&\qquad -\E[X_0^{p-k} X_{i_1} \cdots X_{i_{k}} ] \E[X_0^{q-\ell} X_{j_1} \cdots X_{j_{\ell}} ]        \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}] \E[X_{i_1} \cdots X_{i_{k}} X_{j_1} X_{j_2}\cdots X_{j_{\ell}} ]\\
     &\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\}.    \nonumber 
    \end{aligned}$$ Depending on the values of $k$ and $\ell$, the following two cases arise.\
**Case I.** **Either $k=p, \ell \leq q$ or $\ell=q, k \leq p$ :** Since in this case, we always get $\E[ X_0^{p+q-k-\ell}]= \E[X_0^{p-k}] \E[X_0^{q-\ell}]$. Therefore, if $\{i_1,i_2,\ldots,i_{k}\}\cap \{j_1,j_2,\ldots,j_{\ell}\}=\emptyset$ then from independence of $X_i$’s, we get $$\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}} ] -\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}} ]  =0.$$ Hence in this case, we can get non-zero contribution from (\[eqn:T\_1+T\_2\_SC\]) only when there is at least one cross-matching among $\{i_1,\ldots,i_{k}\}$ and $\{j_1,\ldots,j_{\ell}\}$, i.e., $\{i_1,i_2,\ldots,i_{k}\}\cap \{j_1,j_2,\ldots,j_{\ell}\}\neq\emptyset$ for some $k =0, 1, \ldots, p$ and $\ell=0, 1, \ldots, q$. So from the above observation, (\[eqn:T\_1+T\_2\_SC\]) can be written as $$\begin{aligned}
 \label{eqn:T_k,l_SC}
    \lim_{n\to\infty}  \Cov\big(w_p,w_q\big)  &=\lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{k, \ell =0}^{p, q}\binom{p}{k} \binom{q}{\ell}  \sum_{m=1}^{ \min \{k,\ell\} } \sum_{I_m} \Big\{\E[ X_0^{p+q-k-\ell}] \E[X_{i_1} \cdots X_{i_{k}} X_{j_1} \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\}    \nonumber 