the expression of the limiting moment above, we are able to derive that the Stieltjes transform of $F$: $$s(z):=\int \frac{1}{x-z}dF(x)$$ satisfies the following equation: $$\begin{aligned}
\label{se}
y^2z^2s^3(z)+y^2zs^2(z)-yzs^2(z)-zs(z)-1=0~,\end{aligned}$$ which coincides with an earlier result in [@li] found by using the Stieltjes transform method.

Indeed, by the series expansion of the function $\frac{1}{1-x}$, the Stieltjes transform of a LSD can be expanded using its moments: $$\begin{aligned}
s(z)&=\int \frac{1}{x-z}dF(x)=-\frac 1z-\sum_{i=1}^{\infty}\frac{1}{z^{i+1}}\cdot m_i\\
&=-\frac 1z-\frac 1z\cdot\sum_{i=1}^{\infty}\frac{m_i}{z^{i}}~.\end{aligned}$$ Let $h(z)$ be the moment generating function of $m_i$: $$h(z)=\sum_{i=0}^{\infty} m_iz^i~,$$ then the part $\sum_{i=1}^{\infty}m_i/z^{i}$ equals to $h(1/z)-1$. Therefore, we have the relationship between the Stieltjes transform $s(z)$ and the moment generating function $h(z)$: $$\begin{aligned}
\label{sh}
s(z)=-\frac 1zh\left(\frac 1z\right)~.\end{aligned}$$ In the proof of Theorem \[th1\], we will see that $h$ satisfies the equation: $$\begin{aligned}
xy^2h^3(x)+x(y-y^2)h^2(x)-h(x)+1=0~,\end{aligned}$$ which is detailed in Appendix, see . Let $x=1/z$ in it and combine with leads to .

(of Theorem \[th1\]) After truncation and centralization, see Appendix A in [@li], we may assume in all the following that $$\begin{aligned}
|\varepsilon_{ij}|\leq \delta p^{1/4}:=M, ~{\mathbb{E}}(\varepsilon_{ij})=0, ~{\mathop{\text{Var}}}(\varepsilon_{ij})=1~.\end{aligned}$$ With a little bit calculation, we have $$\begin{aligned}
m_k(A_T)&= \frac 1p\sum_{{\bf i}=1}^T\sum_{{\bf j}=1}^p\frac{1}{T^{2k}}[\varepsilon_{j_1\,i_1}\varepsilon_{j_1\,i_2}\varepsilon_{j_2\,s+i_2}\varepsilon_{j_2\,s+i_3}\varepsilon_{j_3\,i_3}\varepsilon_{j_3\,i_4}\varepsilon_{j_4\,s+i_4}\varepsilon_{j_4\,s+i_5}\nonumber\\
&\quad\quad\quad\quad\quad\quad\cdots \varepsilon_{j_{2k-1}\,i_{2k-1}}\varepsilon_{j_{2k-1}\,i_{2k}}\varepsilon_{j_{2k}\,s+i_{2k}}\varepsilon_{j_{2k}\,s+i_1}]\nonumber\\
&=\frac{1}{pT^{2k}}\sum_{\bf i,j}E_{Q({\bf i,j})}~,\end{aligned}$$ where the summation runs over all $Q(i,j)$-graph of length $4k$ as defined in 