SlowGuess/NeurIPS_2023_OCR huggingface dataset:
A.1 Broader impacts
Automated systems that use human feedback are being used in an increasing number of contexts, spanning everything from predicting user preferences to finetuning language models. It is important to ensure that such systems are as accurate as possible; this naturally requires humans to respond in an accurate and consistent manner. Using the perceptual adjustment query to collect data in such settings could lead to more expressive responses without heavy cognitive burdens on users. Furthermore, providing a user the additional context of a continuous spectrum of items may result in more self-consistent responses. The downstream effects of collecting more expressive and self-consistent human responses could lead to improved models or entirely new paradigms of model development for a myriad of problem settings. With these advantages come associated risks as well. Due to how expressive the responses to PAQs are, the effects of adversarial responses may be magnified. That is, if an adversary purposely chooses to respond in an antagonistic manner, models trained with PAQs may be trained poorly or in opposition to the stated goal. Mitigating such effects likely requires a holistic approach from both the query design and robust model design perspectives.
From a data collection perspective, the perceptual adjustment query requires access to a continuous space where each point corresponds to an item. In many applications, assuming access to this continuous space is reasonable. For example, if we use PAQs to characterize color blindness, then a natural continuous space is the RGB color space. In general, we situate our data collection within the latent space of a generative model, such as a GAN. While GANs are capable of producing extremely high fidelity images, these images are not always free of semantically meaningful artifacts. Our query design and modeling assumptions do not explicitly consider the case where a portion of the continuous spectrum of items may be corrupted. Furthermore, because our work is an initial exploration into low-rank matrix estimation from inverted measurements, we have not considered scenarios such as unbounded noise or heavier-tailed sensing vectors, and we have not established information-theoretic lower bounds for the inverted measurement paradigm. We hope that further exploration of the inverted measurement paradigm will lead to a rich line of follow-up work.
B Simulation details
In this section, we provide details for the simulation results presented in Figure 2. For our experiments, we adopt a normalized version of the setup of [30] and form the metric  $\pmb{\Sigma}^{\star}$  by  $\pmb{\Sigma}^{\star} = \pmb{L}\pmb{L}^{\top} / \| \pmb{L}\pmb{L}^{\top}\|_{F}$ , where  $\pmb{L}$  is a  $50\times 10$  matrix with i.i.d. Gaussian entries. We sweep the number of query responses  $N$ , estimate the metric with  $\widehat{\Sigma}$ , and report the normalized estimation error  $\| \pmb{\Sigma}^{\star} - \widehat{\pmb{\Sigma}}\|_{F} / \| \pmb{\Sigma}^{\star}\|_{F}$  averaged over 10 independent trials. For each query response, items are drawn i.i.d. from a standard multivariate normal distribution, similar to [45].
Pairwise comparison setup. For pairwise comparisons, we use value of  $y = 10$  to denote the squared distance at which items become dissimilar, following our distance-based model for human perception (see Section 2.1). For the  $i$ -th pairwise comparisons, we draw two items  $\boldsymbol{x}_1^{(i)}, \boldsymbol{x}_2^{(i)}$  i.i.d. from a standard multivariate normal distribution. We record the pairwise comparison outcome  $\epsilon_i \in \{-1, +1\}$  as  $\epsilon_i = \mathrm{sign}(\| \boldsymbol{x}_1^{(i)} - \boldsymbol{x}_2^{(i)} \|_{\Sigma^*}^2 - y)$ . To estimate the metric from pairwise comparisons, we utilize a nuclear-norm regularized hinge loss and solve the following optimization problem:
$$
\widehat {\boldsymbol {\Sigma}} _ {\mathrm {P C}} \in \underset {\boldsymbol {\Sigma} \succeq \mathbf {0}} {\arg \min} \frac {1}{N} \sum_ {i = 1} ^ {N} \max \{0, y - \epsilon_ {i} \| \boldsymbol {x} _ {1} ^ {(i)} - \boldsymbol {x} _ {2} ^ {(i)} \| _ {\boldsymbol {\Sigma}} ^ {2} \} + \lambda_ {\mathrm {P C}} \| \boldsymbol {\Sigma} \| _ {*}.
$$
Triplet setup. For the  $i$ -th triplet, we draw three items  $\pmb{x}_1^{(i)},\pmb{x}_2^{(i)},\pmb{x}_3^{(i)}$  i.i.d. from a standard multivariate normal distribution and record the outcome  $\epsilon_{i}\in \{-1, + 1\}$  as  $\epsilon_{i} = \mathrm{sign}(\| \pmb{x}_{1}^{(i)} - \pmb{x}_{2}^{(i)}\|)$ .
