Mean hitting times with 2 absorbing states.

Question

Let the transition matrix be: $$\begin{pmatrix}1 & 0 & 0& 0& 0& 0\\\ 1/2 & 0 & 1/2 & 0& 0& 0 \\\ 1/2 & 0 & 0& 0 & 1/2& 0 \\\ 0 & 1/2 & 0& 0& 0& 1/2 \\\ 0 & 0& 0& 1/2 & 0& 1/2 \\\ 0 & 0& 0& 0& 0& 1\end{pmatrix}$$

Given you start at $P_2$ (the second column) find the mean hitting times.

I have tried to do this by marginalisation, but I am unsure on how to actually apply this here, given there are 2 absorbing states. Is it right to write out all of the mean stopping time, e.g. $MST_{P{_{2, 1000}}} $ = $1+1/2$ $MST_{P{_{5, 1000}}} $ and then do the same the other way around?

Answer 1

I am not totally clear about your notation but you can say something like

• $MST_1=0$
• $MST_2=1+ \frac12MST_1+\frac12MST_3$
• $MST_3=1+ \frac12MST_1+\frac12MST_5$
• $MST_4=1+ \frac12MST_2+\frac12MST_6$
• $MST_5=1+ \frac12MST_4+\frac12MST_6$
• $MST_6=0$

which are six equations in six unknowns and easily solvable by substitution

Answer 2

Hint: There is a positive probability that the walk goes from state 2 immediately to state 1 (an absorbing state) while never visiting any other. Similarly, there is a positive probability that the walk goes from state 2 to 3 to 5 to 6 (the other absorbing state) while never visiting any other.

Thus: for any state $x$, there is a positive probability that the walk never travels from state 2 to $x$. What does that imply about the hitting times?