SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\| (\hat {\Sigma} _ {t} - H _ {t}) y _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} \leq \mathcal {O} \left(\frac {d ^ {3} \log (d T / \delta)}{t}\right)
$$
$$
\left\| x \left(p ^ {\prime}\right) - x \left(p _ {t}\right) \right\| \leq \epsilon , \quad \left\| \hat {x} \left(p ^ {\prime}\right) - \hat {x} \left(p _ {t}\right) \right\| \leq \epsilon , \quad \left\| \hat {\Sigma} \left(p ^ {\prime}\right) - \hat {\Sigma} \left(p _ {t}\right) \right\| _ {F} \leq 7 \epsilon \tag {38}
$$
Thus, for any  $\| y_t\| _2\leq 1$
$$
\begin{array}{l} \left\| \left(\hat {\Sigma} (p _ {t}) - H (p _ {t})\right) y _ {t} \right\| _ {\hat {\Sigma} (p _ {t}) ^ {- 1}} ^ {2} \\ = \left(\| (\hat {\Sigma} (p _ {t}) - H (p _ {t})) y _ {t} \| _ {\hat {\Sigma} (p _ {t}) ^ {- 1}} ^ {2} - \| (\hat {\Sigma} (p ^ {\prime}) - H (p ^ {\prime})) y _ {t} \| _ {\hat {\Sigma} (p ^ {\prime}) ^ {- 1}} ^ {2}\right) + \| (\hat {\Sigma} (p ^ {\prime}) - H (p ^ {\prime})) y _ {t} \| _ {\hat {\Sigma} (p ^ {\prime}) ^ {- 1}} ^ {2} \\ \leq \left(\| (\hat {\Sigma} (p _ {t}) - H (p _ {t})) y _ {t} \| _ {\hat {\Sigma} (p _ {t}) ^ {- 1}} ^ {2} - \| (\hat {\Sigma} (p ^ {\prime}) - H (p ^ {\prime})) y _ {t} \| _ {\hat {\Sigma} (p ^ {\prime}) ^ {- 1}} ^ {2}\right) + \mathcal {O} \left(\frac {d \log (d | \mathbf {P} ^ {\prime} | / \delta)}{t - 1}\right) \\ \end{array}
$$
(by Lemma 14 with an union bound over  $\mathbf{P}'$ )
$$
\begin{array}{l} = \theta_ {t} ^ {\top} \hat {\Sigma} \left(p _ {t}\right) ^ {- 1} \theta_ {t} - \theta^ {\prime \top} \hat {\Sigma} \left(p ^ {\prime}\right) ^ {- 1} \theta^ {\prime} + \mathcal {O} \left(\frac {d \log (d | \mathbf {P} ^ {\prime} | / \delta)}{t - 1}\right) \\ (\text {d e f i n e} \theta_ {t} = (\hat {\Sigma} (p _ {t}) - H (p _ {t})) y _ {t} \text {a n d} \theta^ {\prime} = (\hat {\Sigma} (p ^ {\prime}) - H (p ^ {\prime})) y _ {t}) \\ = \left(\theta_ {t} - \theta^ {\prime}\right) ^ {\top} \hat {\Sigma} \left(p _ {t}\right) ^ {- 1} \theta_ {t} + \theta^ {\prime \top} \left(\hat {\Sigma} \left(p _ {t}\right) ^ {- 1} - \hat {\Sigma} \left(p ^ {\prime}\right) ^ {- 1}\right) \theta_ {t} + \theta^ {\prime \top} \hat {\Sigma} \left(p ^ {\prime}\right) ^ {- 1} \left(\theta_ {t} - \theta^ {\prime}\right) + \mathcal {O} \left(\frac {d \log (d | \mathbf {P} ^ {\prime} | / \delta)}{t - 1}\right) \\ \leq \left(\theta_ {t} - \theta^ {\prime}\right) ^ {\top} \left(\hat {\Sigma} \left(p _ {t}\right) ^ {- 1} \theta_ {t} + \hat {\Sigma} \left(p ^ {\prime}\right) ^ {- 1} \theta^ {\prime}\right) + \theta^ {\prime \top} \hat {\Sigma} \left(p ^ {\prime}\right) ^ {- 1} \left(\hat {\Sigma} \left(p ^ {\prime}\right) - \hat {\Sigma} \left(p _ {t}\right)\right) \hat {\Sigma} \left(p _ {t}\right) ^ {- 1} \theta_ {t} + \mathcal {O} \left(\frac {d \log (d | \mathbf {P} ^ {\prime} | / \delta)}{t - 1}\right) \\ \end{array}
$$
The first two terms above can be bounded by the order of  $\mathcal{O}(\epsilon t^2)$  by Eq. (38). Plugging in the choice of  $\epsilon = \frac{1}{T^3}$  and recalling that  $\log |\mathbf{P}'| = \mathcal{O}(d^2\log (d / \epsilon))$  finishes the proof.
