Difference Expected Hitting Time and Mean Absorption Time Markov Chain

Question

I am a bit confused between the 2 definitions. When you calculate the hitting time of state $j$ from $i$, with $j$ being an absorption state, I'm confused if it is the same or something else of the mean absorption time. And if it is the same, I don't quite see how they are the same.

The hitting time of state $j$ from state $i$ is $H_{ij} = \min \{n\geq 0:X_n=j|X_0=i\}$. Using this definition we can fine expected time to hit a state j starting from i by solving the system of equations (I tried to write a general form but my notation might not be the best): \begin{equation} \begin{cases} H_{1,j} = 1+ p_{1,j}H_{1,j} + p_{1,2}H_{2,j}...+p_{1,j}H_{j,j}\\ H_{2,j} = 1+ p_{2,1}H_{1,j} + p_{2,2}H_{2,2}...+p_{2,j}H_{j,j}\\ ...\\ H_{j,j} = 0 \end{cases}\,. \end{equation} Then the expected time from state $i$ to $j$ is $H_{i,j}$.

Consider the mean absorption time. Then we find the fundamental matrix $(\mathbb{I}-\mathbb{Q})^{-1}$ of our transition matrix. Then multiplying by $$\mathbb{G} = \begin{pmatrix} 1\\ 1\\ ...\\ 1\\ \end{pmatrix}$$ gives every step a "reward" of 1. Thus $(\mathbb{I}-\mathbb{Q})^{-1}\mathbb{G}$ gives the mean time till absorption.

Are both of these definitions and ways of getting an expected absorption time the same, or is there something different that I am overlooking?

Answer 1

I realized the later is just a particular case of the former so it's pretty trivial . But anyways , by referencing wiki I deduce the following :

Assume time homogenous markov chain . Suppose $j$ is the only absorbing state and state space $I$ such that $|I|=n$ . I suppose the mean absorption time starting from state $j$ : $A_{jj} = 0 $ .

We have $n-1$ transient states and $1$ absorbing state $j$ . $Q_{(n-1)\times (n-1)}$ describes the probability of transitioning from some transient state to another . Since $H_{jj} = 0 $ , $H_{ij} , i\neq j$ is independent of $H_{jj} $ . The vector of mean hitting time $H_{(n-1)\times 1} = (H_{ij}) , i\neq j $ can be solved from $$ (I-Q)H= G $$ if $(I-Q)^{-1}$ exists , let the vector of mean absorption time be $A_{(n-1)\times 1} = (A_{ij}), i\neq j $ , we have from the equation you provided $A = (I-Q)^{-1} G = H $ . Furthermore , $A_{jj}=H_{jj}=0$ .

So if the set of all hitting states equals the set of all absorbing state (in this case both are j) , I think mean hitting time is equal to mean absorption time .