SlowGuess/NeurIPS_2023_OCR huggingface dataset:
AutoML Experiment. In our experiments on automated machine learning (AutoML), we use our Q-GP-UCB algorithm to tune the hyperparameters of an SVM for a classification task. Here we consider a Gaussian noise with a variance of  $\sigma^2$  since it is more practical and more commonly used in real-world experiments. The results in Fig. 1 (c) show that our Q-GP-UCB again significantly outperforms the classical GP-UCB. Fig. 1 (d) additionally shows the comparisons with LinUCB and Q-LinUCB. The results again corroborate that BO/kernelized bandit algorithms are considerably superior in real-world applications with highly non-linear reward functions in both the classical and quantum settings. These results here further demonstrate the potential of our Q-GP-UCB to lead to quantum speedup in practical applications.
More Realistic Experiments. We have additionally tested the performance of our Q-GP-UCB algorithm after accounting for the effect of quantum noise, by incorporating into our Qiskit simulations a noise model based on the actual performance of real IBM quantum computers. The results (Fig. 3 in App. J) show that although the addition of quantum noise slightly deteriorates the performance of our Q-GP-UCB, it is still able to significantly outperform classical GP-UCB. Notably, we have additionally performed an experiment using a real quantum computer (details in App. J), in which the performance of our Q-GP-UCB, although further worsened compared to the experiment using simulated noise, is still considerably better than classical GP-UCB (Fig. 4, App. J). Also note that the work of [38] has not shown that Q-LinUCB outperforms LinUCB in the presence of simulated quantum noise. Therefore, the consistently superior performances of our Q-GP-UCB over GP-UCB with both simulated quantum noise and a real quantum computer serve as important new support for the potential practical advantages of quantum bandit algorithms.
Discussion. In our experimental results (e.g., Fig. 1), our Q-GP-UCB usually has relatively larger regrets in the initial stages but quickly achieves zero regret thereafter (i.e., the curve plateaus), which has also been observed in [38] for the Q-LinUCB algorithm. This is because in the initial stages, our Q-GP-UCB explores a smaller number of unique arms than GP-UCB. However, after the initial exploration, our Q-GP-UCB quickly starts to perform reliable exploitation, because the accurate reward observations achieved thanks to our QMC subroutines allow us to rapidly learn the reward function and hence find the optimal arm.
We have introduced the first quantum BO algorithm, named Q-GP-UCB. Our Q-GP-UCB achieves a regret upper bound of  $\mathcal{O}(\mathrm{poly}\log T)$ , which is significantly smaller than the classical regret lower bound of  $\Omega (\sqrt{T})$ . A limitation of our work is that the regret upper bound of our Q-GP-UCB:  $\mathcal{O}((\log T)^{3(d + 1) / 2})$  (ignoring polylog factors) has a worse dependency on the input dimension  $d$  than the regret of the classical GP-UCB:  $\mathcal{O}((\log T)^{d + 1}\sqrt{T})$ . A similar limitation is shared by Q-LinUCB, since its regret has a dependency of  $\mathcal{O}(d^{3 / 2})$  in contrast to the  $\mathcal{O}(d)$  of LinUCB. It is an interesting future work to explore potential approaches to remove this extra dependency on  $d$ . Another limitation of our work is that for the Matérn kernel, we require the reward function to be smooth enough in order to achieve a sub-linear regret upper bound (Sec. 5.7). So, another important future work is to further tighten the regret upper bound of our Q-GP-UCB for the Matérn kernel when the reward function is less smooth, as well as for other kernels encompassing non-smooth functions such as the neural tangent kernel which has been adopted by recent works on neural bandits [13, 42, 44]. Moreover, another interesting future work is to derive regret lower bounds in our setting of quantum kernelized bandits, which can help evaluate the tightness of our regret upper bounds.
This research/project is supported by the National Research Foundation, Singapore under its AI Singapore Programme (AISG Award No: AISG-PhD/2023-01-039J). DesCartes: this research is supported by the National Research Foundation, Prime Minister's Office, Singapore under its Campus for Research Excellence and Technological Enterprise (CREATE) programme.
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A Analysis of the Weighted GP Posterior (3)
Here we show an alternative view of the weighted GP posterior distribution (3) in the RKHS feature space. The notations in this section will be extensively used in the theoretical analyses in the subsequent sections.
