Expected time until different starting states will reach the same state in a Markov chain Consider a Markov chain with $n$ states and transition probabilities $p_{ij}$. For a given pair of states $s_1$ and $s_2$, how can I express the expected time until a chain starting in $s_1$ and a chain starting in $s_2$ will reach the same state?
I take it that the probability that they are in the same state in the next period is given as: $$\sum_{j=1}^n p_{1j}p_{2j}.$$
Based on this expression, I take it that the expected time until they reach the same state is given by the $k$ such that:
$$\sum_{m=1}^k \sum_{j=1}^n p_{1j}^{(m)}p^{(m)}_{2j} = 1.$$
I have two questions. First, is this expression correct? Second, can it be expressed in a more convenient way?
 A: Here is a way to implement the idea saulspatz described in the comments.  I assume the chain starting in $\ s_1\ $ and the one starting in $\ s_2\ $ are independent of each other.  If $\ s_1\ $ and $\ s_2\ $ are in different communicating classes, or they both belong to a communicating class that contains more than one absorbing state, then the expected time for the two chains to reach the same state will be infinite, because there is a positive probability that they will never do so. I'll therefore assume that the original chain comprises a single communicating class with no absorbing states.  If the original chain does have an absorbing state, it can be dealt with by a similar procedure.
Consider the Markov chain with $\ (n-1)^2\ $ non-absorbing states $\ (i,j)\ $ with $\ i\ne j\ $, one absorbing state $\ a\ $, and transition matrix $\ \overline{P}\ $ given by
\begin{align}
\overline{P}_{\left(i_1,i_2\right) \left(j_1,j_2\right)}=&p_{i_1j_1}p_{i_2j_2}\ \ \text{ for }\ \ i_1\ne i_2, j_1\ne j_2\\
\overline{P}_{\left(i_1,i_2\right)a}=&\sum_{j=1}^np_{i_1j}p_{i_2j}\ \ \text{ for }\ \ i_1\ne i_2\\
\overline{P}_{aa}=&1\ .
\end{align}
The expected time you wish to find is the expected time, $ e_{\left(s_1s_2\right)}\ $, to absorption in this chain from the starting state $\ \left(s_1,s_2\right)\ $, which must satisfy the linear equations
$$
e_{s_1s_2}=1+\sum_{j_1=2}^n\sum_{j_2=1}^{j_1-1}p_{s_1j_1}p_{s_2j_2}e_{j_1j_2}+\sum_{j_1=1}^{n-1}\sum_{j_2=j_1+1}^n p_{s_1j_1}p_{s_2j_2}e_{j_1j_2}\ .
$$
If $\ \hat{P}\ $ is the $\ (n-1)^2\times(n-1)^2\ $ submatrix of $\ \overline{P}\ $, obtained from it by deleting the row and column corresponding to the absorbing state $\ a\ $, then these equations can be written in matrix form as
$$
e=\left(I_{(n-1)^2\times(n-1)^2}-\hat{P}\right)^{-1}\mathbb{1}_{(n-1)^2}\ .
$$
