\subsubsection{$\Etwo{n}$ tail bound}\label{appendix:Etwo_bound}

% \subsubsection{Asymptotic variance}

% In this section, we show concentration for our estimator by controlling the tails of the terms in the decomposition provided in Lemma~\ref{lemma:oja_error_decomposition}.

% \paragraph{ ($\Etwo{n}$)} :
% \rd PS: How does this relate to $\Etwo{n}$?\bk
We start by providing a tail bound on higher order terms in the Hoeffding decomposition of $B_{n}-\E\bbb{B_n}$, which may be of independent interest. Let $\mathcal{S}_{n,k}:=\{\{i_1,\dots, i_k\}:1\leq i_1<\dots<i_k\leq n\}$.
    Consider a general product of $n$ matrices, where all but $k$ of the matrices are constant, and $k$ indexed by the subset $S$ are mean zero independent random matrices. With slight abuse of notation, let $M_{S,i}$ denote a constant matrix $M_i$ with $\|M_i\|=:m_i$ when $i\not\in S$ and $W_i$ when $i\in S$, $EW_i=0$, $W_i,i=1,\dots,n$ are mutually independent.
    \begin{align}
    T_{n,k}:=\sum_{S\in \mathcal{S}_{n,k}}\prod_{i=1}^n M_{S,n+1-i}
    \end{align} 
    % \begin{theorem}
    % \label{theorem:higher-order}
    % Let $m_{p,k}$ be an a.s. upper bound on $\max_{S\in \mathcal{S}_{n,k}} \|\prod_i M_{S,i}\|_p$.  We have  $\norm{{T}_{n,k}}_{p,q}\leq C  m_{p,k}(np)^{k/2}$.
    % \end{theorem}
    % \begin{proof}
    % We will prove this by induction on $k$.
    % Note that for $k=1$,
    % \bas{
    % %T_{n,1}=\eta \sum_i (I+\eta\Sigma)^{i-1}(A_i-\Sigma)(I+\eta\Sigma)^{n-i}
    % T_{n,1}=\sum_{j=1}^n \left(\prod_{i=1}^{j-1} M_{i}\right)W_j\left(\prod_{i=j}^{n} M_{i}\right)
    % }
    % Using ~\cite{huang2022matrix} Prop~4.3,
    % \bas{
    % f_{n,1}&=\vertiii{T_{n,1}}_{p,q}^2\leq f_{n-1,1}+C_p m_{p,q,1}^2\leq C np m_{p,q,1}^2
    % % f_{n,1}&=\vertiii{T_{n,1}}_{p,q}^2\leq f_{n-1,1}+C_p \eta^2 \vertiii{(I+\eta\Sigma)^{2(n-1)}(A_i-\Sigma)}_{p,q}^2\\
    % % &\leq f_{n-1,1}+C_p \eta^2 \prod_{i=1}^{n-1} (1+\eta_i \Sigma_i)^2M^2\leq C np\underbrace{\eta^2(1+\eta \Sigma)^{2(n-1)}M^2}_{m^2}
    % }
    % which shows that the base step is correct.
    % Now for the general $T_{n,k}$ we will break it up and write it as a sum of martingale differences and then apply Huang et al's prop 4.3.
    
    % Now define $\Delta_{n,k}$ as all size $k$ subsets containing $n$, i.e.  $\Delta_{n,k}=\mathcal{S}_{n,k}\setminus \mathcal{S}_{n-1,k}$.
    % \bas{
    % T_{n,k}=\sum_{S\in \mathcal{S}_{n-1,k}} \prod_{i=1}^n M_{S,i}+ \sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}
    % }
    % Note that conditioned on the first term, the expectation of the second term is zero, which is why this is essentially a sum of martingale differences. So now we apply prop 4.3 to get:
    % \bas{
    % \vertiii{T_{n,k}}_{p,q}^2\leq \vertiii{\sum_{S\in \mathcal{S}_{n-1,k}} \prod_{i=1}^n M_{S,i}}_{p,q}^2+ C_p \vertiii{\sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}}_{p,q}^2
    % }
    % Note that conditioned on $W_n$, the second term is simply
    % \bas{
    % \vertiii{\sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}}_{p,q}^2 &= \bb{\E \left\|\sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}\right\|_p^q}^{2/q}\\
    % &=\bb{\E_{W_n} \E\bbb{\left.\left\|\sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}\right\|_p^q\right|W_n}}^{2/q}\\
    % &\leq Cp((n-1)p)^{k-1}m_{p,k}^2
    % }
    % \rd [Purna:] I think this is  $m_{p,k}$, but need to check. \bk
    
    % Now the last step holds by the inductive hypothesis since conditioned on $W_n$, $\sum_{S\in \Delta_{n,k}}\prod_{i=1}^n M_{S,i}$ is of the same form as $T_{n,k-1}$.
    
