Let $\{X\_t, t \geq 0\}$ and $\{X\_t', t \geq 0\}$ denote two markov chains on the same state space $\{1, ..., n+1\}$ with transition probability matrices $P$ and $P'$ respectively. Suppose that both chains have one absorbing state, namely state $n+1$ (the same for both chains). Let $Q$ and $Q'$ denote the transient parts of $P$ and $P'$ respectively. Suppose that $Q$ and $Q'$ satisfy $\sum\limits\_{k=1}^j Q\_{ik} \geq \sum\limits\_{k=1}^j Q'\_{ik}$ for each $i,j = 1, ..., n$. I want to show that for each state $i \in \{1, ..., n\}$, the expected times until absorption for both of the chains, denoted by $\mathbb{E}T\_i$ and $\mathbb{E}T\_i'$ respectively satisfy $\mathbb{E}T\_i \geq \mathbb{E}T\_i'$.


I think this is true, but struggle with the proof. My idea was to use the Neumann series expression of the expected hitting times:


\begin{equation}
\mathbb{E}[T] = (I-Q)^{-1}[1, 1 \dots 1]^T= (\sum\limits\_{k=0}^{\infty} Q^k)[1, 1, \dots, 1]^T
\end{equation}


and


\begin{equation}
\mathbb{E}[T'] = (I-Q')^{-1}[1, 1 \dots 1]^T= (\sum\limits\_{k=0}^{\infty} Q'^k)[1, 1, \dots, 1]^T
\end{equation}


I think it is possible to show that the row sums of $Q'^k$ are upper bounded by those of $Q^k$ for each $k$, but at this point I'm getting stuck.


Ideas are very much appreciated! Thanks in advance.



Comparison of time until absorption for two absorbing Markov chains
This conjecture is false in general. E.g., suppose that $n=3$,
$$Q=\frac1{10}
\begin{bmatrix}
 3 & 5 & 2 \\
 2 & 2 & 3 \\
 4 & 5 & 1 \\
\end{bmatrix},\qquad 
Q'=\frac1{10}
\begin{bmatrix}
 2 & 0 & 7 \\
 1 & 1 & 4 \\
 4 & 2 & 4 \\
\end{bmatrix}$$
(so that
$$P=\frac1{10}
\begin{bmatrix}
 3 & 5 & 2&0 \\
 2 & 2 & 3&3 \\
 4 & 5 & 1&0 \\
 0 & 0 & 0&10 \\
\end{bmatrix},\qquad 
P'=\frac1{10}
\begin{bmatrix}
 2 & 0 & 7&1 \\
 1 & 1 & 4&4 \\
 4 & 2 & 4&0 \\
 0 & 0 & 0&10 \\
\end{bmatrix}.)$$


Then all the conditions on $Q$ and $Q'$ hold. However,
$$ET\_1=\frac{442}{51}\not\ge\frac{615}{51}=ET'\_1.$$




---


Here is an informal explanation for this example: We have two Markov chains: a $P$-chain, with transition matrix $P$, and a $P'$-chain, with transition matrix $P'$. For both chains, the state space is $\{1,2,3,4\}$ and the absorbing state is $4$. In both chains, only from state $2$ it is possible to get to the absorbing state $4$ (with a nonzero probability) in one step. Moreover, in the $P'$-chain it is impossible to get to $2$ from $1$ in one step, whereas in the $P$-chain this can be done with probability $5/10$. So, in the $P'$-chain, we can only get to $2$ from $1$ by first visiting state $3$. Also, in the $P'$-chain, the probability of getting to $2$ from $3$ in one step is only $2/10$, whereas this probability for the $P$-chain is $5/10$, which latter is substantially greater than $2/10$. All this suggests that it is harder (and hence takes longer on the average) to get from $1$ to $2$ (and hence from $1$ to the absorbing state $4$) in the $P'$-chain than to do so in the $P$-chain.


The above example just shows that the comparison described informally in the last paragraph is compatible with the conditions imposed in the OP.


