SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\frac {\partial R (a , \theta)}{\partial \theta} = \frac {h}{\theta^ {2}} - a (b + h) \frac {e ^ {- a \theta}}{\theta} - (b + h) \frac {e ^ {- a \theta}}{\theta^ {2}} = \frac {1}{\theta^ {2}} \left(h - (b + h) e ^ {- a \theta} (1 + a \theta)\right) \leq \frac {h}{\theta^ {2}}. \tag {60}
$$
The result follows immediately, since  $\sup_{\theta \in \Theta} \frac{\partial R(a, \theta)}{\partial \theta} \leq \frac{h}{T^2}$ .
Proof. Proof of Lemma B.1
It follows from Lemma C.10 that  $d_L^{NV}(\theta, \theta_0)$  for any  $\theta \in \Theta = [T, \infty)$  and  $\theta_0$  lying the interior of  $\Theta$ , satisfies Assumption C.1 with
$$
K _ {1} = \left[ \frac {\left(\frac {h}{\theta_ {0}} - \frac {h}{T}\right) ^ {2} + (b + h) ^ {2} \left(\frac {e ^ {- \underline {{a}} T}}{T} - \frac {e ^ {- \underline {{a}} \theta_ {0}}}{\theta_ {0}}\right) ^ {2}}{d _ {H} ^ {2} (T , \theta_ {0})} \right] ^ {1 / 2} := K _ {1} ^ {N V}
$$
. Similarly, it follows from Lemma and C.12 that for sufficiently small  $T > 0$ ,  $d_{L}^{NV}(\theta, \theta_{0})$  satisfies Assumption C.2 with  $K_{2} = h / T^{2} \coloneqq K_{2}^{NV}$ . Now using similar arguments as used in Lemma C.8 and Lemma 2.1, for a given  $\epsilon_{n} > 0$  and every  $\epsilon > \epsilon_{n}$ , such that  $n\epsilon_{n}^{2} \geq 1$ , it can be shown that,  $L_{n}^{NV}(\theta, \theta_{0}) = n\left(\sup_{a \in \mathcal{A}} |R(a, \theta) - R(a, \theta_{0})|\right)^{2}$  satisfies
$$
\mathbb {E} _ {P _ {0} ^ {n}} \left[ \phi_ {n} \right] \leq C _ {0} \exp \left(- C n \epsilon^ {2}\right), \tag {61}
$$
$$
\sup  _ {\left\{\theta \in \Theta : L _ {n} ^ {N V} \left(\theta , \dot {\theta} _ {0}\right) \geq C _ {1} n \epsilon^ {2} \right\}} \mathbb {E} _ {P _ {\theta} ^ {n}} \left[ 1 - \phi_ {n} \right] \leq \exp \left(- C n \epsilon^ {2}\right), \tag {62}
$$
where  $C_0 = 20$  and  $C = \frac{C_1}{2(K_1^{NV})^2}$  for a constant  $C_1 > 0$ .
Proof. Proof of Lemma B.2:
First, we write the Rényi divergence between  $P_0^n$  and  $P_{\theta}^{n}$
$$
\begin{array}{l} D _ {1 + \lambda} \left(P _ {0} ^ {n} \| P _ {\theta} ^ {n}\right) = \frac {1}{\lambda} \log \int \left(\frac {d P _ {0} ^ {n}}{d P _ {\theta} ^ {n}}\right) ^ {\lambda} d P _ {0} ^ {n} = n \frac {1}{\lambda} \log \int \left(\frac {d P _ {0}}{d P _ {\theta}}\right) ^ {\lambda} d P _ {0} \\ = n \left(\log \frac {\theta_ {0}}{\theta} + \frac {1}{\lambda} \log \frac {\theta_ {0}}{(\lambda + 1) \theta_ {0} - \lambda \theta}\right), \\ \end{array}
$$
when  $((\lambda + 1)\theta_0 - \lambda \theta) > 0$  and  $D_{1 + \lambda}(P_0^n \| P_\theta^n) = \infty$  otherwise. Also, observe that,  $D_{1 + \lambda}(P_0^n \| P_\theta^n)$  is non-decreasing in  $\lambda$  (this also follows from non-decreasing property of the Rényi divergence with respect to  $\lambda$ ). Therefore, observe that
$$