$\pmb{x}_2^{(i)}\|_{\pmb{\Sigma}^\star}^2 -\| \pmb{x}_1^{(i)} - \pmb{x}_3^{(i)}\|_{\pmb{\Sigma}^\star}^2)$ . To estimate the metric from triplet responses, we follow [19] and utilize a nuclear-norm regularized hinge loss and solve the following optimization problem:
$$
\widehat {\pmb {\Sigma}} _ {\mathrm {T}} \in \underset {\pmb {\Sigma} \succeq \mathbf {0}} {\arg \min} \frac {1}{N} \sum_ {i = 1} ^ {N} \max \left\{0, 1 - \epsilon_ {i} \left(\| \pmb {x} _ {1} ^ {(i)} - \pmb {x} _ {2} ^ {(i)} \| _ {\pmb {\Sigma}} ^ {2} - \| \pmb {x} _ {1} ^ {(i)} - \pmb {x} _ {3} ^ {(i)} \| _ {\pmb {\Sigma}} ^ {2}\right) \right\} + \lambda_ {\mathrm {T}} \| \pmb {\Sigma} \| _ {*}.
$$
Ranking- $k$  query setup. For the  $i$ -th ranking query with a reference item  $\pmb{x}_0$  and  $k$  items  $\pmb{x}_1, \dots, \pmb{x}_k$  to be ranked, we draw all items i.i.d. from a standard multivariate normal distribution. For each item  $\pmb{x}_k$ , we compute the squared distance  $\| \pmb{x}_0 - \pmb{x}_k \|_{\pmb{\Sigma}^*}^2$ . To determine the ranking of items, we sort the items based on this squared distance. To estimate the metric, we follow the approach of [14] and decompose the full ranking into its constituent triplets. A ranking consisting of  $k$  items can equivalently be decomposed into  $k(k-1)/2$  triplet responses. To estimate the metric, we decompose each ranking query and use the triplet estimator presented above with regularization parameter  $\lambda_{\mathbb{R}}$  to obtain estimate  $\widehat{\Sigma}_{\mathbb{R}-k}$ .
PAQ setup. For the  $i$ -th PAQ response, we draw the reference item  $\boldsymbol{x}_i$  and query vector  $\boldsymbol{a}_i$  i.i.d. from the standard multivariate normal distribution. We then receive a scaling  $\gamma_i^2$  satisfying  $\gamma_i^2 = y / \boldsymbol{a}_i^\top \boldsymbol{\Sigma}^\star \boldsymbol{a}_i$ , with  $y = 10$ . To perform estimation, we leverage our method presented in Section 3. Our theoretical results indicate that the averaging parameter  $m$  should be set to 1 in the noiseless setting. Furthermore, the truncation threshold  $\tau$  is large relative to our responses  $\gamma_i^2$ , meaning no truncation is employed. As a result, we solve the nuclear-norm regularized trace regression problem
$$
\widehat {\pmb {\Sigma}} _ {\mathrm {P A Q}} \in \underset {\pmb {\Sigma} \succeq \mathbf {0}} {\arg \min} \frac {1}{N} \sum_ {i = 1} ^ {N} \left(\langle \pmb {a} _ {i} \pmb {a} _ {i} ^ {\top}, \pmb {\Sigma} \rangle - \frac {y}{\gamma_ {i} ^ {2}}\right) ^ {2} + \lambda_ {\mathrm {P A Q}} \| \pmb {\Sigma} \| _ {*}.
$$
In all cases above, we solve all optimization problems with cvxpy and normalize the estimated metric  $\widehat{\Sigma}_{\{\mathrm{PC},\mathrm{T},\mathrm{R - }k,\mathrm{PAQ}\}}$  to be unit Frobenius norm to ensure consistent scaling when compared against the true metric  $\pmb{\Sigma}^{\star}$ . We use a value of 0.05 for all regularization parameters  $\lambda_{\{\mathrm{PC},\mathrm{T},\mathrm{R - }k,\mathrm{PAQ}\}}$  and observe similar performance trends for other choices of regularization parameter.
