SlowGuess/NeurIPS_2023_OCR huggingface dataset:
We introduce the following Lemma, which proof is deferred to later in this section.
Lemma 19. The following inequality holds:  $\sum_{g\in [G]}\mathbb{E}_{\pi ,\mathcal{D}^a}[n_{g,T}]KL(\mathcal{D}_g^a ||\mathcal{D}_g^b)\leq \frac{T\nu^2}{2G}.$
By combining both Lemmas 18 and 19 in the inequality stated in Lemma 6, we obtain
$$
\begin{array}{l} \max  \left\{\operatorname {R e g r e t} _ {\infty} (\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {a}), \operatorname {R e g r e t} _ {\infty} (\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {b}) \right\} \geq \frac {\min  \left(\Sigma_ {\infty} ^ {a} , \Sigma_ {\infty} ^ {b}\right)}{2 T} \| \boldsymbol {\lambda} ^ {*} (\boldsymbol {\sigma} ^ {a}) - \boldsymbol {\lambda} ^ {*} (\boldsymbol {\sigma} ^ {b}) \| _ {1} \exp \left(- \frac {T \nu^ {2}}{2 G}\right) \\ = e ^ {- \frac {T \nu^ {2}}{2 G}} \frac {G}{2 T} \left(\frac {1}{G} - \frac {1}{G + \nu} + (G - 1) \left(\frac {1 + \nu}{G + \nu} - \frac {1}{G}\right)\right) \\ = \frac {e ^ {- \frac {T \nu^ {2}}{2 G}}}{2} \frac {G}{T} \times \frac {(G + \nu) - G + (G - 1) (G + G \nu - G - \nu)}{G (G + \nu)} \\ = \frac {e ^ {- \frac {T \nu^ {2}}{2 G}}}{2} \frac {\nu G}{T} \times \frac {1 + (G - 1) ^ {2}}{G (G + \nu)}. \\ \end{array}
$$
By setting  $\nu = \sqrt{\frac{G}{T}} \leq 1$  and  $\mathcal{D}_{\pi} := \operatorname{argmax}\left\{\mathrm{Regret}_{\infty}(\pi, \mathcal{D}^a), \mathrm{Regret}_{\infty}(\pi, \mathcal{D}^b)\right\}$ , the previous inequality implies
$$
\operatorname {R e g r e t} _ {\infty} (\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} _ {\boldsymbol {\pi}}) = \max  \left\{\operatorname {R e g r e t} _ {\infty} (\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {a}), \operatorname {R e g r e t} _ {\infty} (\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {b}) \right\} \geq \frac {e ^ {- 1 / 2}}{2} \frac {G ^ {1 . 5}}{T ^ {1 . 5}} \frac {1 + (G - 1) ^ {2}}{G (G + 1)} \geq \frac {1}{2} G ^ {1. 5} T ^ {- 1. 5}
$$
which completes the proof of Theorem 4, .
We now prove Lemma 19.
Proof. For convenience, we set  $\nu := \sqrt{\frac{G}{T}} < 1$ . The formula for the  $KL$ -divergence of two univariate normal distributions of zero mean implies
where the first step stems from the inequality  $n_T \leq T$ , the second step stems from using the previous two equalities, the third step stems from simplifying the infinite sum term-wise, the fourth step stems from the expansion of a geometric series, and the fifth step stems from  $\mu = \frac{1}{\sqrt{T}} \geq 0$ . Since  $\mathcal{D}^a$  and  $\mathcal{D}^b$  have the same distribution at all coordinates except coordinate  $h$ , we have:
$$
\sum_ {g \in [ G ]} \mathbb {E} _ {\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {a}} \left[ n _ {g, T} \right] K L \left(\mathcal {D} _ {g} ^ {a} | | \mathcal {D} _ {g} ^ {b}\right) = \mathbb {E} _ {\boldsymbol {\pi}, \boldsymbol {\mathcal {D}} ^ {a}} \left[ n _ {h, T} \right] K L \left(\mathcal {D} _ {h} ^ {a} | | \mathcal {D} _ {h} ^ {b}\right) \leq \frac {T \nu^ {2}}{2 G}
$$
where the inequality follows from the minimality of  $\mathbb{E}_{\pmb{\pi},\pmb{D}^a}[n_{h,T}]$  and  $\sum_{g}\mathbb{E}_{\pmb{\pi},\pmb{D}^a}[n_{g,T}] = T$ . This concludes the proof.