    % Thus,
    % \bas{
    % f_{n,k}&:=\vertiii{T_{n,k}}_{p,q}^2\leq f_{n-1,k}+Cp((n-1)p)^{k-1}m_{p,k}^2\\
    % &\leq Cnp((n-1)p)^{k-1}m_{p,k}^2\leq C(np)^{k}m_{p,k}^2
    % }
    % This completes the proof.
    % \end{proof}
    Let $T_{n,k}$ be a scaled version of the $k^{th}$ term in the Hoeffding projection of the matrix product 
    $B_n:=\prod_{i=1}^n (I+\eta_n A_i)$. Let $W_i=A_i-\Sigma$. We want a tail bound for $\sum_{k \ge 2} T_{n,k}$.

    \begin{lemma}
    \label{lemma:higher-order-helper}
    For $S\in\mathcal{S}_{n,k}$, denote a function $M_{S,i}:=\eta_{n}(A_i-\Sigma)$ when $i\in S$ and $I+\eta_{n}\Sigma$ when $i\not\in S$. Suppose $q \ge 2$ and $\mathcal{M}_q$ are such that $\E\bbb{\norm{A_i-\Sigma}^q}^{1/q} \le \mathcal{M}_q$. Then, for any $1 \le j \le n$ and any $p \ge q$,
    \bas{
    \vertiii{\sum_{k \ge j} T_{n,j}}_{p,q} \leq 2 d^{1/p} (1+\eta_{n} \lambda_1)^n \bb{\frac{\eta_{n} \mathcal{M}_q \sqrt{np}}{1+\eta_{n} \lambda_1}}^j,
    }
    as long as $\frac{2\eta_{n} \mathcal{M}_q \sqrt{np}}{1+\eta_{n} \lambda_1} < 1$.
    \end{lemma}

