%\section{Proof of Theorem \ref{thm:adaHedgeG}}\label{sec:Thm6}
\begin{enumerate}
\item[(i)] From Lemma \ref{lem:expYt} and applying Jensen's inequality to \eqref{thm6:eq15} and substituting $\E\left[\sum_{t=1}^T Y_t\right] \leq \log T +1 $, we get the result.
%\item[(ii)] Refer to the supplementary material for the proof.
\jpcol{\item[(ii)] We present the proof for $d = 1$. Proof for $d > 1$ follows similar steps with more involved analysis and is deferred to the Appendix.} We have
\begin{align}
Y_t & =\frac{e^{-\eta L_{t-1} (n_t)}}{\sum_{j \in \mathcal{B}_{t-1}} e^{-\eta L_{t-1} (j)}}. 
\end{align}
Note that $e^{\eta (L_{t-1}(i)-L_{t-1}(j))} \geq 0$ for all $i$, $j$, and $L_{t-1}(i) \geq L_{t-1}(j)$ implies $e^{\eta (L_{t-1}(i)-L_{t-1}(j))} \geq 1$, since $\eta > 0$. Therefore,
\begin{align}
Y_t & = \frac{1}{\begin{matrix}
    \sum\limits_{\substack{j: L_{t-1}(j) > L_{t-1}(n_t)}} e^{\eta (L_{t-1}(n_t) - L_{t-1}(j))} 
    & \\ + \sum\limits_{\substack{j: L_{t-1}(j) \leq L_{t-1}(n_t)}} e^{\eta (L_{t-1}(n_t) - L_{t-1}(j))}
    \end{matrix}} \nonumber 
\\
& \leq \frac{1}{\sum_{j \in \mathcal{B}_{t-1}} \mathds{1}_{\{L_{t-1}(j) \leq L_{t-1}(n_t)\}}}.
\label{eq:dominance}
\end{align}

In round $t$, we define a random variable $Z_t$ such that $Z_t=k^{-1}$, if $X_t$ falls in the $k^{\text{th}}$ best expert, i.e., $\sum_{j \in \mathcal{B}_{t-1}} \mathds{1}_{\{L_{t-1}(j) \leq L_{t-1}(n_t)\}} = k$.
Note that in the presence of ties, we break ties arbitrarily and strictly order the experts. In this case, if $X_t$ falls in the $k^{\text{th}}$ best partition, it can be shown that the denominator in \eqref{eq:dominance} is at least $k + 1$. Thus, $Y_t < Z_t = 1/k$.

From \eqref{eq:dominance}, we have $Y_t \leq Z_t$, for all $t$. From Lemma \ref{lem:1/t+1}, the probability that $X_t$ falls in $k^\text{th}$ best partition is $\frac{1}{t}$, which implies $\P(Z_t=k^{-1})=1/t$. Therefore,
\begin{align}
    &\E[Z_t] = \sum_{k = 1}^{t} \frac{1}{t}\frac{1}{k} \leq \frac{\log t + 1}{t}, \label{eq:expZ}\\
    \implies & \sum_{t=1}^T \E[Z_t] \leq (\log T + 1)^2. \label{eq:expZ1}
\end{align}
{\allowdisplaybreaks
Further, we have
\begin{align}
    & \P\!\! \left(\sum_{t=1}^T Y_t - \sum_{t=1}^T \E[Y_t] > \delta\right) \nonumber \\
     \leq & \P\!\! \left(\sum_{t=1}^T Z_t - \sum_{t=1}^T \E[Y_t] > \delta\right)  \nonumber \\
     \leq & \P \!\!\left(\sum_{t=1}^T Z_t \!-\! \sum_{t=1}^T \E[Z_t] > \delta\! -\! \sum_{t=1}^T \E[Z_t]\!+\! \sum_{t=1}^T \E\left[Y_t\right]\right) \nonumber\\
    \leq & \P \!\!\left(\sum_{t=1}^T Z_t \!-\! \sum_{t=1}^T \E[Z_t] > \delta \!-\! (\log T + 1)^2 \! +\! \log T \right).
    \label{eq:bern)}
    % & \leq \P\sum Z_i - \sum \E[Z_i] > \delta') \text{ where $\delta'=\delta- \sum \E[Z_i]+ \sum \E[Y_i]$}  
\end{align}
}
To get \eqref{eq:bern)}, we use \eqref{eq:expZ}, \eqref{eq:expZ1}, and the fact that $\sum_{t = 1}^T\E\left[Y_t\right] \geq \log T$. Since the $Z_t$s are independent and are upper bounded by one, using Bernstein's inequality, we get
\begin{align}\label{eq:Bernstein_main_paper}
    \P\left(\sum_{t=1}^T Z_t - \sum_{t=1}^T \E[Z_t] > \delta'\right) \leq e^{-\frac{\delta'^2/2}{V_n+\delta'/3}},
\end{align}
where $\displaystyle V_n=\sum_{t=1}^T \text{Var}(Z_t)$, and $\delta'=\delta - (\log T + 1)^2 + \log T$. We have
\begin{align}\label{eq:varZ}
    \text{Var}(Z_t)=\sum_{j=1}^{t} \frac{1}{t}\frac{1}{j^2} - \E[Z_t]^2 \leq \frac{\pi^2}{6t} \nonumber \\ 
    \implies V_n = \sum_{t=1}^T \text{Var}(Z_t) \leq  \frac{\pi^2}{6}(\log T +1).
\end{align}
Choosing $\delta = (\log T+ 1)^2 + 1$ results in $\delta'=\log T + 1$. \color{black} Substituting $\delta'$ and \eqref{eq:varZ} in \eqref{eq:Bernstein_main_paper}, we obtain \color{black}  
\begin{align*}
    \P\left(\sum_{i=1}^T Y_i \!-\! \sum_{i=1}^T \E[Y_i] \!>\! \delta\right)\! & \! \leq e^{-\frac{(\log T + 1)^2/2}{\frac{\pi^2}{6}(\log T + 1) + (\log T + 1)/3}}\\ &\! \leq \! e^\frac{-3 \log (eT)}{\pi^2+2} \! \leq \!(eT)^{-0.25}.
\end{align*}
\begin{comment}
For $\epsilon \geq 0$, choosing $\delta = (\log T+ 1)^2-\log T +(\log T)^{2+\epsilon}$ results in $\delta'=(\log T)^{2+\epsilon}$. Substituting $\delta'$ and \eqref{eq:varZ} in \eqref{eq:Bernstein}, we get 
\begin{align}
    \P\left(\sum_{i=1}^T Y_i - \sum_{i=1}^T \E[Y_i] > \delta\right)\leq e^{-\frac{(\log T)^{4+2 \epsilon}}{\frac{\pi^2}{6}(\log T + 1) + (\log T)^{2+\epsilon}}}\leq e^{-(\log T)^{2+\epsilon}}.
\end{align}
\end{comment}
Substituting the above result in \eqref{thm6:eq15} proves the final result.
\end{enumerate}

