Let (Xn)n≥0 be a Markov chain with the state space S and Transition Kernel P. Let A⊂S be a subset of interest, and TA the first time that the MC hits the set A: TA(ω)=inf{n≥0:Xn(ω)∈A}
1. TA takes values from {0,1,⋯} 2. {TA=n}={X0∈A,⋯,Xn−1∈A,Xn∈A} 3. {TA=n}={(X0,⋯,Xn)∈Bnc}
For each i∈S, denote the probability of hitting A starting from position i by hiA=P(TA<∞∣X0=i)where TA is the hitting time for event A. For simplicity, we will write Pi for P(⋅∣X0=i).
1. The vector of hitting probabilities hA=(hiA,i∈S) is a solution to the equations: xi=1,xi=j∈S∑pijxj\mboxifi∈A\mboxifi∈A(*) 2. hA is the minimal non-negative solution to (*) in the sense that for any other non-negative solution x to (*), hiA≤xi,\mboxforeachi∈S
For i∈S, the expected time of hitting A is κiA=Ei[TA]=E[TA∣X0=i]where TA is the hitting time of event A. By definition we have κi=n<∞∑nPi(TA=n)+∞∗Pi(TA=∞) If the probability of not hitting the set A from state i is positive, then κi=∞. - i.e. if hitting probability hi=1 then κi<∞ or “if we’re for sure hitting A then we have a expected hitting time” - i.e. if hitting probability hi<1 then κi=∞ or “if we’re not hitting A for sure then expected hitting time is infinite”
1. The vector of expected hitting times κA=(κiA,i∈S) is a solution to the equations:xi=0,xi=1+j∈Ac∑pijxj\mboxifi∈A\mboxifi∈A(*) 2. κA is the minimal non-negative solution to (*) in the sense that for any other non-negative solution x to (*), κiA≤xi,\mboxforeachi∈S