\begin{array}{l} \Pi \left(D _ {1 + \lambda} \left(P _ {0} ^ {n} \| P _ {\theta} ^ {n}\right) \leq C _ {3} n \epsilon_ {n} ^ {2}\right) \geq \Pi \left(D _ {\infty} \left(P _ {0} ^ {n} \| P _ {\theta} ^ {n}\right) \leq C _ {3} n \epsilon_ {n} ^ {2}\right) = \Pi \left(0 \leq \log \frac {\theta_ {0}}{\theta} \leq C _ {3} \epsilon_ {n} ^ {2}\right) \\ = \Pi \left(\theta_ {0} e ^ {- C _ {3} \epsilon_ {n} ^ {2}} \leq \theta \leq \theta_ {0}\right). \\ \end{array}
$$
Now, recall that for a set  $A \subseteq \Theta = [T, \infty)$ , we define  $\Pi(A) = \mathrm{Inv} - \Gamma(A \cap \Theta) / \mathrm{Inv} - \Gamma(\Theta)$ . Now, observe that for sufficiently small  $T$  and large enough  $n$ , we have
$$
\Pi \left(\theta_ {0} e ^ {- C _ {3} \epsilon_ {n} ^ {2}} \leq \theta \leq \theta_ {0}\right) \geq \operatorname {I n v} - \Gamma \left(\theta_ {0} e ^ {- C _ {3} \epsilon_ {n} ^ {2}} \leq \theta \leq \theta_ {0}\right)
$$
The cumulative distribution function of inverse-gamma distribution is  $\operatorname{Inv} - \Gamma(\{\theta \in \Theta : \theta < t\}) \coloneqq \frac{\Gamma(\alpha, \frac{\beta}{t})}{\Gamma(\alpha)}$ , where  $\alpha (>0)$  is the shape parameter,  $\beta (>0)$  is the scale parameter,  $\Gamma(\cdot)$  is the Gamma function, and  $\Gamma(\cdot, \cdot)$  is the incomplete Gamma function. Therefore, it follows for  $\alpha > 1$  that
$$
\begin{array}{l} \operatorname {I n v} - \Gamma \left(\theta_ {0} e ^ {- C _ {3} \epsilon_ {n} ^ {2}} \leq \theta \leq \theta_ {0}\right) \\ = \frac {\Gamma (\alpha , \beta / \theta_ {0}) - \Gamma (\alpha , \beta / \theta_ {0} e ^ {C _ {3} \epsilon_ {n} ^ {2}})}{\Gamma (\alpha)} = \frac {\int_ {\beta / \theta_ {0}} ^ {\beta / \theta_ {0} e ^ {C _ {3} \epsilon_ {n} ^ {2}}} e ^ {- x} x ^ {\alpha - 1} d x}{\Gamma (\alpha)} \\ \geq \frac {e ^ {- \beta / \theta_ {0} e ^ {C _ {3} \epsilon_ {n} ^ {2}} + \alpha C _ {3} \epsilon_ {n} ^ {2}}}{\alpha \Gamma (\alpha)} \left(\frac {\beta}{\theta_ {0}}\right) ^ {\alpha} \left[ 1 - e ^ {- \alpha C _ {3} \epsilon_ {n} ^ {2}} \right] \\ \geq \frac {e ^ {- \beta / \theta_ {0} e ^ {C _ {3}}}}{\alpha \Gamma (\alpha)} \left(\frac {\beta}{\theta_ {0}}\right) ^ {\alpha} \left[ e ^ {- \alpha C _ {3} n \epsilon_ {n} ^ {2}} \right] \\ \end{array}
$$
where the penultimate inequality follows since  $0 < \epsilon_{n}^{2} < 1$  and the last inequality follows from the fact that,  $1 - e^{-\alpha C_3\epsilon_n^2}\geq e^{-\alpha C_3n\epsilon_n^2}$ , for large enough  $n$ . Also note that,  $1 - e^{-\alpha C_3\epsilon_n^2}\geq e^{-\alpha C_3n\epsilon_n^2}$  can't hold true for  $\epsilon_{n}^{2} = 1 / n$ . However, for  $\epsilon_{n}^{2} = \frac{\log n}{n}$  it holds for any  $n\geq 2$  when  $\alpha C_3 > 2$ . Therefore, for inverse-Gamma prior restricted to  $\Theta$ ,  $C_2 = \alpha C_3$  and any  $\lambda >1$  the result follows for sufficiently large  $n$ .
