A 4-dimensional Markov chain Consider a 4-dimensional Markov chain with states $(a,b,c,d) \in \mathbb{N}^4$.
The transition rule is the following:
\begin{align}(a,b,c,d) \rightarrow \begin{cases}
(a+1,b,c,d) \;\;\;\; \text{ with probability }p_x &\text{ if }ad>bc\\
(a,b,c+1,d) \;\;\;\; \text{ with probability }1-p_x &\text{ if }ad>bc\\
(a,b+1,c,d) \;\;\;\; \text{ with probability }p_y &\text{ if }ad\leq bc\\
(a,b,c,d+1) \;\;\;\; \text{ with probability }1-p_y &\text{ if }ad\leq bc\\
\end{cases}
\end{align}
Denote $P_t$ the probability distribution at time $t$, and assume $P_0(a,b,c,d) = \delta(1,1,1,1)$, i.e. the initial state is $(1,1,1,1)$ with probability 1.
I'm trying to describe $P_t$ but am having some trouble.
What i've done: I was thinking to define a recursive relation (denoting $(a-1)$ instead of $(a-1,b,c,d)$ for simplicity):
$$P_t(a,b,c,d) = P_{t-1}(ad>bc)\big[p_x P_{t-1}(a-1) + (1-p_x)P_{t-1}(c-1)\big] + P_{t-1}(ad\leq bc) \big[p_y P_{t-1}(b-1) + (1-p_y)P_{t-1}(d-1)\big]$$
but am not sure how to compute $P_t(ad>bc)$.
$f(a,b,c,d) = ad-bc$ seems to me like a Lyapunov function, and I know vaguely that Lyapunov functions are used to determine whether states are transient/recurrent. Is this useful here?
Ultimately, my goal is to describe the usual quantities of interest (stationary states, distribution of hitting times, etc.) for this process. So if $P_t$ isn't the way to go, I'm happy to use other methods.

Q) What are the recurrent states?
Once you leave a state, you can't get back to it. Therefore the states don't communicate. Therefore no state is recurrent, ie all states are transient.
Q) What are the stationary states?
The states are all transient implying no stationary states.
Q) What is the hitting time of a state given by $\{a,b,c,d\}$?
Starting at $\{0,0,0,0\}$, we can move only right or up. Therefore the process only hits parts of the $\mathbb N^4$ grid. Most states are not hit. Therefore can we describe a finite hitting time?