    \begin{proof}
    We start by deriving a recurrence relation for $T_{n,k}$ as follows:
    \bas{
    T_{n,k} &= \sum_{S \in \mathcal{S}_{n,k}} \prod_{i=1}^n M_{S,n+1-i} \\
    &= \sum_{S \in \mathcal{S}_{n,k}, n \notin S} \prod_{i=1}^n M_{S,n+1-i} +  \sum_{S \in \mathcal{S}_{n,k}, n \in S} \prod_{i=1}^n M_{S,n+1-i} \\
    &= \sum_{S \in \mathcal{S}_{n-1,k}}  (I + \eta_{n} \Sigma) \prod_{i=2}^{n} M_{S,n+1-i} + \sum_{S \in \mathcal{S}_{n-1,k-1}} \eta_{n} (A_n - \Sigma) \prod_{i=2}^{n} M_{S,n+1-i} M_{S,n+1-i} \\
    &=  (I + \eta_{n} \Sigma) \bb{\sum_{S \in \mathcal{S}_{n-1,k}} \prod_{i=1}^{n-1} M_{S,n-i}} + \eta_{n} (A_n - \Sigma) \bb{\sum_{S \in \mathcal{S}_{n-1,k-1}} \prod_{i=1}^{n-1} M_{S,n-i}} \\
    &= (I+\eta_{n} \Sigma)T_{n-1,k} + \eta_{n}(A_n - \Sigma)T_{n-1,k-1}.
    }
    Next, we apply Proposition 4.3. of~\cite{huang2022matrix} to bound $\vertiii{T_{n,k}}_{p,q}$. To apply the proposition, we require $\E\bbb{\eta_{n}(A_n - \Sigma)T_{n-1,k-1} | (I+\eta_{n} \Sigma)T_{n-1,k}} = 0$. Indeed, by independence of $A_1, A_2, \dots, A_n$,
    \bas{
    \E\bbb{\eta_{n}(A_n - \Sigma)T_{n-1,k-1} | (I+\eta_{n} \Sigma)T_{n-1,k}} = \E\bbb{\eta_{n}(A_n - \Sigma)} \E\bbb{T_{n-1,k-1} | (I+\eta_{n} \Sigma)T_{n-1,k}} = 0.
    }
    Therefore, the proposition implies that
    \bas{
    \vertiii{T_{n,k}}_{p,q}^2 \le \vertiii{(I+\eta_{n} \Sigma)T_{n-1,k}}_{p,q}^2 + (p-1) \vertiii{\eta_{n}(A_n - \Sigma)T_{n-1,k-1}}_{p,q}^2.
    }
    From Equation~4.1. and Equation~5.3. of~~\cite{huang2022matrix},
    \bas{
    \vertiii{(I+\eta_{n} \Sigma)T_{n-1,k}}_{p,q} &\le \norm{I+\eta_{n} \Sigma}_{\text{op}} \vertiii{T_{n-1,k}}_{p,q}, \text{ and} \\
    \vertiii{\eta_{n}(A_n - \Sigma)T_{n-1,k-1}}_{p,q} &\le \eta_{n} \E\bbb{\norm{A_n-\Sigma}^q}^{1/q} \vertiii{T_{n-1,k-1}}_{p,q}.
    }
    Plugging these bounds into the recurrence yields
    \bas{
    \vertiii{T_{n,k}}_{p,q}^2 \le (1+\eta_{n} \lambda_1)^2 \vertiii{T_{n,k-1}}_{p,q}^2 + \eta_{n}^2 \mathcal{M}_q^2 (p-1) \E\bbb{\norm{A_n-\Sigma}^q}^{2/q} \vertiii{T_{n-1,k-1}}_{p,q}^2.
    }
    Letting $f_{n,k} \defeq \vertiii{T_{n,k}}_{p,q}^2$, we have the following recurrence for all $n \ge k \ge 1$:
    \bas{
    f_{n,k} \le (1+\eta_{n} \lambda_1)^2 f_{n-1,k} + \eta_{n}^2 \mathcal{M}_q^2 (p-1) f_{n-1,k-1}.
    }
    Defining $a_{n,k} \defeq \frac{f_{n,k}}{(1+\eta_{n} \lambda_1)^{2(n-k)} (\eta_{n}^2 \mathcal{M}_q^2 (p-1))^k}$, we recover an inequality resembling Pascal's identity:
    \bas{
    a_{n,k} \le a_{n-1,k} + a_{n-1,k-1}.
    }
    Moreover, $a_{n,k} = 0$ for all $n < k$ and $a_{n,0} = (1+\eta_{n} \lambda_1)^{-2n} \vertiii{(I+\eta_{n} \Sigma)^{n}}_{p,q}^2 \le d^{2/p}$. Inducting on $n$ and $k$ shows
    \bas{
    a_{n,k} \le d^{2/p} \binom{n}{k}.
    }
    Translating this back to the bound on the norm of $T_{n,k}$, we conclude
    \bas{
    \vertiii{T_{n,k}}_{p,q} \le \sqrt{(1+\eta_{n} \lambda_1)^{2(n-k)} \bb{\eta_{n}^2 \mathcal{M}_q^2 (p-1)}^k d^{2/p} \binom{n}{k}} \le d^{1/p} (1+\eta_{n} \lambda_1)^{n-k} \bb{\eta_{n} \mathcal{M}_q \sqrt{np}}^k 
    }
    Since norms are sub-additive and $\frac{\eta_{n} \mathcal{M}_q \sqrt{np}}{1+\eta_{n} \lambda_1} < \frac{1}{2}$,
    \bas{
    \vertiii{\sum_{k \ge j} T_{n,k}}_{p,q} &\le \sum_{k=j}^n d^{1/p} (1+\eta_{n} \lambda_1)^{n-k} \bb{\eta_{n} \mathcal{M}_q \sqrt{np}}^k \\
    &= d^{1/p} (1+\eta_{n} \lambda_1)^n \sum_{k=j}^n \bb{\frac{\eta_{n} \mathcal{M}_q \sqrt{np}}{1+\eta_{n} \lambda_1}}^k \\
    &\le 2 d^{1/p} (1+\eta_{n} \lambda_1)^n \bb{\frac{\eta_{n} \mathcal{M}_q \sqrt{np}}{1+\eta_{n} \lambda_1}}^j.
    }
    \end{proof}

