How does the mean displacement of a random walk on a $d$-dimensional integer lattice (or $d$-dimensional hexagonal lattice) scale with the number of steps $N$ in the walk? Is the displacement always $\approx N^{1/2}$? My confusion is stemming from the fact that, in some dimension $d$, the number of reachable points on a lattice scales differently with displacement from the origin (more specifically, it scales as $\approx c*r^d$ where $c$ is some dimension specific constant: http://en.wikipedia.org/wiki/Volume_of_an_n-ball).
Presumably we can extrapolate the explanation here to the continuum limit where we might talk about a Brownian process?
Indeed the displacement is still of order $\sqrt{N}$. A formal statement is that $E[\|X_N\|^2]=cN$ for some constant $c$ which depends on the specifics of the lattice and of the distribution of the steps of the random walk $(X_N)_{N\geqslant0}$. The volume of the balls would be relevant when studying the number of steps needed to visit every vertex in some ball, which is an altogether different problem.
To compute the mean square displacement on the lattice $\mathbb Z^d$, assume that $X_N=U_1+\cdots+U_N$ for every $N\geqslant0$, for some i.i.d. centered square-integrable $(U_N)_{N\geqslant1}$ with values in $\mathbb Z^d$. Then, $$ \|X_N\|^2=\sum\limits_{n=1}^N\|U_n\|^2+\sum\limits_{1\leqslant k\ne n\leqslant N}U_k^*U_n. $$ For every $k\ne n$, $\mathbb E[U_k^*U_n]=\mathbb E[U_k]^*\mathbb E[U_n]=0^*0=0$ by independence, hence $\mathbb E[\|X_N\|^2]=c\cdot N$ with $c=\mathbb E[\|U_1\|^2]$. Thus, $\sqrt{\mathbb E[\|X_N\|^2]}\propto\sqrt{N}$.