SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \stackrel {\text {E q .}} {=} ^ {(1 4)} \frac {1}{2} \left(\operatorname {T r} (\Phi (\theta) \Phi^ {- 1} (z)) - d + \log \left(\frac {\det  (\Phi^ {- 1} (\theta))}{\det  (\Phi^ {- 1} (z))}\right) + (\theta - z) ^ {\top} \Phi (\theta) (\theta - z)\right) \\ = \frac {1}{2} \left(\operatorname {T r} \left(\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z)\right) - d + \log \left(\frac {\det  \left(\Phi^ {- 1} (\theta)\right)}{\det  \left(\Phi^ {- 1} (z)\right)}\right) + (\theta - z) ^ {\top} \Phi (\theta) (\theta - z)\right) \\ = \frac {1}{2} \sum_ {i = 1} ^ {d} \lambda_ {i} \left(\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z)\right) - \frac {1}{2} d + \frac {1}{2} \log \left(\frac {1}{\det \left(\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z)\right)}\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ = \frac {1}{2} \sum_ {i = 1} ^ {d} \left(\lambda_ {i} (\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z)) - 1 + \log \left(\frac {1}{\lambda_ {i} (\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z))}\right)\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ \leq \frac {1}{2} \sum_ {i = 1} ^ {d} \left(\lambda_ {i} (\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z)) + \frac {1}{\lambda_ {i} (\Phi^ {- \frac {1}{2}} (z) \Phi (\theta) \Phi^ {- \frac {1}{2}} (z))} - 2\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ \stackrel {\text {E q . (1 3)}}{\leq} \frac {1}{2} \sum_ {i = 1} ^ {d} \left(\max  _ {t \in \left[ \left(1 - \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)}\right) ^ {2}, \left(1 + \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)}\right) ^ {2} \right]} t + \frac {1}{t} - 2\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ = \frac {1}{2} \sum_ {i = 1} ^ {d} \left(\max  _ {t \in \left[ - \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)}, \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)} \right]} (1 + t) ^ {2} + \frac {1}{(1 + t) ^ {2}} - 2\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ \leq \frac {1}{2} \sum_ {i = 1} ^ {d} \left(\max  _ {t \in \left[ - \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)}, \alpha^ {1 / 2} \| \theta - z \| _ {\Phi (\theta)} \right]} 6 t ^ {2}\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ \stackrel {\text {c o n v e x i t y}} {\leq} \frac {1}{2} \left(\sum_ {i = 1} ^ {d} 6 \alpha \| \theta - z \| _ {\Phi (\theta)} ^ {2}\right) + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2} \\ = 3 d \alpha \| \theta - z \| _ {\Phi (\theta)} ^ {2} + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2}, \\ \end{array}
$$
where the first inequality is Pinsker's inequality, the second inequality holds because  $\log \left(\frac{1}{t}\right) \leq \frac{1}{t} - 1$  for all  $t > 0$ , the fourth inequality holds because  $(1 + t)^2 + \frac{1}{(1 + t)^2} - 2 \leq 6t^2$  for all  $t \in \left[-\frac{1}{4}, \frac{1}{4}\right]$ , and the fifth inequality holds since  $t^2$  is convex for  $t \in \mathbb{R}$ .
For every  $j \in \mathbb{N}$ , let  $m_j = m + \lfloor \alpha \eta^{-1} j^2 \rfloor d$ . Consider the matrices  $A^j = [A^\top, I_d, \ldots, I_d]^\top$  where  $A$  is concatenated with  $\frac{m_j - 1}{d}$  copies of the identity matrix  $I_d$ . And consider the vectors  $b^j = (b^\top, j\mathbf{1}^\top, \ldots, j\mathbf{1}^\top)$ , where  $b$  is concatenated with  $\frac{m_j - 1}{d}$  copies of the vector  $j\mathbf{1}$  where  $\mathbf{1} = (1, \ldots, 1)^\top \in \mathbb{R}^d$  is the all-ones vector. Then  $K = \{\theta \in \mathbb{R}^d : A^j\theta \leq b^j\}$ , and the Hessian of the corresponding log-barrier functions is
$$
H _ {j} (w) := \sum_ {i = 1} ^ {m ^ {j}} \frac {a _ {i} ^ {j} (a _ {i} ^ {j}) ^ {\top}}{((a _ {i} ^ {j}) ^ {\top} w - b _ {i} ^ {j}) ^ {2}}.
