## Appendix B Derivation of Type 2 Limits  
Consider a particle's possible horizontal locations after one time step,  
\[x_{i}(N)=x_{i}(N-1)+h_{xi}(N), \tag{101}\]  
We use induction to derive the second equation in (7).  
When \(N=0\), we obtain  
\[\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}(0)h_{xi}(M)=0,\ \ M>0. \tag{102}\]  
When \(N=1\), we have,  
\[\begin{split}\lim_{K\to\infty}&\frac{2}{K}\sum_{i=1 }^{K}x_{i}(1)h_{xi}(M),\ \ M>1\\ &=\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}(x_{i}(0)+h_{xi}(1))h _{xi}(M)\\ &=\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}h_{xi}(1)h_{xi}(M)=0.\end{split} \tag{103}\]  
For \(N>1\), we employ mathematical induction, first assuming that  
\[\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}(N)h_{xi}(M)=0\ \ \text{when}\ M>N, \tag{104}\]  
holds for \(N>1\). We now show that it also holds for \(N+1>2\) via the following steps:  
\[\begin{split}\lim_{K\to\infty}&\frac{2}{K}\sum_{i=1 }^{K}x_{i}(N+1)h_{xi}(M)\\ &=\frac{2}{K}\sum_{i=1}^{K}(x_{i}(N)+h_{xi}(N+1))h_{xi}(M)\\ &=\frac{2}{K}\sum_{i=1}^{K}x_{i}(N)h_{xi}(M)+\frac{2}{K}\sum_{i= 1}^{K}h_{xi}(N+1)h_{xi}(M).\end{split} \tag{105}\]  
Using the inductive hypothesis in Equation (104), we obtain the result  
\[\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}(N+1)h_{xi}(M)=0\ \ \text{ for}\ M>N, \tag{106}\]  
which corresponds to the second relation in (7).  
Lastly, we use mathematical induction to show that  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(N)=Np_{x}(\Delta x)^{2}. \tag{107}\]  
When \(N=1\), Equation (107) becomes  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(1)=\lim_{K\to\infty}\frac {1}{K}\sum_{i=1}^{K}(x_{i}(0)+h_{i}(1))^{2}=p_{x}(\Delta x)^{2}. \tag{108}\]When \(N=2\), Equation (A22) becomes  
\[\begin{split}\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(2)& =\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}(x_{i}(1)+h_{i}(2))^{2} \\ &=\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}(x_{i}^{2}(1)+2x_{i}(1 )h_{i}(2)+h_{i}^{2}(2))\\ &=\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(1)+\lim_{K \to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}(1)h_{i}(2)\\ &\quad+\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{i}^{2}(2). \end{split}\] (A24)  
We can rewrite this as  
\[\begin{split}\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}( 2)&=2p_{x}(\Delta x)^{2}+\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{ K}x_{i}(1)h_{i}(2)\\ &=2p_{x}(\Delta x)^{2}+0.\end{split}\] (A25)  
We now consider the inductive step, assuming that for \(N>2\),  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(N)=Np_{x}(\Delta x)^{2},\] (A26)  
holds for \(N\). We now show that it also holds for \(N+1\) via the following steps:  
\[\begin{split}\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}( N+1)&=\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}(x_{i}(N)+h_{i}(N+1))^ {2}\\ &=\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}\Bigl{(}x_{i}^{2}(N) +2x_{i}(N)h_{i}(N+1)+h_{i}^{2}(N+1)\Bigr{)}\\ &=\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(N)+\lim_{K \to\infty}\frac{2}{K}\sum_{i=1}^{K}x_{i}(N)h_{i}(N+1)\\ &\quad+\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}h_{i}^{2}(N+1). \end{split}\] (A27)  
Using the inductive hypothesis in Equation (A26), we obtain,  
\[\begin{split}\lim_{K\to\infty}&\frac{1}{K}\sum_{i= 1}^{K}x_{i}^{2}(N+1)\\ &=Np_{x}(\Delta x)^{2}+\lim_{K\to\infty}\frac{2}{K}\sum_{i=1}^{K} x_{i}(N)h_{i}(N+1)+p_{x}(\Delta x)^{2}\\ &=(N+1)p_{x}(\Delta x)^{2}+\lim_{K\to\infty}\frac{2}{K}\sum_{i=1} ^{K}x_{i}(N)h_{i}(N+1).\end{split}\] (A28)  
The last term above is zero, so that  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(N+1)=(N+1)p_{x}(\Delta x)^ {2}.\] (A29)  
Hence,  
\[\lim_{K\to\infty}\frac{1}{K}\sum_{i=1}^{K}x_{i}^{2}(N+1)=Np_{x}(\Delta x)^{2},\ \ \text{for all}\ N\geq 1.\] (A30)