## Appendix A Overview and Derivation of Type 1 Limits  
We summarize several useful limits in evaluating Equation (6) (Section 2.2.2). We delineate these limits as those involving only the relative displacements (Type 1) and as those involving the position of the particle (Type 2). Direct evaluation of the Type 1 limits yields the following results associated with case (i):  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}^{N}=0, \tag{11}\] \[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}^{N}h_{xi}^{M}=0 \;\;\text{when}\;N\
eq M,\] (12) \[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}(h_{xi}^{N})^{2}=p_{x}( \Delta x)^{2}\;\;\text{\it i.e.,}\;\text{\it when}N=M. \tag{13}\]  
Analogous results also hold for the vertical transition case (ii) and are derived using the same steps shown in this appendix. The Type 2 limits can also be evaluated directly (see Appendix B) and yield the following results:  
\[\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}^{N}h_{xi}^{M}=0 \;\;\text{when}\;M>N, \tag{14}\] \[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}(x_{i}^{N})^{2}=Np_{x}( \Delta x)^{2}. \tag{15}\]  
The above limits are for the horizontal transition case (\(x_{i}\)), but there are similar analogous results that can be derived using the same steps for the vertical transition case. Derivation of Type 1 Limits  
Evaluating Equation (11). Consider the average displacement in the horizontal direction, over all realizations, as \(K\to\infty\),  
\[\lim_{K\to\infty}\sum_{i=1}^{K}h_{xi}(N). \tag{16}\]  
Let \(k_{x1}\), \(k_{x2}\) and \(k_{x3}\) be defined as follows:  
(i) \(k_{x1}\): the number of times that \(h_{x1}(N)=0\),  
(ii) \(k_{x2}\): the number of times that \(h_{x1}(N)=\Delta x\),  
(iii) \(k_{x3}\): the number of times that \(h_{x1}(N)=-\Delta x\). We can then write  
\[ \begin{split}\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}(N) &=\lim_{K\to\infty}\frac{1}{K}[k_{x2}(\Delta x)+k_{x3}(-\Delta x )]\\ &=\lim_{K\to\infty}(\Delta x)\Big{(}\frac{k_{x2}}{K}-\frac{k_{x3} }{K}\Big{)}.\end{split}  \tag{17}\]For the equally spaced obstacles problem, moving left or right is equally likely. Hence, for very large values of \(K\), it follows that \(\frac{k_{x2}}{K}=\frac{p_{y}}{2}\) and that \(\frac{k_{x3}}{K}=\frac{p_{y}}{2}\). We then write  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}(N)=\lim_{K\to\infty}(\Delta x) \bigg{(}\frac{k_{x2}}{K}-\frac{k_{x3}}{K}\bigg{)}=(\Delta x)\bigg{(}\frac{p_{x }}{2}-\frac{p_{x}}{2}\bigg{)}=0. \tag{10}\]  
Similarly, in the vertical direction, by analogous steps we obtain the result  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{yi}(N)=0. \tag{11}\]  
Evaluating Equation (10). We have three cases for \(h_{xi}(N)\) and \(h_{xi}(M)\):  
1. \(h_{xi}(N)=\Delta x\) and \(h_{xi}(M)=-\Delta x\) (or vice versa),
2. \(h_{xi}(N)=h_{xi}(M)=\Delta x\) or \(h_{xi}(N)=h_{xi}(M)=-\Delta x\),
3. \(h_{xi}(N)=0\) and/or \(h_{xi}(M)=0\). To describe the frequency of these cases, we define \(k_{x4}\), \(k_{x5}\) and \(k_{x6}\), respectively, as follows:  
1. \(k_{x4}\): the number of times that \(h_{xi}(N)=\Delta x\) and \(h_{xi}(M)=-\Delta x\) (or vice versa),
2. \(k_{x5}\): the number of times that \(h_{xi}(N)=h_{xi}(M)=\Delta x\) or \(h_{xi}(N)=h_{xi}(M)=-\Delta x\),
3. \(k_{x6}\): the number of times that \(h_{xi}(N)=0\) and/or \(h_{xi}(M)=0\). Consider the limit  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}(N)h_{xi}(M)=\lim_{K\to\infty} (\Delta x)^{2}\bigg{[}\frac{k_{x5}}{K}-\frac{k_{x4}}{K}\bigg{]}. \tag{12}\]  
We have equal probabilities for events corresponding with \(k_{x4}\) and \(k_{x5}\), and we expect for very large values of \(K\) that \(\frac{k_{x4}}{K}=\frac{k_{x5}}{K}\). Hence,  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{xi}(N)h_{xi}(M)=\lim_{K\to\infty} (\Delta x)^{2}\bigg{[}\frac{k_{x5}}{K}-\frac{k_{x4}}{K}\bigg{]}=0. \tag{13}\]  
Similarly, in the vertical direction, by analogous steps we obtain the result  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{yi}(N)h_{yi}(M)=0, \tag{14}\]  
when \(N\
eq M\). Evaluating Equation (7).