    % \begin{lemma}
    % \label{lemma:higher-order-opnorm}
    % Under the setting of Lemma~\ref{lemma:higher-order-helper},
    % \bas{
    % \E\bbb{\norm{\sum_{k \ge j} T_{n,k}}^2} \le 6(1+\eta \lambda)^{2n} \bb{\frac{2\eta \mathcal{M}_2 \sqrt{n \log d}}{1+\eta \lambda_1}}^{2j}.
    % }
    % \end{lemma}
    % \begin{proof}
    % Since the Schatten $p$-norm is at least the operator norm, for any $p \ge 2$ we have
    % \bas{
    % \E\bbb{\norm{\sum_{k \ge j} T_{n,k}}^2} \le \E\bbb{\norm{\sum_{k \ge j} T_{n,k}}_p^2} = \vertiii{\sum_{k \ge j} T_{n,k}}_{p,2}^2 \le 4d^{2/p} (1+\eta \lambda_1)^{2n} \bb{\frac{\eta \mathcal{M}_2 \sqrt{np}}{1+\eta \lambda_1}}^{2j}
    % }
    % The lemma now follows from Lemma~\ref{lemma:higher-order-helper} by setting $p = 4 \log d$.
    % \end{proof}
    
    \begin{lemma}
    \label{lemma:higher-order-norm}
        For $S\in\mathcal{S}_{n,k}$, denote a function $M_{S,i}:=\eta_n(A_i-\Sigma)$ when $i\in S$ and $I+\eta_n\Sigma$ when $i\not\in S$. Then, for any $1 \le j \le n$, and $2 \le q \le 4 \log d$,
    \bas{
    P\bb{\norm{\sum_{k \ge j} T_{n,k}} \ge 
    \frac{3 (1+\eta_n \lambda_1)^n \bb{\eta_n \mathcal{M}_q \sqrt{4n \log d}}^j}{\delta^{\frac{1}{4\log d}}}} \le \delta,
    }
    as long as $4 \eta_n \mathcal{M}_q \sqrt{n \log d} < 1$.
    \end{lemma}
    \begin{proof}

    Let $p = 4 \log d$; note that the assumption $\frac{2 \eta_n \mathcal{M}_q \sqrt{np}}{1+\eta_n \lambda_1} < 1$ holds. By Markov's inequality, Equation~4.2. of~\cite{huang2022matrix}, and Lemma~\ref{lemma:higher-order-helper},
    \bas{
    P\bb{\norm{\sum_{k \ge j} T_{n,k}} \ge (1+\eta_n \lambda_1)^n t} &\le \inf_{p' \ge 2} \bb{(1+\eta_n \lambda_1)^n t}^{-p'} \E\bbb{\norm{\sum_{k \ge j} T_{n,k}}^{p'}} \\
    &\le \inf_{p' \ge 2} \bb{(1+\eta_n \lambda_1)^n t}^{-p'} \E\bbb{\vertiii{\sum_{k \ge j}  T_{n,k}}_{p',q}^{p'}} \\
    &\le \bb{\frac{2 d^{1/p} \bb{\frac{\eta_n \mathcal{M}_q \sqrt{np}}{1+\eta_n \lambda_1}}^j}{t}}^p \le \bb{\frac{3 \bb{\eta_n \mathcal{M}_q \sqrt{4n \log d}}^j}{t}}^{4\log d}.
    }
    for all $t > 0$. The lemma follows by setting $t = 3 \bb{\eta_n \mathcal{M}_q \sqrt{4n \log d}}^j{\delta^{\frac{-1}{4\log d}}}$.
    % (\rd It is possible to pick $t = (1+\epsilon) \times \text{numerator}$ for $\epsilon$ such that $(1-\epsilon)^{\Theta(\log d)}$ is sufficiently small \bk)
    % \bas{
    % P\bb{\|T_{n,k}\|\geq m_{p,k} t} &\leq \inf_{p\geq 2}(m_{p,k}t)^{-p} \E\bbb{\|T_{n,k}\|^p} \leq \inf_{p\geq 2}(m_{p,k}t)^{-p} \E\bbb{\|T_{n,k}\|^p} \leq \frac{\vertiii{T_{n,k}}_{p,p}^p}{(m_{p,k})^pt^p}\leq C\frac{(np)^{pk/2}  }{t^p}\\
    % }
    % Set $t^p=n^2(np)^{pk/2}$, i.e. $t=n^{2/p}(np)^{k/2}$ to see that
    % \bas{
    % P\bb{\max_k\|T_{n,k}\|\geq m_{p,k}n^{2/p}(np)^{k/2p}}\leq 1/n
    % }
    
    
    % Thus in this setting,
    % \bas{
    % m_{p,k}=\eta^k(1+\eta\lambda_1)^{n-k}M^kd^{1/p}
    % }
    
    % %For simplicity, I will assume all $\eta_i=\eta$.
    
