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Consider an irreducible, aperiodic Markov chain with stationary distribution $\pi$. We will use $E_{\pi} T_j$ to be the hitting time of node $j$ when the initial distribution is $\pi$, and $E_i T_j$ to be the same hitting time when the chain is started at node $i$. If $$ f(i,j) = E_i T_j - E_{\pi} T_j$$ then $$ f(i,j) = f(j,i).$$ Is there a simple proof of this? This is a byproduct of a number of identities in Aldous & Fill textbook, Chapter 2, which are proved using (what seems to me) to be advanced ideas.

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Let me doubt very much that Aldous and Fill are saying that. For a counterexample of the assertion in your post, consider the random walk on a discrete circle of size $n\geqslant3$, with steps $+1$ and $0$ with probabilities $p$ and $1-p$.

Then $\pi$ is uniform and $\mathbb E_\pi(T_i)$ does not depend on $i$ hence $f(i,j)=f(j,i)$ would mean that $\mathbb E_i(T_j)=\mathbb E_j(T_i)$ for every $(i,j)$. But, when $p\to1$, $\mathbb E_i(T_{i+1})\to1$ and $\mathbb E_{i+1}(T_i)\to n-1$.

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Thank you for the answer. I believe I omitted the requirement that the chain should be reversible. – pierre Feb 9 '13 at 22:22

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