Comparison of hitting probability of two Markov chains both with only one absorbing state

Question

Let $N_n:=\{1,2,\cdots,n\}$. Given two finite states Markov chains $\big(X^{(j)}_i\in N_n\}\big)_{i=0}^\infty$ for $j\in\{1,2\}$, both of which have one absorbing state $1$. Pr$(X^{(1)}_{i+1}=1|X_i=1)=$Pr$(X^{(2)}_{i+1}=1|X_i=1)=1, \,\forall a\in N_n$. $$\text{Pr}\big(X^{(1)}_{i+1}=b|X_i=a\big)>\text{Pr}\big(X^{(2)}_{i+1}=b|X_i=a\big)>0, \,\forall 1<a<b, a,b\in N_n,$$ $$0<\text{Pr}\big(X^{(1)}_{i+1}=b|X_i=a\big)\le \text{Pr}\big(X^{(2)}_{i+1}=b|X_i=a\big), \,\forall a\ge b, a>1, a,b\in N_n.$$ Is the following true? $$\text{Pr}\big(X^{(1)}\text{ reaches } b|X^{(1)}_0=a\big)>\text{Pr}\big(X^{(2)}\text{ reaches }b|X^{(2)}_0=a\big), \,\forall a<b,$$ and $$\text{Pr}(X^{(1)}\text{ reaches }b|X^{(1)}_0=a)\le\text{Pr}(X^{(2)}\text{ reaches }b|X^{(2)}_0=a), \,\forall a \ge b.$$

This math.stackexchange.com answer states that it is not true for an irreducible transition probability matrix compared to a reducible transition probability matrix. But is it true for two chains each of which has only one absorbing state?

Would a coupling argument help to resolve this problem?

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Here is the version 2 of this question under a more stringent condition.

Answer 1

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains, such that \begin{gather}p_{11}=q_{11}=1,\\ p_{ij}q_{ij}>0\text{ if }i>1,\\ p_{ij}>q_{ij}\text{ if }1<i<j,\\ p_{ij}\le q_{ij}\text{ if }i>1\text{ and }1\le j\le i. \end{gather} The conjecture was that then \begin{gather}f_{P;ij}>f_{Q;ij}\text{ if }1<i<j,\\ f_{P;ij}\le f_{Q;ij}\text{ if }i>1\text{ and }1\le j\le i, \end{gather} where $f_{P;ij}$ is the probability that the first chain ever reaches $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This conjecture is false in general. E.g., suppose that $n=3$, $$P=\frac1{16} \left( \begin{array}{ccc} 16 & 0 & 0 \\ 1 & 3 & 12 \\ 4 & 4 & 8 \\ \end{array} \right),\quad Q=\frac1{16}\left( \begin{array}{ccc} 16 & 0 & 0 \\ 4 & 4 & 8 \\ 4 & 4 & 8 \\ \end{array} \right). $$ Then $$f_{P;22}=\frac9{16}\not\le\frac8{16}=f_{Q;22}.$$