Proof. Proof of Lemma B.3: Recall,
$$
\frac {\partial R (a , \theta)}{\partial \theta} = \frac {h}{\theta^ {2}} - a (b + h) \frac {e ^ {- a \theta}}{\theta} - (b + h) \frac {e ^ {- a \theta}}{\theta^ {2}} = \frac {1}{\theta^ {2}} \left(h - (b + h) e ^ {- a \theta} (1 + a \theta)\right). \tag {63}
$$
Using the above equation the (finite) critical point  $\theta^{*}$  must satisfy,  $h - (b + h)e^{-a\theta^{*}}(1 + a\theta^{*}) = 0$  Therefore,
$$
R (a, \theta) \geq R (a, \theta^ {*}) = h \left(a - \frac {1}{\theta^ {*}} + \frac {1}{\theta^ {*} (1 + a \theta^ {*})}\right) = \frac {h a ^ {2} \theta^ {*}}{(1 + a \theta^ {*})}.
$$
Since  $h, b > 0$  and  $a\theta^{*} > 0$ , hence
$$
R (a, \theta) \geq \frac {h a ^ {2} \theta^ {*}}{(1 + a \theta^ {*})},
$$
where  $\underline{a} \coloneqq \min \{a \in \mathcal{A}\}$  and  $\underline{a} > 0$ .
Proof. Proof of Lemma B.4:
First, observe that  $R(a,\theta)$  is bounded above in  $\theta$  for a given  $a\in \mathcal{A}$
$$
\begin{array}{l} R (a, \theta) = h a - \frac {h}{\theta} + (b + h) \frac {e ^ {- a \theta}}{\theta} \\ \leq h a + \frac {b}{\theta}. \\ \end{array}
$$
Using the above fact and the Cauchy-Schwarz inequality, we obtain
$$
\begin{array}{l} \int_ {\left\{e ^ {\gamma R (a, \theta)} > e ^ {C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} \right\}} e ^ {\gamma R (a, \theta)} \pi (\theta) d \theta \\ \leq \left(\int e ^ {2 \gamma R (a, \theta)} \pi (\theta) d \theta\right) ^ {1 / 2} \left(\int \mathbb {1} _ {e ^ {\gamma R (a, \theta)} > e ^ {C _ {4} (\gamma) n \epsilon_ {n} ^ {2}}} \pi (\theta) d \theta\right) ^ {1 / 2} \\ \leq \left(\int e ^ {2 \gamma \left(h a + \frac {b}{\theta}\right)} \pi (\theta) d \theta\right) ^ {1 / 2} \left(\int \mathbb {1} _ {\left\{e ^ {\gamma \left(h a + \frac {b}{\theta}\right)} > e ^ {C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} \right\}} \pi (\theta) d \theta\right) ^ {1 / 2} \\ \leq e ^ {- C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} \left(\int e ^ {2 \gamma (h a + \frac {b}{\theta})} \pi (\theta) d \theta\right), \tag {64} \\ \end{array}
$$
where the last inequality follows from using the Chebyshev's inequality.
Now using the definition of the prior distribution, which is an inverse gamma prior restricted to  $\Theta = [T,\infty)$ , we have
$$
\begin{array}{l} \int_ {\left\{e ^ {\gamma R (a, \theta)} > e ^ {C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} \right\}} e ^ {\gamma R (a, \theta)} \pi (\theta) d \theta \leq e ^ {- C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} \left(\int e ^ {2 \gamma \left(h a + \frac {b}{\theta}\right)} \pi (\theta) d \theta\right) \\ \leq e ^ {- C _ {4} (\gamma) n \epsilon_ {n} ^ {2}} e ^ {2 \gamma \left(h \overline {{a}} + \frac {b}{T}\right)}, \\ \end{array}
$$
where  $\overline{a} := \max \{a \in \mathcal{A}\}$  and  $\overline{a} > 0$ . Since  $n\epsilon_n^2 \geq 1$ , we must fix  $C_4(\gamma)$  such that  $e^{C_4(\gamma)} > e^{2\gamma (h\overline{a} + \frac{b}{T})}$ , that is  $C_4(\gamma) > 2\gamma (h\overline{a} + \frac{b}{T})$  and  $C_5(\gamma) = C_4(\gamma) - 2\gamma (h\overline{a} + \frac{b}{T})$ .