$$
Lemma E.7 For all  $w \in \operatorname{int}(K)$  we have that
$$
\lim  _ {j \rightarrow \infty} H _ {j} (w) = \alpha \Phi (w), \tag {15}
$$
uniformly in  $w$ , and that
$$
\lim  _ {j \rightarrow \infty} (H _ {j} (w)) ^ {- 1} = \alpha^ {- 1} (\Phi (w)) ^ {- 1}, \tag {16}
$$
uniformly in  $w$ . Moreover, for any  $\theta \in \operatorname{int}(K)$ , any  $z \in \frac{1}{2} D_{\theta}$  and any sequence  $\{z_j\}_{j=1}^{\infty} \subseteq \frac{1}{2} D_{\theta}$  such that  $\lim_{j \to \infty} z_j = z$  uniformly in  $z$ , we have that
$$
\lim  _ {j \rightarrow \infty} H _ {j} \left(z _ {j}\right) = \alpha \Phi (z), \tag {17}
$$
uniformly in  $z$ , and that
$$
\lim  _ {j \rightarrow \infty} \frac {\det  \left(H _ {j} \left(z _ {j}\right)\right)}{\det  \left(H _ {j} (\theta)\right)} = \frac {\det  (\Phi (z))}{\det  (\Phi (\theta))}, \tag {18}
$$
uniformly in  $z$
$$
\begin{array}{l} H _ {j} (w) = \sum_ {i = 1} ^ {m} \frac {a _ {i} a _ {i} ^ {\top}}{(a _ {i} ^ {\top} w - b _ {i}) ^ {2}} + \lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor \sum_ {i = 1} ^ {d} \frac {e _ {i} e _ {i} ^ {\top}}{(e _ {i} ^ {\top} w - j) ^ {2}} \\ = H (w) + \left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor \sum_ {i = 1} ^ {d} \frac {e _ {i} e _ {i} ^ {\top}}{\left(e _ {i} ^ {\top} w - j\right) ^ {2}}. \tag {19} \\ \end{array}
$$
Now, since  $w \in K \subseteq B(0, R)$ , we have that
$$
(R - j) ^ {2} \leq \left(e _ {i} ^ {\top} w - j\right) ^ {2} \leq (- R - j) ^ {2}. \tag {20}
$$
$$
\lim  _ {j \rightarrow \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{\left(e _ {i} ^ {\top} w - j\right) ^ {2}} \leq \lim  _ {j \rightarrow \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(R - j) ^ {2}} = \alpha \eta^ {- 1}, \tag {21}
$$
$$
\lim  _ {j \rightarrow \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{\left(e _ {i} ^ {\top} w - j\right) ^ {2}} \geq \lim  _ {j \rightarrow \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(- R - j) ^ {2}} = \alpha \eta^ {- 1}. \tag {22}
$$
Thus, (19) (21), and (22) together imply that
$$
\begin{array}{l} \lim _ {j \to \infty} H _ {j} (w) = H (w) + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \lim _ {j \to \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(e _ {i} ^ {\top} w - j) ^ {2}} \\ = H (w) + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \alpha \eta^ {- 1} \\ = H (w) + \alpha \eta^ {- 1} I _ {d} \\ = \alpha \Phi (w), \tag {23} \\ \end{array}
$$
where Inequalities (21) and (22) imply that the convergence to the limit is uniform in  $w$ . This proves Equation (15).
