SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathbb {E} \left(\left| Z ^ {k} \right| / \left(k! \sigma^ {k}\right)\right) \leq \mathbb {E} \left(e ^ {\left| Z \right| / \sigma}\right) = \int_ {0} ^ {\infty} e ^ {t} \mathbb {P} \left(\left| Z \right| / \sigma > t\right) d t \\ \leq \int_ {0} ^ {\infty} 2 \exp \left(t - \frac {t ^ {2}}{2}\right) d t = 2 \sqrt {e} \int_ {0} ^ {\infty} \exp (- (t - 1) ^ {2} / 2) d t \\ \leq 2 \sqrt {e} \int_ {- \infty} ^ {\infty} \exp (- t ^ {2} / 2) d t = 2 \sqrt {2 \pi e} \leq 9 \\ \end{array}
$$
Lemma D.5 For any integer  $k \geq 2$  it holds  $\frac{1}{k^2} - \frac{4}{(k + 1)^3} \leq \frac{1}{(k + 1)^2}$ .
Proof. For any  $k \geq 2$  it holds
$$
\frac {(2 k + 1) (k + 1)}{2 k ^ {2}} = \left(1 + \frac {1}{2 k}\right) \left(1 + \frac {1}{k}\right) \leq \left(1 + \frac {1}{4}\right) \left(1 + \frac {1}{2}\right) <   2
$$
Multiplying  $2 / (k + 1)^3$  in the display above, we have
$$
\frac {2 k + 1}{k ^ {2} (k + 1) ^ {2}} <   \frac {4}{(k + 1) ^ {3}}
$$
The proof is complete by noting that  $\frac{2k + 1}{k^2(k + 1)^2} = \frac{1}{k^2} -\frac{1}{(k + 1)^2}$
Lemma D.6 Let  $[a,b]$  be a sub-interval of [0,1], and  $c\in (a,b)$ . Let  $h$  be a function on  $[a,b]$  such that  $h$  is differentiable on  $(a,c)$  and  $(c,b)$ , but can be discontinuous at  $c$ . Denote  $\Delta h(c)\coloneqq \lim_{t\to c + }h(t) - \lim_{t\to c - }h(t)$ . Suppose
$$
\Delta h (c) > 4 \max  \left\{\int_ {a} ^ {c} | h ^ {\prime} (t) | \mathrm {d} t, \int_ {c} ^ {b} | h ^ {\prime} (t) | \mathrm {d} t \right\} \tag {112}
$$
Then it holds
$$
\inf  _ {w \in \mathbb {R}} \int_ {a} ^ {b} (h (t) - w) ^ {2} \mathrm {d} t \geq \min  \{c - a, b - c \} (\Delta h (c)) ^ {2} / 1 6
$$
Proof. We assume that  $h$  is not continuous at  $c$ , since otherwise, the conclusion holds true trivially. We use the notation  $h(c+) := \lim_{t \to c+} h(t)$  and  $h(c-) := \lim_{t \to c-} h(t)$ . Without loss of generality, assume  $h(c+) > h(c-)$ .
For  $w \geq (1/2)(h(c+) + h(c-))$ , it holds  $w - h(c-) \geq (1/2)\Delta h(c)$ . By (112), we know that for any  $t \in (a, c)$ ,
$$
\left| h (t) - h (c -) \right| \leq \int_ {a} ^ {c} \left| h ^ {\prime} (\tau) \right| d \tau \leq \frac {1}{4} \Delta h (c)
$$
Hence for all  $t\in (a,c)$
$$
w - h (t) = w - h (c -) + h (c -) - h (t) \geq \frac {1}{2} \Delta h (c) - \frac {1}{4} \Delta h (c) = \frac {1}{4} \Delta h (c)
$$
$$
\int_ {a} ^ {b} (h (t) - w) ^ {2} \mathrm {d} t \geq \int_ {a} ^ {c} (h (t) - w) ^ {2} \mathrm {d} t \geq (c - a) (\Delta h (c)) ^ {2} / 1 6 \tag {113}
$$
For  $w < (1/2)(h(c+) + h(c-))$ , similarly, we can prove
$$
\int_ {a} ^ {b} (h (t) - w) ^ {2} \mathrm {d} t \geq (b - c) (\Delta h (c)) ^ {2} / 1 6 \tag {114}
$$
The proof is complete by combining (113) and (114).