Theorem \[thm:symcircovar\]. We shall use trace formula of $SC_n$ to prove \[thm:symcircovar\]. Since for odd and even value of $n$, we have different trace formula, therefore we shall prove \[thm:symcircovar\] in two steps. In Step 1, we calculate limit of $\Cov\big(w_p,w_q\big)$ as $n\to\infty$ with odd $n$ and in Step 2, we calculate limit of $\Cov\big(w_p,w_q\big)$ as $n\to\infty$ with even $n$. We shall show that for both the cases, even and odd value of $n$, limit of $\Cov\big(w_p,w_q\big)$ is same.

Since $\E(w_p)=\E(w_q)=0$, therefore we get $$\begin{aligned}
    \Cov\big(w_p,w_q\big)&=\E[w_p w_q]
    = \frac{1}{n} \Big\{ \E[\Tr(SC_n)^{p}\Tr(SC_n)^{q} ]- \E[\Tr(SC_n)^{p}]\E[\Tr(SC_n)^{q}]   \Big\}. 
    \end{aligned}$$ First we suppose $\Cov\big(w_p,w_q\big)$ for odd value of $n$.\
**Step 1.** Suppose $n$ is odd, then by the trace formula (\[trace formula\_SC\_odd\]), we get $$\begin{aligned}
    \E[\Tr(SC_n)^{p}]&= \E\Big[n\sum_{k=0}^{p}\binom{p}{k}X_0^{p-k}\sum_{A_{k}}\frac{X_{i_1}}{ \sqrt{n}} \cdots \frac{X_{i_{k}}}{ \sqrt{n}} \Big]
    = \frac{1}{ n^ {\frac{p}{2}-1}} \E\Big[ \sum_{k=0}^{p}\binom{p}{k}X_0^{p-k}\sum_{A_{k}} X_{i_1} \cdots X_{i_{k}}\Big].  
    \end{aligned}$$ Therefore $$\begin{aligned}
 \label{eqn:T_1+T_2_SC}
    \Cov\big(w_p,w_q\big)&=\E[w_p w_q] \nonumber\\
    &= \frac{1}{n^{\frac{p+q}{2}-1}} \Big[ \E\Big\{ \Big( \sum_{k=0}^{p}\binom{p}{k} X_0^{p-k}\sum_{A_{k}}  X_{i_1} \cdots X_{i_{k}} \Big) \Big( \sum_{\ell=0}^{q}\binom{q}{\ell}X_0^{q-\ell}\sum_{A_{\ell}} X_{j_1} \cdots X_{j_{\ell}} \Big)  \Big\}\nonumber \\
     &\qquad  - \E \Big( \sum_{k=0}^{p}\binom{p}{k}X_0^{p-k}\sum_{A_{k}}  X_{i_1} \cdots X_{i_{k}} \Big) \E \Big( \sum_{\ell=0}^{q}\binom{q}{\ell}X_0^{q-\ell}\sum_{A_{\ell}} X_{j_1} \cdots X_{j_{\ell}} \Big)   \Big] \nonumber\\
%    &=\frac{1}{n^{\frac{p+q}{2}-1}} \sum_{k, \ell =0}^{p, q}\binom{p}{k} \binom{q}{\ell}  \sum_{A_{k}, A_{\ell}} \Big\{\E[ X_0^{p+q-k-\ell} X_{i_1} \cdots X_{i_{k}} X_{j_1} X_{j_2}\cdots X_{j_{q}} ]\nonumber \\
%    