Moreover, since  $\{z_k\}_{k=1}^{\infty} \subseteq \frac{1}{2} D_\theta$ , and  $D_\theta \subseteq K$ , we also have that
$$
\left| a _ {i} ^ {\top} z _ {j} - b _ {i} \right| \geq \frac {1}{2} \left| a _ {i} ^ {\top} \theta - b _ {i} \right|. \tag {24}
$$
$$
\begin{array}{l} \lim _ {j \to \infty} H _ {j} (z _ {j}) = H (z _ {j}) + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \lim _ {j \to \infty} \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(e _ {i} ^ {\top} z _ {j} - j) ^ {2}} \\ = \sum_ {i = 1} ^ {m} \frac {a _ {i} a _ {i} ^ {\top}}{\left(a _ {i} ^ {\top} z _ {j} - b _ {i}\right) ^ {2}} + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \lim  _ {j \rightarrow \infty} \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{\left(e _ {i} ^ {\top} z _ {j} - j\right) ^ {2}} \\ = \sum_ {i = 1} ^ {m} \frac {a _ {i} a _ {i} ^ {\top}}{\left(a _ {i} ^ {\top} z _ {j} - b _ {i}\right) ^ {2}} + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \alpha \eta^ {- 1} \\ \stackrel {\text {E q .}} {=} ^ {(2 4)} \sum_ {i = 1} ^ {m} \frac {a _ {i} a _ {i} ^ {\top}}{\left(a _ {i} ^ {\top} z - b _ {i}\right) ^ {2}} + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \alpha \eta^ {- 1} \\ = H (z) + \alpha \eta^ {- 1} I _ {d} \\ = \alpha \Phi (z), \\ \end{array}
$$
where the convergence of the limit in the fourth equality holds uniformly in  $z$  by (24) and the fact that  $\lim_{j\to \infty}z_j = z$ . Thus, we have that
$$
\lim  _ {j \rightarrow \infty} H _ {j} \left(z _ {j}\right) = \alpha \Phi (z) \tag {25}
$$
uniformly in  $z$ . This proves Equation (17).
Moreover, since the determinant is a polynomial in the entries of the matrix, Inequality (23) implies that
$$
\lim  _ {j \rightarrow \infty} \det  \left(H _ {j} (w)\right) = \det  (\alpha \Phi (w)), \tag {26}
$$
uniformly in  $w$
We also have, plugging Inequality (20) into Equation (19),
$$
H (w) + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} \preceq H _ {j} (w) \preceq H (w) + \sum_ {i = 1} ^ {d} e _ {i} e _ {i} ^ {\top} \times \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(- R - j) ^ {2}}
$$
and, hence, that
$$
v ^ {\top} \left[ H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d} \right] v \leq v ^ {\top} H _ {j} (w) v \leq v ^ {\top} \left[ H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(- R - j) ^ {2}} I _ {d} \right] v \quad \forall v \in \mathbb {R} ^ {d}. \tag {27}
$$
Thus, Inequality (27) implies that
$$
H (w) + \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(R - j) ^ {2}} I _ {d} \preceq H _ {j} (w) \preceq H (w) + \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(- R - j) ^ {2}} I _ {d} \quad \forall j \in \mathbb {N}. \tag {28}
$$
Thus, Inequality (28) implies that
$$
\left(H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(- R - j) ^ {2}} I _ {d}\right) ^ {- 1} \preceq \left(H _ {j} (w)\right) ^ {- 1} \preceq \left(H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d}\right) ^ {- 1} \quad \forall j \in \mathbb {N}. \tag {29}
$$
Thus, since  $H(w)$  is positive definite, Inequality (29) together with inequalities (21) and (22) imply that
$$
\begin{array}{l} \lim  _ {j \rightarrow \infty} (H _ {j} (w)) ^ {- 1} = (H (w) + \alpha \eta^ {- 1} I _ {d}) ^ {- 1} \\ = \alpha^ {- 1} \Phi (w) ^ {- 1} \quad \forall w \in \operatorname {I n t} (K) \\ \end{array}
$$
uniformly in  $w$ . This proves Equation (16).
Moreover, Inequality (28) implies that
$$
\begin{array}{l} \det  (H _ {j} (w)) \leq \det  \left(H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d}\right) \\ \leq \left(\lambda_ {\max } \left(H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d}\right)\right) ^ {d} \\ \leq \left(\lambda_ {\max } (H (w)) + \lambda_ {\max } \left(\frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d})\right)\right) ^ {d} \\ \leq \left(\lambda_ {\max } (H (w)) + 3 \alpha \eta^ {- 1}\right) ^ {d} \\ = c _ {1} \quad \forall j \geq 3 R, \tag {30} \\ \end{array}
$$
for some  $c_{1} > 0$  which does not depend on  $j$ .