C Scale equivariance
In this section, we verify the scale-equivariance of our derived theoretical bounds (13) and (14). Specifically, we denote by  $\Sigma^{\star}$  and  $\widehat{\Sigma}$  the ground-truth and the estimated matrices corresponding to value  $y$ . We denote by  $\Sigma_{c}^{\star}$  and  $\widehat{\Sigma}_{c}$  the ground-truth and estimated matrices corresponding to value  $c_{\mathrm{scale}}y$  for any  $c_{\mathrm{scale}} > 0$ . By definition, we have  $\Sigma_{c}^{\star} = c_{\mathrm{scale}}\Sigma^{\star}$ , and it can be verified that solving the optimization program (8) yields  $\widehat{\Sigma}_{c} = c_{\mathrm{scale}}\widehat{\Sigma}$ . Hence, one expects the error bound to scale as  $c_{\mathrm{scale}}$ . To verify this linear scaling in  $c_{\mathrm{scale}}$ , we confirm that the noise  $\eta$  scales as  $c_{\mathrm{scale}}$ .
Under the ground-truth metric  $\pmb{\Sigma}^{\star}$ , if the user responds with an item that is a distance  $y + \eta$  away from the reference item, then that same item is a distance  $c_{\mathrm{scale}}(y + \eta)$  away from the reference under the scaled setting. As a result, the noise scales as a result of the choice of  $y$ . Therefore, the following values in the upper bound (13) can be written as scaled versions of their corresponding "ground-truth" values.
$\eta = c_{\mathrm{scale}}\eta_{\star}$
Noise upper bound
Noise variance
$\eta^{\uparrow} = c_{\mathrm{scale}}\eta_{\star}^{\uparrow}$
$\nu_{\eta}^{2} = c_{\mathrm{scale}}^{2}$ $\nu_{\eta ,\star}$
Noise median
Boundary upper bound
Singular values
$\mu_{y} = c_{\mathrm{scale}}\mu_{y}^{\star}$
$$
y ^ {\uparrow} = c _ {\text {s c a l e}} \left(y _ {\star} + \eta_ {\star} ^ {\uparrow}\right)
$$
$$
\sigma_ {k} = c _ {\text {s c a l e}} \sigma_ {k} ^ {\star}, k = 1, \dots , r
$$
Substituting these scaled expressions into the upper bounds (13) and (14), we have
$$
\| \widehat {\boldsymbol {\Sigma}} _ {c} - \boldsymbol {\Sigma} _ {c} ^ {\star} \| _ {F} \leq c _ {\text {s c a l e}} C ^ {\prime} \frac {\left(\sigma_ {1} ^ {\star}\right) ^ {2}}{\sigma_ {r} ^ {\star}} \frac {\left(y _ {\star} ^ {\uparrow}\right) ^ {4 / 3} \left(\nu_ {\eta , \star} ^ {2}\right) ^ {1 / 3}}{\left(\mu_ {y} ^ {\star}\right) ^ {2}} r ^ {3 / 2} \left(\frac {d}{N}\right) ^ {1 / 3}
$$
in the high-noise regime and
$$
\| \widehat {\boldsymbol {\Sigma}} _ {c} - \boldsymbol {\Sigma} _ {c} ^ {\star} \| _ {F} \leq c _ {\text {s c a l e}} C ^ {\prime} \frac {\left(\sigma_ {1} ^ {\star}\right) ^ {2}}{\sigma_ {r} ^ {\star}} \left(\frac {y _ {\star} ^ {\uparrow}}{\mu_ {y} ^ {\star}}\right) ^ {2} r ^ {3 / 2} \left(\frac {d}{N}\right) ^ {1 / 2}
$$
in the low-noise regime. Note that the constant  $C'$  is independent of  $c_{\mathrm{scale}}$ .
D Preliminaries and notation
In this section, we provide an overview of the key tools that are utilized in our proofs. We first introduce notation which is used throughout our proofs.