Proof. Proof of Lemma B.5: Since family  $\mathcal{Q}$  contains all shifted-gamma distributions, observe that  $\{q_n(\cdot)\in \mathcal{Q}\} \forall n\geq 1$ . By definition,  $q_{n}(\theta) = \frac{n^{n}}{\theta_{0}^{n}\Gamma(n)} (\theta -T)^{n - 1}e^{-n\frac{(\theta - T)}{\theta_{0}}}$ . Now consider the first term; using the definition of the KL divergence it follows that
$$
\operatorname {K L} \left(q _ {n} (\theta) \| \pi (\theta)\right) = \int_ {T} ^ {\infty} q _ {n} (\theta) \log \left(q _ {n} (\theta)\right) d \theta - \int_ {T} ^ {\infty} q _ {n} (\theta) \log (\pi (\theta)) d \theta . \tag {65}
$$
Substituting  $q_{n}(\theta)$  in the first term of the equation above and expanding the logarithm term, we obtain
$$
\begin{array}{l} \int_ {T} ^ {\infty} q _ {n} (\theta) \log (q _ {n} (\theta)) d \theta \\ = (n - 1) \int_ {T} ^ {\infty} \log (\theta - T) \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta - n + \log \left(\frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)}\right) \\ = - \log \theta_ {0} + (n - 1) \int_ {T} ^ {\infty} \log \frac {\theta - T}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta - n + \log \left(\frac {n ^ {n}}{\Gamma (n)}\right) \tag {66} \\ \end{array}
$$
Now consider the second term in the equation above. Substitute  $\theta = \frac{t\theta_0}{n} + T$  into the integral, we have
$$
\begin{array}{l} \int_ {T} ^ {\infty} \log \frac {\theta - T}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta = \int_ {0} ^ {\infty} \log \frac {t}{n} \frac {1}{\Gamma (n)} t ^ {n - 1} e ^ {- t} d t \\ \leq \int \left(\frac {t}{n} - 1\right) \frac {1}{\Gamma (n)} t ^ {n - 1} e ^ {- t} d t = 0. \tag {67} \\ \end{array}
$$
Substituting the above result into (66), we get
$$
\begin{array}{l} \int_ {T} ^ {\infty} q _ {n} (\theta) \log (q _ {n} (\theta)) d \theta \leq - \log \theta_ {0} - n + \log \left(\frac {n ^ {n}}{\Gamma (n)}\right) \\ \leq - \log \theta_ {0} - n + \log \left(\frac {n ^ {n}}{\sqrt {2 \pi n} n ^ {n - 1} e ^ {- n}}\right) \\ = - \log \sqrt {2 \pi} \theta_ {0} + \frac {1}{2} \log n, \tag {68} \\ \end{array}
$$
where the second inequality uses the fact that  $\sqrt{2\pi nn^n} e^{-n} \leq n\Gamma(n)$ . Recall  $\pi(\theta) = \frac{\beta^\alpha}{\Gamma(\alpha)}\theta^{-\alpha - 1}e^{-\frac{\beta}{\theta}}$ . Now consider the second term in (65). Using the definition of inverse-gamma prior and expanding the logarithm function, we have
$$
\begin{array}{l} - \int_ {T} ^ {\infty} q _ {n} (\theta) \log (\pi (\theta)) d \theta \\ = - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + (\alpha + 1) \int_ {T} ^ {\infty} \log \theta \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta + \beta \frac {n}{(n - 1) \theta_ {0}} \\ = - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \int_ {T} ^ {\infty} \log \frac {\theta}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta \\ + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0} \\ \leq - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \int_ {T} ^ {\infty} \frac {\theta - T}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta \\ + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0} \\ = - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0}, \tag {69} \\ \end{array}
$$
where the first inequality is due to fact that  $\mathbb{E}_{q_n}[\beta /\theta ]\leq \mathbb{E}_{q_n}[\beta /(\theta -T)]$  for any  $\theta >T$  and the penultimate inequality follows from the observation in (67) and the fact that  $\log \frac{\theta}{\theta_0}\leq \frac{\theta}{\theta_0} -1\leq \frac{\theta}{\theta_0} -\frac{T}{\theta_0}$  for any  $\theta_0 > T$ . Substituting (69) and (68) into (65) and dividing either sides by  $n$ , we obtain
$$
