In p.88 of Wand & Jones (1995), they asked to show the following result.
Let $X_1,\ldots,X_n$ be a set of i.i.d. random variables and define $$U=2n^{-2}\sum_{i=1}^{n-1} \sum_{j=i+1}^n S(X_i - X_j)$$ where the function $S$ is symmetric about zero. Show that \begin{align} \operatorname{Var}(U)= {} & 2n^{-3}(n-1) \operatorname{Var} \{S(X_1-X_2)\} \\ & {} +4n^{-3}(n-1)(n-2) \operatorname{Cov} \{S(X_1-X_2),S(X_2-X_3)\}. \end{align}
I've started with $$\operatorname{Var}(U)=4n^{-4}\sum_{i=1}^{n-1} \sum_{j=i+1}^n \sum_{l=1}^{n-1} \sum_{w=l+1}^n \operatorname{Cov}\{S(X_i-X_j),S(X_l-X_w)\}$$ and then isolating the case where $l=i$ and $w=j$, I could obtain the variance term
\begin{align} & \sum_{i=1}^{n-1} \sum_{j=i+1}^n \operatorname{Cov}\{S(X_i-X_j),S(X_i-X_j)\} \\[8pt] = {} & \left(\sum_{k=1}^{n-1}k\right) \operatorname{Var}\{S(X_1-X_2)\} \\[8pt] = {} & \frac{n(n-1)}{2} \operatorname{Var} \{S(X_1-X_2)\}, \end{align}
using the fact that the random variables are identically distributed.
I'm struggling to show the second term of $\operatorname{Var}(U)$, where the covariance appears. I would appreciate if someone could give me some advice.
**Update**
Let me be more specific. When I expand the covariance summations I get
\begin{align} 4^{-1}n^4 \operatorname{Var}(U) = {} & \sum_{i=1}^{n-1} \sum_{j>i}^n \operatorname{Var} \{S(X_i-X_j)\} \\ & {} + \sum_{i=1}^{n-2} \sum_{j>i}^n \sum_{w>i,w\neq j}^n \operatorname{Cov}\{S(X_i-X_j),S(X_i-X_w)\} \\ & {} +\sum_{i=1}^{n-1} \sum_{l\neq i,j>l}^{n-1} \sum_{j>i}^n \operatorname{Cov} \{S(X_i-X_j),S(X_l-X_j)\}, \end{align}
since $\operatorname{Cov}\{S(X_i-X_j),S(X_l-X_w)\}=0,\forall i\neq l,j \neq w$. And then using the identical distribution assumption, I obtained
\begin{align} 4^{-1}n^4 \operatorname{Var}(U)= {} & \frac{n(n-1)}{2} \operatorname{Var}\{S(X_1-X_2)\} \\[8pt] & {} + \frac{(n-2)(n-1)n}{3} \operatorname{Cov}\{S(X_1-X_2),S(X_1-X_3)\} \\[8pt] & {} + \frac{(n-2)(n-1)n}{3} \operatorname{Cov}\{S(X_1-X_2),S(X_3-X_2)\}. \end{align}
For each $i, i=1,\ldots,n-2$, I have $(n-i)(n-i-1)$ covariance possibilities of the form $\operatorname{Cov}\{S(X_i-X_j),S(X_i-X_w)\},(j,w)>i,j\neq w$, and then summing along $i$ gives $\sum_{k=1}^{n-2}k(k+1)=\frac{(n-2)(n-1)n}{3}$. Analogously, for each $j, j=3,\ldots,n$, I have $(j-1)^2-(j-1)=(j-1)(j-2)$ covariance possibilities of the form $\operatorname{Cov}\{S(X_i-X_j),S(X_l-X_j)\},(i,l)<j,l\neq i$, and then summing along $j$ yields also $\frac{(n-2)(n-1)n}{3}$. Thus, if $\operatorname{Cov}\{S(X_1-X_2),S(X_1-X_3)\}=\operatorname{Cov}\{S(X_1-X_2),S(X_2-X_3)\}$, then
\begin{align} \operatorname{Var}(U) = {} & 2n^{-3}(n-1) \operatorname{Var}\{S(X_1-X_2)\} \\[6pt] & {} + \frac{8}{3} n^{-3} (n-1)(n-2) \operatorname{Cov}\{S(X_1-X_2),S(X_2-X_3)\}. \end{align}
Did I miss something?
Thanks in advance.