We denote the kernel  $k$  by  $k(x,x^{\prime}) = \phi (x)^{\top}\phi (x^{\prime})$ , in which  $\phi (x)$  is the potentially infinite-dimensional feature mapping of  $x$  in the RKHS of  $k$ . Define  $\Phi_s\triangleq [\phi (x_1),\ldots ,\phi (x_s)]^\top$  which is an  $s\times \infty$  matrix. Recall that we have defined  $W_{s}\triangleq \mathrm{diag}(\frac{1}{\epsilon_{1}^{2}},\dots,\frac{1}{\epsilon_{s}^{2}})$  which is an  $s\times s$  diagonal matrix. Define  $V_{s}\triangleq \lambda I + \sum_{\tau = 1}^{s}\frac{1}{\epsilon_{\tau}^{2}}\phi (x_{\tau})\phi (x_{\tau})^{\top} = \lambda I + \Phi_{s}^{\top}W_{s}\Phi_{s}$  which is an  $\infty \times \infty$  matrix, and let  $V_{0}\triangleq \lambda I$ . With these notations, it can be easily verified that  $K_{s} = \Phi_{s}\Phi_{s}^{\top} = [k(x_{\tau},x_{\tau '})]_{\tau ,\tau ' = 1,\ldots ,s}$  and  $\widetilde{K}_s = W_s^{1 / 2}\Phi_s\Phi_s^\top W_s^{1 / 2} = W_s^{1 / 2}K_sW_s^{1 / 2}$ . Moreover, we can also easily show that  $k_{s}(x) = \Phi_{s}\phi (x) = [k(x,x_{\tau})]_{\tau = 1,\ldots ,s}$ ,  $\widetilde{k}_{s}(x) = W_{s}^{1 / 2}\Phi_{s}\phi (x) = W_{s}^{1 / 2}k_{s}(x)$ , and  $\widetilde{Y}_s = W_s^{1 / 2}Y_s$ .
We start by deriving an alternative expression for the weighted GP posterior mean  $\widetilde{\mu}_s(x)$ . To begin with, we have that
$$
\begin{array}{l} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} (\widetilde {K} _ {s} + \lambda I) = \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} (W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda I) \\ = \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} + \lambda \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \\ = \left(\Phi_ {s} ^ {\top} W _ {s} \Phi_ {s} + \lambda I\right) \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \\ = V _ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2}. \\ \end{array}
$$
Now we multiply both sides by  $V_{s}^{-1}$  on the left and by  $(\tilde{K}_s + \lambda I)^{-1}$  on the right, which leads to
$$
V _ {s} ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} = \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} (\widetilde {K} _ {s} + \lambda I) ^ {- 1}
$$
Now we can show that
$$
\begin{array}{l} \widetilde {\mu} _ {s} (x) = \phi (x) ^ {\top} V _ {s} ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} Y _ {s} = \phi (x) ^ {\top} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} (\widetilde {K} _ {s} + \lambda I) ^ {- 1} W _ {s} ^ {1 / 2} Y _ {s} \\ = \widetilde {k} _ {s} ^ {\top} (x) \left(\widetilde {K} _ {s} + \lambda I\right) ^ {- 1} \widetilde {Y} _ {s}. \\ \end{array}
$$
Next, we derive an alternative expression for the weighted GP posterior variance  $\widetilde{\sigma}_s^2 (x)$ . To begin with, we have that
$$
\begin{array}{l} \left(W _ {s} ^ {- 1} + \frac {1}{\lambda} \Phi_ {s} \Phi_ {s} ^ {\top}\right) ^ {- 1} = \left(W _ {s} ^ {- 1 / 2} \left(I + \frac {1}{\lambda} W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2}\right) W _ {s} ^ {- 1 / 2}\right) ^ {- 1} \\ = W _ {s} ^ {1 / 2} \left(I + \frac {1}{\lambda} W _ {s} ^ {1 / 2} \Phi_ {s} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2}\right) ^ {- 1} W _ {s} ^ {1 / 2} \tag {5} \\ = \lambda W _ {s} ^ {1 / 2} \left(\lambda I + \tilde {K} _ {s}\right) ^ {- 1} W _ {s} ^ {1 / 2}. \\ \end{array}