Inequality (27) also implies that
$$
\begin{array}{l} \det  (H _ {j} (w)) \geq \det  \left(H (w) + \frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d}\right) \\ \geq \left(\lambda_ {\min } \left(H (w) + \frac {\left\lfloor \alpha \eta^ {- 1} j ^ {2} \right\rfloor}{(R - j) ^ {2}} I _ {d}\right)\right) ^ {d} \\ \geq \left(\max  \left(\lambda_ {\min } (H (w)), \lambda_ {\min } \left(\frac {\lfloor \alpha \eta^ {- 1} j ^ {2} \rfloor}{(R - j) ^ {2}} I _ {d}\right)\right)\right) ^ {d} \\ \geq \left(\frac {1}{3} \alpha \eta^ {- 1}\right) ^ {d} \\ = c _ {2} \quad \forall j \geq 3 R, \tag {31} \\ \end{array}
$$
for some  $c_{2} > 0$  which does not depend on either  $j$  or  $w$ .
Thus, Inequalities (26), (30), and (31) together imply that for any  $\theta, z \in \operatorname{int}(K)$  we have that
$$
\begin{array}{l} \lim  _ {j \rightarrow \infty} \min  \left(\frac {\det  (H _ {j} (z))}{\det  (H _ {j} (\theta))}, 1\right) \stackrel {{\text {E q .}}} {{=}} ^ {(3 0), (3 1)} \min  \left(\frac {\lim  _ {j \rightarrow \infty} \det  (H _ {j} (z _ {j}))}{\lim  _ {j \rightarrow \infty} \det  (H _ {j} (\theta))}, 1\right) \\ \stackrel {\text {E q .}} {=} \stackrel {(2 6), (2 5)} {=} \min  \left(\frac {\det  (\alpha \Phi (z))}{\det  (\alpha \Phi (\theta))}, 1\right) \\ = \min  \left(\frac {\det  (\Phi (z))}{\det  (\Phi (\theta))}, 1\right), \\ \end{array}
$$
where the limit converges uniformly in  $z$ . This proves Equation (18).
Lemma E.8 Let  $\xi \sim N(0, I_d)$  and let  $\theta \in \operatorname{int}(K)$ . Then with probability at least  $1 - \frac{1}{100} \hat{\delta}$  we have that
$$
\theta + \alpha^ {\frac {1}{2}} H ^ {- 1 / 2} (\theta) \xi \in \frac {1}{2} D _ {\theta}
$$
$$
\| \xi \| _ {2} \leq 1 0 \sqrt {d}.
$$
Proof: Let  $z = \theta + \alpha^{\frac{1}{2}} H^{-\frac{1}{2}}(\theta) \xi$ . Then since  $\xi \sim N(0, I_d)$ , by the Hanson-Wright concentration inequality [22, 40], we have that
$$
\mathbb {P} (\| \xi \| _ {2} > t) \leq e ^ {- \frac {t ^ {2} - d}{8}} \quad \forall t > \sqrt {2 d}.
$$
And hence, since  $\alpha^{-1 / 2}\| z - \theta \|_{H(\theta)} = \| H^{\frac{1}{2}}(\theta)H^{-\frac{1}{2}}(\theta)\xi \| _2 = \| \xi \| _2$  , that
$$
\mathbb {P} \big (\left\| z - \theta \right\| _ {H (\theta)} > \alpha^ {1 / 2} t \big) \leq e ^ {- \frac {t ^ {2} - d}{8}} \quad \forall t > \sqrt {2 d}.
$$
$$
\mathbb {P} (\| z - \theta \| _ {H (\theta)} > \alpha^ {1 / 2} 1 0 \sqrt {d}) \leq \frac {1}{1 0 0}.
$$
Thus, since  $\alpha \leq \frac{\log\left(\frac{1}{\delta}\right)}{100d}$  we have that
$$
\mathbb {P} \left(z \in \frac {1}{2} D _ {\theta}\right) = \mathbb {P} \left(\| z - \theta \| _ {H (\theta)} \leq \frac {1}{2}\right) \geq 1 - \frac {1}{1 0 0} \hat {\delta}.
