(My original question still has not been answered. I have added further clarifications.)
When analyzing random walks (on undirected graphs) by viewing the random walk as a Markov chain, we require the graph to be non-bipartite so that the fundamental theorem of Markov chains applies.
What happens if the graph $G$ is instead bipartite? I am specifically interested in the hitting time $h_{i,j}$, where there is an edge between $i$ and $j$ in $G$. Say the bipartite graph $G$ has $m$ edges. We can add a self-loop to an arbitrary vertex in the graph to make the resulting graph $G'$ non-bipartite; applying the fundamental theorem of Markov chains to $G'$ we then get that $h_{i,j} < 2m+1$ in $G'$, and this is clearly also an upper bound for $h_{i,j}$ in $G$.
Question: Is it true that the stronger claim $h_{i,j} < 2m$ holds in $G$? (It have seen this claimed in analyses of the random walk algorithm for 2SAT.) Or do we really have to go through this extra step of adding the self-loop?