Comparison of hitting probability of two Markov chains both with only one absorbing state version 3

Question

Let $N_n:=\{1,2,\cdots,n\}$. Given two finite states Markov chains $\big(X^{(j)}_t\in N_n\}\big)_{t=0}^\infty$ for $j\in\{1,2\}$, both of which have two absorbing states at $1$ and $n$. Define $p_{i,j}(t):=\text{Pr}\big(X^{(1)}_{t+1}$ and $q_{i,j}(t):=\text{Pr}\big(X^{(2)}_{t+1}=j|X^{(2)}_t=i\big), \,\forall i,j\in N_n$. Dropping the variable $t$ for the brevity of notaion, we stipulate that $$p_{1,1}=p_{n,n}=q_{1,1}=q_{n,n}=1;$$ $$p_{i,j}>q_{i,j}, \forall 1<i<j, i,j\in N_n;$$ $$p_{i,j}<q_{i,j}, \forall n>i>j, i>1, i,j\in N_n;$$ $$p_{i,i}=q_{i,i}, \forall n>i>1.$$

Are the following inequalities true? $$\text{Pr}\big(X^{(1)}\text{ reaches } b \text{ or above}|X^{(1)}_0=a\big)>\text{Pr}\big(X^{(2)}\text{ reaches }b\text{ or above}|X^{(2)}_0=a\big), \,\forall 1<a<b,$$ and $$\text{Pr}(X^{(1)}\text{ reaches }b\text{ or below}|X^{(1)}_0=a)<\text{Pr}(X^{(2)}\text{ reaches }b\text{ or below}|X^{(2)}_0=a), \,\forall n>a>b.$$

This mathoverflow.net answer demonstrates a counterexample for a stronger claim.

Would a coupling argument help to prove the inequalities if they are true?

Answer 1

Let $P=(p_{ij})$ and $Q=(q_{ij})$ be the $n\times n$ transition matrices for the two respective Markov chains. Take $n=5$, $$P=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 1 & 241 & 260 & 38 & 460 \\ 22 & 75 & 283 & 448 & 172 \\ 389 & 67 & 103 & 158 & 283 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right), $$ $$Q=\frac1{1000}\left( \begin{array}{ccccc} 1000 & 0 & 0 & 0 & 0 \\ 4 & 241 & 259 & 37 & 459 \\ 23 & 531 & 283 & 1 & 162 \\ 390 & 197 & 194 & 158 & 61 \\ 0 & 0 & 0 & 0 & 1000 \\ \end{array} \right). $$

Then $f_{P;21}=\dfrac{30684666}{198426719}=0.154\ldots\not\le0.054\ldots= \dfrac{4510572}{83295877}=f_{Q;21}$, where $f_{P;ij}$ is the probability that the first chain ever reaches a state $j$ from $i$, and $f_{Q;ij}$ is defined similarly.

This disproves your conjecture, because there is no state to the left of $1$.

Remark. Using the formula $f=\lim_{t\uparrow1}(I-tR)^{-1}g$, we see that, for $n=5$, $$f_{P;21}=\big[-p_{24} \left(\left(1-p_{33}\right) p_{41}+p_{31} p_{43}\right)-p_{23} \left(p_{34} p_{41}+p_{31} \left(1-p_{44}\right)\right)+p_{21} \left(-p_{44} p_{33}+p_{33}+p_{34} p_{43}+p_{44}-1\right)\big] \\ \big/\big[p_{23} p_{32}+p_{24} p_{43} p_{32}-p_{23} p_{44} p_{32}+p_{33}+p_{24} p_{42}-p_{24} p_{33} p_{42}+p_{23} p_{34} p_{42}+p_{34} p_{43}-p_{33} p_{44}+p_{44}-p_{22} \left(p_{34} p_{43}-p_{33} \left(p_{44}-1\right)+p_{44}-1\right)-1\big]. $$ This expression depends on the $p_{ij}$'s in a very complicated way, which makes simple comparisons such as the one in the OP unlikely to hold.