$$
Lemma E.9 Consider any  $\theta \in \mathrm{int}(K)$ , and  $\xi \sim N(0, I_d)$ . Let  $z = \theta + (\Phi(\theta))^{-\frac{1}{2}}\xi$ . Then
$$
\mathbb {P} \left(\frac {\det (\Phi (z))}{\det (\Phi (\theta))} \geq \frac {4 8}{5 0}\right) \geq 1 - \frac {1}{1 0 0} \hat {\delta}, \tag {32}
$$
$$
\mathbb {P} \left(\| z - \theta \| _ {\Phi (z)} ^ {2} - \| z - \theta \| _ {\Phi (\theta)} ^ {2} \leq \frac {2}{5 0}\right) \geq \frac {9 8}{1 0 0}. \tag {33}
$$
Proof: Let  $z_{j} = \theta +\alpha^{\frac{1}{2}}H_{j}^{-\frac{1}{2}}(\theta)\xi$  for all  $j\in \mathbb{N}$ . Since  $H_{j}(\theta)\succeq H(\theta)$  for all  $j\in \mathbb{N}$ , we have that  $z_{j} = \theta +\alpha^{\frac{1}{2}}H_{j}^{-\frac{1}{2}}(\theta)\xi \in \frac{1}{2} D_{\theta}$  whenever  $\theta +\alpha^{\frac{1}{2}}H(\theta)^{-\frac{1}{2}}\xi \in \frac{1}{2} D_{\theta}$ . Let  $E$  be the event that  $\| \xi \| _2\leq 10\sqrt{d}$  and that  $\{z_j\}_{j = 1}^{\infty}\subseteq \frac{1}{2} D_\theta$ . Thus, by Lemma E.8, we have that
$$
\mathbb {P} (E) \geq \frac {9 9}{1 0 0}. \tag {34}
$$
Moreover, by Equation (16) of Lemma E.7 we have that  $\lim_{j\to \infty}H_j^{-1}(\theta) = \alpha^{-1}\Phi^{-1}(\theta)$ , which implies that
$$
\begin{array}{l} \lim  _ {j \rightarrow \infty} z _ {j} = \lim  _ {j \rightarrow \infty} \theta + \alpha^ {\frac {1}{2}} H _ {j} ^ {- \frac {1}{2}} (\theta) \xi \\ = \theta + \Phi^ {- \frac {1}{2}} (\theta) \xi \\ = z \tag {35} \\ \end{array}
$$
uniformly in  $\xi$  (and hence uniformly in  $z = \theta +\Phi^{-\frac{1}{2}}(\theta)\xi$ ) whenever the event  $E$  occurs. Therefore, by Equation (18) of Lemma E.7 we have that
$$
\lim  _ {j \rightarrow \infty} \frac {\det  \left(H _ {j} \left(z _ {j}\right)\right)}{\det  \left(H _ {j} (\theta)\right)} = \frac {\det  (\Phi (z))}{\det  (\Phi (\theta))}, \tag {36}
$$
uniformly in  $z$  whenever the event  $E$  occurs.
Since, for each  $j \in \mathbb{N}$ ,  $H_{j}$  is the Hessian of a log-barrier function for  $K$ , by last line in the proof of Proposition 6 in [41], we have that, for all  $t \in (0, \frac{1}{2}]$  and all  $\gamma \in (0, 1]$ ,
$$
\mathbb {P} _ {\xi \sim N (0, I _ {d})} \left(\log \det (H _ {j} (\theta + \frac {\gamma}{\sqrt {d}} H _ {j} (\theta) ^ {- \frac {1}{2}} \xi)) - \log \det (H _ {j} (\theta)) \geq - \gamma \sqrt {2 \log (\frac {1}{t})}\right) \geq 1 - t.
$$
Setting  $t = \frac{1}{100}\hat{\delta}$ , we have
$$
\mathbb {P} _ {\xi \sim N (0, I _ {d})} \left(\log \det (H _ {j} (\theta + \alpha^ {\frac {1}{2}} H _ {j} (\theta) ^ {- \frac {1}{2}} \xi)) - \log \det (H _ {j} (\theta)) \geq - \frac {1}{1 0 0}\right) \geq 1 - \frac {1}{1 0 0} \hat {\delta},
$$
since  $\alpha \leq \frac{1}{400^2 d \log\left(\frac{1}{\delta}\right)}$ .
