\subsubsection{$\Efour{n}$ tail bound}
\label{appendix:Efour_bound}

\begin{lemma}\label{lemma:en4_norm} Let $\Efour{n}$ be defined as in Lemma~\ref{lemma:oja_error_decomposition} for $u_{0} = g/\norm{g}_{2}$ with $g \sim \mathcal{N}(0, \id_d)$. Let $\eta_n$ be set according to Lemma~\ref{lemma:learning_rate_choice}. For any $\delta \in \bb{0,1}$, with probability at least $1-\delta$,
\bas{
    \norm{\Efour{n}} \leq \frac{1}{\delta^{3/2}}\bb{d\exp\bb{-2\eta_{n}n\bb{\lambda_{1}-\lambda_{2}} + \eta_{n}^{2}n\bb{\lambda_{1}^{2}+\Mtwo^{2}}} + \frac{e\eta_{n}^{3}n\Mtwo^{4}\bb{1+2\log\bb{d}}}{2\bb{\lambda_{1}-\lambda_{2}} + \eta_{n} \bb{\lambda_{1}^{2}-\lambda_{2}^{2}-\Mtwo^{2}}}}^{1/2}.
}
\end{lemma}

\begin{proof}
Recall that
\bas{
\norm{\Efour{n}} &= \frac{\norm{\vp \vp^{\top} B_{n} \vp \vp^{\top} u_0}}{|v_1^{\top} u_0| (1+\eta_n \lambda_1)^n} =  \frac{\norm{\vp \vp^{\top} B_{n} \vp \vp^{\top} g}}{|v_1^{\top} g| (1+\eta_n \lambda_1)^n}.
}
To bound this quantity, we will bound its square instead. Using Markov's inequality, with probability at least $1-\delta/2$, 
\bas{
\norm{\vp \vp^{\top} B_{n} \vp \vp^{\top} g}^2 
&\le \frac{2}{\delta} E\bbb{\norm{\vp \vp^{\top} B_{n} \vp \vp^{\top} g}^2} \\
&= \frac{2}{\delta} \Tr\bb{E\bbb{\bb{\vp \vp^{\top} B_{n} \vp \vp^{\top} g}\bb{\vp \vp^{\top} B_{n} \vp \vp^{\top} g}^{\top}}} \\
&= \frac{2}{\delta} \E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp \vp^{\top}B_{n}^{\top}\vp}}.
}
By Lemma B.3 of \cite{lunde2021bootstrapping},
\bas{\frac{\E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp}}}{(1+\eta_{n}\lambda_{1})^{2n}} &\leq d\exp\bb{-2\eta_{n}n\bb{\lambda_{1}-\lambda_{2}} + \eta_{n}^{2}n\bb{\lambda_{1}^{2}+\Mtwo^{2}}} + \frac{e\eta_{n}^{3}n\Mtwo^{4}\bb{1+2\log\bb{d}}}{2\bb{\lambda_{1}-\lambda_{2}} + \eta_{n} \bb{\lambda_{1}^{2}-\lambda_{2}^{2}-\Mtwo^{2}}}. 
}
Also, with probability at least $1-\delta/2$, $|v_1^{\top} g| \ge \delta/2$ (see Proposition 7 from \cite{lunde2021bootstrapping} for anticoncentration of gaussians). Combining the two bounds yields the result.
\end{proof}


\begin{lemma}\label{lemma:en4_tail_bound} Let $\Efour{n}$ be defined as in Lemma~\ref{lemma:oja_error_decomposition} for $u_{0} = g/\norm{g}_{2}$ with $g \sim \mathcal{N}(0, \id_d)$. Let $\eta_n$ be set according to Lemma~\ref{lemma:learning_rate_choice}. Let $\left\{\Efour{n}^{(i)}\right\}_{i \in [m]}$ and $\left\{g^{(i)}\right\}_{i \in [m]}$ be $m$ $\iid$ instances of $\Efour{n}$ and $g$ respectively. Fix $\delta \in \bb{0,1}$. Then, conditioned on $\mathcal{E}$, with probability at least $1-\delta$, 
\bas{
    &\frac{\sum_{i \in [m]}\bb{e_{k}^{\top}\Efour{n}^{(i)}}^{2}}{m} \\
    &\quad\quad\quad \leq \frac{b_{k}^{2}m^{2}}{\delta^{3}(1-\delta)}\bb{d\exp\bb{-2\eta_{n}n\bb{\lambda_{1}-\lambda_{2}} + \eta_{n}^{2}n\bb{\lambda_{1}^{2}+\Mtwo^{2}}} + \frac{e\eta_{n}^{3}n\Mtwo^{4}\bb{1+2\log\bb{d}}}{2\bb{\lambda_{1}-\lambda_{2}} + \eta_{n} \bb{\lambda_{1}^{2}-\lambda_{2}^{2}-\Mtwo^{2}}}}
}
for all $k \in [d]$, where $b_k := \norm{\vp^{\top}e_k}_{2}$.
