# Given N1 (time to termination) for an absorbing markov chain find a possible P

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 $$N$$?

• Just build a chain where every main state transitions either to itself or to an absorbing state, and design the probabilities so the expected time to reach an absorbing state is as desired. Commented Aug 2, 2022 at 20:19

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 $$P=I-N^{-1}$$.