I am trying to find the power of the random graph's adjacency matrix.
Consider the $2N$ number of nodes, where $N \geq 2$ is a fixed natural number.
WLOG, let nodes with index $1$ to $N$ belong to Group 1, and nodes with index $N + 1$ to $2N$ belong to Group 2.
Then, for a node pair within the same group, a pair of nodes form an edge with the probability of $p$ (i.e., $P(\{v_{i},v_{j}\} \in \mathcal{E}) = p$ if $v_{i}$ and $v_{j}$ belong to the same group.)
Else, for a node pair in the different groups, a pair of nodes form an edge with the probability of $1-p$ (i.e., $P(\{v_{i},v_{j}\} \in \mathcal{E}) = 1-p$ if $v_{i}$ and $v_{j}$ belong to the different groups.)
We do not allow the self-loop.
You may think this is a variant of the Erods-Renyi random graph, where there are two disjoint node groups. Note that this is an undirected graph.
Let $A$ be an adjacency matrix of this random graph.
My question is: What is the closed-form solution to the expectation of 4-power and 3-power of $A$? That is: $\mathbb{E}_{A}[A^{3}]$ and $\mathbb{E}_{A}[A^{4}]$.
In other words, what would be the expected number of length-3 and length-4 paths between node $v_{i}$ and $v_{j}$ (this would differ whether $v_{i}$ and $v_{j}$ belong to the same group or not)?
Thank you.