Technical question about random walks (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 $G&#39;$ non-bipartite; applying the fundamental theorem of Markov chains to $G&#39;$ we then get that $h_{i,j} &lt; 2m+1$ in $G&#39;$, and this is clearly also an upper bound for $h_{i,j}$ in $G$. 

Question: Is it true that the stronger claim $h_{i,j} &lt; 2m$ 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?
This answer proved something different from what the questioner was actually interested in. Leaving it here so others won't repeat the same mistake.


  In most cases, one can formally justify the intuitive notion that
  "self loops can only slow down the walk" by a coupling argument.  In
  this case for example, one can couple the walk with the self loops
  (let's call it $A$) and the one without the self loops (let's call
  that $B$) so that $A$ takes the same steps as $B$, but delayed in
  time.  This can for example be done as follows:  Suppose that $B$
  starts at $u = x_0$ and goes through $x_i: i =1,2,\ldots,k$. Now, we
  implement $A$ as follows: $A$ also goes through the same vertices as
  $B$, except that vertex $x_i$, it waits for Geometric($p_i$) time
  where $p_i$ is the self loop probability at $x_i$.  Note that this is
  a correct implementation of $A$ (all transition probabilities are
  correct), and the form of the coupling ensures that $A$ never reaches
  any vertex before $B$, that is, we have coupled $H_t^A$ and $H_t^B$
  (the random hitting times in the two walks) so that $H_t^A \geq H_t^B$
  with probability $1$. Thus, the inequality for the expected hitting
  time follows.