Expected time to get from bottom left to top right in a square Consider a two dimensional random walk starting at the bottom left hand corner of an $n$ by $n$ square. At each step you take one step up, down, left or right distance $1$. Each choice has equal and independent probability of $1/4$. The only exception is if the $x$ or $y$ coordinate is $0$ or $n$, that is you are touching a wall of the square. As you never walk outside the square, if you are touching a wall you choose from the available options again with equal probability. 

What is the expected time to reach the top right hand corner of the
  square asymptotic to?

Intuitively it seems that it should take something like $n^3$ time asymptotically but I don't know how to derive this mathematically.
 A: By symmetry, this is $\frac12C_n$ where $C_n$ denotes the commute time between two opposite corners of the discrete square of size $n\times n$. Exact asymptotics of $C_n$ when $n\to\infty$ might be unknown but, for every graph, the commute time $C_{uv}$ between some vertices $u$ and $v$ is $$C_{uv}=2mR_{uv},$$ where $m$ is the total number of edges and $R_{uv}$ the effective resistance between $u$ and $v$. Also, $R_{uv}\leqslant L_{uv}$ where $L_{uv}$ is the length of the shortest path from $u$ to $v$.
In the present case $m=2n^2$ and $L_{uv}=2n$ hence $$C_n\leqslant8n^3.$$ On the other hand, for every graph on $k$ vertices, $C_{u,v}\geqslant k\log k\,(1+o(1))$ hence, in the present case, $$C_n\geqslant2n^2\log n\,(1+o(1)).$$
For this and much more, see Random walks on graphs: a survey by László Lovász.
Edit: A careful analysis of the eigenvalues and eigenvectors of the associated Laplacian (see there) shows that, for every $n$, the effective resistance $R_n$ between two opposite corners of the square lattice of size $n\times n$ is such that $$R_n\lt6+4\log n,$$ hence $$C_n\leqslant16n^2\log n+O(n^2).$$
