If $X$ and $Y$ are independent Markov chains, such that $a$ is the absorbing state of $X$ and $b$ is the absorbing state of $Y$. We know the expected hitting time of $X$ goes to $a$ and expected hitting time of $Y$ goes to $b$. I want to known that the expected hitting time that $X$ goes to $a$ or $Y$ goes to $b$.
I think the answer will be the minimum of the 2 expected hitting time but I am not sure about it and cannot find the proof.
As I suspected, the answer to your question is no, but it's worth understanding why the "intuitive" idea that would suggest the answer is the minimum of the hitting times, fails.
To see the counterexample, let $S = \{1,2\}$, and consider the transition matrices $T_X = \begin{pmatrix}1-p & p \\ 0&1\end{pmatrix}$ and $T_Y = \begin{pmatrix}1-p' & p' \\ 0&1\end{pmatrix}$, so that one can think of the Markov chains $X_n$ and $Y_n$ on $S$ behaving according to the transition matrices $T_X$ and $T_Y$ respectively.
After thinking for a short while, you can convince yourself that $X_n$ and $Y_n$ are in fact modelling geometric random variables, since state $1$ is a failure state, so the systems try to move from state $1$ to state $2$ which is the success state, and once you come back at state $1$, every next try is independent of the previous tries. Therefore, the hitting distribution of state $2$ is $\mbox{Geo}(p)$ for $X_n$ and $\mbox{Geo}(p')$ for $Y_n$.
What about for the probability that either $X_n$ or $Y_n$ hits $2$ at a specific time? The answer to that is that if either hits state $2$ then that particular geometric random variable has recorded a success. In other words, the hitting probability is the minimum of the independent random variables $\mbox{Geo}(p)$ and $\mbox{Geo}(p')$. What is the expectation of this random variable?
As explained here , the answer is $\frac 1{1-(1-p)(1-p')}$, which clearly is not equal to $\min(\frac 1{p},\frac 1{p'})$.
So what's going wrong, and why is it going wrong?
The basic premise to seeing what goes wrong in the example , is the fundamental idea that the minimum does not commute with the expectation, even for independent random variables.
The false intuition that leads us to believe this , is the fact that if the two random variables are independent, then we may as well think of them separately and therefore proceed to answer questions which involve a union of events for both as if we were treating them separately, taking the union event into that sample space and then seeing which one comes first.
This is in fact a false notion , for the union of two events, especially when they are independent, imposes a constraint on both events, since the union not happening entails both not happening simultaneously. Therefore, the simultaneous running of the two chains means that separate treatment of each chain is out of the question.
The simultaneity is expressed in the fact that the event $A \cup B$ is an event concerning a coupling of the random variables $X_n$ and $Y_n$ (even if that's an independent coupling) and therefore one can't correlate this event to any event in the individual sample spaces of $X_n$ and $Y_n$ respectively.
With that behind us, we then realize that a statement that concerns exchange of minimum/maximum must then occur only if , regardless of coupling, one of the random variables (in this case, stopping times) always lags behind the other or stays ahead of the other.
This nice notion is made precise in the MathOverflow answer here. Essentially, the only situation in which a maximum/minimum and an expectation can commute is the "obvious" case where one of the random variables always rises above all the others. For example, since the hitting times of $2$ for the chains $X_n$ and $Y_n$ don't dominate each other (i.e. with non-zero probabilities, each can happen before the other in at least one coupling) it follows that here the minimum and expectation won't commute and so the expectation for the first hitting time of $2$ for either $X$ or $Y$, won't equal the minimum of the two hitting time expectations.
