\sum_{A_{p-1},A_{q-1}} X_{I_{p-1}} X_{J_{q-1}}] & = O(n^{[\frac{p-1}{2}]+ [\frac{q-1}{2} ]}), \\
   \E [ X_0 X_{\frac{n}{2}} \sum_{ \tilde{A}_{p-1}, A_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} ] & = O(n^{[\frac{p-2}{2}]+ [\frac{q-1}{2} ]}), \nonumber \\
   \E[ X_0 X_{\frac{n}{2}}  \sum_{A_{p-1}, \tilde{A}_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} & = O(n^{[\frac{p-2}{2}]+ [\frac{q-1}{2} ]}),  \nonumber  \\
   \E[ (X_{\frac{n}{2}})^2 \sum_{ \tilde{A}_{p-1}, \tilde{A}_{q-1}}  X_{I_{p-1}} X_{J_{q-1}} ] & = O(n^{[\frac{p-2}{2}]+ [\frac{q-2}{2} ]}).   \nonumber 
    \end{aligned}$$ Now using (\[eqn:cov\_k&lt;p,even\]) and the fact that $\E(X_i)=0$ for each $i=1,2, \ldots$ in (\[eqn:cov,even,k&lt;p\]), we get $$\begin{aligned}
 \label{eqn:cov,k<p,even1}
    &\lim_{n\to\infty}  \Cov\big(w_p,w_q\big) \\
       &= \left\{\begin{array}{ll}   \nonumber 
            \displaystyle{\lim_{n\to\infty}} \frac{pq}{n^{\frac{p+q}{2}-1}}  \sum_{A_{p-1}, A_{q-1}} \E[ X_0^{2}] \E[X_{i_1} \cdots X_{i_{p-1}} X_{j_1} X_{j_2}\cdots X_{j_{q-1}} ] & \text{if}\ p, \mbox{} q \mbox{ both are odd},\\\\
            0 & \text{otherwise}.        
              \end{array}\right.    
     \end{aligned}$$ Note that right hand side of (\[eqn:cov,k&lt;p,even1\]) is same as (\[eqn:cov\_k&lt;p\_odd\]). Therefore (\[eqn:cov,k&lt;p,even1\]) will be

$$\begin{aligned}
 \label{eqn:cov_k<p_even}
    & \lim_{n\to\infty}  \Cov\big(w_p,w_q\big)   \\ 
         &= \left\{\begin{array}{ll}\nonumber 
            \displaystyle pq \binom{p-1}{(p-1)/2}\binom{q-1}{(q-1)/2} \l((p-1)/2\r)! \l((q-1)/2\r)!\frac{1}{2^{(\frac{p+q}{2}-1)}} & \text{if}\ p, \mbox{} q \mbox{ both are odd},\\\\
             0 & \text{otherwise}.       
              \end{array}\right.    
     \end{aligned}$$

Now combining both the Cases, from (\[eqn:cov\_k=p\_even\]), (\[eqn:cov\_k&lt;p\_even\]), we get $$\begin{aligned}
 \label{eqn:cov_even,2}
&\lim_{ \substack{{n\to\infty},\\ {n \mbox{ even} } }}  \Cov\big(w_p,w_q\big) \nonumber \\ 
&=  \left\{\begin{array}{ll}     
         \displaystyle  \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},\\\\
         \displaystyle  