$$
This allows us to show that
$$
\begin{array}{l} \widetilde {\sigma} _ {s} ^ {2} (x) = \lambda \phi (x) ^ {\top} V _ {s} ^ {- 1} \phi (x) \\ = \lambda \phi (x) ^ {\top} \left(\lambda I + \Phi_ {s} ^ {\top} W _ {s} \Phi_ {s}\right) ^ {- 1} \phi (x) \\ \stackrel {(a)} {=} \lambda \phi (x) ^ {\top} \left(\frac {1}{\lambda} I - \frac {1}{\lambda} \Phi_ {s} ^ {\top} \left(W _ {s} ^ {- 1} + \Phi_ {s} \frac {1}{\lambda} \Phi_ {s} ^ {\top}\right) ^ {- 1} \Phi_ {s} \frac {1}{\lambda}\right) \phi (x) \\ = \phi (x) ^ {\top} \phi (x) - \phi (x) ^ {\top} \Phi_ {s} ^ {\top} \left(W _ {s} ^ {- 1} + \frac {1}{\lambda} \Phi_ {s} \Phi_ {s} ^ {\top}\right) ^ {- 1} \Phi_ {s} \frac {1}{\lambda} \phi (x) \\ \stackrel {(b)} {=} \phi (x) ^ {\top} \phi (x) - \phi (x) ^ {\top} \Phi_ {s} ^ {\top} W _ {s} ^ {1 / 2} \left(\lambda I + \widetilde {K} _ {s}\right) ^ {- 1} W _ {s} ^ {1 / 2} \Phi_ {s} \phi (x) \\ = k (x, x) - \widetilde {k} _ {s} ^ {\top} (x) (\widetilde {K} _ {s} + \lambda I) ^ {- 1} \widetilde {k} _ {s} (x), \\ \end{array}
$$
in which we have used the matrix inversion lemma in step  $(a)$ , and used equation (5) in step  $(b)$ .
To summarize, we have derived the following alternative expressions for the weighted GP posterior distribution (3):
$$
\widetilde {\mu} _ {s} (x) = \phi (x) ^ {\top} V _ {s} ^ {- 1} \Phi_ {s} ^ {\top} W _ {s} Y _ {s} = \widetilde {k} _ {s} ^ {\top} (x) (\widetilde {K} _ {s} + \lambda I) ^ {- 1} \widetilde {Y} _ {s}, \tag {6}
$$
$$
\widetilde {\sigma} _ {s} ^ {2} (x) = \lambda \phi (x) ^ {\top} V _ {s} ^ {- 1} \phi (x) = k (x, x) - \widetilde {k} _ {s} ^ {\top} (x) (\widetilde {K} _ {s} + \lambda I) ^ {- 1} \widetilde {k} _ {s} (x).
$$
To begin with, we can lower-bound the total number of iterations (i.e., the total number of queries to the quantum oracles) in  $m$  stages by
$$
\sum_ {s = 1} ^ {m} \frac {C _ {1}}{\epsilon_ {s}} \log (\frac {2 \bar {m}}{\delta}) \geq \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s}} \geq \sqrt {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}. \tag {7}
$$
The first inequality follows since  $C_1 > 1$  (Lemma 1) and  $\log \left(\frac{2\overline{m}}{\delta}\right) \geq \log \left(\frac{2}{\delta}\right) \geq 1$  because we have chosen  $\delta \in (0, 2/e]$ . Suppose that  $\sum_{s=1}^{m} \frac{1}{\epsilon_s^2} > T^2$ , then we have 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
$$
\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \leq T ^ {2}. \tag {8}
$$
Here we prove the growth rate of  $\widetilde{\gamma}_m$ , i.e., the weighted information gain. Recall that  $W_m = \mathrm{diag}\left(\frac{1}{\epsilon_1^2}, \frac{1}{\epsilon_2^2}, \ldots, \frac{1}{\epsilon_m^2}\right)$ , and  $\widetilde{\gamma}_m = \frac{1}{2} \log \det\left(I + \lambda^{-1} \widetilde{K}_m\right)$ . Our proof here follows closely the proof of Theorem 3 from the work of [35]. We will omit the subscript  $m$  as long as the context is clear.