D Regret Analysis
Consider the regret decomposition in Section 3.5.
$$
\begin{array}{l} \operatorname {R e g} (u) = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle a _ {t} - u ^ {\mathcal {A} _ {t}}, y _ {t} \right\rangle \right] = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {t}} - \boldsymbol {U} ^ {\mathcal {A} _ {t}}, \gamma_ {t} \right\rangle \right] = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} \right\rangle \right] \\ \leq \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}} , \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right]} _ {\text {B i a s}} + \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}} , \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right]} _ {\text {B o u n s}} + \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}} , \hat {\gamma} _ {t} - \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right]} _ {\text {F T R L - R e g}} \\ \end{array}
$$
where  $\mathcal{A}_0$  is drawn from  $D$  and is independent from the interaction between the learning and the environment. Recall that our algorithm is FTRL:
$$
\boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} = \operatorname * {a r g m i n} _ {\boldsymbol {H} \in \mathcal {H} ^ {\mathcal {A} _ {0}}} \left\{\sum_ {s = 1} ^ {t - 1} \left\langle \boldsymbol {H}, \hat {\gamma} _ {s} - \alpha_ {s} \hat {\boldsymbol {\Sigma}} _ {s} ^ {- 1} \right\rangle + \frac {F (\boldsymbol {H})}{\eta_ {t}} \right\}.
$$
The FTRL-Reg term can be handled by the standard FTRL analysis (Lemma 5). In order to deal with the issue that  $F$  can be unbounded on the boundary of  $\mathcal{H}^{\mathcal{A}_0}$ , we apply Lemma 5 with the regret comparator  $\overline{U}^{\mathcal{A}_0}$  defined as
$$
\overline {{U}} ^ {\mathcal {A} _ {0}} = \left(1 - \frac {1}{T ^ {2}}\right) U ^ {\mathcal {A} _ {0}} + \frac {1}{T ^ {2}} H _ {*} ^ {\mathcal {A} _ {0}}
$$
where  $\pmb{H}_{*}^{\mathcal{A}_{0}}\triangleq \operatorname{argmin}_{\pmb{H}\in \mathcal{H}^{\mathcal{A}_{0}}}F(\pmb{H})$ . Thus,
FTRL-Reg
$$
\begin{array}{l} \leq \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \overline {{\boldsymbol {U}}} ^ {\mathcal {A} _ {0}}, \hat {\gamma} _ {t} - \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right] + \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \overline {{\boldsymbol {U}}} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \hat {\gamma} _ {t} - \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right] \\ \leq \underbrace {\mathbb {E} \left[ \frac {F (\overline {{\boldsymbol {U}}} ^ {\mathcal {A} _ {0}}) - \operatorname* {m i n} _ {\boldsymbol {H} \in \mathcal {H} ^ {\mathcal {A} _ {0}}} F (\boldsymbol {H})}{\eta_ {T}} \right]} _ {\text {P e n a l t y}} + \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \max  _ {\boldsymbol {H} \in \mathcal {H} ^ {\mathcal {A} _ {0}}} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {H}, \hat {\gamma} _ {t} \right\rangle - \frac {D (\boldsymbol {H} , \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}})}{2 \eta_ {t}} \right]} _ {\text {S t a b i l i t y - 1}} \\ + \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \max  _ {\boldsymbol {H} \in \mathcal {H} ^ {\mathcal {A} _ {0}}} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {H}, - \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle - \frac {D (\boldsymbol {H} , \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}})}{2 \eta_ {t}} \right]} _ {\text {S t a b i l i t y - 2}} + \underbrace {\mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \bar {\boldsymbol {U}} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \hat {\gamma} _ {t} - \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right]} _ {\text {E r r o r}} \tag {39} \\ \end{array}
$$
In the rest of this section, we bound the following terms individually: Bias, Bonus, Penalty, Stability-1, Stability-2, Error.