\end{lemma}
\begin{proof}
Note that 
\bas{
\bb{e_{k}^{\top}\Efour{n}}^{2} \le \norm{\vp^{\top}e_k}_{2}^{2} \underbrace{\bb{\frac{\norm{\vp^{\top}B_{n}\vp\vp^{\top}u_{0}}_{2}}{\Abs{v_{1}^{\top}u_{0}}(1+\eta_{n}\lambda_{1})^{n}}}^{2}}_{\Phi_{n}} 
}
Let $\Phi_{n}^{(i)}$ correspond to the $i^{\text{th}}$ instance of the random variable $\Phi_{n}$.
Then, for any $k \in [d]$, 
    \ba{
        \frac{1}{m}\sum_{i \in [m]}\bb{e_{k}^{\top}\Efour{n}^{(i)}}^{2} &\leq \frac{\norm{\vp^{\top}e_k}_{2}^{2}}{m}\sum_{i \in [m]} \Phi_{n}^{(i)}. \label{eq:en4_cs}
    }
    Define the event $\mathcal{E} := \left\{|v_{1}^{\top}g| \geq \frac{\delta}{m}\right\}$ and let $\mathcal{E}^{(i)}$, $i\in [m]$ be the $i^{th}$ instance of this event. First, observe that:
    \ba{
    \E[\Phi_{n} | \mathcal{E}] = \E\bbb{\bb{\frac{\norm{\vp^{\top}B_{n}\vp\vp^{\top}u_{0}}_{2}}{\Abs{v_{1}^{\top}u_{0}}(1+\eta_{n}\lambda_{1})^{n}}}^{2}\bigg|\mathcal{E}}
    \notag &= \E\bbb{\frac{\vp^{\top}B_{n}\vp\vp^{\top}gg^{\top}\vp\vp^{\top}B_{n}^{\top}\vp}{\bb{v_{1}^{\top}g}^{2}(1+\eta_{n}\lambda_{1})^{2n}}\bigg|\mathcal{E}} \notag \\
       &\leq \frac{m^{2}}{\delta^2(1+\eta_{n}\lambda_{1})^{2n}}\E\bbb{\vp^{\top}B_{n}\vp\vp^{\top}gg^{\top}\vp\vp^{\top}B_{n}^{\top}\vp\bigg|\mathcal{E}} \notag \\
       &\leq \frac{m^{2}}{\delta^{2}\mathbb{P}(\mathcal{E})}\frac{\E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp}}}{(1+\eta_{n}\lambda_{1})^{2n}} \label{eq:en4_exp}
    }
    Now, using Markov's inequality conditioned on $\bigcap_{i \in [m]}\mathcal{E}^{(i)}$, we have with probability at least $1-\Prob(\bigcap_{i \in [m]}\mathcal{E}^{(i)})$, 
     \ba{\frac{1}{m}\sum_{i \in [m]}\Phi_{n}^{(i)}&\leq \frac{1}{\delta}\E\bbb{\Phi_n^{(i)}\bigg|\bigcap_{j \in [m]}\mathcal{E}^{(j)}} \notag \\
     \mbox{(By $\iid$ nature of the instances)}&=\frac{1}{\delta}\E\bbb{\Phi_n^{(i)}\bigg|\mathcal{E}^{(i)}} 
     = \frac{1}{\delta}\E[\Phi_n | \mathcal{E}] \notag \\
       %&= \frac{1}{\delta}\E\bbb{\frac{\vp^{\top}B_{n}\vp\vp^{\top}gg^{\top}\vp\vp^{\top}B_{n}^{\top}\vp}{\bb{v_{1}^{\top}g}^{2}(1+\eta_{n}\lambda_{1})^{2n}}\bigg|\bigg|\bigcap_{i \in [m]}\mathcal{E}^{(i)}} \notag \\
       %&\leq \frac{m^{2}}{\delta^{3}(1+\eta_{n}\lambda_{1})^{2n}}\E\bbb{\vp^{\top}B_{n}\vp\vp^{\top}gg^{\top}\vp\vp^{\top}B_{n}^{\top}\vp\bigg|\mathcal{E}} \notag \\
       &\leq \frac{m^{2}}{\delta^{3}\mathbb{P}(\mathcal{E})}\frac{\E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp}}}{(1+\eta_{n}\lambda_{1})^{2n}} \label{eq:en4_markov}
    }
    The last step uses Eq~\ref{eq:en4_exp}.
