_s^{1/2}\Phi_s\Phi_s^\top W_s^{1/2}$  (App. A), and  $\widetilde{\gamma}_m = \frac{1}{2}\log \det(I + \lambda^{-1}\widetilde{K}_m)$  (Sec. 5.1).
According to Corollary 1, for the linear kernel, we have that there exists an absolute constant  $C > 0$  such that  $\widetilde{\gamma}_m \leq C d \log \left( \sum_{s=1}^{m} \frac{1}{\epsilon_s^2} \right)$ . Combining equations (25) and (26), we can show that
$$
m \log 2 = \log \frac {\det  \left(V _ {m}\right)}{\det  \left(V _ {0}\right)} = 2 \widetilde {\gamma} _ {m} \leq 2 C d \log \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right). \tag {27}
$$
$$
\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \geq (2 ^ {m}) ^ {1 / (2 C d)}. \tag {28}
$$
Note that in every stage  $s$ , we query the quantum oracle for  $\frac{C_1}{\epsilon_s} \log \left( \frac{2\overline{m}}{\delta} \right)$  times, in which  $C_1 > 1$ ,  $\delta \in (0, 2 / e]$  and  $\overline{m}$  is an upper bound on the total number of stages. An immediate consequence is that  $\log \left( \frac{2\overline{m}}{\delta} \right) \geq \log \left( \frac{2}{\delta} \right) \geq 1$ . Therefore, the total number of iterations can be analyzed as
$$
\sum_ {s = 1} ^ {m} \frac {C _ {1}}{\epsilon_ {s}} \log \left(\frac {2 \bar {m}}{\delta}\right) \geq \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s}} \geq \sqrt {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}} \geq \sqrt {\left(2 ^ {m}\right) ^ {1 / (2 C d)}}. \tag {29}
$$
Now we derive an upper bound on  $m$  by contradiction. Suppose  $m > \frac{2Cd}{\log 2}\log T$ , this immediately implies that
$$
\sqrt {\left(2 ^ {m}\right) ^ {1 / \left(2 C d\right)}} > T. \tag {30}
$$
This equation, combined with (29), implies that  $\sum_{s=1}^{m} \frac{C_1}{\epsilon_s} \log \left( \frac{2\overline{m}}{\delta} \right) > T$ , which is a contradiction. Therefore, we have that
$$
m \leq \frac {2}{\log 2} C d \log T = \mathcal {O} (d \log T). \tag {31}
$$
That is, for the linear kernel, our algorithm runs for at most  $\mathcal{O}(d\log T)$  stages, which matches the upper bound on the total number of stages for the quantum linear UCB (Q-LinUCB) algorithm, as proved by Lemma 2 from the work of [38].
Next, for the SE kernel, Corollary 1 shows that there exists an absolute constant  $C > 0$  such that  $\widetilde{\gamma}_m \leq C \log^{d+1} \left( \sum_{s=1}^m \frac{1}{\epsilon_s^2} \right)$ . Again combining equations (25) and (26), we can show that
$$
m \log 2 = \log \frac {\det  (V _ {m})}{\det  (V _ {0})} = 2 \widetilde {\gamma} _ {m} \leq 2 C \log^ {d + 1} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right). \tag {32}
$$
$$
\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \geq \exp \left(\left(\frac {m \log 2}{2 C}\right) ^ {1 / (d + 1)}\right). \tag {33}
$$
Now again we derive an upper bound on  $m$  by contradiction. Suppose  $m > \frac{2C}{\log 2} (2\log T)^{d + 1}$ , then this implies that
$$
\sqrt {\exp \left(\left(\frac {m \log 2}{2 C}\right) ^ {1 / (d + 1)}\right)} > T. \tag {34}
$$
This, combined with (29), implies that  $\sum_{s=1}^{m} \frac{C_1}{\epsilon_s} \log \left( \frac{2\overline{m}}{\delta} \right) > T$  which is a contradiction. Therefore
$$
m \leq \frac {2 C}{\log 2} (2 \log T) ^ {d + 1} = \mathcal {O} ((\log T) ^ {d + 1}) \tag {35}
$$
for the SE kernel. This completes the proof.