For any  $t = 2, \dots, T$ , let  $\mathcal{E}_{t-1}$  be the event that the high-probability event in Lemma 17, Lemma 18, and Lemma 19 happens for all  $1, \dots, t-1$  and  $\overline{\mathcal{E}_{t-1}}$  be the opposite event of  $\mathcal{E}_{t-1}$  (i.e. any of these three lemmas fails for any  $1, \dots, t-1$ ). We have  $\mathcal{P}[\mathcal{E}_{t-1}] = 1 - \mathcal{O}(\delta)$  and  $\mathcal{P}[\overline{\mathcal{E}_{t-1}}] = \mathcal{O}(\delta)$ . Let  $\mathbb{E}[\cdot | \mathcal{E}_{t-1}]$  be the conditional expectation that event  $\mathcal{E}_{t-1}$  happens and let  $\mathbb{E}_t^\mathcal{E} = \mathbb{E}[\cdot | \mathcal{F}_{t-1}, \mathcal{E}_{t-1}]$
D.1 Bounding the Bias term
Lemma 20.
$$
\mathbf {B i a s} = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right] \leq \frac {1}{4} \sum_ {t = 1} ^ {T} \alpha_ {t} \| x _ {t} - u \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} + \mathcal {O} \left(\delta T ^ {2} + \sum_ {t = 1} ^ {T} \frac {d ^ {3} \log (T / \delta)}{\alpha_ {t} t}\right)
$$
Proof. For any  $t$ , we have
$$
\begin{array}{l} \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} [ \langle \boldsymbol {H} _ {t} - \boldsymbol {U}, \gamma_ {t} - \hat {\gamma} _ {t} \rangle ] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} [ \langle x _ {t} - u, y _ {t} - \hat {y} _ {t} \rangle ] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \left(x _ {t} - u\right) ^ {\top} \left(y _ {t} - \hat {\Sigma} _ {t} ^ {- 1} \left(a _ {t} - \hat {x} _ {t}\right) a _ {t} ^ {\top} y _ {t}\right) \right] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \left(y _ {t} - \hat {\Sigma} _ {t} ^ {- 1} (a _ {t} - \hat {x} _ {t}) (a _ {t} - \hat {x} _ {t}) ^ {\top} y _ {t}\right) \right] - \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} (a _ {t} - \hat {x} _ {t}) \hat {x} _ {t} ^ {\top} y _ {t} \right] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \left(I - \hat {\Sigma} _ {t} ^ {- 1} \mathbb {E} _ {\mathcal {A} \sim \mathcal {D}} \mathbb {E} _ {a _ {t} \sim p _ {t} ^ {\mathcal {A}}} \left[ (a _ {t} - \hat {x} _ {t}) (a _ {t} - \hat {x} _ {t}) ^ {\top} \right]\right) y _ {t} \right] \\ - \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} \left(\mathbb {E} _ {\mathcal {A} \sim \mathcal {D}} \mathbb {E} _ {a _ {t} \sim p _ {t} ^ {\mathcal {A}}} [ a _ {t} ] - \hat {x} _ {t}\right) \hat {x} _ {t} ^ {\top} y _ {t} \right] (\text {t a k i n g e x p e c t a t i o n o v e r} \mathcal {A} _ {t} \text {a n d} a _ {t}) \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} (\hat {\Sigma} _ {t} - H _ {t}) y _ {t} \right] - \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} (x _ {t} - \hat {x} _ {t}) \hat {x} _ {t} ^ {\top} y _ {t} \right] \\ (b y \\ \leq \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} (\hat {\Sigma} _ {t} - H _ {t}) y _ {t} \right] + \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \left| (x _ {t} - u) ^ {\top} \hat {\Sigma} _ {t} ^ {- 1} (x _ {t} - \hat {x} _ {t}) \right| \right] \quad \left(\left| \hat {x} _ {t} ^ {\top} y _ {t} \right| \leq 1\right) \\ \leq \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \| x _ {t} - u \| _ {\hat {\Sigma} _ {t} ^ {- 1}} \left(\| (\hat {\Sigma} _ {t} - H _ {t}) y _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} + \| x _ {t} - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}}\right) \right] (Cauchy-Schwarz) \\ \leq \mathcal {O} \left(\sqrt {\frac {d ^ {3} \log (T / \delta)}{t}} \| x _ {t} - u \| _ {\hat {\Sigma} _ {t} ^ {- 1}}\right) \quad (\text {L e m m a 1 9 a n d L e m m a 1 8 g i v e n} \mathcal {E} _ {t - 1}) \\ \leq \frac {\alpha_ {t}}{4} \| x _ {t} - u \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} + \mathcal {O} \left(\frac {d ^ {3} \log (T / \delta)}{\alpha_ {t} t}\right) (AM-GM inequality) \\ \end{array}
$$
On the other hand, since  $\hat{\Sigma}_t\succeq \frac{1}{t} I\succeq \frac{1}{T} I$ , for any  $t = 1,\dots ,T$
$$
\| \hat {y} _ {t} \| _ {2} = \| \Sigma_ {t} ^ {- 1} (a _ {t} - \hat {x} _ {t}) a _ {t} ^ {\top} y _ {t} \| _ {2} \leq \| \Sigma_ {t} ^ {- 1} (a _ {t} - \hat {x} _ {t}) \| _ {2} \leq \mathcal {O} (T)
$$
Thus, we have trivial bound
$$
\mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \mid \overline {{\mathcal {E} _ {t - 1}}} \right] = \mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} - \boldsymbol {U}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \mid \overline {{\mathcal {E} _ {t - 1}}} \right] = \mathbb {E} _ {t} \left[ \left\langle x _ {t} - u, y _ {t} - \hat {y} _ {t} \right\rangle \mid \overline {{\mathcal {E} _ {t - 1}}} \right] \leq \mathcal {O} (T)
$$
$$
\begin{array}{l} \mathbf {B i a s} = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right] \\ = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right] \right] \\ = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \mid \mathcal {E} _ {t - 1} \right] \mathbb {I} \{\mathcal {E} _ {t - 1} \} \right] + \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \gamma_ {t} - \hat {\gamma} _ {t} \right\rangle \right| \overline {{\mathcal {E} _ {t - 1}}} \right] \mathbb {I} \{\overline {{\mathcal {E} _ {t - 1}}} \} \\ \leq \frac {1}{4} \sum_ {t = 1} ^ {T} \alpha_ {t} \| x _ {t} - u \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} + \mathcal {O} \left(\sum_ {t = 1} ^ {T} \frac {d ^ {3} \log (T / \delta)}{\alpha_ {t} t} + \delta T ^ {2}\right) \\ \end{array}
$$
D.2 Bounding the Bonus term
We first prove the following useful technique lemma to bound the inner product of lifted matrices.