$$
\mathbb {P} _ {\xi \sim N \left(0, I _ {d}\right)} \left(\frac {\operatorname* {d e t} \left(H _ {j} \left(\theta + \alpha^ {\frac {1}{2}} H _ {j} (\theta) ^ {- \frac {1}{2}} \xi\right)\right)}{\operatorname* {d e t} \left(H _ {j} (\theta)\right)} \geq \frac {4 9}{5 0}\right) \geq 1 - \frac {1}{1 0 0} \hat {\delta}. \tag {37}
$$
Inequalities (34) and (37) together imply that
$$
\mathbb {P} _ {\xi \sim N (0, I _ {d})} \left(\left\{\frac {\operatorname* {d e t} \left(H _ {j} \left(z _ {j}\right)\right)}{\operatorname* {d e t} \left(H _ {j} (\theta)\right)} \geq \frac {4 9}{5 0} \right\} \cap E\right) \geq 1 - \frac {1}{1 0 0} \hat {\delta}. \quad \forall j \in \mathbb {N}. \tag {38}
$$
Moreover, Equation (36) implies that there exists some number  $N \in \mathbb{N}$  such that
$$
\left\{\frac {\det \left(H _ {N} \left(z _ {N}\right)\right)}{\det \left(H _ {N} (\theta)\right)} \geq \frac {4 9}{5 0} \right\} \cap E \subseteq \left\{\frac {\det \left(\Phi (z)\right)}{\det \left(\Phi (\theta)\right)} \geq \frac {4 8}{5 0} \right\} \cap E. \tag {39}
$$
$$
\begin{array}{l} 1 - \frac {1}{1 0 0} \hat {\delta} ^ {\text {E q .} (3 8)} \leq \mathbb {P} _ {\xi \sim N (0, I _ {d})} \left(\left\{\frac {\det  \left(H _ {N} \left(z _ {N}\right)\right)}{\det  \left(H _ {N} (\theta)\right)} \geq \frac {4 9}{5 0} \right\} \cap E\right) \\ \leq^ {(3 9)} \mathbb {P} _ {\xi \sim N (0, I _ {d})} \left(\left\{\frac {\det (\Phi (z))}{\det (\Phi (\theta))} \geq \frac {4 8}{5 0} \right\} \cap E\right). \tag {40} \\ \end{array}
$$
This proves Inequality (32).
Moreover, since, for each  $j \in \mathbb{N}$ ,  $H_{j}$  is the Hessian of a log-barrier function for  $K$ , by Proposition 7 in [41], for all  $t \in (0, \frac{1}{2}]$ , all  $\alpha > 0$  such that  $\sqrt{\alpha d} \leq \frac{t}{20} \log \left( \frac{11}{t} \right)^{-\frac{3}{2}}$ , and all  $j \in \mathbb{N}$ , we have that
$$
\mathbb {P} \left(\left\| z _ {j} - \theta \right\| _ {H _ {j} \left(z _ {j}\right)} ^ {2} - \left\| z _ {j} - \theta \right\| _ {H _ {j} (\theta)} ^ {2} \geq 2 t \alpha\right) \leq 1 - t \quad \forall j \in \mathbb {N}. \tag {41}
$$
Thus, Equations (41) and (34) imply that
$$
\mathbb {P} \left(\left\{\| z _ {j} - \theta \| _ {H _ {j} \left(z _ {j}\right)} ^ {2} - \| z _ {j} - \theta \| _ {H _ {j} (\theta)} ^ {2} \leq \frac {1}{5 0} \alpha \right\} \cap E\right) \geq \frac {9 8}{1 0 0} \quad \forall j \in \mathbb {N}, \tag {42}
$$
since  $\alpha \leq \frac{1}{10^5d}$ .
