variables in $(i_1,i_2,\ldots,i_{p})$ with opposite sign, we get $(\binom{p}{p-2r}\binom{p-2r}{\frac{p-2r}{2}} (\frac{p-2r}{2})!)$ factor. Similarly from $(j_1,j_2,\ldots,j_{q})$, we get $(\binom{q}{q-2r}\binom{q-2r}{\frac{q-2r}{2}} (\frac{q-2r}{2})!)$ factor. Now from , we get

$$\begin{aligned}
\lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{I_{2r}}  \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ]
& = \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \lim_{n\to\infty} \frac{1}{n^{2r-1}}  \sum_{A_{2r}, A_{2r}}  \E[X_{i_1} \cdots X_{i_{2r}} X_{j_1} \cdots X_{j_{2r}} ] \\
&=\frac{a_r}{2^{\frac{p+q-4r}{2}}  } \lim_{n\to \infty}\frac{1}{n^{2r-1}}\sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! |A_{2r}'^{(s)}|\\
&= \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! \lim_{n\to \infty}\frac{|A_{2r}^{(s)}|}{n^{2r-1}}. \end{aligned}$$

The factor $\binom{2r}{s}^2$ appeared because in $\binom{2r}{s}$ ways we can choose $s$ many $+1$ from $\{\epsilon_1,\ldots,\epsilon_{2r}\}$ in one $A_{2r}'$. The factor $(s!(2r-s)!)$ appeared because for each choice of $(i_1,\ldots,i_{2r})$ we have $(s!(2r-s)!)$ many choice for $(j_1,\ldots,j_{2r})$. Now using Result \[result:def\_h\] in right side of the last above equality, we get $$\begin{aligned}
\label{eqn:p,q_even1}
\lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{I_{2r}}  \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ]
&= \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! \ h_{2r}(s),
%\frac{1}{(2k)!}\sum_{s=-\lceil\frac{2k+1-\ell}{2}\rceil}^{\lfloor\frac{\ell}{2}\rfloor}\sum_{q=0}^{2s+2k+1-\ell}\binom{2k+1}{q}\l(\frac{2s+2k+1-\ell-q}{2}\r)^{2k}.\nonumber\end{aligned}$$ where $h_{2r}(s)$ is defined in Result \[result:def\_h\].

Now we calculate first part of (\[eqn:lim\_cov\_SC\]) ($p,q$ both even) for $r=1$. Note that, if $r=1$ in this Case, then from (\[eqn:constraint\_d\_even\]), we get $d_1=d_2$, and hence $$\begin{aligned}
\label{eqn:p,q_even2}
\lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{I_{2}}  \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}} ]
&= \frac{a_1}{2^{\frac{p+q-4}{2}}  } (\E X^4_1- (\E X^2_1)^2) \nonumber \\