Lemma 21. Let  $\mathbf{G} = \left[ \begin{array}{cc} G + gg^{\top} & g \\ g^{\top} & 1 \end{array} \right]$ ,  $\mathbf{H} = \left[ \begin{array}{cc} H + hh^{\top} & h \\ h^{\top} & 1 \end{array} \right]$  where  $G$  and  $H$  are positive semi-
definite, and  $\pmb{H}^{\prime} = \pmb{H} + vv^{\top}$  where  $v = \left[ \begin{array}{c}0\\ \sqrt{\beta} \end{array} \right]\in \mathbb{R}^{d + 1}$ . Then we have
1.  $\operatorname{Tr}\left(\boldsymbol{H}^{-1}\boldsymbol{G}\right) = \operatorname{Tr}(H^{-1}G) + \| g - h\|_{H^{-1}}^2 + 1$
2.  $\operatorname{Tr}\left((\pmb{H}')^{-1}\pmb{G}\right) \geq \frac{1}{2\left(1 + \frac{\beta}{1 + \beta} \|h\|_{H^{-1}}^{2}\right)}\|g - h\|_{H^{-1}}^{2} - \frac{\beta^{2}}{(1 + \beta)^{2}}\|h\|_{H^{-1}}^{2}$
Proof. From Theorem 2.1 of Lu and Shiou [2002], for any block matrix  $R = \begin{bmatrix} A & B \\ C & D \end{bmatrix}$  if  $A$  is invertible and its Schur complement  $S_A = D - CA^{-1}B$  is invertible, then
$$
R ^ {- 1} = \left[ \begin{array}{c c} A ^ {- 1} + A ^ {- 1} B S _ {A} ^ {- 1} C A ^ {- 1} & - A ^ {- 1} B S _ {A} ^ {- 1} \\ - S _ {A} ^ {- 1} C A & S _ {A} ^ {- 1} \end{array} \right]
$$
Using above equation, for the first equation, Since  $(H + hh^{\top})^{-1} = H^{-1} - \frac{H^{-1}hh^{\top}H^{-1}}{1 + h^{\top}H^{-1}h}$ . The inverse Schur complement of  $H + hh^{\top}$  is  $1 + h^{\top}H^{-1}h$ . Thus
$$
\boldsymbol {H} ^ {- 1} = \left[ \begin{array}{c c} (I + H ^ {- 1} h h ^ {\top}) (H + h h ^ {\top}) ^ {- 1} & - H ^ {- 1} h \\ - h ^ {\top} H ^ {- 1} & 1 + h ^ {\top} H ^ {- 1} h \end{array} \right] = \left[ \begin{array}{c c} H ^ {- 1} & - H ^ {- 1} h \\ - h ^ {\top} H ^ {- 1} & 1 + h ^ {\top} H ^ {- 1} h \end{array} \right]
$$
$$
\begin{array}{l} \operatorname {T r} \left(\boldsymbol {H} ^ {- 1} \boldsymbol {G}\right) = \operatorname {T r} \left(H ^ {- 1} G + H ^ {- 1} g g ^ {\top} - H ^ {- 1} h g ^ {\top}\right) - h ^ {\top} H ^ {- 1} g + 1 + h ^ {\top} H ^ {- 1} h \\ = \operatorname {T r} (H ^ {- 1} G) + g ^ {\top} H ^ {- 1} g - 2 g ^ {\top} H ^ {- 1} h + h ^ {\top} H ^ {- 1} h + 1 \\ = \operatorname {T r} \left(H ^ {- 1} G\right) + \| g - h \| _ {H ^ {- 1}} ^ {2} + 1. \\ \end{array}
$$
For the second equation, observe that
$$
\boldsymbol {H} ^ {\prime} = \left[ \begin{array}{c c} H + h h ^ {\top} & h \\ h ^ {\top} & 1 + \beta \end{array} \right] = (1 + \beta) \left[ \begin{array}{c c} \frac {1}{1 + \beta} (H + h h ^ {\top}) & \frac {1}{1 + \beta} h \\ \frac {1}{1 + \beta} h ^ {\top} & 1 \end{array} \right] = (1 + \beta) \left[ \begin{array}{c c} H ^ {\prime} + h ^ {\prime} h ^ {\prime \top} & h ^ {\prime} \\ h ^ {\prime \top} & 1 \end{array} \right]
$$
where  $h' = \frac{1}{1 + \beta} h$  and  $H' = \frac{1}{1 + \beta} H + \left( \frac{1}{1 + \beta} - \frac{1}{(1 + \beta)^2} \right) hh^\top = \frac{1}{1 + \beta} H + \frac{\beta}{(1 + \beta)^2} hh^\top \succeq 0.$
Applying the first equality, we have
$$
\operatorname {T r} \left(\left(\boldsymbol {H} ^ {\prime}\right) ^ {- 1} \boldsymbol {G}\right) = \frac {1}{1 + \beta} \left(\operatorname {T r} \left(\left(H ^ {\prime}\right) ^ {- 1} G\right) + \left\| g - h ^ {\prime} \right\| _ {H ^ {\prime} - 1} ^ {2} + 1\right) \geq \frac {1}{1 + \beta} \left\| g - h ^ {\prime} \right\| _ {H ^ {\prime} - 1} ^ {2}.