By Equation (17) of Lemma E.7, Equation (35) implies that
$$
\lim  _ {j \rightarrow \infty} H _ {j} \left(z _ {j}\right) = \alpha \Phi (z), \tag {43}
$$
uniformly in  $z$ , whenever the event  $E$  occurs. Thus, Equation (43) implies that
$$
\begin{array}{l} \lim  _ {j \rightarrow \infty} \| z _ {j} - \theta \| _ {H _ {j} (z _ {j})} ^ {2} - \| z _ {j} - \theta \| _ {H _ {j} (\theta)} ^ {2} = \lim  _ {j \rightarrow \infty} (z _ {j} - \theta) ^ {\top} H _ {j} (z _ {j}) (z _ {j} - \theta) - (z _ {j} - \theta) ^ {\top} H _ {j} (\theta) (z _ {j} - \theta) \\ \stackrel {\text {E q .}} {=} ^ {(4 3)} \lim  _ {j \rightarrow \infty} (z - \theta) ^ {\top} \alpha \Phi (z) (z - \theta) - (z - \theta) ^ {\top} \alpha \Phi (\theta) (z - \theta) \\ = \alpha \| z - \theta \| _ {\Phi (z)} ^ {2} - \alpha \| z - \theta \| _ {\Phi (\theta)} ^ {2} \tag {44} \\ \end{array}
$$
uniformly in  $z$  (and hence in  $\xi = \Phi^{\frac{1}{2}}(\theta)(z - \theta)$ ) whenever event  $E$  occurs. Thus, Equation (44) implies that there exists a number  $M\in \mathbb{N}$  such that
$$
\left\{\| z _ {M} - \theta \| _ {H _ {M} (z _ {M})} ^ {2} - \| z _ {M} - \theta \| _ {H _ {M} (\theta)} ^ {2} \leq \frac {1}{5 0} \alpha \right\} \cap E \subseteq \left\{\alpha \| z - \theta \| _ {\Phi (z)} ^ {2} - \alpha \| z - \theta \| _ {\Phi (\theta)} ^ {2} \leq \frac {2}{5 0} \alpha \right\} \cap E. \tag {45}
$$
$$
\begin{array}{l} \frac {9 8}{1 0 0} \stackrel {\text {E q . (4 2)}} {\leq} \mathbb {P} \left(\left\{\| z _ {M} - \theta \| _ {H _ {M} (z _ {M})} ^ {2} - \| z _ {M} - \theta \| _ {H _ {M} (\theta)} ^ {2} \leq \frac {1}{5 0} \alpha \right\} \cap E\right) \\ \stackrel {\text {E q . (4 5)}} {\leq} \mathbb {P} \left(\left\{\alpha \| z - \theta \| _ {\Phi (z)} ^ {2} - \alpha \| z - \theta \| _ {\Phi (\theta)} ^ {2} \leq \frac {2}{5 0} \alpha \right\} \cap E\right). \\ \end{array}
$$
This proves Inequality (33).
E.4 Bounding the conductance
Lemma E.10 For any  $\theta, z \in \operatorname{int}(K)$  we have that
$$
\| P _ {\theta} - P _ {z} \| _ {\mathrm {T V}} \leq \frac {2 9}{3 0}, \quad \text {w h e n e v e r} \quad \| \theta - z \| _ {\Phi (\theta)} \leq 1.
$$
Proof: First, we note that,
$$
\begin{array}{l} \| P _ {\theta} - P _ {z} \| _ {\mathrm {T V}} \leq \| P _ {\theta} - N (\theta , \Phi^ {- 1} (\theta)) \| _ {\mathrm {T V}} \\ + \| N (\theta , \Phi^ {- 1} (\theta)) - N (z, \Phi^ {- 1} (z)) \| _ {\mathrm {T V}} + \| P _ {z} - N (z, \Phi^ {- 1} (z)) \| _ {\mathrm {T V}}. \tag {46} \\ \end{array}
$$
By Lemma E.6, the middle term on the RHS of (46) satisfies
$$
\| N (\theta , \Phi^ {- 1} (\theta)) - N (z, \Phi^ {- 1} (z)) \| _ {\mathrm {T V}} \leq \sqrt {3 d \alpha \| \theta - z \| _ {\Phi (\theta)} ^ {2} + \frac {1}{2} \| \theta - z \| _ {\Phi (\theta)} ^ {2}} \tag {47}
$$
Plugging in our choice of hyperparameter  $\alpha = \frac{1}{10^5d}$ , Inequality (47) simplifies to
$$
\left\| N \left(\theta , \Phi^ {- 1} (\theta)\right) - N \left(z, \Phi^ {- 1} (z)\right) \right\| _ {\mathrm {T V}} \leq \sqrt {\frac {5 3}{1 0 0}} \| \theta - z \| _ {\Phi (\theta)}. \tag {48}
$$
Thus, if we can show that  $\| P_{\theta} - N(\theta, \Phi^{-1}(\theta)) \|_{\mathrm{TV}} \leq \frac{1}{5}$  for all  $\theta \in \operatorname{int}(K)$ , we will have that  $\| P_{\theta} - P_{z} \|_{\mathrm{TV}} \leq \frac{29}{30}$  whenever  $\| \theta - z \|_{\Phi(\theta)} \leq \frac{1}{2}$ , as desired.