    Using Lemma B.3 from \cite{lunde2021bootstrapping}, we have
    \ba{\frac{\E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp}}}{(1+\eta_{n}\lambda_{1})^{2n}} &\leq d\exp\bb{-2\eta_{n}n\bb{\lambda_{1}-\lambda_{2}} + \eta_{n}^{2}n\bb{\lambda_{1}^{2}+\Mtwo^{2}}} + \frac{e\eta_{n}^{3}n\Mtwo^{4}\bb{1+2\log\bb{d}}}{2\bb{\lambda_{1}-\lambda_{2}} + \eta_{n} \bb{\lambda_{1}^{2}-\lambda_{2}^{2}-\Mtwo^{2}}} \label{eq:en4_expectation_bound}
    }
    Finally, we note that using Proposition 7 from \cite{lunde2021bootstrapping}, we have 
    \ba{
        \forall i \in [m], \Prob\bb{\mathcal{E}^{(i)}} \geq 1 - \frac{\delta}{m} \implies  \mathbb{P}\bb{\bb{\bigcap_{i \in [m]}\mathcal{E}^{(i)}}^{\complement}} \leq \sum_{i \in [m]}\Prob\bb{\mathcal{E}_i^{\complement}} \leq \sum_{i \in [m]}\frac{\delta}{m} = \delta \label{eq:good_event_union_bound}
    }
    The result follows by substituting \eqref{eq:en4_expectation_bound} in \eqref{eq:en4_markov} and then using \eqref{eq:en4_cs}, along with the union-bound provided in \eqref{eq:good_event_union_bound}.
\end{proof}

% \rd Does our Lemma B.3 apply in your subG setting? We need that.\bk
% \begin{lemma}\label{lemma:en4_fourth_moment_bound}
%     Let $E_{n,4} := \frac{\vp\vp^{\top}B_{n}\vp\vp^{\top}u_{0}}{\Abs{v_{1}^{\top}u_{0}}(1+\eta_{n}\lambda_{1})^{n}}$ be as defined in Lemma~\ref{lemma:oja_error_decomposition} for $u_{0} = g/\norm{g}_{2}$ with $g \sim \mathcal{N}(0, \id_d)$. Assume $Cn\eta_{n}^2\lambda_{1}\Tr\bb{\Sigma} \leq 1$ and $\eta_{n} L^2\sigma^{2}\leq \frac{1}{4}\min\left\{\frac{1}{\lambda_{1}}, \frac{1}{\Tr\bb{\Sigma}}, \frac{1}{\sqrt{\lambda_{1}\Tr\bb{\Sigma}}}\right\}$ for a sufficiently large universal constant $C > 0$. For any $\delta \in (0,1)$, define the event $\mathcal{E} := \left\{\Abs{v_{1}^{\top}g} \geq \delta\right\}$. Then, for any $k \in [d]$, \bas{\E\bbb{(e_{k}^{\top}E_{n,4})^{4}|\mathcal{E}} \leq \frac{C''}{\delta^{4}\Prob(\mathcal{E})}\bb{\norm{\vp^{\top}e_{k}}_{2}^{2}\bb{\frac{\gamma_{2}}{\gamma_{1}}}^{n} + \eta_{n}^{5}n^{2}\log^{2}(n)\frac{L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}}{\lambda_{1}-\lambda_{2}} }
%     }
%     where $C'' > 0$ is a universal constant,  $\gamma_2:=1+4\eta_{n}\lambda_2+150\eta_{n}^2\log(n)\lambda_1\Tr \bb{\Lambda_2}$ and $\gamma_1:=1+4\eta_{n}\lambda_1$.