$$
Below, we continue to lower bound this term. By the same formula above, we have
$$
H ^ {\prime - 1} = \left(\frac {1}{1 + \beta} H + \frac {\beta}{(1 + \beta) ^ {2}} h h ^ {\top}\right) ^ {- 1} = (1 + \beta) H ^ {- 1} - \frac {\beta H ^ {- 1} h h ^ {\top} H ^ {- 1}}{1 + \frac {\beta}{1 + \beta} h ^ {\top} H ^ {- 1} h}.
$$
$$
\begin{array}{l} \frac {1}{1 + \beta} \| g - h ^ {\prime} \| _ {H ^ {\prime} - 1} ^ {2} \\ \geq \frac {1}{2 (1 + \beta)} \| g - h \| _ {H ^ {\prime} - 1} ^ {2} - \frac {1}{1 + \beta} \| h - h ^ {\prime} \| _ {H ^ {\prime} - 1} ^ {2} \quad (\text {u s i n g} \| a + b \| ^ {2} \leq 2 \| a \| ^ {2} + 2 \| b \| ^ {2}) \\ = \frac {1}{2} (g - h) ^ {\top} \left(H ^ {- 1} - \frac {\frac {\beta}{1 + \beta} H ^ {- 1} h h ^ {\top} H ^ {- 1}}{1 + \frac {\beta}{1 + \beta} h ^ {\top} H ^ {- 1} h}\right) (g - h) - (h - h ^ {\prime}) ^ {\top} \left(H ^ {- 1} - \frac {\frac {\beta}{1 + \beta} H ^ {- 1} h h ^ {\top} H ^ {- 1}}{1 + \frac {\beta}{1 + \beta} h ^ {\top} H ^ {- 1} h}\right) (h - h ^ {\prime}) \\ \geq \frac {1}{2} \| g - h \| _ {H ^ {- 1}} ^ {2} - \frac {\frac {\beta}{1 + \beta} ((g - h) ^ {\top} H ^ {- 1} h) ^ {2}}{2 \left(1 + \frac {\beta}{1 + \beta} \| h \| _ {H ^ {- 1}} ^ {2}\right)} - \frac {\beta^ {2}}{(1 + \beta) ^ {2}} \| h \| _ {H ^ {- 1}} ^ {2} \quad (\text {u s i n g} h - h ^ {\prime} = \frac {\beta}{1 + \beta} h) \\ \geq \frac {1}{2} \| g - h \| _ {H ^ {- 1}} ^ {2} - \frac {\frac {\beta}{1 + \beta} \| h \| _ {H ^ {- 1}} ^ {2}}{2 \left(1 + \frac {\beta}{1 + \beta} \| h \| _ {H ^ {- 1}} ^ {2}\right)} \| g - h \| _ {H ^ {- 1}} ^ {2} - \frac {\beta^ {2}}{(1 + \beta) ^ {2}} \| h \| _ {H ^ {- 1}} ^ {2} \quad (\text {C a u c h y - S c h w a r z}) \\ = \frac {1}{2 \left(1 + \frac {\beta}{1 + \beta} \| h \| _ {H ^ {- 1}} ^ {2}\right)} \| g - h \| _ {H ^ {- 1}} ^ {2} - \frac {\beta^ {2}}{(1 + \beta) ^ {2}} \| h \| _ {H ^ {- 1}} ^ {2}. \\ \end{array}
$$
Using Lemma 21, we are able to show Corollary 22 which bound part of the second term.
Corollary 22.  $\operatorname{Tr}(U\hat{\Sigma}_t^{-1}) \geq \frac{1}{4}\|u - \hat{x}_t\|_{\hat{\Sigma}_t^{-1}}^2 - \frac{1}{4}.$
Proof. From Lemma 21, we have
$$
\operatorname {T r} (\boldsymbol {U} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}) \geq \frac {1}{2 \left(1 + \frac {\beta_ {t}}{1 + \beta_ {t}} \| \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2}\right)} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} - \frac {\beta_ {t} ^ {2}}{(1 + \beta_ {t}) ^ {2}} \| \hat {x} _ {t} \| _ {\Sigma_ {t} ^ {- 1}} ^ {2}.
