For many enough steps the non-diagonal contribution is $$ \begin{equation*} \sum_{i\ne j}^{N} \Delta x_i\Delta x_j=0, \end{equation*} $$ since \( \Delta x_{i,j} = \pm l \). The variance is then $$ \begin{equation} \langle x(n)^2\rangle - \langle x(n)\rangle^2 = l^2n. \tag{7} \end{equation} $$ It is also rather straightforward to compute the variance for \( L\ne R \). The result is $$ \begin{equation*} \langle x(n)^2\rangle - \langle x(n)\rangle^2 = 4LRl^2n. \end{equation*} $$