Denote by  $K_{m} = \Phi_{m}\Phi_{m}^{\top} = [k(x_{i},x_{j})]_{i,j = 1,\dots,m}$  the  $m\times m$  covariance matrix, and define  $\widetilde{K}_m = W_m^{1 / 2}\Phi_m\Phi_m^\top W_m^{1 / 2} = W_m^{1 / 2}K_mW_m^{1 / 2}$ . Following Theorem 1 from [35], our kernel  $k$  can be denoted as  $k(x,x^{\prime}) = \sum_{\tau = 1}^{\infty}\lambda_{\tau}\psi_{\tau}(x)\psi_{\tau}(x^{\prime})$ , in which  $\{\lambda_{\tau}\}_{\tau = 1,\ldots ,\infty}$  and  $\{\psi_{\tau}\}_{\tau = 1,\ldots ,\infty}$  represent, respectively, the eigenvalues and eigenfunctions of the kernel  $k$ . We will make use of the projection onto a  $D$ -dimensional RKHS ( $D < \infty$ ), which consists of the first  $D$  features corresponding to the  $D$  largest eigenvalues of the kernel  $k$ . Specifically, define  $k_{P}$  as the  $D$ -dimensional projection in the RKHS of  $k$ , and  $k_{O}$  corresponds to the orthogonal element such that  $k(\cdot ,\cdot) = k_{P}(\cdot ,\cdot) + k_{O}(\cdot ,\cdot)$ . That is, following Section 3.2 from the work of [35], define
$$
k _ {P} \left(x, x ^ {\prime}\right) = \sum_ {\tau = 1} ^ {D} \lambda_ {\tau} \psi_ {\tau} (x) \psi_ {\tau} \left(x ^ {\prime}\right). \tag {9}
$$
We define  $K_P \triangleq [k_P(x_i, x_j)]_{i,j=1,\dots,m}$  and  $K_O \triangleq [k_O(x_i, x_j)]_{i,j=1,\dots,m}$ . Similarly, define  $\widetilde{K}_P = W_m^{1/2} K_P W_m^{1/2}$  and  $\widetilde{K}_O = W_m^{1/2} K_O W_m^{1/2}$ . This implies that  $K_m = K_P + K_O$  and that  $\widetilde{K}_m = \widetilde{K}_P + \widetilde{K}_O$ .
Here to be consistent with [35], we use  $I_{m}$  to denote the  $m \times m$ -dimensional identity matrix. Denote by  $\widetilde{I}(\mathbf{y}_{m};f)$  the information gain about the function  $f$  from the observations  $\mathbf{y}_{m} = [y_{1},\dots,y_{m}]$ . We have that
$$
\begin{array}{l} \widetilde {I} (\mathbf {y} _ {m}; f) = \frac {1}{2} \log \det  \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {m}\right) \\ = \frac {1}{2} \log \det  \left(I _ {m} + \frac {1}{\lambda} \left(\widetilde {K} _ {P} + \widetilde {K} _ {O}\right)\right) \\ = \frac {1}{2} \log \det  \left(\left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) \left(I _ {m} + \frac {1}{\lambda} \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) ^ {- 1} \widetilde {K} _ {O}\right)\right) \tag {10} \\ = \frac {1}{2} \log \det  \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) + \frac {1}{2} \log \det  \left(I _ {m} + \frac {1}{\lambda} \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) ^ {- 1} \widetilde {K} _ {O}\right). \\ \end{array}
$$
The last line follows because  $\det(AB) = \det(A)\det(B)$ . In the following, we will separately upper-bound the two terms in (10).
