same as (\[eqn:lim\_cov\_SC\]). Therefore from (\[eqn:lim\_cov\_SC\]) and (\[eqn:cov\_k=p\_odd\]), (\[eqn:cov\_even1\]) will be $$\begin{aligned}
 \label{eqn:cov_k=p_even}
    & \lim_{n\to\infty}  \Cov\big(w_p,w_q\big) \nonumber \\ 
    &=  \left\{\begin{array}{ll}     
            \frac{a_1}{2^{\frac{p+q-4}{2}}  } (\E X^4_1- 1) + \sum_{r=2}^{ \min\{ \frac{p}{2},\frac{q}{2} \} } \frac{a_r}{2^{\frac{p+q-4r}{2}}  } \sum_{s=0}^{2r}\binom{2r}{s}^2 s!(2r-s)! \ h_{2r}(s) & \text{if}\ p, \mbox{} q \mbox{ both are even},\\\\
            \sum_{r=0}^{ \min\{ \frac{p-1}{2},\frac{q-1}{2} \} }  \frac{b_r}{2^{\frac{p+q-4r-2}{2}}  } \sum_{s=0}^{2r+1}\binom{2r+1}{s}^2 s!(2r+1-s)! \ h_{2r+1}(s)  & \text{if}\ p, \mbox{} q \mbox{ both are odd},\\\\
            0 & \text{otherwise}.        
              \end{array}\right.                  
    \end{aligned}$$\
**Case II.** **$k=p-1$ and $\ell=q-1$ :** Since for $k=p-1$ and $\ell=q-1, Y_{p-1}=Y_{q-1}= 2X_0$ and $\tilde{Y}_{p-1}=\tilde{Y}_{q-1}=2X_{\frac{n}{2}}$. Therefore in this case, (\[eqn:cov\_even\]) will be $$\begin{aligned}
 \label{eqn:cov,even,k<p}
 & \lim_{n\to\infty}  \Cov\big(w_p,w_q\big)  \\
 &= \lim_{n\to\infty} \frac{1}{4n^{\frac{p+q}{2}-1}} p q\Big[ \E  (4X_0^2 \sum_{A_{p-1},A_{q-1}} X_{I_{p-1}} X_{J_{q-1}}) + \E ( 4 X_0 X_{\frac{n}{2}} \sum_{ \tilde{A}_{p-1}, A_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} ) \nonumber \\
&\qquad  +\E( 4 X_0 X_{\frac{n}{2}}  \sum_{A_{p-1}, \tilde{A}_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} ) + \E( 4 (X_{\frac{n}{2}})^2 \sum_{ \tilde{A}_{p-1}, \tilde{A}_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} ) \nonumber \\
& \qquad  - \E[(2X_0\sum_{A_{p-1}} X_{I_{p-1}} + 2X_{\frac{n}{2}} \sum_{ \tilde{A}_{p-1}}  X_{I_{p-1}} ]  \E[ 2X_0\sum_{A_{q-1}} X_{J_{q-1}} + 2X_{\frac{n}{2}} \sum_{ \tilde{A}_{q-1}}  X_{J_{q-1}}] \Big]. \nonumber 
 \end{aligned}$$ In this case, we get non-zero contribution when $\{i_1,i_2,\ldots,i_{p-1}\}\cap \{j_1,j_2,\ldots,j_{q-1}\}=\emptyset$ and $(i_1,i_2,\ldots,i_{p-1}), (j_1,j_2,\ldots,j_{q-1})$ are [*opposite sign pair matched*]{}. Since from (\[eqn:condition\]) we have that all moments are bounded, therefore $$\begin{aligned}
 \label{eqn:cov_k<p,even}
    \E[ X_0^2 