Hitting probabilities for a velocity-switching Markov ladder ($\vartheta =\mu (\lambda +\alpha )/[\lambda (\mu +\beta )]$)

Consider states $C_i,W_i$ for integer levels $i\ge 0$ and strictly positive rates $\lambda ,\mu ,\alpha ,\beta$. From $C_i$ the chain moves to $C_{i+1}$ at rate $\lambda$ and to $W_i$ at rate $\alpha$. From $W_i$, $i\ge 1$, it moves to $W_{i-1}$ at rate $\mu$ or to $C_i$ at rate $\beta$. At the reflecting boundary $W_0$ there is only the switch to $C_0$; no negative level is allowed. All rates are bounded, so the chain is nonexplosive.

Let $h$ be the probability of hitting $W_0$, with $h_{W_0}=1$. First-step equations arewith the first equation also valid at $i=0$. Put $\vartheta =\mu (\lambda +\alpha )/[\lambda (\mu +\beta )]$. If $\mu \alpha <\lambda \beta$, then $\vartheta <1$ andSubstitution verifies both equations and the boundary. Eliminating the $C$ variables gives a second-order recurrence with modes $1,{\vartheta }^i$. The displayed decaying solution is the minimal nonnegative harmonic solution and hence the Hitting probability.

When $\mu \alpha \ge \lambda \beta$, the growing mode is excluded by boundedness; at equality the second mode is linear and is likewise excluded. The only bounded solution with the boundary value is one, proving almost-sure hitting. Since the specified reflected chain is irreducible, these cases classify it as recurrent or transient respectively. Specifying the boundary rule is essential for that recurrence statement; making $W_0$ absorbing would be a different chain.