We start by upper-bounding the first term in equation (10). Denote by  $\psi_{D}(x)$  the  $D$ -dimensional feature vector corresponding to the first  $D$  principle features of the kernel  $k$  (i.e., the features corresponding to the largest  $D$  eigenvalues). Define  $\Psi_{m,D} \triangleq [\psi_{D}(x_{1}), \ldots, \psi_{D}(x_{m})]^{\top}$  which is an  $m \times D$  matrix. Denote by  $\Lambda_{D} = \mathrm{diag}(\lambda_{1}, \ldots, \lambda_{D})$  a  $D \times D$ -diagonal matrix whose diagonal entries consist of the largest  $D$  eigenvalues of the kernel  $k$  in descending order. With this definition, we have that  $K_{P} = \Psi_{m,D} \Lambda_{D} \Psi_{m,D}^{\top}$  and
$$
\tilde {K} _ {P} = W _ {m} ^ {1 / 2} \Psi_ {m, D} \Lambda_ {D} \Psi_ {m, D} ^ {\top} W _ {m} ^ {1 / 2}. \tag {11}
$$
Also define the gram matrix:
$$
\widetilde {G} _ {m} = \Lambda_ {D} ^ {1 / 2} \Psi_ {m, D} ^ {\top} W _ {m} \Psi_ {m, D} \Lambda_ {D} ^ {1 / 2}. \tag {12}
$$
Based on these definitions, we have that
$$
\det  \left(I _ {D} + \frac {1}{\lambda} \widetilde {G} _ {m}\right) = \det  \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right), \tag {13}
$$
which follows from the Weinstein-Aronszajn identity:  $\operatorname*{det}\left(I_D + AA^\top\right) = \operatorname*{det}\left(I_m + A^\top A\right)$  for a  $D\times m$  matrix  $A$ , and we have plugged in  $A = \frac{1}{\sqrt{\lambda}}\Lambda_{D}^{1 / 2}\Psi_{m,D}^{\top}W_{m}^{1 / 2}$ . In addition, the following inequality will also be useful in the subsequent proof:
$$
\left\| \Lambda_ {D} ^ {1 / 2} \psi_ {D} (x) \right\| _ {2} ^ {2} = \sum_ {\tau = 1} ^ {D} \lambda_ {\tau} \psi_ {\tau} ^ {2} (x) = k _ {P} (x, x) \leq k (x, x) \leq 1, \tag {14}
$$
in which we have made use of the definition of  $k_{P}$  from (9), and our assumption (w.l.o.g.) that  $k(x,x^{\prime})\leq 1,\forall x,x^{\prime}$  (Sec. 3.1). In the following, we will denote the trace of a matrix  $A$  by  $\operatorname {Tr}(A)$  Next, we have that
$$
\begin{array}{l} \operatorname {T r} \left(I _ {D} + \frac {1}{\lambda} \widetilde {G} _ {m}\right) = D + \frac {1}{\lambda} \operatorname {T r} \left(\widetilde {G} _ {m}\right) \\ = D + \frac {1}{\lambda} \operatorname {T r} \left(\Lambda_ {D} ^ {1 / 2} \Psi_ {m, D} ^ {\top} W _ {m} \Psi_ {m, D} \Lambda_ {D} ^ {1 / 2}\right) \\ \stackrel {(a)} {=} D + \frac {1}{\lambda} \operatorname {T r} \left(\Lambda_ {D} ^ {1 / 2} \left[ \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \psi_ {D} (x _ {s}) \psi_ {D} ^ {\top} (x _ {s}) \right] \Lambda_ {D} ^ {1 / 2}\right) \\ = D + \frac {1}{\lambda} \operatorname {T r} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \Lambda_ {D} ^ {1 / 2} \psi_ {D} (x _ {s}) \psi_ {D} ^ {\top} (x _ {s}) \Lambda_ {D} ^ {1 / 2}\right) \\ = D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \operatorname {T r} \left(\Lambda_ {D} ^ {1 / 2} \psi_ {D} \left(x _ {s}\right) \psi_ {D} ^ {\top} \left(x _ {s}\right) \Lambda_ {D} ^ {1 / 2}\right) \tag {15} \\ \stackrel {(b)} {=} D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \operatorname {T r} \left(\boldsymbol {\psi} _ {D} ^ {\top} (x _ {s}) \Lambda_ {D} ^ {1 / 2} \Lambda_ {D} ^ {1 / 2} \boldsymbol {\psi} _ {D} (x _ {s})\right) \\ = D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \boldsymbol {\psi} _ {D} ^ {\top} (x _ {s}) \Lambda_ {D} ^ {1 / 2} \Lambda_ {D} ^ {1 / 2} \boldsymbol {\psi} _ {D} (x _ {s}) \\ = D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \left\| \Lambda_ {D} ^ {1 / 2} \psi_ {D} (x _ {s}) \right\| _ {2} ^ {2} \\ \stackrel {(c)} {\leq} D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}, \\ \end{array}
$$
in which  $(a)$  follows because  $\Psi_{m,D}^{\top}W_{m}\Psi_{m,D} = \sum_{s = 1}^{m}\frac{1}{\epsilon_{s}^{2}}\pmb{\psi}_{D}(x_{s})\pmb{\psi}_{D}^{\top}(x_{s})$ ,  $(b)$  has made use of the cyclic property of the trace operator, and  $(c)$  follows from (14) above.
