Consider the first $2n$ steps of a simple random walk on the integers, starting at the origin. A simple binomial argument shows that regardless of $n$, the origin gets visited the most (in expectation). 


**Question 1:** Does this fact generalize? that is, suppose we have a symmetric random walk on $\mathbb{Z}$ starting at $0$. Is it still true that the origin gets visited the most (in expectation)? 


**Question 2:** Suppose we have two random walks (not necessarily symmetric) $X=(X\_j)\_j$ and $Y=(Y\_j)\_j$ on $\mathbb{Z}\_{\geq 0}$ that can be coupled such that $X\_j\leq Y\_j$ for all $j$, and such that the integer most visited by $Y$ is $0$. Does this imply the same for $X$? that is, is it necessarily true that the integer most visited by $X$ is $0$?


Thanks!



Expected visits to the origin by a symmetric random walk on the integers
Here we prove that if steps are independent and identically symmetrically distributed, then Q1 has positive answer.


Let $p(k)$ be probability of step $k$, $p(-k)=p(k)$. Denote by $f(t)=\sum\_k p(k)t^k$ the generating function. Then average number of visits of 0 during first $n$ steps is $[1] g(t)$, where $g(t)=1+f(t)+\dots+f^{n-1}(t)$, $[1]g$ means constant term of $g$. Average number of visits of a point $w$ is $[t^w] g(t)=[t^{-w}] g(t)=[1] \frac{t^w+t^{-w}}2g(t)$. Next, we use the formula $[1] h(t)=\frac1{2\pi}\int\_0^{2\pi} h(e^{ix})dx$. Note that $f(e^{ix})=p\_0+\sum\_{k>0} 2p\_k \cos kx\in [-1,1]$. Thus $$g(e^{ix})=\frac{1-f^n(e^{ix})}{1-f(e^{ix})}\geqslant 0.$$
Hence $$[1]g=\frac1{2\pi}\int\_0^{2\pi} g(e^{ix})dx\geqslant \frac1{2\pi}\int\_0^{2\pi} g(e^{ix})\cos (wx)dx=[1] \frac{t^w+t^{-w}}2g(t)$$
as desired.