Our proof here follows closely the proof of Theorem 2 from [9]. Denote by  $\zeta_s$  the observation noise for the observation in stage  $s$ :  $y_s = f(x_s) + \zeta_s$ . Define  $\zeta_{1:s} \triangleq [\zeta_1, \dots, \zeta_s]^\top$  which is an  $s \times 1$  vector. Denote by  $\mathcal{F}_s$  the  $\sigma$ -algebra:  $\mathcal{F}_s = \{x_1, \dots, x_{s+1}, \zeta_1, \dots, \zeta_s\}$ . With this definition, we have that  $x_s$  is  $\mathcal{F}_{s-1}$ -measurable, and  $\zeta_s$  is  $\mathcal{F}_s$ -measurable. Note that we have used  $\phi(\cdot)$  to denote the RKHS feature map of the kernel  $k$ :  $k(x,x') = \phi(x)^\top \phi(x')$ , and here we let  $\phi(x) = k(x,\cdot)$ . Then using the reproducing property, we have that  $f(x) = \langle \phi(x), f \rangle_k = \phi(x)^\top f$ , in which the inner product denotes the inner product induced by the RKHS of  $k$ .
To begin with, we have that
$$
\Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \left(W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I\right) ^ {- 1} = \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2}, \tag {36}
$$
which follows from noting that  $(\Phi_s^\top W_s\Phi_s + \lambda I)\Phi_s^\top W_s^{1 / 2} = \Phi_s^\top W_s^{1 / 2}(W_s^{1 / 2}\Phi_s\Phi_s^\top W_s^{1 / 2} + \lambda I)$ , and then multiplying by  $(\Phi_s^\top W_s\Phi_s + \lambda I)^{-1}$  on the left and by  $(W_{s}^{1 / 2}\Phi_{s}\Phi_{s}^{\top}W_{s}^{1 / 2} + \lambda I)^{-1}$  on the right.
Now let  $f_{s} \triangleq [f(x_{k})]_{k=1,\ldots,s}^{\top} = \Phi_{s}f$  which is an  $s \times 1$ -dimensional vector, and  $\widetilde{f}_{s} \triangleq W_{s}^{1/2}f_{s} = W_{s}^{1/2}\Phi_{s}f$ . Next, we can prove that
$$
\begin{array}{l} | f (x) - \widetilde {k} _ {s} ^ {\top} (x) (\widetilde {K} _ {s} + \lambda I) ^ {- 1} \widetilde {f _ {s}} | = | f (x) - \phi (x) ^ {\top} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \left(W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I\right) ^ {- 1} \widetilde {f _ {s}} | \\ \stackrel {(a)} {=} | f (x) - \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} W _ {s} ^ {1 / 2} \Phi_ {s} f | \\ = | \phi (x) ^ {\top} f - \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} f | \\ = | \lambda \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} f | \\ \leq \| f \| _ {k} \left\| \lambda \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \right\| _ {k} \tag {37} \\ = \| f \| _ {k} \sqrt {\lambda \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \lambda I \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \phi (x)} \\ \leq \| f \| _ {k} \sqrt {\lambda \phi (x) ^ {\top} (\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I) ^ {- 1} (\lambda I + \Phi_ {s} ^ {\top} W _ {s} \Phi_ {s}) (\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I) ^ {- 1} \phi (x)} \\ \leq \| f \| _ {k} \sqrt {\lambda \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \phi (x)} \\ \begin{array}{l} \stackrel {(b)} {\leq} B \widetilde {\sigma} _ {s} (x), \end{array} \\ \end{array}
$$
in which step  $(a)$  has made use of (36), and step  $(b)$  follows from our assumption that  $\| f\| _k\leq B$  (Sec. 3.1) and the expression for the weighted GP posterior variance (6). Next, we have that
