Question:
Let $K_{m,n}$ be a weighted complete bipartite graph with weights $w_{ij}$ for each edge $(i,j)$. What is the effective resistance $R_{i,j}$ between any two nodes of the graph? I am interested in using its expression to compute the pseudoinverse of the graph Laplacian.
Some context:
A possible starting point may be this question, which I would like to extend to the weighted graph case. I believe (I may be completely off track) that following the answer (3rd solution) from @Misha Lavrov it is possible to find the resistance between any two points belonging to a different partition of the graph. My attempt is as follows.
Let any edge $(i,j)$ of the graph have weight $w_{ij}$, i.e. resistance $r_{ij} := 1/w_{ij}$. Let $s_i := \sum_j{w_{ij}}$ for each node $i$. We can define a random walk on the edges of the graph using the transition matrix $P_{ij} := \frac{w_{ij}}{s_{i}}$ (note that even if the weight metrix is symmetric in general $P_{ij} \neq P_{ji}$). Then the effective resistance then can be measured as the probability that a random walk starting at $u$ ends up at $v$ by taking the edge $uv$, that should be:
$$ R_{u,v} := P_{uv} + (1-P_{uv})P_{vu} $$
with the interpretation: $P_{uv}$ ''$=$'' probability of going from $u$ to $v$ when doing the first step $+$ $(1-P_{uv})$ ''$=$'' probability of jumping to another node as the first step, and then $P_{vu}$ "=" probability of reaching $v$ from $u$ every time we are on the same side of $u$.
Here I get stuck:
Is the random walk ''approach'' applicable also for two nodes $u,v$ belonging to the same partition? What is the interpretation of the effective resistance then? (since there is no edge $(u,v)$). Is there another approach to find $R_{u,v}$ using other tools? (e.g. 1st and 2nd solution of the answer from @Misha Lavrov).
Thanks in advance!