The following problem seems to be a basic exercise in probability theory, but I fear the elementary setting is misleading. References and partial proofs will be appreciated!
Suppose $x = \{x_t : t \geq 0 \}$ is a Markov chain on $\mathbb{Z}$ with $x_0 = 1$ and single-step transition probabilities $$p(x,y) = \left\{ \begin{array}{cc} p & \text{ if } ~~y = x+1, \\ 1 - p & \text{ if }~~ y = x-1, \\ 0 & \text{ otherwise, } \end{array} \right. \text{ for } x > 1, \text{ and } ~~ p(x,y) = p(1,y) = \delta_{2,y} \text{ when } x = 1.$$ We suppose $p \neq 1/2$. This random walk has a drift towards or away from zero, and is reflected upon reaching $x = 1$.
I would like an analytic description of the $n$-step transition probabilities; that is, a formula for $P(x_n = k | x_0 = 1)$. Literature searches turned up empty and the bias (drift) makes using reflection techniques impractical (referring to arguments in Lawler & Limic's text). Messy, messy formulas for the transitions can be derived by considering paths composed of $k$ positive excursions (paths starting and ending at a given height $\ell$, which never drop below $\ell$) of length $\tau_{\ell}$, $\ell = 1, 2, \dots, k$. One finds \begin{eqnarray*} P(x_n = k) & = & \sum_{t_1 + \dots + t_k = n-k} P(\tau_1 = t_1) P(\tau_2 = t_2) \cdots P(\tau_k = t_k)p^{k-1} \\ & = & p^{k-1} \sum_{t_1+\dots+t_k = n-k} \prod_{i = 1}^k C_{t_i} p^{t_i/2} (1-p)^{t_i/2} \\ & = & p^{(n+k)/ 2 - 1}(1-p)^{(n-k)/2} \sum_{t_1+\dots+t_k = n-k} \prod_{i = 1}^k C_{t_i} \end{eqnarray*} where $C_m = {1 \over m+1} {2m \choose m}$ is the $m^{\rm th}$ Catalan number. For example, a path that starts at 1 and ends at 2 after $n$ steps must have some time $\tau_1 < n$ such that $\tau_1 = \max \{0 \leq t \leq n : x_t = 1 \}$, after which the walk increases to 2, never to return to 1. Then the walk experiences another excursion of length $\tau_2$ during which it passes from 2 to 2 without dipping back to 1 or below. These three events occur with probability $C_{\tau_1}p^{\tau_1/2}(1-p)^{\tau_1/2}$, $p$, and $C_{\tau_2} p^{\tau_2} (1-p)^{\tau_2}$, respectively, and we must have $$ \tau_1 + 1 + \tau_2 = n.$$The Catalan numbers $C_{\tau_i}$ describe the number of positive excursions of length $\tau_i$. However, I need a closed form solution so this argument seems to be a dead end.
The continuum limit of the process is simply a Brownian motion with drift $2p-1$ and reflection at zero, which might inform one of the general behavior, but no asymptotic information will be sufficient either, as it cannot describe $P(x_n = k | x_0 = 1)$ exactly.