situation, the cardinality of $I_{2r+1}$ will be $o(n^{p+q-1}).$ Since each entries of $ \{d_1, d_2, \ldots, d_{2r+1}\}$ are distinct, as (ii) holds. Therefore $$\E[ X_{i_1} \cdots X_{i_{p}} ] \E[ X_{j_1} \cdots X_{j_{q}} ] =0.$$ Now by the similar calculations as we have done in Case I, second part of (\[eqn:lim\_cov\_SC\]) ($p,q$ both odd) will be $$\begin{aligned}
 \label{eqn:p,q_odd}
%   \lim_{n\to\infty} &\Cov\big(w_p,w_q\big)=
    &\lim_{n\to\infty} \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{r=0}^{ \min\{ \frac{p-1}{2},\frac{q-1}{2} \} } \sum_{I_{2r+1}} \big( \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ] -\E[ X_{i_1} \cdots X_{i_{p}} ] \E[ X_{j_1} \cdots X_{j_{q}} ]  \big) \nonumber\\
    &  = \lim_{n\to\infty}  \frac{1}{n^{\frac{p+q}{2}-1}} \sum_{r=0}^{ \min\{ \frac{p-1}{2},\frac{q-1}{2} \} } \sum_{I_{2r+1}}  \E[X_{i_1} \cdots X_{i_{p}} X_{j_1} \cdots X_{j_{q}} ]  \nonumber\\
    &  = \sum_{r=0}^{ \min\{ \frac{p-1}{2},\frac{q-1}{2} \} } \frac{b_r}{2^{\frac{p+q-4r-2}{2}}  } \sum_{s=0}^{2r+1}\binom{2r+1}{s}^2 s!(2r+1-s)! \ h_{2r+1}(s) ,
    \end{aligned}$$ where $h_{2r+1}(s)$ is defined in Result \[result:def\_h\] and $b_r=  \binom{p}{p-2r-1}\binom{p-2r-1}{\frac{p-2r-1}{2}} (\frac{p-2r-1}{2})!\binom{q}{q-2r-1}\binom{q-2r-1}{\frac{q-2r-1}{2}} (\frac{q-2r-1}{2})!.$ Now, after combining both the sub-cases I and II, using (\[eqn:p,q\_even\]) and (\[eqn:p,q\_odd\]) in (\[eqn:lim\_cov\_SC\], we get $$\begin{aligned}
 \label{eqn:cov_k=p_odd}
    &\lim_{n\to\infty} \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\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 