$$
\begin{array}{l} | \widetilde {k} _ {s} ^ {\top} (x) (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1: s} | = | \phi (x) ^ {\top} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \left(W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I\right) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1: s} | \\ \stackrel {(a)} {=} | \phi (x) ^ {\top} (\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} W _ {s} ^ {1 / 2} \zeta_ {1: s} | \\ = | \phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} \zeta_ {1: s} | \\ \leq \left\| (\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I) ^ {- 1 / 2} \phi (x) \right\| _ {k} \left\| (\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I) ^ {- 1 / 2} \Phi_ {s} ^ {\top} W _ {s} \zeta_ {1: s} \right\| _ {k} \\ = \sqrt {\phi (x) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \phi (x)} \sqrt {\left(\Phi_ {s} ^ {\top} W _ {s} \zeta_ {1 : s}\right) ^ {\top} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} \zeta_ {1 : s}} \\ = \frac {1}{\sqrt {\lambda}} \widetilde {\sigma} _ {s} (x) \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} \Phi_ {s} \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} W _ {s} ^ {1 / 2} \zeta_ {1 : s}} \\ \stackrel {(b)} {=} \frac {1}{\sqrt {\lambda}} \widetilde {\sigma} _ {s} (x) \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} (W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1 : s}} \\ = \frac {1}{\sqrt {\lambda}} \widetilde {\sigma} _ {s} (x) \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} ^ {1 / 2} W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \left(W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I\right) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1 : s}} \\ = \frac {1}{\sqrt {\lambda}} \widetilde {\sigma} _ {s} (x) \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} ^ {1 / 2} \widetilde {K} _ {s} (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1 : s}}, \\ \end{array}
$$
in which  $(a)$  and  $(b)$  follow from (36). Since  $\widetilde{\mu}_s(x) = \widetilde{k}_s^\top (x)(\widetilde{K}_s + \lambda I)^{-1}\widetilde{Y}_s$  and  $\widetilde{Y}_s = \widetilde{f}_s + W_s^{1 / 2}\zeta_{1:s}$ , then the two equations above combine to tell us that
$$
\begin{array}{l} \left| f (x) - \widetilde {\mu} _ {s} (x) \right| \leq \widetilde {\sigma} _ {s} (x) \left(B + \frac {1}{\sqrt {\lambda}} \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} ^ {1 / 2} \widetilde {K} _ {s} (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1 : s}}\right) \tag {39} \\ \leq \widetilde {\sigma} _ {s} (x) \left(B + \sqrt {\zeta_ {1 : s} ^ {\top} W _ {s} ^ {1 / 2} \widetilde {K} _ {s} (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1 : s}}\right), \\ \end{array}
$$
in which the second inequality follows since  $\lambda = 1 + \eta > 1$ . Again following [9], we apply a few more steps of transformations. Note that for an invertible  $K$ , we have that  $K(K + I)^{-1} = ((K + I)K^{-1})^{-1} = (I + K^{-1})^{-1}$ . Substituting  $K = \widetilde{K}_s + \eta I$ , we have that
$$
(\widetilde {K} _ {s} + \eta I) (\widetilde {K} _ {s} + (\eta + 1) I) ^ {- 1} = \left(I + (\widetilde {K} _ {s} + \eta I) ^ {- 1}\right) ^ {- 1}.