Notation. For two real numbers  $a$  and  $b$ , let  $a \wedge b = \min \{a, b\}$ . Given a vector  $\boldsymbol{x} \in \mathbb{R}^d$ , denote  $\| \boldsymbol{x} \|_1$  and  $\| \boldsymbol{x} \|_2$  as the  $\ell_1$  and  $\ell_2$  norm, respectively. Denote  $\mathcal{S}^{d-1} := \{x \in \mathbb{R}^d : \| x \|_2 = 1\}$  to be the set of vectors with unit  $\ell_2$  norm. Given a matrix  $\boldsymbol{A} \in \mathbb{R}^{d_1 \times d_2}$ , denote  $\| \boldsymbol{A} \|_F$ ,  $\| \boldsymbol{A} \|_*$ , and  $\| \boldsymbol{A} \|_{\mathrm{op}}$  as the Frobenius norm, nuclear norm, and operator norm, respectively. Denote  $\mathbb{S}^{d \times d} = \{\boldsymbol{A} \in \mathbb{R}^{d \times d} : \boldsymbol{A} = \boldsymbol{A}^\top\}$  to be the set of symmetric  $d \times d$  matrices. Denote  $\boldsymbol{A} \succeq \boldsymbol{0}$  to mean  $\boldsymbol{A}$  is symmetric positive semi-definite. For  $\boldsymbol{A} \succeq \boldsymbol{0}$ , define the (pseudo-) inner product  $\langle x, y \rangle_A = x^\top A y$  and the associated (pseudo-) norm  $\| x \|_A = \sqrt{x^\top A x}$ . For matrices  $\boldsymbol{A}, B \in \mathbb{R}^{d_1 \times d_2}$ , denote  $\langle A, B \rangle = \operatorname{tr}(A^\top B)$  as the Frobenius inner product.
We use the notation  $f(x) \lesssim g(x)$  to denote that there exists some universal positive constant  $c > 0$ , such that  $f(x) \leq c \cdot g(x)$ , and use the notation  $f(x) \gtrsim g(x)$  when  $g(x) \lesssim f(x)$ .
We define random matrices
$$
\bar {A} = \bar {\gamma} ^ {2} \boldsymbol {a} \boldsymbol {a} ^ {\top} = \frac {y + \bar {\eta}}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \boldsymbol {a} \boldsymbol {a} ^ {\top} \tag {15}
$$
$$
\widetilde {\boldsymbol {A}} = \widetilde {\gamma} ^ {2} \boldsymbol {a} \boldsymbol {a} ^ {\top} = \left(\frac {y + \bar {\eta}}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \wedge \tau\right) \boldsymbol {a} \boldsymbol {a} ^ {\top} \tag {16}
$$
as the sensing matrix formed with the  $m$ -averaged responses  $\bar{\gamma}$  and truncated responses  $\widetilde{\gamma}$ , respectively.
D.1 Inverted measurement sensing matrices result in estimation bias
Recall from Equation (4) that the random sensing matrix  $A^{\mathrm{inv}}$  takes the form
$$
\boldsymbol {A} ^ {\text {i n v}} = \frac {y + \eta}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \boldsymbol {a} \boldsymbol {a} ^ {\top}.
$$
Standard trace regression analysis assumes that for some sensing matrix  $\mathbf{A}$  and measurement noise  $\eta$ ,  $\mathbb{E}[\eta \mathbf{A}] = \mathbf{0}$ . Specifically, it is often typically assumed that  $\eta$  is zero-mean conditioned on the sensing matrix  $\mathbf{A}$ . The following lemma shows that for the inverted measurements, we have  $\mathbb{E}[\eta \mathbf{A}^{\mathrm{inv}}] \neq 0$ , resulting in bias in estimation.
Lemma 1. Let  $A^{\mathrm{inv}}$  be the random matrix defined in Eq. (4) and  $\eta$  be the measurement noise. Then
$$
\mathbb {E} \left[ \eta A ^ {\text {i n v}} \right] \neq 0.
$$
The proof of Lemma 1 is provided in Appendix D.6.1. Hence, utilizing established low-rank matrix estimators for inverted measurements results in biased estimation.
D.2 Sub-exponential random variables
Our analysis utilizes properties of sub-exponential random variables, a class of random variables with heavier tails than the Gaussian distribution.
Lemma 2 (Moment bounds for sub-exponential random variables [46, Proposition 2.7.1(b)]). If  $X$  is a sub-exponential random variable, then there exists some constant  $c$  (only dependent on the distribution of the random variable  $X$ ) such that for all integers  $p \geq 1$ ,
$$
\left(\mathbb {E} | X | ^ {p}\right) ^ {1 / _ {p}} \leq c p.
$$
D.3 Bernstein's inequality
In our proofs, we use Bernstein's inequality to bound the sums of independent sub-exponential random variables.
