$\sum_{k=1}^{q-1} \epsilon_k=0$ in $A_{q-1}$. $\l(\frac{p-1}{2}\r)!$ appeared because for each free choice of $(p-1)/2$ variables among $\{i_1,\ldots,i_{p-1}\}$ with positive sign, we can choose rest of the $(p-1)/2$ variables with negative sign in $\l(\frac{p-1}{2}\r)!$ ways to have pair matching. Using same argument for $\{j_1,\ldots,j_{q-1}\}$, we get another $\l(\frac{q-1}{2}\r)!$ factor. $\frac{1}{2^{(\frac{p+q}{2}-1)}}$ arises because $1\leq i_k,j_k\leq n/2$.

Now combining both the Cases, from (\[eqn:cov\_k=p\_odd\]), (\[eqn:cov\_k&lt;p\_odd\]), we get $$\begin{aligned}
 \label{eqn:cov_odd}
    \lim_{ \substack{{n\to\infty} \\ {n \mbox{ odd} } }} & \Cov\big(w_p,w_q\big) \nonumber \\ 
    &=  \left\{\begin{array}{ll}     
            \displaystyle\frac{a_1}{2^{\frac{p+q-4}{2}}  } (\E X^4_1- 1) + \sum_{r=2}^{ \min\{ \frac{p}{2},\frac{q}{2} \} } \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! \ h_{2r}(s) & \text{if}\ p, \mbox{} q \mbox{ both are even},\\\\
            \displaystyle\sum_{r=0}^{ \min\{ \frac{p-1}{2},\frac{q-1}{2} \} }  \frac{b_r}{2^{\frac{p+q-4r-2}{2}}  } \sum_{s=0}^{2r+1}\binom{2r+1}{s}^2 s!(2r+1-s)! \ h_{2r+1}(s) \\         
        \displaystyle + pq \binom{p-1}{(p-1)/2}\binom{q-1}{(q-1)/2} \l((p-1)/2\r)! \l((q-1)/2\r)!\frac{1}{2^{(\frac{p+q}{2}-1)}}     & \text{if}\ p, \mbox{} q \mbox{ both are odd},\\\\
            0 & \text{otherwise}.        
              \end{array}\right.                  
    \end{aligned}$$

**Step 2.** Suppose $n$ is even, then by the trace formula (\[trace formula\_SC\_even\]), we get $$\begin{aligned}
 \label{eqn:cov_even}
    & \lim_{n\to\infty}  \Cov\big(w_p,w_q\big) \nonumber \\
      &= \lim_{n\to\infty} \frac{1}{4n^{\frac{p+q}{2}-1}} \sum_{k, \ell =0}^{p, q}\binom{p}{k} \binom{q}{\ell}   \Big[ \E \{ (Y_k  \sum_{A_k} X_{I_{k}} + \tilde{Y}_k \sum_{ \tilde{A}_k}  X_{I_{k}}) ( Y_\ell  \sum_{A_\ell} X_{J_{\ell}} + \tilde{Y}_\ell \sum_{ \tilde{A}_\ell}  X_{J_{\ell}} ) \}  \nonumber\\
      &\qquad -   \E[(Y_k  \sum_{A_k} X_{I_{k}} + \tilde{Y}_k \sum_{ \tilde{A}_k}  X_{I_{k}} 