]  \E[Y_\ell  \sum_{A_\ell} X_{J_{\ell}} + \tilde{Y}_\ell \sum_{ \tilde{A}_\ell}  X_{J_{\ell}}] \Big] \nonumber \\
&= \lim_{n\to\infty} \frac{1}{4n^{\frac{p+q}{2}-1}} \sum_{k, \ell =0}^{p, q}\binom{p}{k} \binom{q}{\ell}   \Big[ \E  (Y_k Y_\ell \sum_{A_k,A_\ell} X_{I_{k}} X_{J_{\ell}}) + \E (\tilde{Y}_k Y_\ell \sum_{ \tilde{A}_k, A_\ell}  X_{I_{k}}  X_{J_{\ell}}) +\E( Y_k \tilde{Y}_\ell  \sum_{A_k, \tilde{A}_\ell} X_{I_{k}}  X_{J_{\ell}})  \\
&\qquad + \E( \tilde{Y}_k \tilde{Y}_\ell  \sum_{ \tilde{A}_k, \tilde{A}_\ell}  X_{I_{k}} X_{J_{\ell}} ) - \E[(Y_k  \sum_{A_k} X_{I_{k}} + \tilde{Y}_k \sum_{ \tilde{A}_k}  X_{I_{k}} ]  \E[Y_\ell  \sum_{A_\ell} X_{J_{\ell}} + \tilde{Y}_\ell \sum_{ \tilde{A}_\ell}  X_{J_{\ell}}] \Big]. \nonumber 
    \end{aligned}$$ By the similar arguments as we have done in Step 1, we can show that right hand side of (\[eqn:cov\_even\]) has non-zero contribution only when $k=p, \ell=q$ with $\{i_1,i_2,\ldots,i_{k}\}\cap \{j_1,j_2,\ldots,j_{\ell}\} \neq\emptyset$ and $k=p-1, \ell=q-1$ with $\{i_1,i_2,\ldots,i_{k}\}\cap \{j_1,j_2,\ldots,j_{\ell}\}=\emptyset$.\
**Case I.** **$k=p$ and $\ell=q$ :** First recall $Y_k$ and $\tilde{Y}_k$ from (\[eqn:Y\_k\]) $$Y_k  = {(X_0 +  X_{\frac{n}{2}})}^{p-k} + {(X_0 -  X_{\frac{n}{2}})}^{p-k}, \ \ \tilde{Y}_k = {(X_0 +  X_{\frac{n}{2}})}^{p-k} - {(X_0 -  X_{\frac{n}{2}})}^{p-k}.$$ Since for $k=p$ and $\ell=q, Y_p=Y_q= 2$ and $\tilde{Y}_p=\tilde{Y}_q=0$. Therefore in this case, (\[eqn:cov\_even\]) will be $$\begin{aligned}
 \label{eqn:cov_even1}
 \lim_{n\to\infty} \Cov\big(w_p,w_q\big) &= \lim_{n\to\infty} \frac{1}{4n^{\frac{p+q}{2}-1}} \Big[ \E  (4\sum_{A_p,A_q} X_{I_{p}} X_{J_{q} }) - \E[ 2\sum_{A_p} X_{I_{p}}] \E[ 2\sum_{A_q} X_{J_{q}}] \Big] \nonumber \\
 &= \lim_{n\to\infty} \frac{1}{n^{\frac{p+q}{2}-1}}  \sum_{A_p,A_q} \E[X_{I_{p}} X_{J_{q}}] - \E[ X_{I_{p}}] \E[X_{J_{q}}].
 \end{aligned}$$ Note that (\[eqn:cov\_even1\]) is 