Lemma 3 (Bernstein's inequality, adapted from [47, Theorem 2.10]). Let  $X_{1},\ldots ,X_{n}$  be independent real-valued random variables. Assume there exist positive numbers  $u_{1}$  and  $u_{2}$  such that
$$
\mathbb {E} \left[ X _ {i} ^ {2} \right] \leq u _ {1} \quad a n d \quad \mathbb {E} \left[ | X _ {i} | ^ {p} \right] \leq \frac {p !}{2} u _ {1} u _ {2} ^ {p - 2} f o r a l l i n e g e r s p \geq 2,
$$
Then for all  $t > 0$
$$
\mathbb {P} \left(\left| \frac {1}{n} \sum_ {i = 1} ^ {n} \left(X _ {i} - \mathbb {E} \left[ X _ {i} \right]\right) \right| \geq \sqrt {\frac {2 u _ {1} t}{n}} + \frac {u _ {2} t}{n}\right) \leq 2 \exp (- t).
$$
D.4 Moments of the ratios of quadratic forms
The quadratic term  $\boldsymbol{a}^{\top}\boldsymbol{\Sigma}^{\star}\boldsymbol{a}$  appears in the denominator of our sensing matrices, so we use the following result to quantify the moments of the ratios of quadratic forms.
Lemma 4. There exists an absolute constant  $c > 0$  such that the following is true. Let  $\pmb{a} \sim \mathcal{N}(\pmb{0},\pmb{I}_d)$ ,  $\pmb{\Sigma}^{\star} \in \mathbb{R}^{d\times d}$  be any PSD matrix with rank  $r$ , and  $\pmb{U} \in \mathbb{R}^{d\times d}$  be an arbitrary symmetric matrix.
(a) Suppose that  $r > 8$ . Then we have
$$
\mathbb {E} \left(\frac {1}{\boldsymbol {a} ^ {T} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) ^ {4} \leq \frac {c}{\sigma_ {r} ^ {4} r ^ {4}}.
$$
(b) Suppose that  $r > 2$ . Then we have
$$
\mathbb {E} \left(\frac {\boldsymbol {a} ^ {\top} \boldsymbol {U} \boldsymbol {a}}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) \leq \frac {c}{\sigma_ {r} r} \| \boldsymbol {U} \| _ {*}.
$$
The proof of Lemma 4 is presented in Appendix D.6.2.
D.5 A fourth moment bound for  $\bar{\gamma}^2$
Recall from Equation (6) that the averaged measurement  $\bar{\gamma}^2$  takes the form
$$
\bar {\gamma} _ {i} ^ {2} = \frac {1}{m} \sum_ {j = 1} ^ {m} \frac {y + \eta_ {i} ^ {(j)}}{\boldsymbol {a} _ {i} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a} _ {i}} = \frac {y + \bar {\eta} _ {i}}{\boldsymbol {a} _ {i} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a} _ {i}}.
$$
Throughout our analysis, we utilize the fact that  $\bar{\gamma}^2$  has a bounded fourth moment, as characterized in the following lemma.
Lemma 5. Assume  $r > 8$ . Then there exists a universal constant  $c > 0$ , such that
$$
\mathbb {E} \left(\bar {\gamma} ^ {2}\right) ^ {4} \leq c \left(\frac {y + \eta^ {\uparrow}}{\sigma_ {r} r}\right) ^ {4},
$$
where  $\sigma_r$  is the smallest non-zero singular value of  $\pmb{\Sigma}^{\star}$ .
The proof of Lemma 5 is presented in Appendix D.6.3. For notational simplicity of the proofs, we denote  $M = c\left(\frac{y + \eta^{\uparrow}}{\sigma_{r}r}\right)^{4}$ .
D.6 Proofs of preliminary lemmas
In this section, we present proofs for preliminary lemmas from Appendices D.1, D.4, and D.5.