\begin{array}{l} \frac {1}{n} \mathrm {K L} \left(q _ {n} (\theta) \| \pi (\theta)\right) \\ \leq \frac {1}{n} \left(- \log \sqrt {2 \pi} \theta_ {0} + \frac {1}{2} \log n - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0}\right) \\ = \frac {1}{2} \frac {\log n}{n} + \beta \frac {1}{(n - 1) \theta_ {0}} + \frac {1}{n} \left(- \log \sqrt {2 \pi} - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + (\alpha) \log \theta_ {0}\right). \tag {70} \\ \end{array}
$$
Now consider the second term in the assertion of the lemma. Since  $\xi_i, i \in \{1, 2 \dots n\}$  are independent and identically distributed, we obtain
$$
\frac {1}{n} \mathbb {E} _ {q _ {n} (\theta)} \left[ \mathrm {K L} \left(d P _ {0} ^ {n} \| p (\tilde {X} _ {n} | \theta)\right) \right] = \mathbb {E} _ {q _ {n} (\theta)} \left[ \mathrm {K L} \left(d P _ {0} \| p (\xi | \theta)\right) \right]
$$
Now using the expression for KL divergence between the two exponential distributions, we have
$$
\begin{array}{l} \frac {1}{n} \mathbb {E} _ {q _ {n} (\theta)} \left[ \mathrm {K L} \left(d P _ {0} ^ {n} \| p (\tilde {X} _ {n} | \theta)\right) \right] = \int_ {T} ^ {\infty} \left(\log \frac {\theta_ {0}}{\theta} + \frac {\theta}{\theta_ {0}} - 1\right) \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} (\theta - T) ^ {n - 1} e ^ {- n \frac {\theta - T}{\theta_ {0}}} d \theta \\ \leq \frac {n}{n - 1} + 1 - 2 = \frac {1}{n - 1}, \tag {71} \\ \end{array}
$$
where second inequality uses the fact that  $\log x\leq x - 1\leq x - \frac{T}{\theta_0}$  for  $\theta_0 > T$ . Combined together (71) and (70) for  $n\geq 2$  implies that
$$
\begin{array}{l} \frac {1}{n} \left[ \mathrm {K L} \left(q _ {n} (\theta) \| \pi (\theta)\right) + \mathbb {E} _ {q _ {n} (\theta)} \left[ \mathrm {K L} \left(d P _ {0} ^ {n}\right) \| p (\tilde {X} _ {n} | \theta)\right) \right] \Bigg ] \\ \leq \frac {1}{2} \frac {\log n}{n} + \frac {1}{n} \left(2 + \frac {2 \beta}{\theta_ {0}} - \log \sqrt {2 \pi} - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \alpha \log \theta_ {0}\right) \leq C _ {9} \frac {\log n}{n}. \tag {72} \\ \end{array}
$$
where  $C_9 \coloneqq \frac{1}{2} + \max \left(0, 2 + \frac{2\beta}{\theta_0} - \log \sqrt{2\pi} - \log \left(\frac{\beta^\alpha}{\Gamma(\alpha)}\right) + \alpha \log \theta_0\right)$  and the result follows.
Proof. Proof of Lemma B.5: Since family  $\mathcal{Q}$  contains all gamma distributions, observe that  $\{q_n(\cdot)\in$ $\mathcal{Q}\} \forall n\geq 1$  . By definition,  $q_{n}(\theta) = \frac{n^{n}}{\theta_{0}^{n}\Gamma(n)}\theta^{n - 1}e^{-n\frac{\theta}{\theta_{0}}}$  . Now consider the first term; using the definition of the KL divergence it follows that
$$
\operatorname {K L} \left(q _ {n} (\theta) \| \pi (\theta)\right) = \int q _ {n} (\theta) \log \left(q _ {n} (\theta)\right) d \theta - \int q _ {n} (\theta) \log (\pi (\theta)) d \theta . \tag {73}
$$
$$
\begin{array}{l} \int q _ {n} (\theta) \log (q _ {n} (\theta)) d \theta = (n - 1) \int \log \theta \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta - n + \log \left(\frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)}\right) \\ = - \log \theta_ {0} + (n - 1) \int \log \frac {\theta}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta - n + \log \left(\frac {n ^ {n}}{\Gamma (n)}\right) \tag {74} \\ \end{array}
$$
Now consider the second term in the equation above. Substitute  $\theta = \frac{t\theta_0}{n}$  into the integral, we have
$$
\begin{array}{l} \int \log \frac {\theta}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta = \int \log \frac {t}{n} \frac {1}{\Gamma (n)} t ^ {n - 1} e ^ {- t} d t \\ \leq \int \left(\frac {t}{n} - 1\right) \frac {1}{\Gamma (n)} t ^ {n - 1} e ^ {- t} d t = 0. \tag {75} \\ \end{array}
$$
Substituting the above result into (74), we get
$$
\begin{array}{l} \int q _ {n} (\theta) \log (q _ {n} (\theta)) d \theta \leq - \log \theta_ {0} - n + \log \left(\frac {n ^ {n}}{\Gamma (n)}\right) \\ \leq - \log \theta_ {0} - n + \log \left(\frac {n ^ {n}}{\sqrt {2 \pi n} n ^ {n - 1} e ^ {- n}}\right) \\ = - \log \sqrt {2 \pi} \theta_ {0} + \frac {1}{2} \log n, \tag {76} \\ \end{array}
$$
where the second inequality uses the fact that  $\sqrt{2\pi nn} n^{n}e^{-n}\leq n\Gamma (n)$ . Recall  $\pi (\theta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)}\theta^{-\alpha -1}e^{-\frac{\beta}{\theta}}$ . Now consider the second term in (73). Using the definition of inverse-gamma prior and expanding the logarithm function, we have
$$
\begin{array}{l} - \int q _ {n} (\theta) \log (\pi (\theta)) d \theta \\ = - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + (\alpha + 1) \int \log \theta \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta + \beta \frac {n}{(n - 1) \theta_ {0}} \\ = - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + (\alpha + 1) \int \log \frac {\theta}{\theta_ {0}} \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta \\ + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0} \\ \leq - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0}, \tag {77} \\ \end{array}
$$
where the last inequality follows from the observation in (75). Substituting (77) and (76) into (73) and dividing either sides by  $n$ , we obtain
$$
\begin{array}{l} \frac {1}{n} \mathrm {K L} \left(q _ {n} (\theta) \| \pi (\theta)\right) \\ \leq \frac {1}{n} \left(- \log \sqrt {2 \pi} \theta_ {0} + \frac {1}{2} \log n - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \beta \frac {n}{(n - 1) \theta_ {0}} + (\alpha + 1) \log \theta_ {0}\right) \\ = \frac {1}{2} \frac {\log n}{n} + \beta \frac {1}{(n - 1) \theta_ {0}} + \frac {1}{n} \left(- \log \sqrt {2 \pi} - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + (\alpha) \log \theta_ {0}\right). \tag {78} \\ \end{array}
$$
Now, consider the second term in the assertion of the lemma. Since,  $\xi_{i}, i \in \{1, 2 \dots n\}$  are independent and identically distributed, we obtain
$$
\frac {1}{n} \mathbb {E} _ {q (\theta)} \left[ \mathrm {K L} \left(d P _ {0} ^ {n} \| p (\tilde {X} _ {n} | \theta)\right) \right] = \mathbb {E} _ {q _ {n} (\theta)} \left[ \mathrm {K L} \left(d P _ {0} \| p (\xi | \theta)\right) \right]
$$
$$
\begin{array}{l} \frac {1}{n} \mathbb {E} _ {q (\theta)} \left[ \mathrm {K L} \left(d P _ {0} ^ {n} \| p (\tilde {X} _ {n} | \theta)\right) \right] = \int \left(\log \frac {\theta_ {0}}{\theta} + \frac {\theta}{\theta_ {0}} - 1\right) \frac {n ^ {n}}{\theta_ {0} ^ {n} \Gamma (n)} \theta^ {n - 1} e ^ {- n \frac {\theta}{\theta_ {0}}} d \theta \\ \leq \frac {n}{n - 1} + 1 - 2 = \frac {1}{n - 1}, \tag {79} \\ \end{array}
$$
where second inequality uses the fact that  $\log x\leq x - 1$ . Combined together (79) and (78) for  $n\geq 2$  implies that
$$
\begin{array}{l} \frac {1}{n} \left[ \operatorname {K L} (q (\theta) \| \pi (\theta)) + \mathbb {E} _ {q (\theta)} \left[ \operatorname {K L} \left(d P _ {0} ^ {n}\right) \| p (\tilde {X} _ {n} | \theta)\right) \right] \\ \leq \frac {1}{2} \frac {\log n}{n} + \frac {1}{n} \left(2 + \frac {2 \beta}{\theta_ {0}} - \log \sqrt {2 \pi} - \log \left(\frac {\beta^ {\alpha}}{\Gamma (\alpha)}\right) + \alpha \log \theta_ {0}\right) \leq C _ {9} \frac {\log n}{n}. \tag {80} \\ \end{array}
$$
C.7 Multi-product Newsvendor problem
In the multi-dimensional newsvendor problem, we fix  $n^{-1/2} \sqrt{L_n^{MNV}(\theta, \theta_0)} = (\sup_{a \in \mathcal{A}} |R(a, \theta) - R(a, \theta_0)|)$ , where  $R(a, \theta) = \sum_{i=1}^{d} \left[ (h_i + b_i)a_i\Phi(a_i) - b_i a_i + \theta_i(b_i - h_i) \right]$ .
$$
\left. + \sigma_ {i i} \left[ h \frac {\phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)}{\Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)} + b \frac {\phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)}{1 - \Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)} \right] \right].
$$
For brevity, we denote  $d_L^{MNV}(\theta ,\theta_0) = n^{-1 / 2}\sqrt{L_n^{MNV}(\theta,\theta_0)}$  . First, we show that
Lemma C.13. For any compact decision space  $\mathcal{A}$  and compact model space  $\Theta$ ,
$$
d _ {L} ^ {M N V} (\theta , \theta_ {0}) \leq K \| \theta - \theta_ {0} \|,
$$
for a constant  $K$  depending on compact sets  $\mathcal{A}$  and  $\Theta$  and given  $b, h$  and  $\Sigma$ .
$$
\begin{array}{l} \partial_ {\theta_ {i}} R (a, \theta) \\ = \left(b _ {i} - h _ {i}\right) + \left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i} \phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right) \left[ \frac {h}{\Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)} + \frac {b}{1 - \Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)} \right] \\ + \sigma_ {i i} \phi \left(\frac {(a _ {i} - \theta_ {i})}{\sigma_ {i i}}\right) \left[ \frac {h \phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i})}{\sigma_ {i i} \Phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i}) ^ {2}} - \frac {b \phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i})}{\sigma_ {i i} (1 - \Phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i})) ^ {2}} \right] \\ = (b _ {i} - h _ {i}) + (a _ {i} - \theta_ {i}) / \sigma_ {i i} \phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i}) \left[ \frac {h}{\Phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i})} + \frac {b}{1 - \Phi ((a _ {i} - \theta_ {i}) / \sigma_ {i i})} \right] \\ + \phi \left(\frac {\left(a _ {i} - \theta_ {i}\right)}{\sigma_ {i i}}\right) \left[ \frac {h \phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)}{\Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right) ^ {2}} - \frac {b \phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)}{\left(1 - \Phi \left(\left(a _ {i} - \theta_ {i}\right) / \sigma_ {i i}\right)\right) ^ {2}} \right]. \tag {81} \\ \end{array}
$$
Since,  $\mathcal{A}$  and  $\Theta$  are compact sets, therefore  $\{(a_i - \theta_i) / \sigma_{ii}\}_{i = 1}^d$  lie in a compact set. Consequently,  $\phi ((a_i - \theta_i) / \sigma_{ii})$  and  $\Phi ((a_i - \theta_i) / \sigma_{ii})$  also lie in bounded subset of  $\mathbb{R}$  and thus  $\sup_{\mathcal{A},\Theta}\| \partial_{\theta_i}R(a,\theta)\| \leq K$  for a given  $b,h$  and  $\Sigma$ . Since, the norm of the derivative of  $R(a,\theta)$  is bounded on  $\Theta$  for any  $a\in \mathcal{A}$ , therefore,  $d_L^{MNV}(\theta ,\theta_0)$  is uniformly Lipschitz in  $\mathcal{A}$  with Lipschitz constant  $K$ , that is
$$
d _ {L} ^ {M N V} (\theta , \theta_ {0}) \leq K \| \theta - \theta_ {0} \|.
$$
Next, we show that the  $P_{\theta}$  satisfies Assumption 2.1, for distance function  $L_{n}^{MNV}(\theta ,\theta_{0})$
Proof. Proof of Lemma B.6:
First consider the following test function, constructed using  $\tilde{X}_n = \{\xi_1,\xi_2,\dots ,\xi_n\}$
$$
\phi_ {n, \epsilon} := \mathbb {1} _ {\left\{\tilde {X} _ {n}: \left\| \hat {\theta} _ {n} - \theta_ {0} \right\| > \sqrt {C \epsilon^ {2}} \right\}},
$$