% %    \rd[PS:] Can we also add $\E[\|E_{n,4}\|^4|\mathcal{E}]$?\bk
%  \end{lemma}
% \begin{proof}
%  We have
%     \bas{
%         \E\bbb{(e_{k}^{\top}E_{n,4})^{4}|\mathcal{E}} &= \E\bbb{\bb{\frac{e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}u_{0}}{\Abs{v_{1}^{\top}u_{0}}(1+\eta_{n}\lambda_{1})^{n}}}^{4}\bigg|\mathcal{E}} \\
%         &= \frac{1}{(1+\eta_{n}\lambda_1)^{4}}\E\bbb{\frac{(e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}g)^{4}}{(v_{1}^{\top}g)^{4}}\bigg|\mathcal{E}} \\
%         &\leq \frac{1}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}}\E\bbb{(e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}g)^{4}\bigg|\mathcal{E}} \\
%         &\leq \frac{1}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}\Prob(\mathcal{E})}\E\bbb{(e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}g)^{4}}
%     }
%     Let $M := \vp\vp^{\top}B_{n}\vp\vp^{\top}$. Consider the random variable $r_{n} := e_{k}^{\top}Mg$. Then, over the random of $g$, $r_{n}$ is a scalar gaussian random variable with mean zero and variance 
%     \bas{
%         \E\bbb{r_n^{2}} &= \E\bbb{(e_{k}^{\top}Mg)^{2}} \\
%         &= e_{k}^{\top}M\E\bbb{gg^{\top}}M^{\top}e_{k} \\
%         &= e_{k}^{\top}MM^{\top}e_{k} \\
%         &= e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}\vp\vp^{\top}B_{n}^{\top}\vp\vp^{\top}e_{k} \\
%         &= e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp\vp^{\top}e_{k}
%     }
%     Therefore, using standard properties of a scalar gaussian random variable 
%     \bas{
%          \E\bbb{r_{n}^{4}} &= 3(e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}B_{n}\vp\vp^{\top}e_{k})^{2}
%     }
%     Therefore, 
%     \ba{
%         \E\bbb{(e_{k}^{\top}E_{n,4})^{4}|\mathcal{E}} &\leq \frac{3}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}\Prob(\mathcal{E})}\E\bbb{(e_{k}^{\top}\vp\vp^{\top}B_{n}\vp\vp^{\top}B_{n}^{\top}\vp\vp^{\top}e_{k})^{2}} \notag \\
%         &= \frac{3}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}\Prob(\mathcal{E})}\E\bbb{\Tr\bb{\vp^{\top}B_{n}^{\top}\vp\vp^{\top}e_{k}e_{k}^{\top}\vp\vp^{\top}B_{n}\vp}^{2}} \notag \\
%         &= \frac{3}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}\Prob(\mathcal{E})}\E\bbb{\Tr\bb{\vp^{\top}B_{n}\vp\vp^{\top}e_{k}e_{k}^{\top}\vp\vp^{\top}B_{n}^{\top}\vp}^{2}} \label{eq:fourth_moment_conditional_bound}
%     }
%     where in the last line, we used the fact that $B_{n}$ and $B_{n}^{\top}$ are identically distributed. Let $\beta_{n} := \E\bbb{\Tr\bb{\vp^{\top}B_{n}UU^{\top}B_{n}^{\top}\vp}^{2}}$ and $\alpha_{n} := \E\bbb{\bb{v_{1}^{\top}B_{n}UU^{\top}B_{n}^{\top}v_{1}}^{2}}$ for $U := \vp\vp^{\top}e_{k}$. Using Lemma A.2.10 from \cite{kumar2024oja}, we have
