# Expected time of absorption for time-changed Markov chain

Let $S=\{1,...,s\}$ be a finite number of states, where $s$ is the absorbing state. The transition probability from state $i$ to $j$ is $P_{ij}$, but the time it takes from state $i$ to $j$ is given by $t_{ij}$.

When all $t_{ij}=1$, we can calculate the expected time (i.e. number of steps) needed from state $1$ to $s$ by taking the standard matrix: $E(T) = (1\;0\;0\cdots0) (I-Q)^{-1} (1\;1\;1\cdots1)^T$ where $Q=(P_{ij})_{1\leq i,j\leq s-1}$.

Is there an analogue of such formula for non-constant times $t_{ij}$?

Simply adapt the classical proof, namely, consider $v_i=E_i(T)$ for every $i$, then $v_s=0$ and $$v_i=\sum_jP_{ij}(t_{ij}+v_j)$$ for every $i\ne s$, that is, $v=(v_i)_{i\ne s}$ solves $$v=w+Qv$$ where $Q$ is the matrix in your post and $$w_i=\sum_jP_{ij}t_{ij}$$ Thus, $$v=(I-Q)^{-1}w$$ and, in particular, $$E_1(T)=v_1=(1,0,\ldots,0)(I-Q)^{-1}w$$ When $t_{ij}=1$ for every $(i,j)$, $w=(1,1,\ldots,1)^T$ hence one recovers the formula in your post.