To bound the other two terms on the RHS of (46), we observe that
$$
\left\| P _ {\theta} - N \left(\theta , \Phi^ {- 1} (\theta)\right) \right\| _ {\mathrm {T V}} = 1 - \mathbb {E} _ {z \sim N \left(\theta , \Phi^ {- 1} (\theta)\right)} [ q (\theta , z) ], \tag {49}
$$
$$
q (\theta , z) := \min \{1, \frac {\pi (z)}{\pi (\theta)} \frac {\sqrt {\det \Phi (z)}}{\sqrt {\det \Phi (\theta)}} \exp (\| z - \theta \| _ {\Phi (\theta)} ^ {2} - \| \theta - z \| _ {\Phi (z)} ^ {2}) \times \mathbb {1} \{z \in K \} \}
$$
is the acceptance ratio.
By Lemma E.5, we have that  $\mathbb{P}_{z\sim N(\theta ,\Phi^{-1}(\theta))}\left(\frac{\pi(z)}{\pi(\theta)}\geq \frac{99}{100}\right)\geq \frac{99}{10}$ . Therefore,
$$
\begin{array}{l} \mathbb {P} _ {z \sim N (\theta , \Phi^ {- 1} (\theta))} [ q (\theta , z) \geq \frac {9}{1 0} ] \\ \stackrel {\text {L e m m a}} {\geq} 1 - \mathbb {P} _ {z \sim N (\theta , \Phi^ {- 1} (\theta))} \left(\frac {\pi (z)}{\pi (\theta)} <   \frac {9 9}{1 0 0}\right) - \mathbb {P} _ {z \sim N (\theta , \Phi^ {- 1} (\theta))} \left(\frac {\sqrt {\det  (\Phi (z))}}{\sqrt {\det  (\Phi (\theta))}} <   \sqrt {\frac {4 8}{5 0}}\right) \\ - \mathbb {P} _ {z \sim N (\theta , \Phi^ {- 1} (\theta))} \left(e ^ {\| z - \theta \| _ {\Phi (\theta)} ^ {2} - \| \theta - z \| _ {\Phi (z)} ^ {2}} <   0. 9 6\right) - \mathbb {P} _ {z \sim N (\theta , \Phi^ {- 1} (\theta))} (z \notin K) \\ \geq 1 - \frac {1}{1 0 0} - \frac {2}{1 0 0} - \frac {2}{1 0 0} - \frac {1}{1 0 0} \\ \geq \frac {9}{1 0}, \\ \end{array}
$$
where the term with the exponent is bounded by Inequality (33) of Lemma E.9, and the other two terms are bounded by Lemmas E.8 and E.9. Therefore,
$$
\left\| P _ {\theta} - N \left(\theta , \Phi^ {- 1} (\theta)\right) \right\| _ {\mathrm {T V}} = 1 - \mathbb {E} _ {z \sim N \left(\theta , \Phi^ {- 1} (\theta)\right)} [ q (\theta , z) \geq \frac {9}{1 0} ] \geq 1 - \frac {9}{1 0} \times \frac {9}{1 0} \geq \frac {4}{5}. \tag {50}
$$
Plugging Inequalities (48) and (50) into (46), we obtain that  $\| P_{\theta} - P_{z}\|_{\mathrm{TV}}\leq \frac{29}{30}$  whenever  $\| \theta -z\|_{\Phi (\theta)}\leq \frac{1}{2}$ , as desired.
We recall the following isoperimetric inequality for a log-concave distribution on a convex body, which uses the cross-ratio distance: