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Tight Non-asymptotic Inference via Sub-Gaussian Intrinsic Moment Norm
\maketitle
\begin{abstract}
In non-asymptotic learning, variance-type parameters of sub-Gaussian distributions are of paramount importance. However, directly estimating these parameters using the empirical moment generating function (MGF) is infeasible. To address this, we suggest using the sub-Gaussian intrinsic moment norm [Buldygin and Kozachenko (2000), Theorem 1.3] achieved by maximizing a sequence of normalized moments. Significantly, the suggested norm can not only reconstruct the exponential moment bounds of MGFs but also provide tighter sub-Gaussian concentration inequalities. In practice, we provide an intuitive method for assessing whether data with a finite sample size is sub-Gaussian, utilizing the sub-Gaussian plot. The intrinsic moment norm can be robustly estimated via a simple plug-in approach. Our theoretical findings are also applicable to reinforcement learning, including the multi-armed bandit scenario.
\end{abstract}
\begin{keywords}
non-asymptotic inference; uncertainty quantification; sub-Gaussian; concentration inequality; multi-armed bandit.
\end{keywords}
\begin{AMS}
68Q25, 68R10, 68U05
\end{AMS}
\section{Introduction}
Establishing rigorous error bounds and related finte--sample inferences for specific learning procedures is a popular topic in machine learning, statistics and econometrics communities, such as \cite{wainwright2019high,zhang2020concentration,yang2020non, romano2000finite,chassang2009non,arlot2010some,yang2020non,horowitz2020inference,armstrong2021finite,zheng2021finite,lucas2008rigorous,owhadi2013optimal,wang2020sub}. In particular, concentration-based inferences have been developed when data is unbounded \cite{romano2000finite,auer2002finite,maurer2009empirical,hao2019bootstrapping,wang2021two,shiu2021efficient} or Gaussian \cite{arlot2010some,duy2021exact,bettache2021fast,feng2021nonasymptotic}.
Notably, \cite{phan2021towards,austern2022efficient} derived a result that is more refined than Hoeffding's inequality for bounded data, while \cite{bettache2021fast} provided a concentration of the correlations for multivariate Gaussian data.
However, in reality, it is hard to learn the boundedness or distribution of data. If we mistakenly apply Hoeffding's inequality \cite{hoeffding1963probability} to unbounded data, we will obtain very in-consistent confidence intervals (CIs); see Appendix SM1.1. Compromisely, it is a common practice to assume that data follows a sub-Gaussian distribution \cite{kahane1960}. In this case, the Cramér–Chernoff method\footnote{For simplicity, we focus on zero-mean random variables (r.v.) throughout this paper when referencing all sub-Gaussian r.v.s.} \cite{boucheron2013concentration} gives
\[
\mathrm{P} (X \geq t) \leq \inf_{s > 0} [ \exp\{-st\} \mathrm{E} \{\exp (s X)\}]
\]
for all $t \ge 0$. Therefore, a tight control of the upper bound for the moment generating function (MGF) \(\mathrm{E} \{\exp (s X)\}\) is crucial in statistical inferences. This consideration leads to the concept of optimal variance proxy \cite{chow1966some,buldygin2000metric} of sub-Gaussian distribution; see Appendix for more introductions of sub-Gaussianity.
\begin{definition}\label{defsubG}
A r.v. $X$ is sub-Gaussian (sub-G) with a {variance proxy} $\sigma^2$ {\rm{[}}denoted as $X \sim \operatorname{subG}(\sigma^{2})${\rm{]}} if its MGF satisfies $\mathrm{E} \{\exp(tX)\} \leq \exp(\sigma^2 t^2 / 2)$ for all $t \in \mathbb{R}.$ The {optimal variance proxy} is defined as:
\begin{align}\label{eq:opvar}
\sigma^2_{opt}(X):= \inf \big\{ \sigma^2 > 0 : \mathrm{E} \exp(tX) \leq \exp({\sigma^2 t^2}/{2}), t \in \mathbb{R} \big\}=2 \sup\limits_{t \in \mathbb{R}} \frac{ \log \{\mathrm{E} \exp(t X)\}}{{t^{2}}}.
\end{align}
\end{definition}
The optimal proxy variance has a conservative lower bound:
$\sigma^2_{opt}(X) \ge \operatorname{var}(X)$, as detailed in the Appendix SM1.2. When \( \sigma_{opt}^2(X) = \operatorname{var}(X) \), it denotes strict sub-Gaussianity \cite{arbel2020strict}. Drawing from Theorems 1.5 in \cite{buldygin2000metric}, one has
$\mathrm{P} \left( X \ge t \right) \leq e^{{t^2}/\{{2 \sigma^2_{opt}(X)}\}}$
and
\begin{align}\label{eq:Sub-Gaussian}
\mathrm{P} \left(|\begin{matrix}\sum_{i=1}^{n}\end{matrix}X_{i}| \ge t \right)\le 2 \exp \left[-t^{2}/\left\{2\begin{matrix}\sum_{i=1}^{n}\end{matrix}\sigma_{opt}^2(X_i)\right\}\right].
\end{align}
for independent sub-G r.v.s \( \{ X_{i}\}_{i = 1}^n \). By the Cramér–Chernoff method, (\ref{eq:Sub-Gaussian}) offers the tightest upper bound in the form of \( \mathrm{P} (X > t) \leq e^{-C t^2}\) [or \( \mathrm{P} (|\sum_{i = 1}^n X_i | > t) \leq e^{-C t^2} \)], where $c$ a constant.
Given the unbounded data \( \{X_i\}_{i = 1}^n \stackrel{\rm{i.i.d.}} \sim \operatorname{subG}(\sigma^2_{opt}(X)) \), a direct application of (\ref{eq:Sub-Gaussian}) yields a non-asymptotic \( 100(1-\alpha)\% \) CI:
\begin{equation}\label{eq:ci}
\mathrm{E} X = 0 \in \left[{\overline X_n} \pm \sigma_{opt}(X)\{{2}{n^{-1}}\log ({2}/{\alpha})\}^{1/2}\right].
\end{equation}
A plug-in estimate\footnote{We note an inconsistent estimator, \(2 \inf_{t \in \mathbb{R}} \log [n^{-1}\sum_{i = 1}^n e^{t X_i}]/t^2\), from the statistical physics literature \cite{wang2020sub}. It's important to stress that \(\widehat\sigma_{opt}^2(X)\) should take the supremum over \( {t \in \mathbb{R}} \), not the infimum.} for \( \sigma_{opt}^2(X) \) is given by
$$\widehat\sigma_{opt}^2(X) := 2 \sup\limits_{t \in \mathbb{R}}{t^{-2}}\log M_n(t),$$
where $M_n(t):=n^{-1}\sum\nolimits_{i = 1}^n \exp(t X_i)$ is the empirical MGF. However, $\widehat\sigma_{opt}^2(X)$ is not practically useful for at least two reasons:
(i) optimization outcome is unstable due to potential non-convexity of the objective function; (ii) variance of logarithm empirical MGF explodes as $t$ diverges, i.e.,
\[
\mathrm{var}\left\{\frac{\log (M_n(t))}{t^2}\right\}\approx \frac{\exp(2t\mu){\mathrm{var}\{X_1\}}}{nt^2M^2(t)}\to \infty\;\text{as}~t \to \infty,~\mu \ne 0;
\]
see Appendix SM1.3 for the proof. These concerns are empirically supported by the subsequence simulations in Sections \ref{sec:rb} \& \ref{sec_MAB}, e.g.,
Figure \ref{4_est}.
On the other hand, several other forms of variance-type parameters exist in the literature. For example, \cite{vaart1996weak} introduced the Orlicz norm, defined as \(\|X\|_{w_2}:=\inf\{c>0: \mathrm{E} \{\exp (|X|^2/c^2) \}\leq 2\}\); \cite{vershynin2010introduction} proposed a norm based on the scaling of moments, expressed as \(\|X\|_{\psi_2} := \max_{k \geq 2} k^{- 1/ 2} \mathrm{E} \{( |X|^k )^{1 / k}\}\). However, as illustrated in Table \ref{tab2} and further discussed in Appendix SM1.4, these two norms often fall short in providing {nice} probability bounds even for strict sub-Gaussian distributions, such as the standard Gaussian and symmetric Beta distributions.
\begin{table*}[!ht]\scriptsize
\centering
\caption{Comparison of sub-Gaussian norms $\|\cdot\|_{*}$ for centralized and symmetric $X$.}
\label{tab2}
\begin{tabular}{@{}ccccc@{}}
\toprule
$\|\cdot\|_{*}$-norm & sharp tail bound & sharp MGF bound & {half length } & easy to \\
& of $\mathrm{P} (|X| \ge t)$ & of $\mathrm{E} e^{tX}$& {of $(1 - \delta)$-CI} & learn \\\toprule
${\sigma_{opt}(X)}$ & Yes $2{e^{ \frac{-t^2}{2}/\sigma^2_{opt}(X)}}$ & Yes ${e^{\sigma^2_{opt}(X)\frac{t^2}{2}}}$ & $ \sqrt{2 \log (2 / \delta)}\sigma_{opt}(X)$ & No \\ \hline
{$ \|X\|_{w_2}$ } & Yes $2e^{\frac{-t^2}{2}/(\frac{\|X\|_{w_2}}{\sqrt 2})^2}$& No $e^{({\sqrt{5/2}}\|X\|_{w_2})^2\frac{t^2}{2}}$ & $ \sqrt{{2 \log (2 / \delta) }\cdot 5/2}\| X\|_{w_2}$ & No \\ \hline
{$\|X\|_{\psi_2} $ } & No $2{e^{ \frac{-t^2}{2}/(2e\|X\|_{\psi_2}^2)}}$& No $e^{{{[\sqrt{10e}\|X\|_{\psi_2}]^2}\frac{t^2}{2}}}$ & $\sqrt{2 \log (2 / \delta)\cdot 10e}\| X\|_{\psi_2}$ & Yes \\ \hline
{${\| X \|_{G}} $}~(Def. \ref{def:TypeIX}) & Yes $2{e^{ \frac{-t^2}{2}/{\| X \|_{G}^2}}}$&Yes ${e^{{\| X \|_{G}^2}\frac{t^2}{2}}}$& $\sqrt{2 \log (2 / \delta) }\| X \|_G $ & Yes \\\bottomrule
\end{tabular}
\end{table*}
Recently, \cite{horowitz2020inference} derived confidence intervals for structural parameters in econometrics using a class of optimization problems based on Berry-Esseen (B-E) bounds. When compared to the traditional normal approximation underpinned by B-E bounds, our results offer notable advantages, especially for datasets with extremely small sample sizes. For instance, in certain experimental science measurements, the sample size ranges merely from \( n=4 \) to \( n=8 \) \cite{rousseeuw2002robust}. We showcase this by examining Bernoulli observations, comparing two types of confidence intervals: one based on the B-E-corrected Central Limit Theorem (CLT) and the other on Hoeffding's inequality, as illustrated in Figure \ref{toy}.
\begin{figure*}[h]
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-210818-fig1.pdf}
\endminipage\hfill
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-210818-fig1.pdf}
\endminipage\hfill
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-210818-fig1.pdf}
\endminipage
\par
\caption{CIs via Hoeffding's inequality (red line) and B-E-corrected CLT (blue line). This illustration points out the deficiencies of the B-E-corrected CLT {under small sample cases}. Interestingly, it suggests that a straightforward application of Hoeffding's inequality might even offer better performance. }\label{toy}
\end{figure*}
\subsection{Contributions}
In light of previous discussions, we advocate the use of the intrinsic moment norm (see Definition \ref{def:TypeIX}) in the construction of tight non-asymptotic CIs for two reasons: (i) it approximately recovers tight inequalities \eqref{eq:Sub-Gaussian}; (ii) it can be easily estimated in a robust manner.
\begin{definition}[Page 6 in \cite{buldygin2000metric}]\label{def:TypeIX}
Let \( (2k-1)!!:=(2k-1) \cdot (2k-3) \cdot \ldots \cdot 3\cdot1 \). The intrinsic moment norm is given by
\begin{center}
${\left\| X \right\|_{G}} :=\max\limits_{k \ge 1} {\left[ {\frac{{{2^k}k!}}{{(2k)!}}\mathrm{E} (X^{2k})}\right]^{1/(2k)}} = \max\limits_{k \ge 1} {\left[ {\frac{{1}}{{(2k - 1)!!}}\mathrm{E} (X^{2k})}\right]^{1/(2k)}}.$
\end{center}
\end{definition}
Note that \(\| X\|_G < \infty\) implies \(\sigma_{opt}(X) < \infty\) based on Theorem 2.6 in \cite{wainwright2019high}. Thus, a random variable with a finite intrinsic moment norm is ineeded sub-Gaussian.
In this paper, our contributions are summarized as
\begin{itemize}
\item[1.] Using \({\left\| X \right\|_{G}}\), we derive a more refined Hoeffding-type inequality, especially for asymmetric distributions, as detailed in Theorem \ref{eq:sgnew}(b);
\item[2.] We estimate \({\left\| X \right\|_{G}}\) and study the finite sample properties of the estimates;
\item[3.] We introduce a \textit{sub-Gaussian plot} to assess whether data is sub-Gaussian or not.
\item[4.] We introduce an \textit{optimal lower variance proxy} (as defined in Definition \ref{def:TypeIXlow}) that yields a sharp reverse Chernoff inequality (refer to Lemma \ref{eq:low}) alongside sharp lower tails for sub-Gaussian maxima (outlined in Theorem \ref{eq:uplow});
\item[5.] We apply the intrinsic moment norm estimation to the Bootstrapped UCB-algorithm in multi-armed bandits.
\end{itemize}
\subsection{Outline}
This article is organized as follows: In Section 2, we present upper and lower bounds for the tail probabilities of sub-Gaussian variables and introduce the sub-Gaussian plot to determine whether data follows a sub-Gaussian distribution. In Section 3, we delve into the finite sample properties of the intrinsic moment norm, exploring its inherent characteristics and the concentration for summations' sharper bounds. In Section 4, we propose estimation techniques for this norm, including the plug-in estimator, the robust mean-of-median estimation, and methods for extremely small sample sizes, along with a discussion of their theoretical properties. In Section 5, we apply the insights derived from this norm to the non-asymptotic analysis of problems found in the multi-armed bandit. Detailed proofs and supplementary information can be found in the Appendix.
\section{Sub-Gaussian plot}
Before estimating $\|X\|_G$, the first step is to verify the data $X$ is indeed sub-G given its i.i.d. copies $\{X_i\}_{i=1}^n$. For bounded $X$, it is naturally sub-Gaussian by using Hoeffding's inequality. For unbounded $X$, Corollary 7.2 (b) in \cite{zhang2020concentration} shows for r.v.s $X_i \sim \operatorname{subG}(\sigma_{opt}^2(X))$ (without independence assumption)
\begin{equation}\label{eq:maxU}
\mathrm{P} \left\{ \max\limits_{1 \leq i \leq j} X_i \leq \sigma_{opt}(X) \sqrt{2 (\log j + t)}\right\} \geq 1-\exp(-t),
\end{equation}
which implies $\max_{1 \leq i \leq j} X_i =O_{\mathrm{P}}(\sqrt{\log j})$. Moreover, we will show the rate $\sqrt{\log j}$ is indeed sharp for a class of unbound sub-G r.v.s characterized by the \textit{lower intrinsic moment norm} below.
\begin{definition}[Lower intrinsic moment norm]\label{def:TypeIXlow}
Define ${\left\| \cdot \right\|_{\tilde G}}$-norm for a sub-G $X$ as
\begin{center}
${\left\| X \right\|_{\tilde G}} := \min\limits_{k \ge 1} [ {{{\{(2k - 1)!!\}^{-1}}}\mathrm{E} (X^{2k})}]^{1/(2k)}.$
\end{center}
\end{definition}
By the method in Theorem 1 of \cite{zhang2020non}, we obtain the following tight rate result with a lower bound.
\begin{theorem}\label{eq:uplow}
{(a)}. If ${\left\| X \right\|_{\tilde G}}>0$ for i.i.d. symmetric sub-G r.v.s $\{ X_i\}_{i = 1}^n \sim X$, then with probability at least $1-\delta$
\begin{align*}
L(X)\left\{\log [nC^{2}(X)]-\log \log \left(\frac{2}{ \delta}\right)\right\}^{1/2}\leq \max _{1 \leq i \leq n}\frac{X_{i}}{\left\| X \right\|_{G}}\leq \left\{2\log n + 2\log \left(\frac{2}{ \delta}\right)\right\}^{1/2},
\end{align*}
where $L(X):={\frac{{\left\| X \right\|_{\tilde G}}/{(2\left\| X \right\|_{G})}}{[{2{\left\| X \right\|_{G}^2}}/{\left\| X \right\|_{\tilde G}^2}-1]^{1/2}}}$ and
\[C(X):=\frac{{\left\| X \right\|_{\tilde G}^2}}{{4{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}}}\left( \frac{{4{\left\| X \right\|_{G}^2}-2{\left\| X \right\|_{\tilde G}^2}}}{4{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}}\right)^{({4\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}})-2}\in (0,1);\]
{(b)} if $X$ is bounded, then ${\left\| X \right\|_{\tilde G}}=0$.
\end{theorem}
For instance, a mixture of two independent Gaussian variables satisfies the condition ${\left\| X \right\|_{\tilde G}}>0$ in Theorem \ref{eq:uplow}(a). Refer to Example SM1.5 in Appendix for verification. For a Gaussian variable $X$, ${\left\| X \right\|_{G}}={\left\| X \right\|_{\tilde G}}$ and the constant $L(X)$ takes on the value of $1 / 2$ in the aforementioned lower bound. The upper bound follows from the proof of \eqref{eq:maxU} similarly. The proof of lower bound relies on the following sharp reverse Chernoff inequality, which is derived from Paley--Zygmund inequality (see \cite{paley1932note}).
\begin{lemma}[A reverse Chernoff inequality]\label{eq:low}
Suppose ${\| X \|_{\tilde G}}>0$ for a symmetric sub-G variable $X$. {Then}
\begin{center}
$\mathrm{P}(X \ge t) \geq C^2(X) \exp \{-4{[2{\| X \|_{G}^2}/{\| X \|_{\tilde G}^4}-\| X \|_{\tilde G}^{-2}]t^2}\},~\forall~t \geq 0$.
\end{center}
\end{lemma}
Theorem 1 of \cite{zhang2020non} does not optimize the constant in Paley-Zygmund inequality. In contrast, our Lemma \ref{eq:low} has an optimal constant; see Section \ref{se:key}.
Motivated by Theorem \ref{eq:uplow}, we propose a novel \textit{sub-Gaussian plot} to check whether i.i.d data $\{ X_i\}_{i = 1}^n$ follow a sub-G distribution.
\textbf{Sub-Gaussian plot under unbounded assumption}. Suppose that for each $j$, $\{ X_i^*\}_{i = 1}^j$ are independently sampled from $\mathbb{F}_n(x) = \frac{1}{n} \sum_{i = 1}^n \mathds{1}(X_i \leq x)$ as the empirical distribution of $\{ X_i\}_{i = 1}^n$. Specifically, we plot the order statistics $\{ \max_{1 \leq i \leq j} X_i^*\}_{j = 1}^n$ on the plane coordinate axis, where $x$ axis represents $\sqrt{\log j + 1}$ and $y$ axis the value of $ \max_{1 \leq i \leq j} X_i^*$. We check whether those points have a linear tendency at the boundary: the closer they are to the tendency of a straight line, the more we can trust the data are sub-Gaussian.
\begin{figure}[h]
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/cg-220624-fig1-new-new-new.pdf}
\endminipage\hfill
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/cg-220624-fig2-new-new-new.pdf}
\endminipage
\caption{
Sub-Gaussian plot for standard Gaussian and standard exponential random variables with \( n = 1000 \).
(Left: The two dotted lines denote a triangular region where the points are likely to fall with high probability.
{The data points corresponding to the exponential random variable predominantly occupy a curved triangular region, thereby exhibiting quadratic behavior.})
}
\label{sub-G-plot}
\end{figure}
The Figure \ref{sub-G-plot} illustrates
\textit{sub-Gaussian plot} of $N(0, 1)$ and $\operatorname{Exp}(1)$. It can be seen that sub-Gaussian plot of $N(0, 1)$ shows a linear behavior near the boundary, while $\operatorname{Exp}(1)$ shows quadratic tendency at the boundary. For the quadratic tendency, we note that if $\{ X_i\}_{i = 1}^n$ have heavier tails such as sub-exponentiality, then $\max_{1 \leq i \leq j} X_i =O_{\mathrm{P}}( {\log j})$ instead of the order $O_{\mathrm{P}}(\sqrt{\log j})$; see Corollary 7.3 in \cite{zhang2020concentration}.
Sub-Gaussian plots provide an insightful way to visualize sub-Gaussian distributions, serving a similar role to how Q-Q plots represent Gaussian distributions. While both types of plots exhibit linear trends, they achieve this in different ways: Q-Q plots rely on the actual distribution of the data, whereas sub-G plots make use of the MGF bound. To the best of the author's knowledge, this paper represents the first formal attempt to introduce a methodology for determining whether a dataset follows a sub-Gaussian distribution. It is worth emphasizing that the utility of sub-G plots is most pronounced for datasets with a large enough sample size. For small values of $n$, the data is approximately treated as bounded random variables, and there is no need to use a sub-G plot.
Another consideration is the data that exhibits heavy-tailed outliers, the next section will discuss employing mean-of-median estimation. This approach allows for the exclusion of outliers, ensuring that the inliers conform to a sub-Gaussian distribution. This will particularly suitable if the inliers are approximately bounded, a scenario often encountered in real-world datasets.
\section{Finite Sample Properties of Intrinsic Moment Norm}
In this section, we characterize two important properties of the intrinsic moment norm that are used in constructing non-asymptotic confidence intervals.
\subsection{Basic Properties}
Lemma \ref{thm_arg_k} below establishes that the intrinsic moment norm is estimable.
\begin{lemma}\label{thm_arg_k}
Let $2 \mathbb{N}:=\{2 ,4, \cdots\}$ is the even number set. For sub-G $X$, we have
\[\arg \max_{m \in 2 \mathbb{N}} \left[ \frac{\mathrm{E} (X^m)}{(m - 1)!!} \right]^{1 / m} < \infty.\]
\end{lemma}
Lemma \ref{thm_arg_k} ensure that for any sub-G variable $X$, its intrinsic moment norm can be computed as
\[\|X \|_G := \max_{m \in 2 \mathbb{N}} \left[ \frac{\mathrm{E} (X^{m})}{{(m - 1)!!}} \right]^{1 / m} = \max_{1 \leq k \leq k_{X}} \left[ \frac{\mathrm{E} (X^{2k})}{{(2k - 1)!!}} \right]^{1 / (2k)}\]
with some finite $k_X < \infty$.
This is an important property that other norms may not have. The $\|X\|_{\psi_2} := \max_{k \geq 2} k^{- 1/ 2} ( \mathrm{E} |X|^k )^{1 / k}$ for Gaussian $X$ achieves its optimal point at $k = \infty$; see Example \ref{ex:comparingver} in Appendix \ref{se;his2}.
As for $\sigma_{opt}^2(X):=2 \sup_{t \in \mathbb{R}} \frac{ \log \{\mathrm{E} [\exp(t X)]\}}{t^2}$, it is unclear that its value can be achieved at a finite $t$. Note that if $k_X = 1$, one has $\| X\|_G^2 =\operatorname{var}(X)$.
Next, we present an example in calculating the values of $k_X$. Denote $\left. \operatorname{Exp}(1) \right|_{[0, M]}$ as the truncated standard exponential r.v. on $[0, M]$ with the density as $f(x) = \frac{e^{-x}}{\int_0^M e^{-x} \, dx} 1_{\{x \in [ 0, M]\}}(x)$.
\begin{example}\label{eg:1}
a. $X \sim U[-a, a]$, $k_X = 1$ for any $a \in \mathbb{R}$; b. $X \sim \left. \operatorname{Exp}(1) \right|_{[0, 2.75]} - \mathrm{E} \left. \operatorname{Exp}(1) \right|_{[0, 2.75]}$, $k_X = 2$; c. $X \sim \left. \operatorname{Exp}(1) \right|_{[0, 3]} - \mathrm{E} \left. \operatorname{Exp}(1) \right|_{[0, 3]}$, $k_{X} = 3$. Indeed, for any fixed $k_0 \in \mathbb{N}$, we can construct a truncated exponential r.v. $X:=\left. \operatorname{Exp}(1) \right|_{[0, M]}$ such that $k_X = k_0$ by properly adjusting the truncation level $M$.
\end{example}
\subsection{Concentration for summations}
{In the following discussion, we demonstrate an additional property of $\|X \|_G$, illustrating that it nearly recovers the tight MGF bounds as defined in Definition \ref{defsubG}. More significantly, this property allows us to derive the sub-Gaussian version of Hoeffding's inequality, as shown in \eqref{eq:Sub-Gaussian}.}
\begin{theorem}\label{eq:sgnew}
Assume $\{X_{i}\}_{i=1}^n$ are independent r.v.s with $ {{\max }_{i \in [n]}\left\| X_i \right\|_{G}} <\infty$. We have
{\rm{(a)}}. If $X_i$ is {symmetric about zero}, then
\begin{center}
$\mathrm{E} \{\exp(t X_i)\} \le \exp ( {{t^2\left\| X_i \right\|_{G}^2/2}})$ for any $t \in \mathbb{R}$,
\end{center}
and
\begin{center}
$\mathrm{P}\left(|\sum_{i = 1}^n {{X_i}} | \ge s \right) \le 2\exp \left\{-{s^2}/[2\sum_{i = 1}^n {\| X_i \|_{G}^2}]\right\}$ ~~~for~~~ $s \ge 0$.
\end{center}
{\rm{(b)}}. If $X_i$ is {not symmetric}, then
\begin{center}
$\mathrm{E} \{\exp(t X_i)\} \leq \exp [(17/12){{t^2\left\| X_i \right\|_{G}^2/2}}]$ for any $t \in \mathbb{R}$,
\end{center}
and
\begin{center}
$\mathrm{P}\left(|\sum_{i = 1}^n {{X_i}} | \ge s \right) \le 2 \exp \left\{-\frac{12s^2}{17}/ [2\sum_{i = 1}^n {\| X_i \|_{G}^2} ]\right\}$ ~~~for~~~ $s \ge 0$.
\end{center}
\end{theorem}
Theorem \ref{eq:sgnew}(a) can be found in Theorem 2.6 of \cite{wainwright2019high}. As for Theorem \ref{eq:sgnew}(b), it's noteworthy that \((17/12)^{1/2}\) is approximately 1.19. In contrast, Lemma 1.5 in \cite{buldygin2000metric} provides the bound $\mathrm{E} \{\exp\{tX_i\}\}\le \exp\big\{ {{\frac{t^2}{2}({3.1}^{1/4}{{\| X_i\|}_{G}})^2}}\big\}$ for \(t \in \mathbb{R}\), where \({3.1}^{1/4}\) is approximately 1.32.
{Essentially, $(17/12)^{1/2}>1$ appears for asymmetric variables, since ${\| \cdot\|}_{G}$ is defined by comparing a Gaussian variable $G$ that is symmetric}. A technical improvement in our approach is due to the fact that \(\| \cdot\|_G\) bypasses the need for Stirling's approximation, thereby providing a more precise MGF bound when expanding the exponential function using Taylor's expansion.
To further demonstrate the precision of Theorem \ref{eq:sgnew}(b), Figure \ref{HvsBE} offers a comparative analysis with various parameters such as \(\sigma_{o p t}(X)\), \((17/12)^{1/2}\| X \|_{G}\), \((10e)^{1/2}\|X\|_{\psi_2}\), \((5/2)^{1/2}\|X\|_{w_2}\), and \((\operatorname{var}X)^{1/2}\). The comparisons specifically explore the confidence lengths {under asymmetric Bernoulli and Beta distributions,} as detailed in Table \ref{tab2}.
\begin{figure}[!htb]
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/cg-211014-fig1-new-new-new.pdf}
\caption*{Bernoulli}
\endminipage\hfill
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/cg-211014-fig2-new-new-new.pdf}
\caption*{$\operatorname{Beta}(\alpha, \beta)$}
\endminipage
\caption{
Half-length of the \(1 - \delta\) confidence intervals, normalized by \([2 \log (2 / \delta)]^{1/2}\) to eliminate the influence of \(\delta\), across different norms.}
\label{HvsBE}
\end{figure}
\section{Estimation of the intrinsic moment norm}
\subsection{Sub-Gaussian Norm Estimation: Plug-in Estimators}\label{se:naive}
Ideally, suppose we have i.i.d. mean-zero sample generating from $X$. Then a straightforward approach to estimate $\| X\|_G$ is by the plug-in approach. Although $k_X$ is proven to be finite in Lemma \ref{thm_arg_k}, its (possibly large) exact value is still unknown in practice. Hence, we suggest a feasible plug-in estimator with a non-decreasing \textit{index sequence} $\{\kappa_n\}$ to replace $k_X$ {as follows:}
\begin{equation}\label{DE}
\widehat{\| X \|}_G = \max_{1 \leq k \leq \kappa_n} {\left[ {\frac{{1}}{{(2k - 1)!!}} \frac{1}{n} \sum_{i = 1}^n {X_i^{2k}}}\right]^{1/(2k)}}.
\end{equation}
For example, $\kappa_n=\lceil \log n \rceil$ or a large constant.
Deriving the non-asymptotic property of the $\widehat{\|X\|}_G$ is not an easy task: the maximum point
\[\hat k(\kappa_n):=\arg \max_{1 \leq k \leq \kappa_{n}} \left[ {\frac{{1}}{{(2k - 1)!!}} \frac{1}{n} \sum\limits_{i = 1}^n {X_i^{2k}}}\right]^{1/(2k)}\]
will change with the sample size $n$ even $\kappa_n$ is fixed. To resolve this, we first examine the oracle estimator defined as
\begin{center}
$\widetilde{\| X\|}_G = \left[ {\frac{{1}}{{(2k_X - 1)!!}} \frac{1}{n} \sum_{i = 1}^n {X_i^{2k_X}}} \right]^{{1}/{(2 k_X})}.$
\end{center}
Here, based on Orlicz norm $\|Y\|_{\psi_{\theta}}:=\inf \{t>0: \mathrm{E} \exp (|Y|^{\theta} / t^{\theta}) \leq 2 \}$ of sub-Weibull r.v. $Y$ with $\theta>0$ \cite{hao2019bootstrapping}, we present the non-asymptotic concentration of $\widetilde{\| X\|}_G$ {around it ture value $\|X \|_G$.}
\begin{proposition}\label{pro1}
Suppose $\{X_i\}_{i = 1}^n \stackrel{\rm{i.i.d.}} \sim X$ with $\| X\|_{\psi_{1/k_X}}<\infty$, then for any $t > 0$, with probability at least $1- 2{\exp( - t)}$
\begin{equation}\label{eq:non-robust}
\Big| \widetilde{\|X \|}_G^{2 k_X} - \| X\|_G^{2 k_X} \Big| \leq 2 e \| X\|_{\psi_{1/k_X}} C(k_X^{-1})\left\{\left(\frac{t}{n}\right)^{1/2} +\gamma^{2 k_X} A(k_X^{-1}) \frac{t^{k_X}}{n} \right\},
\end{equation}
where $\gamma \approx 1.78$, and the constants $C(\cdot)$ and $A(\cdot)$ are defined in Appendix \ref{app_proof}.
\end{proposition}
\subsection{Sub-Gaussian Norm Estimation: Robust Estimators}\label{sec:rb}
In practice, datasets often contain a significant number of outliers; see \cite{hsu2016loss}. The exponential-moment condition $\| X\|_{\psi_{1/k_X}}<\infty$ is too strong for the error bound of $ \widetilde{\|X \|}_G^{2 k_X} - \| X\|_G^{2 k_X} $ in Proposition \ref{pro1}. {Additionally, the constants in \eqref{eq:non-robust} may potentially be large, leading to overly loose CIs when the sample size $n$ is small. }
Apart from the direct plug-in estimator, we utilize the median-of-means (MOM) as described on Page 244 of \cite{nemirovskij1983problem} as a robust alternative. Given a positive integer \( m \) and \( b \) such that \( n=mb \), let \( B_{1}, \ldots, B_{b} \) partition the set \( [n] \) into blocks of equal size \( m \). For every \( s \in[b] \), and define \( \mathbb{P}_m^{B_{s}} X \) as \( m^{-1} \sum_{i \in B_{s}} X_{i} \) for the independent dataset \( \{X_{i}\}_{i=1}^n \). We can then express the MOM-based intrinsic moment norm estimator as:
\begin{equation}\label{MOM}
{\widehat{\| X \|}_{b,G}}:= \max_{1 \le k \le \kappa_{n} }\mathop{\operatorname{med}}\limits_{s \in [b]}\left[\{(2 k-1) ! !\}^{-1}\mathbb{P}_m^{B_{s}} X^{2k}\right]^{1/(2k)}.
\end{equation}
The naive plug-in estimator in Section \ref{se:naive} can be just reviewed as a special case of MOM estimator with $b=1$. As stated in Proposition \ref{pro1}, the naive plug-in estimator $\widehat{\|X \|}_G = \widehat{\| X \|}_{1,G}$ is not robust. MOM estimators \eqref{MOM} with $b\gg 1$ have two merits: (a) it only needs finite moment conditions, but the exponential concentration bounds are still achieved; (b) it permits some outliers in the data.
For non-asymptotic inference, we need to precisely bound \( \| X\|_{G} \) using a feasible estimator \( {\widehat{\| X \|}_{b,G}} \) while ensuring sharp constants. We proceed to establish a high-probability upper bound for the estimated norm under the following outlier assumptions labeled as \( O \cup I \):
$\bullet$ (M.1) Suppose that the dataset $\{X_{i}\}{i=1}^n$ consists of $n-n_o$ inliers $\{X_{i}\}{i \in I}$, each i.i.d. following a target sub-Gaussian distribution (for the $n_o$ outliers ${X{i}}_{i \in O}$, no distributional assumptions are made).
$\bullet$ (M.2) $b=b_O+b_S$, where $b_O$ is the number of blocks \textit{containing at least one outliers} and $b_S$ is the number of \textit{sane blocks containing no outliers}. Let $\varepsilon:=n_o/n$ be the fraction of the outliers and ${n_o}/{b}<{1}/{2}$. Assume here exists a fraction function $\eta(\varepsilon)$ for sane block such that
$b_S \ge \eta(\varepsilon) b~\text{for a function}~ \eta(\varepsilon)\in (0,1].$
(M.1) is a reasonable assumption in our sub-Gaussian inference, since when the outliers are thrown away with a large proportion such that ${n_o}/{b}<{1}/{2}$ that are allowed in (M.2) the remaining inliers is approximately bounded data that are naturely sub-Gaussian. To serve for error bounds in the presence of outliers, (M.2) considers the specific fraction function of the polluted inputs; see \cite{laforgue2021generalization}.
Define ${{\underline{g}_{k,m}}({\sigma _k})}$ and ${\bar g_{k,m}({\sigma _k})}$ as the sequences for any $m \in \mathbb{N}$ {and} $1 \le k \le \kappa_{n}$:
$ {\overline g_{k,m}({\sigma _k})} : = 1 - \left[ \frac{{{\mathrm{E}}(X^{2k})}}{{(2k - 1)!!}} \right]^{-1/(2k)}\mathop {\max }\limits_{1 \le k \le \kappa_{n}} {\left[ {\frac{{{\mathrm{E}}(X^{2k})}}{{(2k - 1)!!}} -2\{\frac{m}{\eta(\varepsilon)}\}^{-1/2}\frac{\sigma_{k}^{k}}{{\mathrm{E}} (X^{2 k})} } \right]^{1/(2k)}}$
and ${ \underline{g}_{k,m}({\sigma _k})}:=[{2\{m/\eta(\varepsilon)\}^{-1/2}\sigma_{k}^{k}}/{{\mathrm{E}} (X^{2 k})}+1]^{1 /(2 k)}-1.$ A robust and non-asymptotic CI for $\left\| X \right\|_{G}$ is obtained as follow.
\begin{theorem}[Finite sample guaranteed coverage for MOM estimted norms]\label{thm:mom}
Suppose ${\mathrm{var}({X^{2k}}) }\le \sigma _k^{2k}$ for a sequence $\{\sigma _k\}_{k = 1}^{\kappa_{n}}$, for $\kappa_{n} \ge k_{X}$ under (M.1-M.2), we have
\begin{center}
$\mathrm{P} \left\{ {\| X \|_{G} \le {{[1 - {\mathop {\max }\limits_{1 \le k \le \kappa_{n}} \overline g_{k,m}}({\sigma _k})]^{-1}}}{\widehat{\left\| X \right\|}_{b,G}}} \right\} > 1 - \kappa_{n} \cdot {e^{ - 2b\eta(\varepsilon) (1-\frac{3}{4\eta(\varepsilon)})^2}}$;
\end{center}
and $\mathrm{P} \{ \| X \|_{G} \ge [{1 +\max _{1 \le k \le \kappa_{n}} {\underline{g}_{k,m}}({\sigma _k})}]^{-1}{\widehat{\left\| X \right\|}_{b,G}}\} > 1 - \kappa_{n} \cdot {e^{ - 2b\eta(\varepsilon) (1-\frac{3}{4\eta(\varepsilon)})^2}}$.
\end{theorem}
\begin{figure}[h]
\centering
\minipage{0.50\textwidth}
\includegraphics[width=\linewidth]{figures/4_est_cmp_Gau.pdf}
\endminipage\hfill
\minipage{0.50\textwidth}
\includegraphics[width=\linewidth]{figures/4_est_cmp_Gau_Cau.pdf}
\endminipage\\
\minipage{0.50\textwidth}
\includegraphics[width=\linewidth]{figures/4_est_cmp_Rad.pdf}
\endminipage\hfill
\minipage{0.50\textwidth}
\includegraphics[width=\linewidth]{figures/4_est_cmp_Rad_Cau.pdf}
\endminipage
\caption{OP refers to the plug-in estimator for the optimal variance proxy, which employs the computational trick detailed in \cite{lieber2022estimating}. DE represents the naive plug-in estimator as defined in equation \eqref{DE}, while MOM denotes the MOM estimator given in equation \eqref{MOM} with a fixed block size of $b = 5$. MOM\_LOCV is the MOM estimator where the block number $b$ is selected using the leave-one-out method, as outlined in equation \eqref{MOM_LOOCV}. The absolute error vector, averaged over $n \in [250]$, for the estimators (OP, DE, MOM, MOM\_LOCV) is $(1.54, 0.07, 0.06, 0.05)$, $(1.30, 0.25, 0.07, 0.07)$, $(1.49, 0.00, 0.00, 0.00)$, and $(1.78, 0.59, 0.00, 0.00)$, respectively. These vectors correspond to the four different figures, arranged from top to bottom.}
\label{4_est}
\end{figure}
Theorem \ref{thm:mom} ensures the concentration of the estimator $\widehat{\| X\|}_{b,G}$ when $\kappa_{n} \ge k_X$ under enough sample. {If $\eta(\varepsilon)=1$ with $\varepsilon=0$, then the data are i.i.d., which have no outlier, and outlier assumptions in M.1-M.2 can be dropped in Theorem \ref{thm:mom}}. {When the data is i.i.d. Gaussianc variable, Proposition 4.1 in \cite{auer2002finite} also gave a high-probability estimated upper bound for Gaussian standard deviations; our result is for the sub-G intrinsic moment norm and the Gaussian standard deviation is a special case of the sub-G norm.}
In practice, the block number $b$ can be taken by the adaptation method based on the Lepski method \cite{depersin2022robust}. However, they only discuss Lepski's method in a theoretical context, and no simulation studies were conducted due to the method involving a large search radius with a constant of $808\times 1200$ (see Page 514). In practical terms, implementing Lepski's method can be computationally intensive, given the use of a dyadic grid for $b$. As an alternative, we propose the leave-one-out cross-validation (LOCV) method to determine $b$:
\begin{equation}\label{MOM_LOOCV}
\begin{aligned}
b^{\text{LOCV}} = \mathop{\operatorname{argmin}}\limits_{{1\le b \le n}} \max _{1 \leq {k} \leq \kappa_{{n}}} \sum_{j=1}^{n} \left[\left\{\frac{X_{{j}}^{2 {k}}}{(2 {k}-1) ! !} \right\}^{1/(2k)} - \mathop { \operatorname{med}}\limits_{{s} \in[{b}]}\left\{\frac{\sum\nolimits_{{i} \in {B}_{{s}}, \, i \neq {j}} X_{{i}}}{{~m}(2 {k}-1) ! !} \right\}^{{1}/(2k)}\right]^2.
\end{aligned}
\end{equation}
The high probability guarantee in Theorem \ref{thm:mom} requires that the index sequence $\kappa_{n}$ should not be very large for fixed $b$. The larger $\kappa_{n}$ needs larger $b$ in blocks $B_{1}, \ldots, B_{b}$. In the simulation, we will see that an increasing index sequence $\kappa_{n}$ with a slow rate will lead to a good performance.
We compare the estimated optimal variance proxy from \cite{lieber2022estimating} with our estimated intrinsic moment norm. The latter is calculated using the direct method (equation \eqref{DE}), the MOM estimator (equation \eqref{MOM}) with a fixed block number $b \equiv 5$, and the MOM estimator with the block number determined by equation \eqref{MOM_LOOCV}.
We consider variables $X$ that follow standard Gaussian and Rademacher distributions. In both cases, we have $\| X \|_G^2 = \sigma_{opt}^2 (X) = \operatorname{var}(X) = 1$. Figure \ref{4_est} depicts the performance of the three estimators for sample sizes ranging from $n=10$ to $1000$, with $\kappa_n=\lceil \log n \rceil$.
From Figure \ref{4_est}, we observe that both variants of the MOM estimator (fixed block size and LOCV-determined block size) outperform others in terms of accuracy. In comparison, the plug-in estimator from \cite{lieber2022estimating} shows a less consistent performance.
\subsection{Small Sample Leave-one-out Average in Moment Norm Estimations}\label{AppC}
For high-quality datasets with a very limited sample size \( n \), specifically when \( n \) is extremely small, the leave-one-out Hodges-Lehmann method \cite{rousseeuw2002robust} can be employed to effectively augment the sample size. In scenarios where the sample size is small (\( n \leq 20 \)), apart from the direct empirical moment method (DE), there are two alternative techniques to consider.
The first method, a variant of the Bootstrap approach, employs the ($n-1$)-out-of-$n$ Bootstrap technique. In this procedure, \( n - 1 \) Bootstrap estimators are generated from the data by iteratively resampling \( n - 1 \) times while leaving one observation out during each iteration. Once these estimators are constructed, the median of these estimators is taken as the final estimate. This method increases the robustness of the estimator, particularly in the context of small sample sizes.
The second approach, known as the Leave-One-Out Hodges-Lehmann method (LOO-HL), has been introduced by \cite{rousseeuw2002robust}. To elucidate, given a sample \( \{X_i \}_{i = 1}^n \), the LOO-HL empirical mean estimator can be explicitly defined as:
\[
\widehat \mu_{\rm LOO} := \operatorname{med} \left\{ \frac{\operatorname{DE} (X_{(-i)}) + \operatorname{DE} (X_{(-j)})}{2} : 1 \leq i < j \leq n \right\},
\]
Here, the function \( \operatorname{DE} \) computes the empirical mean of the dataset excluding a particular observation. Mathematically, \( \operatorname{DE} (X_{(-i)}) :=\sum_{k \ne i}X_k/(n-1) \) is the empirical mean estimator when data point \( i \) is excluded. This mean is calculated over the set $X_{(-i)} = \{ X_1, \ldots, X_{i - 1}, X_{i + 1}, \ldots, X_n\}$.
Both of these methods aim to enhance the reliability and robustness of estimates derived from datasets with limited observations. By focusing on sub-samples or manipulating the data in a particular way, they mitigate some of the vulnerabilities associated with small sample statistics.
The nuances of the leave-one-out Hodges-Lehmann method differ significantly from the classic Hodges-Lehmann empirical mean estimator. The classic Hodges-Lehmann empirical mean estimator is represented as $\widehat \mu_{\rm HL} :=\operatorname{med}_{i \leq j} \frac{X_i+X_j}{2}$. In essence, this method computes the median of the means of all possible pairs of the sample data. Contrastingly, the leave-one-out version employs the leave-one-out mean, \( \operatorname{DE} (X_{(-i)}) \), as opposed to directly using individual data points \( X_i \). The departure from the classic Hodges-Lehmann method is driven by empirical findings, wherein the traditional method did not offer satisfactory performance under certain conditions.
To challenge these methods further, the underlying data gets perturbed with a \( 1\% \) independent Cauchy$(0, 5)$ noise, introducing outliers and asymmetry. The intention behind this perturbation is to assess the robustness of each method under non-ideal circumstances. The performance evaluation is in Figure \ref{small_sample_subG}: while all three methods manage to yield reasonably good performance, the LOO-HL method seems to be in the lead. Its superiority becomes more pronounced for very small sample sizes, particularly when \( 2 \leq n \leq 4 \). This observation underscores the importance of selecting appropriate statistical techniques, especially the sample is extremely small.
\begin{figure}[!htb]
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-211017-fig1.pdf}
{\caption*{Standard Gaussian}}
\endminipage\hfill
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-211017-fig2.pdf}
\caption*{Centralized bernoulli}
\endminipage\hfill
\minipage{0.333\textwidth}
\includegraphics[width=\linewidth]{figures/cg-211017-fig3.pdf}
\caption*{Uniform on $[-0.5, 0.5]$}
\endminipage
\caption{The relative error of $\| \cdot\|_G$-norm estimation for three kind distributions by using direct empirical moment method (DE), Bootstrap, and Leave-one-out Hodges-Lehmann method (LOO-HL).}
\label{small_sample_subG}
\end{figure}
\section{Application in Multi-armed Bandit Problem}\label{sec_MAB}
In the multi-armed bandit problem (MAB), a player must choose between \( K \) slot machines, termed an \( K \)-armed bandit. Each machine provides a unique, unknown random reward with a random variable set \( \{Y_{k}\}_{k=1}^K \subseteq \mathbb{R} \). Every realization from a specific arm \( k \) is independent and identically distributed. Furthermore, we assume these rewards to be sub-Gaussian:
\begin{equation}\label{eq:REGsubg}
\|Y_{k}-\mu_{k}\|_{G} < \infty, \quad \forall k \in [K].
\end{equation}
The primary objective is to identify the optimal arm with the highest expected reward, denoted as \( Y_{k^{\star}} \), through arm pulls. In every round \( t \in [T] \), the player selects an arm \( A_{t} \in [K] \). Given \( \{A_{t}=k\} \), the observed reward is defined as \( \{Y_{k,t}\}_{t \in [T]}\stackrel{\rm{i.i.d.}} \sim P_k \). The MAB exploration's aim is to minimize the cumulative regret after \( T \) rounds, represented as:
\[
\operatorname{Reg}_{T}(Y,A) := \sum_{t=1}^{T}(\mu_{k^{\star}}-\mu_{A_{t}}).
\]
Without loss of generality, let \( {k^{\star}}=1 \). We seek to evaluate the expected bounds from the decomposition (see Lemma 4.5 in \cite{lattimore2020bandit}),
\begin{equation}\label{eq:REG}
\operatorname{Reg}_{T}:= \mathrm{E} (\operatorname{Reg}_{T}(Y,A) )= \sum_{k = 1}^K \Delta_k \mathrm{E} \left[ \sum_{t=1}^{T} 1 \left\{A_{t}=k\right\} \right],
\end{equation}
{where \( \Delta_{k}:=\mu_1-\mu_k \) is the gap that represents the sub-optimality for arm \( k \) relative to the best arm. Here,\( \mathrm{E} \) is calculated based on the randomness in player's actions \( \{A_{t}\}_{t \in [T]} \).}
For each round \( t \), let \( T_{k}(t) \) denote the number of pulls for arm \( k \) up to time \( t \). The running average of rewards from arm \( k \) at time \( t \) is defined as:
$\overline{Y}_{T_{k}(t)} := \frac{1}{T_{k}(t)} \sum_{\tau \leq t, A_{\tau}=k} Y_{k,\tau}.$
Given a tight concentration inequality, a \( 100(1-\delta)\% \) CI for \( \mu_k \) can be extracted as:
\begin{center}
$ \left[\overline{Y}_{T_{k}(t)}-{c}_{k}(t), \overline{Y}_{T_{k}(t)}+{c}_{k}(t)\right].$
\end{center}
This confidence interval allows us to establish an upper bound for the reward of arm \( k \), \( \overline{Y}_{T_{k}(t)}+{c}_{k}(t) \), and subsequently play arm \( A_{t}=k \) to maximize the reward with a high probability for a given finite \( t \). This strategy is known as the Upper Confidence Bound (UCB) method; see \cite{auer2002finite}.
Many works based on UCB methods appears; see \cite{lattimore2020bandit}. However, many existent bandit algorithms contain unknown norms for the random rewards, they are actually hard to use. For example, \cite{hao2019bootstrapping} use bootstrap method with the second order correction to give an algorithm with the explicit regret bounds for sub-Gaussian rewards. {However, the UCB algorithm in \cite{hao2019bootstrapping} necessitates the use of the unknown Orlicz-norm (or its upper bound) of $\overline{Y}_k - \mu_k$. Consequently, employing this algorithm in practice proves to be infeasible, or it becomes risky to use an arbitrary value for the unknown norm (or its upper bound).}
Fortunately, our estimator can solve this problem. Suppose that $Y_{k}-\mu_{k}$ is symmetric around zero, by one-side version of Theorem \ref{eq:sgnew}, the \eqref{eq:REGsubg} implies that for all $k$ and all $t$,
\[
\mathrm{P} \left(\overline{Y}_{T_{k}(t)} > \mu_{k}+\|Y_{k}-\mu_{k}\|_{G}\sqrt{\frac{2}{T_{k}(t)}\log \frac{1}{\delta}} \right)\le \delta.
\]
Let sub-sample size $m_k$ and block size $b_k$ be positive integers such that $T_k(t) = m_k b_k$ for MOM estimators $\widehat{\|Y_{k}-\mu_{k}\|}_{b_k,G}$ in Section 3. Theorem \ref{thm:mom} {\rm{(a)}} guarantee that true norms can be replaced by MOM-estimated norms such that
\begin{center}
$ {\mathrm{P}}\left\{\overline{Y}_{T_{k}(t)} \le \mu_{k}+\frac{\widehat{\|Y_{k}-\mu_{k}\|}_{b_k,G}}{1-o(1)}\sqrt{\frac{2}{T_{k}(t)}\log \frac{1}{\delta}} \right\}\ge 1-\delta- k_{Y_k} \cdot \exp (- b_k / 8)$
\end{center}
as $\eta(\varepsilon)=1$ with $\varepsilon=0$.
In real-world scenarios, underlying data distribution details are mostly unknown. The multiplier bootstrap is a versatile method for uncertainty qualification, enabling the simulation of the non-asymptotic properties of the target statistic by reweighting its summands. The multiplier bootstrapped quantile for the i.i.d. observation \( \mathbf{Y}_n \) is:
\[
q_{\alpha} (\mathbf{Y}_n - \overline{Y}_n, \mathbf{w}) = \inf \left\{ x \in \mathbb{R} \mid \mathrm{P}_{w} \left( n^{-1} \sum_{i = 1}^n w_i (Y_i - \overline{Y}_n) > x \right) \leq \alpha \right\},
\]
where \( \mathbf{w} \) are bootstrap random weights independent of \( \mathbf{Y}_n \).
{Inspired by \cite{hao2019bootstrapping}, we developed Algorithm \ref{alg1} that utilizes MOM estimators for the UCB. Our innovative regret bounds are articulated in Theorem \ref{thm.UCB}. These bounds assure comparatively minimal regret by incorporating a concentration-based second-order correction, $\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)})$, as elucidated in Theorem \ref{thm:mom}, into the bootstrapped threshold $q_{\alpha / 2} (\mathbf{Y}_{T_k(t)} - \overline{Y}_{T_k(t)}, \mathbf{w})$.}
\begin{theorem}\label{thm.UCB}
Consider a $K$-armed sub-G bandit under \eqref{eq:REGsubg} and suppose that $Y_k - \mu_k$ is symmetric around zero. For any round $T$, according to moment conditions in Theorem \ref{thm:mom} with $\eta(\varepsilon)=1$ and $\varepsilon=0$, choosing $\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)})$ as
\begin{equation}\label{eq:widehatvarphi}
\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)}) = \frac{\{{2 \log(4 / \alpha)}\}^{1/2}}{{T_{k}^{1 / 2}(t)} - 1} {\widehat{\| {Y}_k - \mu_k \|}_{b_k,G}}
\end{equation}
as a re-scaled version of MOM estimator ${\widehat{\| Y_k - \mu_k \|}_{b_k,G}}$ with block number $b_k$ satisfying the moment assumptions \ref{UCB.A1}[UCB1] and \ref{UCB.A2}[UCB2] in Appendix \ref{app_proof}. Fix a confidence level $\alpha = 4 / T^2$, if the player pull an arm $A_{t} \in [K]$ according to Algorithm 1, then we have the problem-dependent regret of Algorithm 1 is bounded by
\begin{center}
$\operatorname{Reg}_{T} \leq 16 (2 + \sqrt{2})^2 \displaystyle\max_{k \in [K]}\|Y_k - \mu_k \|_G^2 \log T \sum_{k = 2}^K \Delta_k^{-1} + ( 4 T^{-1} + 2 T^{- 25 - 16 \sqrt{2}} + 8) \sum_{k = 2}^K \Delta_k$,
\end{center}
where $\Delta_k$ is the sub-optimality gap.
Moreover, let $\mu_1^* := \max_{k_1 \in [K]} \mu_{k_1} - \min_{k_2 \in [K]} \mu_{k_2}$ be the range over the rewards, the problem-independent regret
\begin{center}
$\operatorname{Reg}_{T} \leq 8 (2 + \sqrt{2}) \max_{k \in [K]}\|Y_k - \mu_k \|_G \sqrt{T K \log T} + ( 4 T^{-1} + 2 T^{-25 - 16 \sqrt{2}} + 8 ) K \mu_1^*.$
\end{center}
\end{theorem}
For $\eta(\varepsilon)\in (0,1)$ where $\varepsilon\in(0,1)$, the regret bounds are comparable up to certain constants.
According to Theorem \ref{thm.UCB}, the regret of our approach attains the minimax rate of $\log T$ in a problem-dependent scenario, and $\sqrt{TK}$ for a problem-independent one. This is consistent up to a logarithmic factor, corroborated by \cite{tao2021optimal}, demonstrating the {near-optimal property} of Algorithm \ref{alg1}. In comparison to the traditional vanilla UCB, our technique refines the constants. To illustrate, when $Y_k \sim N(\mu_k, 1)$, the regret bound's constant factor in Theorem 4 of \cite{auer2002finite} is $256$, which is significantly greater than the $16(2+\sqrt{2})^2$ presented in our theorem.
In scenarios where the UCB possesses unknown sub-G parameters, Theorem \ref{thm.UCB} primarily investigates a practical UCB algorithm integrating sub-G parameter estimations. {In contrast, many earlier UCB algorithms operate under the assumption that the upper bound of the sub-G parameter is known beforehand.} One can refer to the algorithm detailed in \cite{hao2019bootstrapping} for an example.
\begin{algorithm*}[!h]\caption{Bootstrapped and Estimated UCB algorithms (BeUCB)}\label{alg1}
\begin{footnotesize}
{\textbf{Input:} $\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)})$ is given by \eqref{eq:widehatvarphi}.}
{\textbf{For} $t = 1, \ldots, K$:}
\quad Pull each arm once to initialize the algorithm.
{\textbf{For} $t = K + 1, \ldots, T$:}
{
\quad Set a confidence level $\alpha \in (0, 1)$.
\quad Calculate the boostrapped quantile $q_{\alpha / 2} (\mathbf{Y}_{T_k(t)} - \overline{Y}_{T_k(t)}, \mathbf{w})$ with the Rademacher bootstrapped weights $\mathbf{w}$ independent with any $Y$. Let
\begin{center}
$\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)}) = \frac{\{{2 \log(4 / \alpha)}\}^{1/2}}{{T_{k}^{1 / 2}(t)} - 1} {\widehat{\| {Y}_k - \mu_k \|}_{b_k,G}}$.
\end{center}
\quad Pull the arm $A_t = \mathop{\operatorname{argmax}}\limits_{k \in [K]} \operatorname{UCB}_k(t - 1)$, where
\begin{center}
$\operatorname{UCB}_k(t - 1) := \overline{Y}_{T_k(t - 1)} + q_{\alpha / 2} (\mathbf{Y}_{T_k(t - 1)} - \overline{Y}_{T_k(t - 1)}, \mathbf{w}) + \left\{\frac{2 \log(4 / \alpha)}{T_{k}(t - 1)}\right\}^{1/2}\widehat{\varphi}_G (\mathbf{Y}_{T_k(t - 1)} ).$
\end{center}
\quad Receive reward $Y_{A_t}$.
}
\end{footnotesize}
\end{algorithm*}
Next, we present a simulation of Algorithm \ref{alg1} for two sub-G cases to assess the performance of the estimated norms. Taking cues from \cite{hao2019bootstrapping,wang2020residual}, we design the four UCB methods:
\textbf{1}. Use our method $\widehat{\varphi}_1(\mathbf{Y}_{T_k(t)})=\widehat{\varphi}_G (\mathbf{Y}_{T_k(t)})$ with \textit{Estimated Norm} in Theorem \ref{thm.UCB} with block number $b_k = 2 \lceil \log T_k(t) \rceil$;
\textbf{2}. Let
\begin{center}
$\widehat{\varphi}_2(\mathbf{Y}_{T_k(t)}) :=\widehat\sigma_{opt}(Y_k) \{2 \log (2 / \alpha) / T_k(t)\}^{1/2}$,
\end{center}
where $\widehat\sigma_{opt}^2(Y_k)$ is estimated using the formula in Remark 24 of \cite{lieber2022estimating}.
\textbf{3}. Use \textit{Asymptotic Naive varphi} $\widehat{\varphi}(\mathbf{Y}_{T_k(t)})$ satisfying
\[\mathrm{P} \big(|\overline{Y}_{T_k(t)} - \mu_k| \leq \widehat{\varphi}_3(\mathbf{Y}_{T_k(t)}) \big) \rightarrow \alpha\]by CLT, i.e.
\begin{center}
$\widehat{\varphi}_3 (\mathbf{Y}_{T_k(t)}) = \widehat{\sigma}_k \Phi^{-1} (1 - \alpha / 2) /\sqrt{{T_k(t)}}$
\end{center}
with $\widehat{\sigma}_k =\{\frac{1}{T_k(t)} \sum_{\tau \leq t, A_{\tau}=k} (Y_{k,\tau} - \overline{Y}_{T_k(t)})^2\}^{1/2}$ as the estimated standard deviation;
\textbf{4}. Regard all the unbounded rewards as bounded r.v. and use Hoeffding's inequality (\textit{wrongly use Hoeffding's inequality}) to construct $\varphi$, i.e.
\[
\widehat{\varphi}_4 (\mathbf{Y}_{T_k(t)}) = \big[\max\{ \mathbf{Y}_{T_k(t)} \} - \min \{\mathbf{Y}_{T_k(t)}\}\big] \left\{\frac{\log (2 / \alpha)}{2T_k(t)}\right\}^{1/2}.
\]
where $\max\{ \mathbf{Y}_{T_k(t)} \}:=\max\{Y_1, Y_2, \ldots, Y_{T_k(t)} \}$.
\begin{figure}[!htb]
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/Gau_MAB.pdf}
\endminipage\hfill
\minipage{0.5\textwidth}
\includegraphics[width=\linewidth]{figures/Mixed_Gau_MAB.pdf}
\endminipage
\caption{The regret of MAB with sub-G rewards under three methods. $x$-axis represents the round number and $y$-axis is the cumulative regret.}
\label{UCB_fig}
\end{figure}
For our MAB simulation, we consider as follows, in each case, the number of arms is assigned as $K = 10$, and the mean reward from $k$-th arm $\mu_k$ is independently drawn from $\operatorname{Exp}(1)$. We consider two types of unbounded reward distributions: EG1. Gaussian $N(\mu_k, 1)$; EG2. Mix of two independent Gaussian
\begin{center}
$p_k \times N(2 \mu_k, 1) + (1 - p_k) \times N \left(\frac{1 - 2p_k}{1 - p_k} \mu_k, 1\right)$
\end{center}
with $p_k \, \sim \, U(0, \frac{1}{2})$.
Simultaneously, we account for potential data contamination due to Cauchy noise. For any given arm, the rewards generated in each round have a 15\% chance of being affected by a $\operatorname{Cauchy}(0,0.01)$ (standard Cauchy distribution with a scale parameter of 1\%).
Additionally, we utilize Thompson Sampling \cite{agrawal2017near} with Gaussian distributions for both rewards and priors. We tune the prior parameters on $\{ 2^{4- k} \}_{k = 1}^6$, using its optimal performance as a robust baseline for this simulation.
Both EG1 and EG2 represent sub-G reward structures. In the simulation, $\mu_k$ may not be bounded, complicating this problem. The simulation results are shown in Figure \ref{UCB_fig}. These results show that our method outperforms the other two methods under unbounded sub-Gaussian rewards and is even comparable to Thompson Sampling when sufficient correct prior knowledge is available. Furthermore, our method demonstrates the strong robustness of our estimated norm method.
\section{Proofs of Key Results}\label{se:key}
\begin{proof}{of Lemma \ref{thm_arg_k}}
Note that $\| X\|_G := \max_{m \in 2 \mathbb{N}} \left[ \frac{\mathrm{E} (X^m)}{\mathrm{E} (Z^m)} \right]^{1 / m}$, where $Z\sim N(0,1)$. By the formula of $2k$-th moment of standard normal distribution [see (18) in \cite{winkelbauer2012moments}], one has
\[
\max_{m = 2k \in 2 \mathbb{N}} \left[ \frac{\mathrm{E} (X^m)}{\mathrm{E} (Z^m)} \right]^{1 / m} = \max_{k \in \mathbb{N}} \left[ \frac{2^{k}}{\sqrt{\pi}} \Gamma \left( k + \frac{1}{2} \right)\right]^{-1 / (2k)} \| X\|_{2k}.
\]
If the maximum above take at $m = \infty$, then $\left[ \frac{2^{k}}{\sqrt{\pi}} \Gamma \left( k + \frac{1}{2} \right)\right]^{-1 / (2k)} \| X\|_{2k}$ is an increasing function for some sub-sequence $\{k_{\ell}\} \subseteq \{ k\}$ such that $\lim_{\ell \rightarrow \infty} k_{\ell} = \infty$ when $\ell$ is large enough.
Therefore,
\[
\left[ \frac{2^{k_{\ell}}}{\sqrt{\pi}} \Gamma \left( k_{\ell} + \frac{1}{2} \right)\right]^{-1 / (2k_{\ell})} \| X\|_{2k_{\ell}} \leq \left[ \frac{2^{k_{\ell} + 1}}{\sqrt{\pi}} \Gamma \left( k_{\ell} + \frac{3}{2} \right)\right]^{-1 / (2k_{\ell} + 2)} \| X\|_{2k_{\ell} + 2}
\]
i.e.
$\frac{2 k_{\ell} + 1}{2} \left[ \frac{\sqrt{\pi}}{\Gamma(k_{\ell} + 1 / 2)}\right]^{1 / k_{\ell}} \leq {\| X\|_{2 k_{\ell} + 2}^{2 k_{\ell} + 2}}/{\| X\|_{2 k_{\ell}}^{2 k_{\ell} + 2}}$
for any $k_{\ell}$ is large enough.
Let $\ell \rightarrow \infty$, we have
\[
\begin{aligned}
1 &= \limsup_{\ell \rightarrow \infty} \frac{\| X\|_{2 k_{\ell} + 2}^{2 k_{\ell} + 2}}{\| X\|_{2 k_{\ell}}^{2 k_{\ell} + 2}} \geq \lim_{\ell \rightarrow \infty} \frac{2 k_{\ell} + 1}{2} \left[ \frac{\sqrt{\pi}}{\Gamma(k_{\ell} + 1 / 2)}\right]^{1 / k_{\ell}} \\
& = \lim_{k_{\ell} \rightarrow \infty} (k_{\ell} + 1 / 2) \left[ \frac{\sqrt{\pi}}{\sqrt{2 \pi} (k_{\ell} + 1 / 2)^{k_{\ell}} e^{- (k_{\ell} + 1 / 2)}}\right]^{1 / k_{\ell}} = \lim_{\ell \rightarrow \infty} \frac{e^{1 + 1 / (2 k_{\ell})}}{\sqrt{2}} = \frac{e}{\sqrt{2}} > 1,
\end{aligned}
\]
which leads to a contradiction, where we use a fact that $\lim_{n \rightarrow \infty} \| X\|_n = \| X\|_{\infty} = \operatorname{esssup} |X|$, and hence
\begin{center}
$\limsup_{\ell \rightarrow \infty}{\| X\|_{2 k_{\ell} + 2}^{2 k_{\ell} + 2}}/{\| X\|_{2 k_{\ell}}^{2 k_{\ell} + 2}} = 1$.
\end{center}
As a result, one must have $\arg \max_{m \in 2 \mathbb{N}} \left[ \frac{\mathrm{E} (X^m)}{\mathrm{E} (Z^m)} \right]^{1 / m} < \infty$.
\end{proof}
\begin{proof}{of Theorem \ref{eq:sgnew}}
If $X_i$ is symmetric around zero, by $\mathrm{E} (X_i^{2k+1})=0$ for $k \in \mathbb{N}_{+}$ then we have
\begin{align*}
\mathrm{E} {e^{tX_i}} = 1 + \sum\limits_{k = 1}^\infty {\frac{{{t^{2k}}\mathrm{E} (X_i^{2k})}}{{(2k)!}}}&\le 1 + \sum\limits_{k = 1}^\infty {\frac{{{t^{2k}}}}{{(2k)!}}} \frac{{(2k)!\left\| X_i \right\|_{G}^{2k}}}{{{2^k}k!}} \\
&= 1 + \sum\limits_{k = 1}^\infty {\frac{{{{({t^2}\left\| X_i \right\|_{G}^2/2)}^k}}}{{k!}}} = \exp \left\{ \frac{{{t^2}\left\| X_i \right\|_{G}^2}}{2}\right\}
\end{align*}
for all $t \in \mathbb{R}$, where the last inequality is by the definition of $\left\| X_i \right\|_{G}<\infty$ such that $\mathrm{E} (X_i^{2k}) \le \frac{{(2k)!}}{{{2^k}k!}}\left\| X_i\right\|_{G}^{2k}$.
Then it proves $X_i \sim \operatorname{subG}({\left\| X_i \right\|_{G}^2})$, which shows case {\rm{(a)}}.
For case {\rm{(b)}}, if $X_i$ has zero mean, then we bound the odd moment by even moments. For $k=1,2,\cdots$ and $c_k>0$, Cauchy's inequality and mean value inequality imply
\[\mathrm{E} (|tX_i|^{2k + 1}) \le \sqrt{ {c_k^{-1}{\mathrm{E} (|tX_i|^{2k}}) c_k\mathrm{E} (|t X_i|^{2k + 2}}) }\le \frac{ c_k^{-1}{{t^{2k}}\mathrm{E} (X_i^{2k})} + c_k{t^{2k + 2}}\mathrm{E} (X_i^{2k + 2})}{2}.\]
So, $\frac{{\mathrm{E} (|tX_i|^3)}}{{3!}} \le \frac{ {c_1^{-1}{t^2}\mathrm{E} (X_i^2) + c_1{t^4}\mathrm{E} (X_i^4)} }{{2 \cdot 3!}}$, $\frac{{\mathrm{E} (|tX_i|^5)}}{{5!}} \le \frac{c_2^{-1} {{t^4}\mathrm{E} (X_i^4) + c_2{t^6}\mathrm{E} (X_i^6)} }{{2 \cdot 5!}}$, and so on, which implies
\begin{align}\label{eq:oddmoments}\small
&~~~~\mathrm{E} {e^{tX_i}}\le 1 + \sum\limits_{k = 2}^\infty {\frac{{{t^k}\mathrm{E} (|X_i|^k)}}{{k!}}} \le 1 + {\frac{{{t^2}\mathrm{E} (X_i^2)}}{{2!}}} +\frac{ {c_1^{-1}{t^2}\mathrm{E} (X_i^2) + c_1{t^4}\mathrm{E} (X_i^4)} }{{2 \cdot 3!}}+ {\frac{{{t^4}\mathrm{E} (X_i^4)}}{{4!}}} \nonumber\\
&+\frac{c_2^{-1} {{t^4}\mathrm{E} (X_i^4) + c_2{t^6}\mathrm{E} (X_i^6)} }{{2 \cdot 5!}}+ {\frac{{{t^6}\mathrm{E} (X_i^6)}}{{6!}}}+\frac{c_3^{-1} {{t^6}\mathrm{E} (X_i^6) + c_3{t^8}\mathrm{E} (X_i^8)} }{{2 \cdot 7!}}+\cdots \nonumber\\
& \le 1 + \left( {1+ \frac{c_1^{-1}}{{ 3!}}} \right){\frac{{{t^2}\mathrm{E} (X_i^2)}}{{2!}}}+\left( {1+ \frac{{4!}c_1}{{2\cdot 3!}}} + \frac{{4!c_2^{-1}}}{2\cdot{ 5!}} \right){\frac{{{t^4}\mathrm{E} (X_i^4)}}{{4!}}}+ \left( {1+ \frac{{6!}c_2}{{2\cdot 5!}}} + \frac{{6!c_3^{-1}}}{2\cdot{ 7!}} \right){\frac{{{t^6}\mathrm{E} (X_i^6)}}{{6!}}}+\cdots\nonumber\\
& \le 1 + \left( {1+ \frac{1}{6c_1}} \right){\frac{{{t^2}\mathrm{E} (X_i^2)}}{{2!}}}+\left( 1+ 2c_1+ \frac{c_2^{-1}}{10} \right){\frac{{{t^4}\mathrm{E} (X_i^4)}}{{4!}}}+\left( 1+ 3c_2+ \frac{c_3^{-1}}{14} \right){\frac{{{t^6}\mathrm{E} (X_i^6)}}{{6!}}}+\cdots \nonumber\\
&\le 1 + \left( {1+ \frac{1}{6c_1}} \right){\frac{{{t^2}\mathrm{E} (X_i^2)}}{{2!}}}+\left( 1+ 2c_1+ \frac{c_2^{-1}}{10} \right){\frac{{{t^4}\mathrm{E} (X_i^4)}}{{4!}}}+\sum\limits_{k = 3}^\infty {\left( {1 + kc_{k-1} + \frac{c_k^{-1}}{{4k + 2}} } \right)} \frac{{t^{2k}}\mathrm{E} (X_i^{2k})}{{(2k)!}}.
\end{align}
To bound \eqref{eq:oddmoments}, we assign $c_k=x^{-1}\cdot \frac{m^{k+1}}{2k+2}$ for $k \ge 2$ and $x, m>0$. Consider the following system of equations:
\begin{equation*}
\left\{ \begin{array}{l}
1 + \frac{1}{{6{c_1}}} = m \\
{1 + 2{c_1} + \frac{{c_2^{ - 1}}}{{10}} = {m^2}}
\end{array} \right..
\end{equation*}
This system with ${c_2} = \frac{1}{6\cdot x}{m^3}$ gives
\[ 1 + 2{c_1} + 0.6x{\left( {1 + \frac{1}{{6{c_1}}}} \right)^{ - 3}} = {\left( {1 + \frac{1}{{6{c_1}}}} \right)^2}\]
which could implies ${c_1}=0.4$ if we set $x=0.9806308$. And then $m={1+ \frac{1}{6c_1}}= \frac{17}{12}$.
Since
\begin{center}
$1 + \frac{{{m^k}}}{{2x}} + \frac{{(k + 1)x}}{{2k + 1}} \cdot {m^{ - k - 1}} \le {m^k}$
\end{center}
with $m = 17/12,k = 3,4,\cdots$. Therefore, \eqref{eq:oddmoments} has a further upper bound
\begin{align}
\mathrm{E} {e^{tX_i}} &\le 1 + m{\frac{{{t^2}\mathrm{E} (X_i^2)}}{{2!}}}+m^2{\frac{{{t^4}\mathrm{E} (X_i^4)}}{{4!}}}+\sum\limits_{k = 3}^\infty {\left( {1 + kc_{k-1} + \frac{c_k^{-1}}{{4k + 2}} } \right)} \frac{{t^{2k}}\mathrm{E} (X_i^{2k})}{{(2k)!}}\nonumber\\
&\le \sum\limits_{k = 0}^\infty \frac{{{(\sqrt m t)^{2k}}\mathrm{E} (X_i^{2k})}}{{(2k)!}}\nonumber\\
&\le 1 + \sum\limits_{k = 1}^\infty {\frac{{{(\sqrt m t)^{2k}}}}{{(2k)!}}} \frac{{(2k)!\left\| X_i \right\|_{G}^{2k}}}{{{2^k}k!}}\nonumber\\
&\le \exp \left\{ \frac{{t^2{(\sqrt{17/12}{{\left\| X_i\right\|}_{G}})^2}}}{2}\right\},
\end{align}
where the last inequality is by the definition of $\left\| X_i \right\|_{G}$.
Thus we show $X_i \sim \operatorname{subG}(17{\left\| X_i \right\|_{G}^2}/12)$.
\end{proof}
\begin{proof}{of Theorem \ref{eq:uplow}}
For {(a)}, it remains to show the lower tail bound. For $t\ge 0$, by the independence of $\{ X_i\}_{i = 1}^n$,
\begin{align*}
\mathrm{P}\{\max _{1 \leq i \leq n}X_{i}\leq t\} &=\mathrm{P}\left(X_{1} \leq t, \cdots,X_{n}\leq t\right) = \prod_{i=1}^{n} \mathrm{P}\left(X_{i} \leq t\right) \\
[\text{Applying~Lemma}~\ref{eq:low}]~& \leq\left(1-C^2(X)\mathrm{e}^{-4{{\left\| X \right\|_{\tilde G}^{-2}}}{[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]t^2}}\right)^{n} \\ &{\leq} \exp \left(-nC^2(X)\mathrm{e}^{-4{\left\| X \right\|_{\tilde G}^{-2}}{[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]t^2}}\right),
\end{align*}
where we use $1-x \leq e^{-x} \text { for all } x \in \mathbb{R}$ in the last inequality.
Let $\delta=\exp \left(-nC^2(X)\mathrm{e}^{-4{\left\| X \right\|_{\tilde G}^{-2}}{[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]t^2}}\right)$ and we get
\[t={\frac{{\left\| X \right\|_{\tilde G}}/{\left\| X \right\|_{G}}}{2\{{2{\left\| X \right\|_{G}^2}}/{\left\| X \right\|_{\tilde G}^2}-1\}^{1/2}}} \left\{\log n-\log C^{-2}(X)-\log \log \left(\frac{2}{ \delta}\right)\right\}^{1/2}.\]
For {(b)}, if $X\le M<\infty$, it shows
\[0\le {\left\| X \right\|_{\tilde G}} \le \min\limits_{k \ge 1} \left[ {\frac{{{2^k}k!}}{{(2k)!}} {M^{2k}}}\right]^{1/(2k)}=M\min\limits_{k \ge 1} \left[ {\frac{{{2^k}k!}}{{(2k)!}} }\right]^{1/(2k)}= 0,\]
So we immediately get $ {\left\| X \right\|_{\tilde G}}=0$.
\end{proof}
\begin{proof}{of Lemma \ref{eq:low}}
The proof is based on Paley--Zygmund inequality
\[\mathrm{P}(Z\ge\theta \mathrm{E}(Z)) \geq(1-\theta)^{2} \frac{\{\mathrm{E}(Z)\}^{2}}{\mathrm{E}(Z^{2})}\]
for a positive r.v. $Z$ with finite variance, where $\theta \in (0,1)$; see Page 47 in \cite{boucheron2013concentration}.
Since $X$ is symmetric around zero, one has $\mathrm{E} (X^{2k+1})=0$ for $k \in \mathbb{N}_{+}$, which gives for all $s \in \mathbb{R}$,
\begin{align}\label{eq:subGlower}
\mathrm{E} {e^{sX}} = 1 + \sum\limits_{k = 1}^\infty {\frac{{{s^{2k}}\mathrm{E} (X^{2k})}}{{(2k)!}}}&\ge 1 + \sum\limits_{k = 1}^\infty {\frac{{{s^{2k}}}}{{(2k)!}}} \frac{{(2k)!\left\| X \right\|_{\tilde G}^{2k}}}{{{2^k}k!}}\nonumber\\
&= 1 + \sum\limits_{k = 1}^\infty {\frac{{{{({s^2}\left\| X \right\|_{\tilde G}^2/2)}^k}}}{{k!}}} = \exp \{ \frac{{{s^2}\left\| X \right\|_{\tilde G}^2}}{2}\},
\end{align}
where the last inequality stems from the definition of $\left\| X \right\|_{\tilde G}<\infty$ such that $\mathrm{E} (X^{2k}) \ge \frac{{(2k)!}}{{{2^k}k!}}\left\| X\right\|_{\tilde G}^{2k}$.
Let $Z:=\exp\{sX\}$. The above Paley-Zygmund inequality and \eqref{eq:subGlower} imply
\begin{align}\label{eq:PZ}
\mathrm{P} (Z \ge t) &= \mathrm{P}(\exp\{sX\} \ge\theta \exp\{{\left\| X \right\|_{\tilde G}^2}s^2 / 2\})\nonumber\\
& \ge \mathrm{P}(\exp\{sX\} \ge\theta \mathrm{E}\exp\{sX\})\nonumber\\
&\geq (1-\theta)^{2} \frac{[\mathrm{E}\exp\{sX\}]^{2}}{\mathrm{E}\exp\{2sX\}}\nonumber\\
& \geq (1-\theta)^{2} \frac{\exp\{{\left\| X \right\|_{\tilde G}^2}s^2\}}{\exp\{2{\left\| X \right\|_{ G}^2} s^2\}}=(1-\theta)^{2}\exp\{-[2{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}] s^2\}
\end{align}
where $t:= {\left\| X \right\|_{\tilde G}^2}\frac{s}{2}+\frac{\log \theta}{s}>0$, and the last inequality is from Theorem \ref{eq:sgnew}(a).
Put $s=\frac{t+\{t^2+2{\left\| X \right\|_{\tilde G}^2}\log({1}/{\theta})\}^{1/2}}{{\left\| X \right\|_{\tilde G}^2}}$, which is solved from equation
$$
\frac{1}{2}{\left\| X \right\|_{\tilde G}}s^2-ts-\log({1}/{\theta})=0,
$$
Then we have
\[
s^2={\left\| X \right\|_{\tilde G}^{-4}}{\left[t+\{t^2+2{\left\| X \right\|_{\tilde G}^2}\log({1}/{\theta})\}^{1/2}\right]^2}\le \frac{4t^2+4{\left\| X \right\|_{\tilde G}^2}\log({1}/{\theta})}{\left\| X \right\|_{\tilde G}^4}.
\]
Substitute this upper bound into \eqref{eq:PZ}, and it leads to
\begin{align}\label{eq:PZ1}
\mathrm{P} (X \ge t) &\geq (1-\theta)^{2}\exp\left\{-\frac{[2{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}][{4t^2+4{\left\| X \right\|_{\tilde G}^2}\log({1}/{\theta})}]}{\left\| X \right\|_{\tilde G}^4}\right\}\nonumber\\
&=(1-\theta)^{2}\exp\{-{[2{\left\| X \right\|_{G}^2}/{\left\| X \right\|_{\tilde G}^4}-\left\| X \right\|_{\tilde G}^{-2}][{4t^2+4{\left\| X \right\|_{\tilde G}^2}\log({1}/{\theta})}]}\}\nonumber\\
&=(1-\theta)^{2}\theta^{4[2{\left\| X \right\|_{G}^2}/{\left\| X \right\|_{\tilde G}^4}-\left\| X \right\|_{\tilde G}^{-2}]}\exp\{-4{[2{\left\| X \right\|_{G}^2}/{\left\| X \right\|_{\tilde G}^4}-\left\| X \right\|_{\tilde G}^{-2}]t^2}\}.
\end{align}
Taking $\sup$ on $\theta \in (0,1)$ over the two sides of \eqref{eq:PZ1}, we have
\begin{align}\label{eq:PZ2}
&\mathrm{P} (X \ge t)\geq \exp\{-4{[2{\left\| X \right\|_{G}^2}/{\left\| X \right\|_{\tilde G}^4}-\left\| X \right\|_{\tilde G}^{-2}]t^2}\} \sup\limits_{\theta \in (0,1)}(1-\theta)^{2}\theta^{4[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]}\nonumber\\
& =\left( \frac{{\left\| X \right\|_{\tilde G}^2}}{{4{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}}}\right)\left( \frac{{4{\left\| X \right\|_{G}^2}-2{\left\| X \right\|_{\tilde G}^2}}}{4{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}}\right)^{2[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]}e^{-4{[2{\left\| X \right\|_{G}^2}/{\left\| X \right\|_{\tilde G}^4}-\left\| X \right\|_{\tilde G}^{-2}]t^2}}
\end{align}
where the $\sup_{\theta \in (0,1)}(1-\theta)^{2}\theta^{4[2{\left\| X \right\|_{G}^2}/{{\left\| X \right\|_{\tilde G}^2}}-1]}$ is attained at $\theta_0=\frac{{4{\left\| X \right\|_{G}^2}-2{\left\| X \right\|_{\tilde G}^2}}}{4{\left\| X \right\|_{G}^2}-{\left\| X \right\|_{\tilde G}^2}}$.
\end{proof}
\noindent \textbf{Proof of Theorem \ref{thm.UCB}:}
Denote $Y_{k,i}$ represents the $i$-th value of reward of arm $k$ in its history, where $i \in [T_k(t)]$ for any round $t$.
It is needed to bound $\mathrm{E} [T_k(t)]$ as one has
\[\mathrm{E} (\operatorname{Reg}_T) \leq \sum_{k = 2}^K \Delta_k \mathrm{E} [T_k(t)],\]
from \eqref{eq:REG}, so we can only focus on some fixed arm. Hence, we can just drop the subscript $k$ in $\{ Y_{k, i} \}_{i = 1}^{T_k(t)}$ as $\{ Y_{i}\}_{i = 1}^{T_k(t)}$.
We first give a lemma, which is crucial in the following proof.
\begin{lemma}\label{bootstrap.lem1}
Let $\{ Y_i \}_{i = 1}^n$ be independent r.v.s with $\mu_i = \mathrm{E}(Y_i)$, and assume that $Y_i - \mu_i$ is symmetric around zero with $\| Y_i - \mu_i \|_{G} \leq C$ and $\{w_i \}_{i = 1}^n $ are i.i.d. Rademacher r.v.s independent of $\{ Y_i \}_{i = 1}^n$. Let $\overline{w}:=\frac{1}{n}\sum_{i=1}^nw_i$. Then,
\begin{equation*}
\mathrm{P} \bigg( \frac{1}{n} \sum_{i = 1}^n (w_i - \overline{w}) (Y_i - \mu_i ) \leq C \left\{\frac{2\log (1 / \alpha)}{n}\right\}^{1/2}\bigg) \geq 1 - \alpha.
\end{equation*}
\end{lemma}
\begin{proof}{of Lemma \ref{bootstrap.lem1}}
From Theorem \ref{eq:sgnew}, we know that under the conditions in the lemma,
\begin{equation}\label{eq:subGweight}
\mathrm{P} \bigg( \frac{1}{n}\sum_{i = 1}^n a_i (Y_i - \mu_i) \leq C \| a \|_2 \frac{\{2\log (1 / \alpha)\}^{1/2}}{n} \bigg) \geq 1 - \alpha
\end{equation}
for any vector $a:=(a_1,\ldots,a_n)^\top \in \mathbb{R}^n$.
On the other hand, we have following inequalities,
\begin{equation}\label{eq:weight}
\| w - \overline{w}\|_2^2 := \sum_{i = 1}^n (w_i - \overline{w})^2 = \sum_{i = 1}^n w_i^2 - n \overline{w}^2 = n (1 - \overline{w}^2) \leq n,~\text{and}~\| w - \overline{w} \|_{\infty} \leq 2.
\end{equation}
Therefore, one has
\begin{equation*}
\begin{aligned}
&~~~~ \mathrm{P} \bigg( \frac{1}{n} \sum_{i = 1}^n (w_i - \overline{w}) (Y_i - \mu_i ) \leq C \left\{\frac{2\log (1 / \alpha)}{n}\right\}^{1/2}\bigg)\\
& = \mathrm{P}_w \mathrm{P}_y \bigg( \frac{1}{n} \sum_{i = 1}^n (w_i - \overline{w}) (Y_i - \mu_i ) \leq C \left\{\frac{2\log (1 / \alpha)}{n}\right\}^{1/2}\bigg) \\
& \geq \mathrm{P}_w \mathrm{P}_y \bigg( \frac{1}{n} \sum_{i = 1}^n (w_i - \overline{w}) (Y_i - \mu_i ) \leq C \| w - \overline{w}\|_2 \left\{\frac{2\log (1 / \alpha)}{n}\right\}^{1/2}\bigg) \geq 1 - \alpha,
\end{aligned}
\end{equation*}
where the second inequality is by \eqref{eq:weight} and the last inequality applies \eqref{eq:subGweight}.
\end{proof}
Based on Lemma \ref{bootstrap.lem1}, next we can prove Theorem \ref{thm.UCB}. We first state the assumptions for Theorem \ref{thm.UCB} in detail.
\begin{itemize}
\item[(UCB1)\def#2{(UCB1)}\label{UCB.A1}] In round $t$, the rewards of $k$-the arm $Y_k$ satisfies ${\mathrm{var}(Y_k^{2\kappa}) }\le \sigma_{k, \kappa}^{2\kappa},~{1 \le \kappa \le k_{Y_k}}$ for a given sequence $\{\sigma_{k, \kappa}\}_{\kappa \ge 1}$;
\item[(UCB2)\def#2{(UCB2)}\label{UCB.A2}]\label{UCB.A2} For any $k \in [K]$
\[
{\overline g_{\kappa,m}({\sigma_{k,\kappa}})} : = 1 - {\left[ {\frac{{{\mathrm{E}}({Y_k^{2\kappa}})}}{{(2\kappa - 1)!!}}} \right]^{{\frac{-1}{2\kappa}}}} \mathop {\max }\limits_{1 \le \kappa \le \kappa_{Y_k}} {\left[ { \frac{{\mathrm{E}}(Y_{k}^{2 \kappa})}{{(2\kappa - 1)!!}}-2m^{-1/2}\frac{\sigma_{k, \kappa}^{\kappa}}{{\mathrm{E}}(Y_{k}^{2 \kappa})}} \right]^{\frac{1}{2\kappa}}}
\]
and
\[
{ \underline{g}_{\kappa,m}({\sigma_{k, \kappa}})}:=[{2m^{-1/2}\sigma_{k, \kappa}^{\kappa}}/\{{\mathrm{E}} (Y_{k}^{2 \kappa})\}+1]^{1 /(2 \kappa)} - 1
\]
are both less than $m^{-1/2}$ for sufficient large $m$.
\end{itemize}
Denote the population version of
\begin{center}
$ \widehat{\varphi}_G (\mathbf{Y}_{n}) =\left\{\frac{2 \log (4 / \alpha)}{n}\right\}^{1/2}\frac{\widehat{\| Y - \mu \|}_{b_k,G}}{1 - {n^{- 1 / 2}}}$ \quad
for \quad $\varphi_G( \mathbf{Y}_n ) := \| Y - \mu \|_{G} \left\{\frac{2 \log (4 / \alpha)}{n}\right\}^{1/2}$
\end{center}
with $\mathrm{E}(Y) = \mu$. Theorem \ref{eq:sgnew} gives
\begin{equation*}
\mathrm{P} \big( | \overline{Y} - \mu | \geq {\varphi}_G(\mathbf{Y}_n) \big) \leq 2 \exp \left\{- \frac{n^2 \varphi_G^2(\mathbf{Y}_n)}{2 n \|Y - \mu \|_{G}^2} \right\} = \alpha / 2,
\end{equation*}
by the fact that $Y - \mu$ is symmetric around zero.
From Theorem \ref{thm:mom}, we take $b \geq 8 \log (k_{Y} / \alpha)$ and define the event $\mathcal{E}_Y$ bellow associated with the MOM estimation of the intrinsic moment norm
\begin{equation*}
\mathcal{E}_Y:=\{ \widehat{\| Y - \mu \|}_{b,G} \geq [ 1 - \max_{1 \leq k \leq k_{Y} / 2} \overline{g}_{k, m} (\sigma_k)] \| Y - \mu \|_{G}\}
\end{equation*}
with probability at least $1 - \alpha / 2$. Note that
\begin{align*}
& ~~~~\mathrm{P} \big( | \overline{Y} - \mu | \geq \widehat{\varphi}_G (\mathbf{Y}_n) \big) \leq \mathrm{P} \big( | \overline{Y} - \mu | \geq \widehat{\varphi}_G (\mathbf{Y}_n), \, \mathcal{E}_Y\big) +\mathrm{P} \big( \mathcal{E}_Y^c\big)\\
& \leq \mathrm{P} \bigg( | \overline{Y} - \mu | \geq \left\{\frac{2 \log (4 / \alpha)}{n}\right\}^{1/2}(1 - n^{- 1 / 2})^{-1} \widehat{\| Y - \mu \|}_{b,G}, \, \mathcal{E}_Y \bigg) + \alpha / 2\\
&\le \mathrm{P} \bigg( | \overline{Y} - \mu | \geq (1 - n^{- 1 / 2})^{-1} [ 1 - \max_{1 \leq k \leq k_{Y} / 2} \overline{g}_{k, m} (\sigma_k)]\sqrt{\frac{2 \log (4 / \alpha)}{n}}\| Y - \mu \|_{G} \bigg) + \alpha / 2\\
&\le \mathrm{P} \big( | \overline{Y} - \mu | \geq \varphi_G (\mathbf{Y}_n)) + \alpha / 2 \leq \alpha.
\end{align*}
where the last inequality is by taking $m$ big enough such that $\max_{1 \leq k \leq k_{Y} / 2} \overline{g}_{k, m}(\sigma_k) \leq 1 / \sqrt{n}$. Then, for $s \in \mathbb{N}_+$,
\begin{equation*}
\mathrm{P} \big( |\overline{Y}_s - \mu_k | \leq \widehat{\varphi}_G (\mathbf{Y}_s)\big) \geq 1 - \alpha.
\end{equation*}
Now for any $k \in [K]$ and fixed $T_k(t) = s$, we know that
\begin{equation*}
\mathrm{P} \big( \overline{Y}_s - \mu_k \geq \widehat{\varphi}_G (\mathbf{Y}_s) \big)\leq \mathrm{P} \big( | \overline{Y}_k - \mu_k | \geq \widehat{\varphi}_G (\mathbf{Y}_n) \big) \leq \alpha.
\end{equation*}
By the non-asymptotic second-order correction [see Theorem 2.2 in \cite{hao2019bootstrapping}] and the assumption that $Y - \mu$ is symmetric around zero, one has
\[
\mathrm{P} \left\{ \mu_k - \overline{Y}_s \geq q_{\alpha / 2} (\mathbf{Y}_s - \overline{Y}_s) + \left\{\frac{2 \log (4 / \alpha)}{s}\right\}^{1/2} \widehat{\varphi}_G(\mathbf{Y}_s) \right\} \leq 2 \alpha,
\]
where $q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) : = q_{\alpha / 2}\left(\mathbf{Y}_s -\overline{Y}_s, \frac{1}{s} \mathbf{1}_s \right)$.
Denote the UCB index $\operatorname{UCB}_k (t) = \overline{Y}_{T_k(t)} + h_{\alpha} (\mathbf{Y}_{T_k(t)})$, and the good event for $k \in [K]$
\begin{equation*}\small
\mathcal{E}_k := \{ \mu_1 < \min_{t \in [T]} \operatorname{UCB}_1 (t)\} \, \cap \, \left\{ \overline{Y}_{B_k} + q_{\alpha / 2} (\mathbf{Y}_s - \overline{Y}_s) + \{{2 \log(4 / \alpha)
\}^{1/2} / {s}} \cdot \widehat{\varphi}_G(\mathbf{Y}_s) < \mu_1 \right\},
\end{equation*}
where $B_k \in [T]$ is a constant to be chosen later. Following from the proof in (B.16)-(B.18) of \cite{hao2019bootstrapping}, we can gives that $T_k(t) \leq B_k$ and
\begin{equation}\label{thm.UCB.n_kt}\small
\mathrm{E} (T_k(t)) \leq B_k + T \Big[ 2 \alpha T + \mathrm{P} \big( \overline{Y}_{B_k} + q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) + \{{2 \log(4 / \alpha)} / {{B_k}}\}^{1/2} \widehat{\varphi}_G(\mathbf{Y}_{B_k}) \geq \mu_1 \big) \Big].
\end{equation}
On the other hand, from Lemma \ref{bootstrap.lem1}, then
\begin{equation*}
\mathrm{P} \bigg( q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) \geq C \{{2 \log(4 / \alpha)} / {{B_k}}\}^{1/2}\bigg) \leq \alpha / 2.
\end{equation*}
Next, we need to the following assumptions for block $\{ b_k\}_{k \in [K]}$ in MOM estimator corresponding to Theorem \ref{thm:mom}. Here we include the subscript $k$ to avoid confusion.
Under $\eta(\varepsilon)=1$ with $\varepsilon=0$, Theorem \ref{thm:mom} ensures for $b_k \geq 8 \log (k_{Y} / \alpha)$ and take $m$ large enough such that $\max_{1 \leq \kappa \leq k_{Y}} \underline{g}_{\kappa, m} (\sigma_\kappa) \leq n^{-1 / 2}$, then
\begin{equation*}
\begin{aligned}
&~~~~ \mathrm{P} \big( (1 + n^{- 1 /2}) C \le {\widehat{\| {Y} - \mu\|}}_{b,G} \big) \\
& \leq \mathrm{P} \big( ({1 + \mathop {\max }\limits_{1 \le \kappa \le k_{Y} / 2}{\underline{g}_{\kappa,m}}({\sigma _{k, \kappa}})})\left\| {Y} - \mu \right\|_{G} \le {\widehat{\| {Y} - \mu \|}}_{b_k,G}\big) \leq k_{Y}{e^{ - b_k/8}} \leq \alpha / 2
\end{aligned}
\end{equation*}
Hence, we have
\begin{align*}\small
& \mathrm{P}\left( q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) + \sqrt{\frac{2\log (4 / \alpha)}{{B_k}}} \widehat{\varphi}_G (\mathbf{Y}_{B_k})\geq \right.\\
& C\left[1 + \left\{\frac{2\log (4 / \alpha)}{{B_k}}\right\}^{1/2} \frac{{1 + {B_k}^{- 1 / 2}}}{1 - {B_k}^{- 1 / 2}}\right] \sqrt{\frac{2\log (4 / \alpha)}{{B_k}}}\bigg)\\
\leq & \mathrm{P} \bigg( q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) + {\frac{2\log (4 / \alpha)}{{B_k}}}\frac{\widehat{\| Y - \mu \|}_{b_k,G}}{1 - {B_k}^{- 1 / 2} }\geq C \sqrt{\frac{2\log (4 / \alpha)}{{B_k}}}+ {\frac{2\log (4 / \alpha)}{{B_k}}}\frac{({1 + {B_k}^{- 1 / 2}})C}{1 - B_k^{- 1 / 2}} \bigg) \\
\leq & \mathrm{P} \bigg( q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) \geq C \sqrt{\frac{2\log (4 / \alpha)}{{B_k}}} \bigg) + \mathrm{P} \big( \widehat{\| Y - \mu \|}_{b_k,G} \ge ({1 + {B_k}^{- 1 / 2}})C \big)
\end{align*}
which implies with probability at least $1 - \alpha$,
\begin{align*}
q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) +\sqrt{\frac{2\log (4 / \alpha)}{{B_k}}} \widehat{\varphi}_G(\mathbf{Y}_{B_k}) & \le C\bigg[1 + \sqrt{\frac{2\log (4 / \alpha)}{{B_k}}}\frac{{1 + {B_k}^{-1 / 2}}}{1 - B_k^{-1 / 2}}\bigg]\sqrt{\frac{2\log (4 / \alpha)}{{B_k}}}\\
& \leq 2 ( 2 + \sqrt{2}) C\sqrt{\frac{2\log (4 / \alpha)}{{B_k}}}
\end{align*}
where $B_k \geq 8 \log ( k_{Y_k}/ \alpha) \vee 2 \log (4 / \alpha)$ and $\max_{1 \leq \kappa \leq k_{Y_k} / 2} \overline{g}_{\kappa, m} (\sigma_{k, \kappa})\leq 1 / \sqrt{{B_k}}$ for each arm $k$.
\vspace{2ex}
Now, define the event
\begin{center}
$\mathcal{B}_k := \{q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) +\{{2 \log(4 / \alpha)} / {{B_k}}\}^{1/2} \cdot \widehat{\varphi}_G(\mathbf{Y}_{B_k}) \leq \Delta_k / 2 \}$
\end{center}
with $\Delta_k := \mu_1 - \mu_k$. Choose $B_k$ as
\begin{equation}\label{choosing.B_k}
B_k = \frac{4^2 (2 + \sqrt{2})^2 C^2}{\Delta_k^2} \log (4 / \alpha)\ge 2,
\end{equation}
we have
\begin{equation*}
\begin{aligned}
\mathrm{P} \big( \mathcal{B}_k^c \big) & = \mathrm{P} \big( q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) + \{{2 \log(4 / \alpha)} / {{B_k}}\}^{1/2}\cdot \widehat{\varphi}_G(\mathbf{Y}_{B_k}) > \Delta_k / 2 \big) \\
& \leq \mathrm{P} \bigg( 2 (2 + \sqrt{2}) C \left\{\frac{ \log (4 / \alpha)}{B_k}\right\}^{1/2} > \Delta_k / 2 \bigg) + \alpha = 0 + \alpha = \alpha.
\end{aligned}
\end{equation*}
Applying Theorem \ref{eq:sgnew} for the concentration of $\overline{Y}_{B_k} - \mu_k$ when $B_k$ is chosen as in \eqref{choosing.B_k},
\begin{equation*}
\begin{aligned}
& \mathrm{P} \big( \overline{Y}_{B_k} + q_{\alpha / 2} (\mathbf{Y}_{B_k} - \overline{Y}_{B_k}) + \{{2 \log(4 / \alpha)} / {{B_k}}\}^{1/2} \cdot \widehat{\varphi}_G(\mathbf{Y}_{B_k}) \geq \mu_1 \big)\\
& \leq \mathrm{P} \big( \overline{Y}_{B_k} - \mu_k \geq \Delta_k / 2\big) + \mathrm{P} \big( \mathcal{B}_k^c\big) \\
& \leq 2 \exp \Big\{ - \frac{(B_k \Delta_k / 2)^2}{2 \cdot B_k C^2}\Big\} + \alpha = 2 \exp \Big\{ - \frac{B_k \Delta_k^2}{8 C^2}\Big\} + \alpha.
\end{aligned}
\end{equation*}
Taking account these results into \eqref{thm.UCB.n_kt}, we get that
\begin{equation*}
\begin{aligned}
& ~~ \mathrm{E}(T_k(t)) \leq B_k + 2 \alpha T^2 + \alpha T + 2 T \exp \Big\{ - \frac{B_k \Delta_k^2}{8 C^2}\Big\} \\
& = \frac{4^2 (2 + \sqrt{2})^2 C^2}{\Delta_k^2} \log (4 / \alpha) + 2 \alpha T^2 + \alpha T + 2 T \exp \Big\{ - 2 (2 + \sqrt{2})^2 \log (4 / \alpha)\Big\} \\
& = \frac{16 (2 + \sqrt{2})^2 C^2}{\Delta_k^2} \log T + \frac{4 }{T} + \frac{2}{T^{25 + 16 \sqrt{2}}} + 8
\end{aligned}
\end{equation*}
by taking $\alpha = 4 / T^2$. Under the problem-dependent case, the regret is bounded by
\begin{align*}
\operatorname{Reg}_T &= \sum_{k = 2}^K \Delta_k \mathrm{E} (T_k(t))\\
&\leq 16 (2 + \sqrt{2})^2 C^2 \log T \sum_{k = 2}^K \Delta_k^{-1} + \left( \frac{4}{T} + \frac{2}{T^{25 + 16 \sqrt{2}}} + 8 \right) \sum_{k = 2}^K \Delta_k.
\end{align*}
To get the problem-independent bound, we let $\Delta > 0$ as an arbitrary threshold, then decompose $\operatorname{Reg}_T$, we get
\begin{equation*}
\begin{aligned}
&~~ \operatorname{Reg}_T = \sum_{\Delta_k : \Delta_k < \Delta} \Delta_k \mathrm{E} (T_k(t)) + \sum_{\Delta_k : \Delta_k \geq \Delta} \Delta_k \mathrm{E} (T_k(t)) \\
& \leq T \Delta + 16 (2 + \sqrt{2})^2 C^2 \log T \sum_{\Delta_k : \Delta_k \geq \Delta} \Delta_k^{-1} + \left( \frac{4}{T} + \frac{2}{T^{25 + 16 \sqrt{2}}} + 8 \right) \sum_{\Delta_k : \Delta_k \geq \Delta} \Delta_k \\
& \leq T \Delta + \frac{16 (2 + \sqrt{2})^2 C^2 K \log T}{\Delta} + \left( \frac{4}{T} + \frac{2}{T^{25 + 16 \sqrt{2}}} + 8 \right) K \mu_1^* \\
& = 8 (\sqrt{2} + 2) C \{T K \log T\}^{1/2} + \left( \frac{4}{T} + \frac{2}{T^{25 + 16 \sqrt{2}}} + 8 \right) K \mu_1^*,
\end{aligned}
\end{equation*}
by taking $\Delta = 8 (2 + \sqrt{2}) C \{(K \log T)/ T\}^{1/2}$. And finally, we take $C = \max_{k \in [K]}\|Y_k - \mu_k \|_G$.
\section*{Acknowledgements}
{H. Zhang's research received support from the National Natural Science Foundation of China under Grant No. 12101630. G. Cheng's research was partly funded by the NSF ¨C SCALE MoDL, Grant No. 2134209.
The authors extend their gratitude to Prof. Hongjun Li for initial discussions, and to Dr. Ning Zhang and Dr. Yanpeng Li for their invaluable insights and suggestions.}
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