\mbox{ both are odd},\\\\
            0 & \text{otherwise}.        
              \end{array}\right.                  
    \end{aligned}$$ **Case II.** **$k<p$ and $\ell<q$ :** First recall $\Cov\big(w_p,w_q\big)$ from (\[eqn:T\_1+T\_2\_SC\]) for $k<p$ and $\ell<q$ $$\begin{aligned}
        \Cov\big(w_p,w_q\big) \mathbb I_{\{k<p,\ell<q\}}&=\frac{1}{n^{\frac{p+q}{2}-1}} \sum_{k, \ell =0}^{p-1, q-1}\binom{p}{k} \binom{q}{\ell}  \sum_{A_{k}, A_{\ell}} \Big\{\E[ X_0^{p+q-k-\ell}] \E[X_{i_1} \cdots X_{i_{k}} X_{j_1} X_{j_2}\cdots X_{j_{\ell}} ]\nonumber \\
     &\qquad -\E[X_0^{p-k}]\E[ X_{i_1} \cdots X_{i_{k}} ] \E[X_0^{q-\ell}]\E[ X_{j_1} \cdots X_{j_{\ell}} ]   \Big\}.\end{aligned}$$ Similar to Case I, we get maximum contribution when $\{i_1,i_2,\ldots,i_{k}\}\cap \{j_1,j_2,\ldots,j_{\ell}\}=\emptyset$ and $(i_1,i_2,\ldots,i_{k}), (j_1,j_2,\ldots,j_{\ell})$ are [*opposite sign pair matched*]{}. Since from (\[eqn:condition\]) we have that all moments are bounded, therefore $$\begin{aligned}
 \label{eqn:cov_k<p}
    \sum_{A_{k}, A_{\ell}} \E[ X_0^{p+q-k-\ell}] \E[ X_{I_{k}}  X_{J_{\ell}} ] -\E[X_0^{p-k}]\E[X_{I_{k}} ] \E[X_0^{q-\ell}]\E[ X_{J_{\ell}} ] = O(n^{[\frac{k}{2}]+ [\frac{\ell}{2} ]}).
    \end{aligned}$$ Now using (\[eqn:cov\_k&lt;p\]) and the fact that $\E(X_i)=0$ for each $i=1,2, \ldots,$ we get $$\begin{aligned}
 \label{eqn:cov_k<p_odd}
    \lim_{n\to\infty} & \Cov\big(w_p,w_q\big) \mathbb I_{\{k<p,\ell<q\}}  \nonumber \\
       &= \left\{\begin{array}{ll}   \nonumber
            \displaystyle\lim_{n\to\infty} \frac{pq}{n^{\frac{p+q}{2}-1}}  \sum_{A_{p-1}, A_{q-1}} \E[ X_0^{2}] \E[X_{i_1} \cdots X_{i_{p-1}} X_{j_1} X_{j_2}\cdots X_{j_{q-1}} ] & \text{if}\ p, \mbox{} q \mbox{ both are odd},\\\\
            0 & \text{otherwise},        
              \end{array}\right.    \\ 
         &= \left\{\begin{array}{ll} 
             \displaystyle pq \binom{p-1}{(p-1)/2}\binom{q-1}{(q-1)/2} \l((p-1)/2\r)! \l((q-1)/2\r)!\frac{1}{2^{(\frac{p+q}{2}-1)}} & \text{if}\ p,  q \mbox{ both are odd},\\\\
             0 & \text{otherwise},       
              \end{array}\right.    
     \end{aligned}$$ where the factor $\binom{p-1}{(p-1)/2} \binom{q-1}{(q-1)/2}$ appeared because in $\binom{p-1}{(p-1)/2}$ many ways $\sum_{k=1}^{p-1} \epsilon_k=0$ in one $A_{p-1}$ and $\binom{q-1}{(q-1)/2}$ many ways 