D.6.1 Proof of Lemma 1
Using the independence of the noise  $\eta$  and the sensing vector  $\mathbf{a}$ , and the assumption that  $\eta$  is zero mean, we have
$$
\begin{array}{l} \mathbb {E} \left[ \eta \boldsymbol {A} ^ {\text {i n v}} \right] = \mathbb {E} \left[ \frac {\eta (y + \eta)}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \boldsymbol {a} \boldsymbol {a} ^ {\top} \right] \\ = \mathbb {E} [ \eta (y + \eta) ] \cdot \mathbb {E} \left[ \frac {1}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \boldsymbol {a} \boldsymbol {a} ^ {\top} \right] \\ = \nu_ {\eta} ^ {2} \mathbb {E} \left[ \frac {1}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}} \boldsymbol {a} \boldsymbol {a} ^ {\top} \right]. \tag {17} \\ \end{array}
$$
The expectation in (17) is non-zero, because the random matrix  $\frac{1}{\pmb{a}^{\top}\pmb{\Sigma}^{\star}\pmb{a}}\pmb{a}\pmb{a}^{\top}$  is symmetric positive definite almost surely. Therefore, we have  $\mathbb{E}\left[\eta \pmb{A}^{\mathrm{inv}}\right] \neq \mathbf{0}$ , as desired.
D.6.2 Proof of Lemma 4
Since  $\Sigma^{\star}$  is symmetric positive semidefinite, it be decomposed as  $Q\Sigma Q^{\top}$ , where  $Q$  is a square orthonormal matrix and  $\Sigma$  is a diagonal matrix with non-negative entries. Multiplying  $\pmb{a}$  by any square orthonormal matrix does not change its distribution. Therefore, without loss of generality, we assume that  $\Sigma^{\star}$  is diagonal with all non-negative diagonal entries. We first note that the moments of the ratios in both parts of Lemma 4 exist, because by [48, Proposition 1], for non-negative integers  $p$  and  $q$ , the quantity  $\mathbb{E}\frac{(a^\top Ua)^p}{(a^\top\Sigma^{\star}a)^q}$  exists if  $\frac{r}{2} > q$ . Furthermore, we use the following expression from [48, Proposition 2]:
$$
\mathbb {E} \frac {\left(\boldsymbol {a} ^ {\top} \boldsymbol {U} \boldsymbol {a}\right) ^ {p}}{\left(\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}\right) ^ {q}} = \frac {1}{\Gamma (q)} \int_ {0} ^ {\infty} t ^ {q - 1} \cdot | \boldsymbol {\Delta} _ {t} | \cdot \mathbb {E} \left(\boldsymbol {a} ^ {\top} \boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t} \boldsymbol {a}\right) ^ {p} \mathrm {d} t, \tag {18}
$$
where  $\pmb{\Delta}_{t} = (\pmb{I}_{d} + 2t\pmb{\Sigma}^{\star})^{-1 / 2}$  and  $|\pmb{\Delta}_t|$  is the determinant of  $\pmb{\Delta}_t$ . To characterize the determinant  $|\pmb{\Delta}_t|$ , we note that  $\pmb{\Delta}_t$  is a diagonal matrix whose  $d$  diagonal entries are
$$
\frac {1}{(1 + 2 t \sigma_ {1}) ^ {1 / 2}}, \dots , \frac {1}{(1 + 2 t \sigma_ {r}) ^ {1 / 2}}, 1, \dots , 1.
$$
Hence, the determinant is the product  $|\pmb{\Delta}_t| = \prod_{i=1}^{r} \frac{1}{(1 + 2t\sigma_i)^{1/2}}$ . Furthermore, this product can be bounded as:
$$
\left| \boldsymbol {\Delta} _ {t} \right| \leq \frac {1}{\left(1 + 2 t \sigma_ {r}\right) ^ {r / 2}}. \tag {19}
$$
We now prove parts (a) and (b) separately.
