# Difference Expected Hitting Time and Mean Absorption Time Markov Chain

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): $$$$\begin{cases} H_{1,j} = 1+ p_{1,j}H_{1,j} + p_{1,2}H_{2,j}...+p_{1,j}H_{j,j}\\ H_{2,j} = 1+ p_{2,1}H_{1,j} + p_{2,2}H_{2,2}...+p_{2,j}H_{j,j}\\ ...\\ H_{j,j} = 0 \end{cases}\,.$$$$ Then the expected time from state $$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?

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 , $$A_{jj}=H_{jj}=0$$ .