be $o(n^{p+q-1}).$ Also note that, as each entries of $ \{d_1, d_2, \ldots, d_{2r}\}$ are distinct, therefore $$\E[ X_{i_1} \cdots X_{i_{p}} ] \E[ X_{j_1} \cdots X_{j_{q}} ] =0.$$ Hence for each fixed $r\geq 2$, first part of (\[eqn:lim\_cov\_SC\]) ($p,q$ both even) will be $$\begin{aligned}
 \label{eqn:I_2k2_SC}
%   \lim_{n\to\infty} &\Cov\big(w_p,w_q\big)=
%  &\lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{I_{2r}}  \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ] -\E[ X_{i_1} \cdots X_{i_{p}} ] \E[ X_{j_1} \cdots X_{j_{q}} ]  \nonumber \\
    &  \lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{I_{2r}}  \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ] \nonumber \\
    & = \lim_{n\to\infty} \frac{1}{n^{\frac{p+q}{2}-1}}  a_r  (n/2)^{\frac{p-2r}{2}+\frac{q-2r}{2}} \sum_{A_{2r}, A_{2r}}  \E[X_{i_1} \cdots X_{i_{2r}} X_{j_1} \cdots X_{j_{2r}} ] \nonumber \\
    & =  \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \lim_{n\to\infty} \frac{1}{n^{2r-1}}  \sum_{A_{2r}, A_{2r}}  \E[X_{i_1} \cdots X_{i_{2r}} X_{j_1} \cdots X_{j_{2r}} ],
    \end{aligned}$$ where $ a_r=    \binom{p}{p-2r}\binom{p-2r}{\frac{p-2r}{2}} (\frac{p-2r}{2})!\binom{q}{q-2r}\binom{q-2r}{\frac{q-2r}{2}} (\frac{q-2r}{2})!$. $a_r$ factor is arising for pair-matching of $(p-2r)$ many variables in $(i_1,i_2,\ldots,i_{p})$ and $(j_1,j_2,\ldots, j_{q})$ both with opposite sign. In $(i_1,i_2,\ldots,i_{p})$, we can choose $(p-2r)$ variables in $\binom{p}{p-2r}$ many ways. Out of $(p-2r)$ variables, $(\frac{p-2r}{2})$ many variables can be chosen with positive sign in $\binom{p-2r}{\frac{p-2r}{2}}$ many ways. After free choice of $(\frac{p-2r}{2})$ variables with positive sign, rest of the $(\frac{p-2r}{2})$ variables with negative sign can be chosen in $(\frac{p-2r}{2})!$ ways. Therefore for pair matching of $(p-2r)$ many 