$$
$$
\begin{array}{l} \zeta_ {1: s} ^ {\top} W _ {s} ^ {1 / 2} \widetilde {K} _ {s} (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1: s} \leq \zeta_ {1: s} ^ {\top} W _ {s} ^ {1 / 2} (\widetilde {K} _ {s} + \eta I) (\widetilde {K} _ {s} + (1 + \eta) I) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1: s} \\ = \zeta_ {1: s} ^ {\top} W _ {s} ^ {1 / 2} \left(\left(\tilde {K} _ {s} + \eta I\right) ^ {- 1} + I\right) ^ {- 1} W _ {s} ^ {1 / 2} \zeta_ {1: s} \\ = \left\| W _ {s} ^ {1 / 2} \zeta_ {1: s} \right\| _ {((\widetilde {K} _ {s} + \eta I) ^ {- 1} + I) ^ {- 1}} ^ {2}. \\ \end{array}
$$
This allows us to further re-write (39) as
$$
\left| f (x) - \widetilde {\mu} _ {s} (x) \right| \leq \widetilde {\sigma} _ {s} (x) \left(B + \left\| W _ {s} ^ {1 / 2} \zeta_ {1: s} \right\| _ {((\widetilde {K} _ {s} + \eta I) ^ {- 1} + I) ^ {- 1}}\right). \tag {40}
$$
Recall that  $W_{s} = \mathrm{diag}\left(\frac{1}{\epsilon_{1}^{2}},\frac{1}{\epsilon_{2}^{2}},\ldots ,\frac{1}{\epsilon_{s}^{2}}\right)$ , therefore,  $W_{s}^{1 / 2}\zeta_{1:s} = [\zeta_{k}\frac{1}{\epsilon_{k}} ]_{k = 1,\dots,s}^{\top}$ . Of note, since we have that  $|f(x_{s}) - y_{s}|\leq \epsilon_{s},\forall s = 1,\ldots ,m$  (with high probability, refer to the beginning of Sec. 5), so, the noise  $\zeta_s = y_s - f(x_s)$  is bounded:  $|\zeta_s|\leq \epsilon_s,\forall s = 1,\dots ,m$ . In other words,  $\zeta_{s}$  is  $\epsilon_{s}$ -sub-Gaussian.
To begin with, note that  $\epsilon_{k}$  is  $\mathcal{F}_{k - 1}$ -measurable. This can be seen recursively: conditioned on  $\{x_{1}\}$ ,  $\epsilon_{1} = \widetilde{\sigma}_{0}(x_{1}) / \sqrt{\lambda}$  is a deterministic constant (i.e., predictable); conditioned on  $\{x_{1}, x_{2}\}$ ,  $\widetilde{\sigma}_{1}(\cdot)$  is predictable because it depends on  $x_{1}$  and  $\epsilon_{1}$  (via the weight  $\frac{1}{\epsilon_1^2}$ ), and hence  $\epsilon_{2} = \widetilde{\sigma}_{1}(x_{2}) / \sqrt{\lambda}$  is predictable conditioned on  $\{x_{1}, x_{2}\}$ . By induction, conditioned on  $\{x_{1}, \ldots, x_{k}\}$ ,  $\epsilon_{k} = \widetilde{\sigma}_{k - 1}(x_{k}) / (\sqrt{\lambda})$  is predictable. Therefore,  $\epsilon_{k}$  is  $\mathcal{F}_{k - 1}$ -measurable, because  $\mathcal{F}_{k - 1} = \{x_{1}, \ldots, x_{k}, \zeta_{1}, \ldots, \zeta_{k - 1}\}$ .
Next, note that since  $\zeta_{k}$  is  $\mathcal{F}_k$ -measurable, we have that  $\frac{\zeta_k}{\epsilon_k}$  is  $\mathcal{F}_k$ -measurable. Moreover, conditioned on  $\mathcal{F}_{k-1}$  (i.e.,  $\epsilon_k$  is a predictable),  $\frac{\zeta_k}{\epsilon_k}$  is 1-sub-Gaussian, because  $\zeta_k$  is  $\epsilon_k$ -sub-Gaussian as discussed above. To summarize, we have that every element  $\frac{\zeta_k}{\epsilon_k}$  of this noise vector  $W_s^{1/2} \zeta_{1:s}$  is  $(a)\mathcal{F}_k$ -measurable and  $(b)$  1-sub-Gaussian conditionally on  $\mathcal{F}_{k-1}$ . Therefore, we can apply Theorem 1 of [9] to 1-sub-Gaussian noise, to show that
$$
\left\| W _ {s} ^ {1 / 2} \zeta_ {1: s} \right\| _ {((\widetilde {K} _ {s} + \eta I) ^ {- 1} + I) ^ {- 1}} \leq \sqrt {2 \log \frac {\sqrt {\det ((1 + \eta) I + \widetilde {K} _ {s})}}{\delta}} \tag {41}
$$
with probability of at least  $1 - \delta$ . When applying Theorem 1 of [9], we have also replaced the original unweighted covariance matrix  $K_{s}$  by the corresponding weighted covariance matrix  $\widetilde{K}_{s} = W_{s}^{1/2} K_{s} W_{s}^{1/2}$ . This is possible because every  $\epsilon_{k}$  is  $\mathcal{F}_{k-1}$ -measurable. More concretely, in the proof of Theorem 1 of [9], when defining the super-martingale  $\{M_{t}\}_{t}$  which is conditioned on  $\mathcal{F}_{\infty}$ , we only need to replace the covariance matrix  $K_{t}$  (for the multivariate Gaussian distribution of  $h(x_{1}), \ldots, h(x_{t})$ ) by  $\widetilde{K}_{t} = W_{t}^{1/2} K_{t} W_{t}^{1/2}$ , because both the original  $K_{t}$  and our  $\widetilde{K}_{t}$  only require conditioning on  $\{x_{1}, \ldots, x_{t}\}$ .