Also note that for a positive definite matrix  $P \in \mathbb{R}^{n \times n}$ , we have that
$$
\log \det  (P) \leq n \log \frac {\operatorname {T r} (P)}{n}. \tag {16}
$$
$$
\begin{array}{l} \log \det  \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) = \log \det  \left(I _ {D} + \frac {1}{\lambda} \widetilde {G} _ {m}\right) \\ \leq D \log \frac {D + \frac {1}{\lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{D} = D \log \left(1 + \frac {1}{\lambda D} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right). \tag {17} \\ \end{array}
$$
The first equality follows from (13), and the inequality results from (16) and (15).
Now we upper-bound the second term in equation (10). To begin with, we have that
$$
\operatorname {T r} \left(\left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) ^ {- 1} \widetilde {K} _ {O}\right) \leq \operatorname {T r} \left(\widetilde {K} _ {O}\right), \tag {18}
$$
which is because the matrix  $\left(I_m + \frac{1}{\lambda}\widetilde{K}_P\right)^{-1}$  is positive semi-definite (PSD) whose largest eigenvalue is upper-bounded by 1, and  $\mathrm{Tr}(P_1P_2)\leq \lambda_{P_1}\mathrm{Tr}(P_2)$  where  $P_{1}$  and  $P_{2}$  are PSD matrices and  $\lambda_{P_1}$  is the largest eigenvalue of  $P_{1}$ . Next, define  $\delta_D\triangleq \sum_{\tau = D + 1}^{\infty}\lambda_{\tau}\psi^{2}$  where  $|\phi_{\tau}(x)|\leq \psi ,\forall x,\tau$ . Then we immediately have that  $k_{O}(x,x^{\prime}) = \sum_{\tau = D + 1}^{\infty}\lambda_{\tau}\phi_{\tau}(x)\phi_{\tau}(x^{\prime})\leq \delta_{D},\forall x,x^{\prime}$ . Therefore, we have that
$$
\operatorname {T r} \left(\widetilde {K} _ {O}\right) = \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} k _ {O} \left(x _ {s}, x _ {s}\right) \leq \delta_ {D} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}. \tag {19}
$$
As a result, we have that
$$
\operatorname {T r} \left(I _ {m} + \frac {1}{\lambda} \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) ^ {- 1} \widetilde {K} _ {O}\right) \leq m + \frac {1}{\lambda} \delta_ {D} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}, \tag {20}
$$
in which the inequality follows from (18) and (19).
Next, again making use of equation (16) and incorporating the upper bound on the trace from (20), we have that
$$
\begin{array}{l} \log \det  \left(I _ {m} + \frac {1}{\lambda} \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {P}\right) ^ {- 1} \widetilde {K} _ {O}\right) \leq m \log \frac {m + \frac {1}{\lambda} \delta_ {D} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{m} \\ \leq m \log \left(1 + \frac {\delta_ {D}}{m \lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \tag {21} \\ \leq m \times \frac {\delta_ {D}}{m \lambda} \sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}} \\ \leq \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \frac {\delta_ {D}}{\lambda}. \\ \end{array}
$$
in which the second last step is because  $\log (1 + z)\leq z,\forall z\in \mathbb{R}$
Finally, combining the upper bounds from 17 and 21, equation (10) can be upper-bounded by
$$
\widetilde {I} (\mathbf {y} _ {m}; f) \leq \frac {1}{2} D \log \left(1 + \frac {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{D \lambda}\right) + \frac {1}{2} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \frac {\delta_ {D}}{\lambda}. \tag {22}
$$
$$
\widetilde {\gamma} _ {m} = \widetilde {I} (\mathbf {y} _ {m}; f) = \frac {1}{2} \log \det  \left(I _ {m} + \frac {1}{\lambda} \widetilde {K} _ {m}\right) \leq \frac {1}{2} D \log \left(1 + \frac {\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}}{D \lambda}\right) + \frac {1}{2} \left(\sum_ {s = 1} ^ {m} \frac {1}{\epsilon_ {s} ^ {2}}\right) \frac {\delta_ {D}}{\lambda}. \tag {23}
$$