%     \ba{
%         \beta_{n} & \leq \bb{1 + 4\eta_{n}\lambda_{2} + 100\eta_{n}^{2}\log\bb{n}L^4\sigma^{4}\lambda_{2}\Tr\bb{\Lambda_{2}}}\beta_{n-1} + 100\eta_{n}^{2}\log\bb{n}L^4\sigma^{4}\lambda_{1}\Tr\bb{\Lambda_{2}}\sqrt{\alpha_{n-1}\beta_{n-1}} \notag \\
%         & \;\;\;\;\; + 600\eta_{n}^{2}\log^{2}\bb{n}L^4\sigma^{4}\lambda_{1}^{2}\alpha_{n-1} \\
%         & \leq \bb{1 + 4\eta_{n}\lambda_{2} + 150\eta_{n}^{2}\log\bb{n}L^4\sigma^{4}\lambda_{1}\Tr\bb{\Sigma}}\beta_{n-1} + 650\eta_{n}^{2}\log^{2}\bb{n}L^4\sigma^{4}\lambda_{1}\Tr\bb{\Sigma}\alpha_{n-1} \label{eq:beta_n_recursion}
%     }
%     Now we bound $\alpha_{n}$ using Theorem~5.1 from~\cite{huang2022matrix}.  For $p>4$, for a universal constant $C > 0$, 
%     \bas{  \alpha_n &= \E\bbb{\bb{v_{1}^{T}B_{n}\vp e_{k}e_{k}^{\top}\vp^{T}B_{n}^{T}v_{1}}^{2}} = \E\bbb{\bb{v_{1}^{T}(B_{n}-\E[B_n])\vp e_{k}e_{k}^{\top}\vp^{T}(B_{n}-\E[B_n])^{T}v_{1}}^{2}} \\
%     &\leq \E\bbb{\normop{B_n-\E[B_n]}^4} \leq \normpq{B_n-\E[B_n]}_{p,4}^4\leq (\exp(Cpn\eta_{n}^2\lambda_{1}\Tr\bb{\Sigma})-1)^2(1+\eta_{n}\lambda_1)^{4n}d^{4/p}
%     }
% Taking $p=2(1+\log d)\geq 4$ and using the bound on $\eta_{n}$, we have for a universal constant $c > 0$,
% \bas{  
%  \alpha_n \leq c(\eta_{n}^2 n\lambda_{1}\Tr\bb{\Sigma}  )^2(1+\eta_{n}\lambda_1)^{4n}(1+2\log d)^4
% }
% Recall $\gamma_2:=1+4\eta_{n}\lambda_2+150\eta_{n}^2\log(n)\lambda_1\Tr \bb{\Lambda_2}$ and $\gamma_1:=1+4\eta_{n}\lambda_1$. Now Eq~\ref{eq:beta_n_recursion} gives, for a universal constant $C' > 0$:
% \bas{
%     \beta_{n} &\leq \gamma_{2}\beta_{n-1} + C'\eta_{n}^{6}n^{2}\log^{2}(n)L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}\gamma_{1}^{n}
% }
% Unrolling the recursion, 
% \bas{
%     \beta_n &\leq \gamma_{2}^{n}\beta_{0} + \bb{C'\eta_{n}^{6}n^{2}\log^{2}(n)L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}\gamma_{1}}\frac{\gamma_1^{n}}{\gamma_{1}-\gamma_{2}}\bb{1- \bb{\frac{\gamma_2}{\gamma_1}}^{n}} \\
%     &\leq \gamma_{2}^{n}\beta_{0} + \bb{C'\eta_{n}^{6}n^{2}\log^{2}(n)L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}\gamma_{1}}\frac{\gamma_1^{n}}{\eta_{n}(\lambda_1 - \lambda_2)}
% }
% Susbtituting in \eqref{eq:fourth_moment_conditional_bound}, 
% \bas{
%     \E\bbb{(e_{k}^{\top}E_{n,4})^{4}|\mathcal{E}} &\leq \frac{3}{\delta^{4}(1+\eta_{n}\lambda_1)^{4}\Prob(\mathcal{E})}\beta_{n} \\
%     &\leq \frac{C''}{\delta^{4}\Prob(\mathcal{E})}\bb{\beta_{0}\bb{\frac{\gamma_{2}}{\gamma_{1}}}^{n} + \eta_{n}^{5}n^{2}\log^{2}(n)\frac{L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}}{\lambda_{1}-\lambda_{2}} }
% }
% The proof follows by noting that $\beta_{0} = e_{k}^{\top}\vp\vp^{\top} e_{k}$.