    
    
    % First see that, for $p=O(\log \max(n,d))$, 
    % \bas{
    % \sum_{k\geq 2} \eta^k (n^2)^{1/p}(np)^{k/2}m_{p,k}&\leq n^{2/p} d^{1/p}(1+\eta\lambda_1)^{n}\sum_{k\geq 2}\bb{\frac{\eta M\sqrt{np}}{1+\eta\lambda_1}}^k \\
    % &\leq (1+\eta\lambda_1)^{n}O\bb{\frac{\eta^2M^2n\log(\max(n,d))}{(1+\eta\lambda_1)^2}},
    % %O(\eta^2 n \log (\max(n,d)))\leq 
    % } as long as $\frac{\eta^2M^2n\log(\max(n,d))}{(1+\eta\lambda_1)^2}< 1$, which is a reasonable assumption since $\eta=O(\log n/n)$.
    % This completes the proof.
    \end{proof}
    % \begin{lemma}
    % \label{lemma:error2-bound}
    % Let $\Eone{n}$ and $\Etwo{n}$ be defined as in Lemma~\ref{lemma:oja_error_decomposition}. Then,
    % \bas{
    % P\bb{\norm{\Etwo{n}} \ge \frac{12 \eta^2 \mathcal{M}_2^2 n \log d} {\delta^{\frac{1}{4\log d}}}} \le \delta \qquad \text{and} \qquad P\bb{\norm{\Eone{n} + \Etwo{n}} \ge \frac{6 \eta \mathcal{M}_2 \sqrt{n\log d}}{\delta^{\frac{1}{4\log d}}}} \le \delta.
    % }
    % \end{lemma}

    % \begin{proof}
    % By Lemma~\ref{lemma:higher-order-norm}, with probability at least $1-\delta$,
    % \bas{
    % \norm{\sum_{k \ge 2} T_{n,k}} <  \frac{12 (1+\eta \lambda_1)^n \eta^2 \mathcal{M}_2^2 n \log d} {\delta^{\frac{1}{4\log d}}}.
    % }
    % Under this event,
    % \bas{
    % \norm{\Etwo{n}} &= \norm{\frac{\vp \vp^\top \sum_{k \ge 2} T_{n,k} v_1 \sign(v_{1}^{\top}u_{0})}{(1+\eta \lambda_1)^n}} \le \frac{\norm{\vp \vp^\top} \norm{\sum_{k \ge 2} T_{n,k}} \norm{v_1}}{(1+\eta \lambda_1)^n} < \frac{12 \eta^2 \mathcal{M}_2^2 n \log d} {\delta^{\frac{1}{4\log d}}}
    % }
    % Similarly,  with probability at least $1-\delta$,
    % \bas{
    % \norm{\sum_{k \ge 1} T_{n,k}} < \frac{6(1+\eta \lambda_1)^n \eta \mathcal{M}_2 \sqrt{n\log d}}{\delta^{\frac{1}{4\log d}}}.
    % }
    % Under this event,
    % \bas{
    % \norm{\Eone{n} + \Etwo{n}} \le \frac{\norm{\vp \vp^\top} \norm{\sum_{k \ge 1} T_{n,k}} \norm{v_1}}{(1+\eta \lambda_1)^n} < \frac{6 \eta \mathcal{M}_2 \sqrt{n\log d}}{\delta^{\frac{1}{4\log d}}}.
    % }
    % \end{proof}

\begin{lemma}\label{lemma:en2_norm}
Let $\Etwo{n}$ be as defined in Lemma~\ref{lemma:oja_error_decomposition} with $u_0 = g/\norm{g}_{2}$. Then, for any $\delta \in (0,1)$,
\bas{
\Prob\bb{\norm{\Etwo{n}} \le \frac{12\eta_{n}^2 \mathcal{M}_2^2 n \log d} {\sqrt{\delta}}} \geq 1-\delta.
}
\end{lemma}

