I am working with an absorbing markov chain as described here: https://en.wikipedia.org/wiki/Absorbing_Markov_chain
Normally you would have a Matrix:
$$P=\begin{pmatrix}Q & R\\ 0 & I_r\end{pmatrix}$$
where $Q$ is a t-by-t matrix, $R$ is a nonzero t-by-r matrix, $0$ is an r-by-t zero matrix, and $I_r$ is the r-by-r identity matrix. Thus, $Q$ describes the probability of transitioning from some transient state to another while $R$ describes the probability of transitioning from some transient state to some absorbing state.
From here you would compute things like $N$ (a t-by-t matrix where each cell is the expected time spent in state j when starting from state i before absorption)
And $N1$ which is a sum over the rows of $N$ and is the expected time to termination from every initial state i
My Question is this:
Given $N1$ can a plausible matrix $P$ be discovered that would result in $N1$?
If not can $P$ be computed given $N$?
We assume that $N_{ij}$ for transient $i,j$ is the expected time (including time $0$) spent in state $j$ when starting from state $i$, before absorption. Extend $N$ to be a square matrix of side $t+r$ as follows: If $i$ is absorbing, let $N_{ii}=1$. If $i,j$ are distinct and at least one of them is absorbing, let $N_{ij}=0$.
If $i,j$ are both transient, then $(PN)_{ij}=\sum_{\ell}p_{i \ell}N_{\ell j}$ is is the expected time (NOT including time $0$) spent in state $j$ when starting from state $i$, before absorption. Thus in this case, $$(N-PN)_{ij}=N_{ij}-\sum_{\ell}p_{i \ell}N_{\ell j}=I_{ij} \,.$$ By inspection, this identity also holds if at least one of $i,j$ is absorbing, so it holds for all $i,j$. Thus $(I-P)N=I$ so $P=I-N^{-1}$.