$$
Since  $\hat{\Sigma}_t\succeq \beta_tI,\hat{\Sigma}_t^{-1}\preceq \frac{1}{\beta_t} I$  . Since  $\| \hat{x}_t\| _2\leq 1$  , we have  $\| \hat{x}_t\|_{\hat{\Sigma}_t^{-1}}^2\leq \frac{1}{\beta_t}$  . Then
$$
\begin{array}{l} \operatorname {T r} \left(\boldsymbol {U} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}\right) \geq \frac {1}{2 \left(1 + \frac {1}{1 + \beta_ {t}}\right)} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} - \frac {\beta_ {t}}{(1 + \beta_ {t}) ^ {2}} \\ \geq \frac {1}{4} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} - \frac {\beta_ {t}}{(2 \sqrt {\beta_ {t}}) ^ {2}} \quad (\beta_ {t} \geq 0) \\ = \frac {1}{4} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} - \frac {1}{4}. \\ \end{array}
$$
Lemma 23.
$$
\begin{array}{l} \mathbf {B o u n s} = \mathbb {E} \left[ \sum_ {t = 1} ^ {T} \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right] \\ \leq 2 (d + 2) \sum_ {t = 1} ^ {T} \alpha_ {t} - \frac {1}{4} \sum_ {t = 1} ^ {T} \alpha_ {t} \| u - x _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} + \mathcal {O} \left(\sum_ {t = 1} ^ {T} \frac {d ^ {3} \alpha_ {t} \log (T / \delta)}{t} + \delta T \sum_ {t = 1} ^ {T} \alpha_ {t}\right). \\ \end{array}
$$
$$
\begin{array}{l} \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \right] \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \operatorname {T r} \left(\alpha_ {t} \left(\boldsymbol {H} _ {t} - \boldsymbol {U}\right) \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}\right) \right] \quad (\text {t a k i n g e x p e c t a t i o n o v e r} \mathcal {A} _ {0}) \\ = \mathbb {E} _ {t} ^ {\mathcal {E}} \left[ \alpha_ {t} \operatorname {T r} \left(\boldsymbol {H} _ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}\right) - \alpha_ {t} \operatorname {T r} \left(\boldsymbol {U} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}\right) \right] \\ \leq \alpha_ {t} \operatorname {T r} \left(\mathbb {E} _ {t} ^ {\varepsilon} \left[ \boldsymbol {H} _ {t} \right] \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1}\right) - \mathbb {E} _ {t} ^ {\varepsilon} \left[ \frac {\alpha_ {t}}{4} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} \right] + \frac {1}{4} \alpha_ {t} (Corollary 22) \\ \leq 2 \alpha_ {t} (d + 2) - \mathbb {E} _ {t} ^ {\varepsilon} \left[ \frac {\alpha_ {t}}{4} \| u - \hat {x} _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} \right] \\ \leq 2 \alpha_ {t} (d + 2) - \mathbb {E} _ {t} ^ {\varepsilon} \left[ \frac {\alpha_ {t}}{4} \| u - x _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} - \frac {\alpha_ {t}}{4} \| \hat {x} _ {t} - x _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} \right] \\ \leq 2 \alpha_ {t} (d + 2) - \frac {\alpha_ {t}}{4} \| u - x _ {t} \| _ {\hat {\Sigma} _ {t} ^ {- 1}} ^ {2} + \mathcal {O} \left(\frac {d ^ {3} \alpha_ {t} \log (T / \delta)}{t}\right) (Lemma 18) \\ \end{array}
$$
On the other hand, since  $\hat{\pmb{\Sigma}}_t\succeq \frac{1}{t}\pmb {I}\succeq \frac{1}{T}\pmb{I}$  , we have trivial bound
$$
\mathbb {E} _ {t} \left[ \left\langle \boldsymbol {H} _ {t} ^ {\mathcal {A} _ {0}} - \boldsymbol {U} ^ {\mathcal {A} _ {0}}, \alpha_ {t} \hat {\boldsymbol {\Sigma}} _ {t} ^ {- 1} \right\rangle \mid \overline {{\mathcal {E} _ {t - 1}}} \right] \leq \mathcal {O} (\alpha_ {t} T)
$$