% \end{proof}


% \begin{lemma}\label{lemma:en4_concentration}
%     Let $\Efour{n} := \frac{\vp\vp^{\top}B_{n}\vp\vp^{\top}u_{0}}{\Abs{v_{1}^{\top}u_{0}}(1+\eta_{n}\lambda_{1})^{n}}$ be as defined in Lemma~\ref{lemma:oja_error_decomposition} for $u_{0} = g/\norm{g}_{2}$ with $g \sim \mathcal{N}(0, \id_d)$. Fix $\delta, \delta' \in (0,1)$. Assume $Cn\eta_{n}^2\lambda_{1}\Tr\bb{\Sigma} \leq 1$ and $\eta_{n} L^2\sigma^{2}\leq \frac{1}{4}\min\left\{\frac{1}{\lambda_{1}}, \frac{1}{\Tr\bb{\Sigma}}, \frac{1}{\sqrt{\lambda_{1}\Tr\bb{\Sigma}}}\right\}$ for a sufficiently large universal constant $C > 0$. Define the event $\mathcal{E} := \left\{\Abs{v_{1}^{\top}g} \geq \delta\right\}$. Let $\left\{\Efour{n}^{(i)}\right\}_{i \in [m]}$ be $m$ independent and identically distributed instances of $\Efour{n}$. Then, conditioned on $\mathcal{E}$, for any $k \in [d]$, with probability at least $1-\delta'$,  \bas{
%     \frac{1}{m}\sum_{i \in [m]}(\bb{e_{k}^{\top}\Efour{n}^{(i)}}^{2} - \E\bbb{(e_{k}^{\top}\Efour{n})^{2}|\mathcal{E}}) \leq \frac{1}{\sqrt{m\delta'}}\sqrt{\frac{C''}{\delta^{4}\Prob(\mathcal{E})}\bb{\norm{\vp^{\top}e_{k}}_{2}^{2}\bb{\frac{\gamma_{2}}{\gamma_{1}}}^{n} + \eta_{n}^{5}n^{2}\log^{2}(n)\frac{L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}}{\lambda_{1}-\lambda_{2}} }}
%     }
%     where $C'' > 0$ is a universal constant,  $\gamma_2:=1+4\eta_{n}\lambda_2+150\eta_{n}^2\log(n)\lambda_1\Tr \bb{\Lambda_2}$ and $\gamma_1:=1+4\eta_{n}\lambda_1$.
%  \end{lemma}
%  \begin{proof}
%      We have using Chebyshev's inequality, conditioned on $\mathcal{E}$, with probability at least $1-\delta'$,  
%      \bas{
%         \frac{1}{m}\sum_{i \in [m]}\bb{e_{k}^{\top}\Efour{n}^{(i)}}^{2} &\leq \frac{\sqrt{\E\bbb{(e_{k}^{\top}E_{n,4})^{4}|\mathcal{E}}}}{\sqrt{m\delta'}} \\
%         &\leq \frac{1}{\sqrt{m\delta'}}\sqrt{\frac{C''}{\delta^{4}\Prob(\mathcal{E})}\bb{\norm{\vp^{\top}e_{k}}_{2}^{2}\bb{\frac{\gamma_{2}}{\gamma_{1}}}^{n} + \eta_{n}^{5}n^{2}\log^{2}(n)\frac{L^{4}\sigma^{4}\lambda_{1}^{3}\Tr\bb{\Sigma}^{3}}{\lambda_{1}-\lambda_{2}} }}
%      }
%      where the last line follows from Lemma~\ref{lemma:en4_fourth_moment_bound}.
