Stationary distribution of random walker on a directed weighted graph I know that on an undirected graph, the stationary distribution has a nice closed form solution as $\frac{d_u}{2m}$. I was wondering if I can generalize this formula for directed weighted graph like below.
$$\pi_u = \frac{ \sum_{v \in N \setminus u} w_{v \rightarrow u}} { \sum_{u \in N} \sum_{v \in N} w_{u \rightarrow v}}$$
where $w_{u \rightarrow v}$ is the probability of jumping from $u$ to $v$, thus $\sum_{v \in N \setminus u}w_{u \rightarrow v} = 1$. Does this hold true even for directed graphs? I will be very grateful if someone can explain it. 
I am concerned about general case and not bipartite graphs in particular. 
 A: Such a formula does not hold and, in general, no "local" formula such as this one can hold because the stationary distribution for random walks on weighted graphs really depends on the global structure of such a graph.
To see a very clear-cut example of this, build a directed graph $G$ consisting of two complete directed graphs $G_1$ and $G_2$ (with arbitrary nonzero weights on all edges if you like) and a single edge $(v,w)$ with $v \in V(G_1)$, $w \in V(G_2)$.
Then the limiting distribution of a random walk on $G$ assigns zero probability to any vertex in $G_1$: even if the random walk starts in $G_1$, it will eventually take edge $(v,w)$ with probability $1$, and then the random walk will get stuck in $G_2$ and never come back. On the other hand, if we reverse the edge $(v,w)$ to $(w,v)$, it's vertices in $G_2$ that have a zero limiting probability.
So any formula for the stationry distribution would have to somehow "realize" that $(v,w)$ is a key edge and vary dramatically depending on its orientation.
(For a strongly connected version of this construction, have edges $(v,w)$ and $(w,v)$ both be present, but with the weight of one $100$ times greater than the weight of the other. The resulting graph will still be sensitive to the behavior of this one edge.)

In general, the reason that there's a nice formula in the undirected case and no good formula for the directed case is that random walks on directed graphs are equivalent to arbitrary Markov chains (which we can't say anything specific about) while random walks on undirected graphs are equivalent to time-reversible Markov chains (which have all kinds of nice properties).