Combining the two equations above allows us to show that
$$
\left| f (x) - \widetilde {\mu} _ {s} (x) \right| \leq \widetilde {\sigma} _ {s} (x) \left(B + \sqrt {2 \log \frac {\sqrt {\det  ((1 + \eta) I + \widetilde {K} _ {s})}}{\delta}}\right) \tag {42}
$$
with probability of  $\geq 1 - \delta$ . Next, we can further upper-bound the log determinant term:
$$
\begin{array}{l} \log \det \left((1 + \eta) I + \widetilde {K} _ {s}\right) \stackrel {(a)} {\leq} s \log (1 + \eta) + \log \det \left(I + \frac {1}{1 + \eta} \widetilde {K} _ {s}\right) \\ \stackrel {(b)} {\leq} \log \det  \left(I + \frac {1}{\lambda} \widetilde {K} _ {s}\right) + s \eta \tag {43} \\ \stackrel {(c)} {=} 2 \widetilde {\gamma} _ {s} + s \eta \\ \stackrel {(d)} {\leq} 2 \widetilde {\gamma} _ {s} + 2, \\ \end{array}
$$
in which  $(a)$  follows since  $\operatorname{det}(AB) = \operatorname{det}(A)\operatorname{det}(B)$  and  $\operatorname{det}(cA) \leq c^s\operatorname{det}(A)$  for a scalar  $c$  and an  $s \times s$ -matrix  $A$ ,  $(b)$  has made use of  $\log(1 + z) \leq z$ ,  $(c)$  has made use of the definition of  $\gamma_s$ , and  $(d)$  follows because  $\eta = 2 / T$  and  $s \leq T$ . This eventually allows us to show that
$$
\begin{array}{l} | f (x) - \widetilde {\mu} _ {s} (x) | \leq \widetilde {\sigma} _ {s} (x) \left(B + \sqrt {2 \log \frac {\sqrt {\det \left((1 + \eta) I + \widetilde {K} _ {s}\right)}}{\delta}}\right) \tag {44} \\ \leq \widetilde {\sigma} _ {s} (x) \left(B + \sqrt {2 \left(\widetilde {\gamma} _ {s} + 1 + \log \left(1 / \delta\right)\right)}\right), \\ \end{array}
$$
Of note, although the work of [15] has also adopted weighted GP regression in a similar way to our (3), our proof technique here cannot be applied in their analysis. This is because the work of [15] has chosen the weight to be  $w_{s} = \eta^{-s}$  for  $\eta \in (0,1)$  and hence  $W_{s} = \mathrm{diag}[w_{1},\ldots ,w_{s}]$ . Therefore, every element of the scaled noise vector  $W_{s}^{1 / 2}\zeta_{1:s}$  (i.e.,  $\zeta_k\sqrt{w_k}$ ) is not guaranteed to be sub-Gaussian. As a result, this makes it infeasible for them to apply the self-normalizing concentration inequality from Theorem 1 of [9], i.e., our (41) above does not hold in their case. The work of [15] has come up with other novel proof techniques suited to their setting. Therefore, our proof here only works in our problem setting of quantum BO.
