all the $Q$-graphs that have at least one single vertical edge.

Category 3 (denoted by $Q_3$) contains all other $Q$-graphs.

The almost sure convergence of the ESD of $A_T$ will result from the following two assertions: $$\begin{aligned}
{\mathbb{E}}(m_k(A_T))&=\frac 1p\sum_{{\bf i}=1}^T\sum_{{\bf j}=1}^p\frac{1}{T^{2k}}{\mathbb{E}}[\varepsilon_{j_1\,i_1}\varepsilon_{j_1\,i_2}\varepsilon_{j_2\,s+i_2}\varepsilon_{j_2\,s+i_3}\varepsilon_{j_3\,i_3}\varepsilon_{j_3\,i_4}\varepsilon_{j_4\,s+i_4}\varepsilon_{j_4\,s+i_5}\nonumber\\
&\quad\quad\quad\quad\quad\quad\cdots \varepsilon_{j_{2k-1}\,i_{2k-1}}\varepsilon_{j_{2k-1}\,i_{2k}}\varepsilon_{j_{2k}\,s+i_{2k}}\varepsilon_{j_{2k}\,s+i_1}]\label{etrak}\\
&=\frac{1}{pT^{2k}}\sum_{{\bf i},{\bf j}}{\mathbb{E}}\left(E_{Q({\bf i},{\bf j})}\right)\nonumber\\
&=\sum_{i=0}^{k-1}\frac 1k \begin{pmatrix}
                               2k \\
                               i \\
                             \end{pmatrix}\begin{pmatrix}
                                            k \\
                                            i+1 \\
                                          \end{pmatrix}y_T^{2k-1-i}+o(1)~,\label{e}\end{aligned}$$ and $$\begin{aligned}
\label{va}
&\quad~{\mathop{\text{Var}}}(m_k(A_T))\nonumber\\
&=\frac{1}{p^2T^{4k}}\sum_{{\bf i_1},{\bf j_1},{\bf i_2},{\bf j_2}}\left[{\mathbb{E}}\left(E_{Q_1({\bf i_1},{\bf j_1})}E_{Q_2({\bf i_2},{\bf j_2})}\right)-{\mathbb{E}}\left(E_{Q_1({\bf  i_1},{\bf  j_1})}\right){\mathbb{E}}\left(E_{Q_2({\bf  i_2},{\bf  j_2})}\right)\right]\nonumber\\
&=O(p^{-2})~.\end{aligned}$$

[**The proof of .**]{} Since ${\mathbb{E}}\varepsilon_{ij}=0$, the only non-vanishing terms in are those for which each edge in the $Q$-graph occurs at least twice. So the contribution of Category 2 is zero.

Next, we only consider those $Q$-graphs that fall in Category 1 and 3. Denote $b_l$ the degree associated to the $I$-vertex $i_l$ $(1\leq l\leq t)$ in its corresponding $M$-pillar, then we have $b_1+\cdots+b_t=4k$, which is the total number of edges. On the other hand, since we glue the $I$-vertex $i_l$ and $i_l+s$ in the definition of $M$-pillar, each degree $b_l$ should be no less than $4$; otherwise, there will be some single vertical edges in the $Q$-graph, which results in Category 2. Therefore, we have $$\begin{aligned}
4k=b_1+\cdots+b_t\geq 4t~,\end{aligned}$$ which 