\begin{proof}
By Lemma~\ref{lemma:higher-order-norm}, with probability at least $1-\delta$,
\bas{
\norm{\sum_{k\geq 2}T_{n,k}} \le \frac{3 (1+\eta_{n} \lambda_1)^n \bb{\eta_{n} \mathcal{M}_2 \sqrt{4n \log d}}^2}{\delta^{\frac{1}{4\log d}}} < \frac{12(1+\eta_{n} \lambda_1)^n \eta_{n}^2 \mathcal{M}_2^2 n \log d} {\sqrt{\delta}}.
}
Conditioned on this event,
\bas{
\norm{\Etwo{n}} &= \frac{\norm{\vp\vp^{\top}(\sum_{k\geq 2}T_{n,k})v_{1}\sign(v_{1}^{\top}u_{0})}}{(1+\eta_{n}\lambda_{1})^{n}} \le \frac{\norm{\vp \vp^{\top}} \norm{\sum_{k\geq 2}T_{n,k}}\norm{v_{1}}}{(1+\eta_{n}\lambda_{1})^{n}} \\
&\le \frac{\norm{\sum_{k\geq 2}T_{n,k}}}{(1+\eta_{n}\lambda_{1})^{n}} \le \frac{12\eta_{n}^2 \mathcal{M}_2^2 n \log d} {\sqrt{\delta}}.
}
\end{proof}

\begin{lemma}\label{lemma:en2_tail_bound} Let $\Etwo{n}$ 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 $\left\{\Etwo{n}^{(i)}\right\}_{i \in [m]}$ and $\left\{g^{(i)}\right\}_{i \in [m]}$ be $m$ $\iid$ instances of $\Etwo{n}$ and $g$ respectively, and let $\delta \in (0,1)$. Then, with probability at least $1-\delta$,
\bas{
    \frac{\sum_{i \in [m]}\bb{e_{k}^{\top}\Etwo{n}^{(i)}}^{2}}{m} \le \frac{144b_k^2 \eta_{n}^4 \mathcal{M}_2^4 n^2 \log^2 d}{\delta},
}
for all $k \in [d]$, where $b_k := \norm{\vp^{\top}e_k}_{2}$.
\end{lemma}

\begin{proof}
We have
\bas{
\Abs{e_k^\top \Etwo{n}} &= \frac{\Abs{e_k^\top \vp\vp^{\top}(\sum_{k\geq 2}T_{n,k})v_{1}\sign(v_{1}^{\top}u_{0})}}{(1+\eta_{n}\lambda_{1})^{n}} = \frac{\Abs{e_k^\top \vp\vp^{\top} \vp \vp^{\top} (\sum_{k\geq 2}T_{n,k})v_{1}\sign(v_{1}^{\top}u_{0})}}{(1+\eta_{n}\lambda_{1})^{n}} \\
&= \Abs{e_k^{\top} \vp^{\top} \vp \Etwo{n}} \le \norm{e_k^\top \vp} \norm{\Etwo{n}} \le \frac{b_k \norm{\sum_{k\geq 2}T_{n,k}}}{(1+\eta_{n}\lambda_{1})^{n}}.
}
By Lemma~\ref{lemma:en2_norm}, for each $i \in [m]$, with probability at least $1-\frac{\delta}{m}$,
\bas{
\Abs{e_k^\top \Etwo{n}^{(i)}} \le \frac{12b_k \eta_{n}^2 \mathcal{M}_2^2 n \log d} {\sqrt{\delta/m}}.
}
By a union bound, the above holds for all $i \in [m]$ with probability at least $1-\delta$. Under this event,
\bas{
\frac{\sum_{i \in [m]}\bb{e_{k}^{\top}\Etwo{n}^{(i)}}^{2}}{m} \le \frac{\sum_{i \in [m]} \bb{\frac{12b_k \eta_{n}^2 \mathcal{M}_2^2 n \log d} {\sqrt{\delta/m}}}^2}{m} = \frac{144b_k^2 \eta_{n}^4 \mathcal{M}_2^4 n^2 \log^2 d}{\delta}.
}
\end{proof}
%    \rd Purna: Note that here, the trivial way to do a time uniform bound on $n$ which adds a extra $\log n$ factor works out, because it is a lower order term than the leading term. So as long as we are careful about the leading term, possibly by using the techniques of Shubanshu we should be okay.\blk