To begin with, we can show that
$$
\begin{array}{l} r _ {s} = f (x ^ {*}) - f (x _ {s}) \stackrel {(a)} {\leq} \widetilde {\mu} _ {s - 1} (x ^ {*}) + \beta_ {s} \widetilde {\sigma} _ {s - 1} (x ^ {*}) - f (x _ {s}) \\ \stackrel {(b)} {\leq} \tilde {\mu} _ {s - 1} \left(x _ {s}\right) + \beta_ {s} \tilde {\sigma} _ {s - 1} \left(x _ {s}\right) - f \left(x _ {s}\right) \tag {45} \\ \stackrel {(c)} {\leq} 2 \beta_ {s} \widetilde {\sigma} _ {s - 1} (x _ {s}) \stackrel {(d)} {=} 2 \beta_ {s} \epsilon_ {s} \sqrt {\lambda}, \\ \end{array}
$$
in which  $(a)$  and  $(c)$  follow from Theorem 3,  $(b)$  results from our policy for selecting  $x_{s}$  (line 2 of Algo. 1), and  $(d)$  follows from our design of  $\epsilon_s = \widetilde{\sigma}_{s - 1}(x_s) / \sqrt{\lambda}$  (line 3 of Algo. 1). As a result, the total regret in stage  $s$  can be upper bounded by:
$$
\frac {C _ {1}}{\epsilon_ {s}} \log \frac {2 \bar {m}}{\delta} \times 2 \beta_ {s} \epsilon_ {s} \sqrt {\lambda} = 2 C _ {1} \beta_ {s} \sqrt {\lambda} \log \frac {2 \bar {m}}{\delta}. \tag {46}
$$
Then an upper bound on the total cumulative regret can be obtained by summing up the regrets from all  $m$  stages:
$$
R _ {T} \leq \sum_ {s = 1} ^ {m} 2 C _ {1} \beta_ {s} \sqrt {\lambda} \log \frac {2 \bar {m}}{\delta} \leq m 2 C _ {1} \beta_ {m} \sqrt {\lambda} \log \frac {2 \bar {m}}{\delta} = \mathcal {O} \left(m \beta_ {m} \log \frac {\bar {m}}{\delta}\right). \tag {47}
$$
For the linear kernel, recall from Theorem 3 and Corollary 1 that  $\beta_{m} = \mathcal{O}(\sqrt{\widetilde{\gamma}_{m - 1} + \log(1 / \delta)}) = \mathcal{O}(\sqrt{d\log T + \log(1 / \delta)})$ . Also recall that Theorem 2 tells us that  $m \leq \overline{m} = \mathcal{O}(d\log T)$ . Therefore, for the linear kernel,
$$
R _ {T} = \mathcal {O} \left(d \log T \sqrt {d \log T + \log (1 / \delta)} \log \frac {d \log T}{\delta}\right) = \mathcal {O} \left(d ^ {3 / 2} \log^ {3 / 2} T \log (d \log T)\right), \tag {48}
$$
in which we have ignored the dependency on  $\log (1 / \delta)$  in the second step to get a cleaner expression.
For the SE kernel, recall from Theorem 3 and Corollary 1 that  $\beta_{m} = \mathcal{O}\bigl (\sqrt{\widetilde{\gamma}_{m - 1} + \log(1 / \delta)}\bigr) =$ $\mathcal{O}(\sqrt{(\log T)^{d + 1} + \log(1 / \delta)})$  . Also recall that Theorem 2 tells us that  $m\leq \overline{m} = \mathcal{O}((\log T)^{d + 1})$  Therefore, for the linear kernel,
$$
\begin{array}{l} R _ {T} = \mathcal {O} \left(\left(\log T\right) ^ {d + 1} \sqrt {\left(\log T\right) ^ {d + 1} + \log (1 / \delta)} \log \frac {(\log T) ^ {d + 1}}{\delta}\right) \tag {49} \\ = \mathcal {O} \left((\log T) ^ {3 (d + 1) / 2} \log ((\log T) ^ {d + 1})\right), \\ \end{array}
$$