%  \end{proof}



% \begin{lemma}[Lemma A.2.10]
%     \label{lemma:fourth_moment_beta}
%     For all $t > 0$, under subgaussianity (Definition \ref{definition:subgaussian}), let $U \in \mathbb{R}^{d \times m}$, $\alpha_{n} := \E{\bb{v_{1}^{T}B_{n}UU^{T}B_{n}^{T}v_{1}}^{2}}$, $\beta_{n} := \E{\Tr\bb{\vp^{T}B_{n}UU^{T}B_{n}^{T}\vp}^{2}}$ and $\eta_{n} L^2\sigma^{2}\leq \frac{1}{4}\min\left\{\frac{1}{\lambda_{1}}, \frac{1}{\Tr\bb{\Lambda_{2}}}, \frac{1}{\sqrt{\lambda_{1}\Tr\bb{\Lambda_{2}}}}\right\}$ then
%     \bas{
%         & \beta_{n} \leq \bb{1 + 4\eta_{n}\lambda_{2} + 100\eta_{n}^{2}\log\bb{n}L^4\sigma^{4}\lambda_{2}\Tr\bb{\Lambda_{2}}}\beta_{n-1} + 100\eta_{n}^{2}\log\bb{n}L^2\sigma^{2}\lambda_{1}\Tr\bb{\Lambda_{2}}\sqrt{\alpha_{n-1}\beta_{n-1}} \\
%         & \;\;\;\;\; + 600\eta_{n}^{2}\log^{2}\bb{n}L^4\sigma^{4}\lambda_{1}^{2}\alpha_{n-1}
%     }
%     where $B_{n}$ is defined in Eq~\ref{definition:Bn}.
% \end{lemma}

% \rd I am using $\sum_i\sigma_i^2\leq Cn\eta_{n}^2\mathcal{V}$.
% \bk
% Note that,

% \ba{\label{eq:beta}
%          \beta_{n} &\leq \bb{1 + 4\eta_{n}\lambda_{2} + C\eta_{n}^{2}\log\bb{n}\lambda_{1}\Tr\bb{\Lambda_{2}}}\beta_{n-1} 
%        + C'\eta_{n}^{2}\log^{2}\bb{n}\lambda_{1}^{2}\alpha_{n-1}
%     }

%     Now we bound $\alpha_{n}$ using Theorem~5.1 from~\cite{huang2022matrix}.  For $p>4$,
%     \bas{  \alpha_n&=\E{\bb{v_{1}^{T}B_{n}\vp\vp^{T}B_{n}^{T}v_{1}}^{2}}=\E{\bb{v_{1}^{T}(B_{n}-\E[B_n])\vp\vp^{T}(B_{n}-\E[B_n])^{T}v_{1}}^{2}}\\
%     &\leq \E\|B_n-\E[B_n]\|^4\leq \|B_n-\E[B_n]\|_{p,4}^4\leq (\exp(Cpn\eta_{n}^2\mathcal{V})-1)^2(1+\eta_{n}\lambda_1)^{4n}d^{4/p}
%     }
%     Taking $p=2(1+\log d)\geq 4$,
% we see,
%  \bas{  \alpha_n\leq c(\eta_{n}^2 n \mathcal{V} )^2(1+\eta_{n}\lambda_1)^{4n}(1+2\log d)^4
%     }
% Let $\gamma_2=1+4\eta_{n}\lambda_2+C\eta_{n}^2\log(n)\lambda_1\Tr \bb{\Lambda_2}$ and $\gamma_1=(1+4\eta_{n}\lambda_1)$.
%     Now Eq~\ref{eq:beta} gives:
%     \bas{
%     \beta_n %d\bb{1 + 4\eta_{n}\lambda_{2} + C\eta_{n}^{2}\log\bb{n}\lambda_{1}\Tr\bb{\Lambda_{2}}}^n+C'(\log n)^2 n^2\eta_{n}^6\mathcal{V}^4\sum_k \bb{1 + 4\eta_{n}\lambda_{2} + C\eta_{n}^{2}\log\bb{n}\lambda_{1}^2\Tr\bb{\Lambda_{2}}}^{k}\bb{1 + 4\eta_{n}\lambda_{2} + C\eta_{n}^{2}\log\bb{n}\lambda_{1}\Tr\bb{\Lambda_{2}}}^{n-k}\\
%     &\leq \gamma_1^n\left(d(\gamma_2/\gamma_1)^n+C'(\log n)^2 n^2\eta_{n}^6\sum_k (\gamma_2/\gamma_1)^{k}\right)\\
%     &\leq \gamma_1^n\left(d(\gamma_2/\gamma_1)^n+C'\frac{(\log n)^2 n^2\eta_{n}^6}{4\eta_{n}(\lambda_1-\lambda_2)+\rd Cn\eta_{n}^2\bk}\right)\\
%     }