Part (a). Using the integral expression (18) with  $p = 0$  and  $q = 4$ , and the upper bound (19) on the determinant, we have
$$
\begin{array}{l} \mathbb {E} \left(\frac {1}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) ^ {4} = \frac {1}{\Gamma (4)} \int_ {0} ^ {\infty} t ^ {3} \cdot | \boldsymbol {\Delta} _ {t} |   \mathrm {d} t \\ \leq \frac {1}{\Gamma (4)} \int_ {0} ^ {\infty} t ^ {3} \frac {1}{\left(1 + 2 t \sigma_ {r}\right) ^ {r / 2}} d t. \\ \end{array}
$$
Denoting  $s \coloneqq 1 + 2t\sigma_r$ , we have
$$
\begin{array}{l} \mathbb {E} \left(\frac {1}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) ^ {4} \leq \frac {1}{2 \Gamma (4) \sigma_ {r}} \int_ {1} ^ {\infty} \left(\frac {s - 1}{2 \sigma_ {r}}\right) ^ {3} \frac {1}{s ^ {r / 2}} d s \\ \lesssim \frac {1}{\sigma_ {r} ^ {4}} \int_ {1} ^ {\infty} \frac {(s - 1) ^ {3}}{s ^ {r / 2}} \mathrm {d} s \\ = \frac {1}{\sigma_ {r} ^ {4}} \int_ {1} ^ {\infty} \left(\frac {s ^ {3}}{s ^ {r / 2}} - 3 \frac {s ^ {2}}{s ^ {r / 2}} + 3 \frac {s}{s ^ {r / 2}} - \frac {1}{s ^ {r / 2}}\right) d s \\ = \frac {1}{\sigma_ {r} ^ {4}} \left(\frac {2}{r - 8} - \frac {6}{r - 6} + \frac {6}{r - 4} - \frac {2}{r - 2}\right) \\ \leq \frac {c}{\sigma_ {r} ^ {4} r ^ {4}}, \\ \end{array}
$$
Part (b). Using the integral expression (18)  $p = q = 1$  and the upper bound (19) on the determinant, we have
$$
\begin{array}{l} \mathbb {E} \left(\frac {\boldsymbol {a} ^ {\top} \boldsymbol {U} \boldsymbol {a}}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) = \frac {1}{\Gamma (1)} \int_ {0} ^ {\infty} | \boldsymbol {\Delta} _ {t} | \cdot \mathbb {E} \left[ \boldsymbol {a} ^ {\top} \boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t} \boldsymbol {a} \right] d t \\ \leq \frac {1}{\Gamma (1)} \int_ {0} ^ {\infty} \frac {1}{\left(1 + 2 t \sigma_ {r}\right) ^ {r / 2}} \mathbb {E} \left[ \boldsymbol {a} ^ {\top} \boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t} \boldsymbol {a} \right] \mathrm {d} t. \tag {20} \\ \end{array}
$$
We now bound the expectation term in (20). Note that for  $\mathbf{a} \sim \mathcal{N}(\mathbf{0},\mathbf{I}_d)$ , we have  $\mathbb{E}\left[\mathbf{a}^\top \mathbf{B}\mathbf{a}\right] = \operatorname {tr}(\mathbf{B})$  for any symmetric matrix  $\mathbf{B}$ . Therefore, we have
$$
\begin{array}{l} \mathbb {E} \left[ \boldsymbol {a} ^ {\top} \boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t} \boldsymbol {a} \right] = \operatorname {t r} \left(\boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t}\right) \\ \stackrel {(i)} {\leq} \left\| \boldsymbol {\Delta} _ {t} \boldsymbol {U} \boldsymbol {\Delta} _ {t} \right\| _ {*} \\ \stackrel {\text {(i i)}} {\leq} \| \boldsymbol {U} \| _ {*}, \tag {21} \\ \end{array}
$$
where (i) the fact that  $\operatorname{tr}(\boldsymbol{B}) \leq \| \boldsymbol{B} \|_*$  for any symmetric matrix  $\boldsymbol{B}$ . Furthermore, (ii) follows from Hölder's inequality for Schatten- $p$  norms, where we have that  $\| \Delta_t U \Delta_t \|_* \leq \| \Delta_t \|_{\mathrm{op}}^2 \cdot \| U \|_*$ . Because  $\Delta_t$  is diagonal and the entries of  $\Delta_t$  are bounded between 0 and 1, we bound the operator norm as  $\| \Delta_t \|_{\mathrm{op}} \leq 1$ . Substituting (21) to (20), we obtain
$$
\begin{array}{l} \mathbb {E} \left(\frac {\boldsymbol {a} ^ {\top} \boldsymbol {U} \boldsymbol {a}}{\boldsymbol {a} ^ {\top} \boldsymbol {\Sigma} ^ {\star} \boldsymbol {a}}\right) \leq \| \boldsymbol {U} \| _ {*} \cdot \int_ {0} ^ {\infty} \frac {1}{(1 + 2 t \sigma_ {r}) ^ {r / 2}} d t \\ \lesssim \frac {1}{\sigma_ {r} r} \cdot \| \boldsymbol {U} \| _ {*}, \\ \end{array}
$$
D.6.3 Proof of Lemma 5