Is there any way to calculate the equilibrium (stationary) distribution for a weighted directed acyclic graph? Some references emphasized adjacency matrix to be symmetric. https://arxiv.org/abs/1012.1211#content Graph Example
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If the graph is acyclic, then either you have sinks (nodes with no edge out), or your graph is infinite.
If you have sinks, then this would indicate that you have a loop of probability $1$ at this node (you have to do something of your probability). In such a configuration, only the sink nodes can have a non-zero stationary probability. And any probability distribution whose support is among the sink nodes is stationnary.
If you don't have sinks, then you have infinite paths, so infinite (discrete) number of nodes. This means discrete probability techniques won't work (no unique node can have a non-zero probability). Then, maybe the results you could have there depend on the $\Sigma$-algebra you use, but my guess is that in most interesting cases you won't have any stationary distribution.
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1$\begingroup$ For graphs that are not acyclic but only directed, you can study the strongly connected components of the graph. Then again, only the sink components will possibly have non-zero probability. The study of strongly connected directed graph is another question, though. $\endgroup$ Commented Mar 13, 2021 at 19:37
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$\begingroup$ Thanks for helpful answer, Graphs I'm referring to have one start node and one end node $\endgroup$ Commented Mar 13, 2021 at 19:46
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$\begingroup$ Then everything would be in the end node. $\endgroup$ Commented Mar 13, 2021 at 19:50
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$\begingroup$ So there is not a Unique stationary distribution? $\endgroup$ Commented Mar 13, 2021 at 20:06
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