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Consider a directed, weighted graph $G$. Let $s$ and $t$ be two distinct vertices of $G$ and consider a walker that starts at $s$ and traverses a random shortest path from $s$ to $t$, chosen uniformly at random from all shortest paths from $s$ to $t$.

Question: What is the transition matrix $P$ that characterises this "random walk" (specifically, choosing a random shortest path) that starts at $s$ and ends at $t$?

Intuitively, it seems clear that $P$ is given by \begin{equation} P_{ij} = \frac{\omega_{ij}}{\sum_k \omega_{ik}}\,, \end{equation} where $\omega_{ij}$ is the number of shortest paths from $s$ to $t$ that traverse the directed edge $(i,j)$.

To see this, consider the $k$ shortest paths from $s$ to $t$. If we let $k$ walkers walk each of the $k$ shortest paths, their empirical transition matrix is given by $P$.

However, this is not a rigorous proof. Can someone provide a rigorous proof or fill in the missing step(s)?

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    $\begingroup$ Of course you must assume a path from $s$ to $t$ exists. Is your graph finite (or do you have some other condition that implies there exist finitely many shortest paths)? Then $P_{ij}$ is the fraction of shortest paths from $i$ to $t$ that start with $(i,j)$, which is clear from its formulation as a conditional probability. $\endgroup$ Jun 13, 2019 at 21:07
  • $\begingroup$ Indeed. Yes, I assume the graph is finite. $\endgroup$ Jun 14, 2019 at 8:28

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The formula you write is correct. Your chain is constrained to go from any node to a node that is one step closer to the sink $t$. This can be thought of as a Doob-transform of the space-time chain $(X_k,k)$. See page 256 in the book https://bookstore.ams.org/mbk-107/ (also available at https://pages.uoregon.edu/dlevin/MARKOV/mcmt2e.pdf ) or http://staff.utia.cas.cz/swart/chain2.pdf

More directly, you can infer your formula from Bayes' theorem.

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