Consider a biased simple random walk $S_{n}$ on $\mathbb{Z}_{\geq 0}$ which moves up with probability $p$ and down with probability $q = 1-p$, $q>p$, except when at $0$, in which case it stays put with probability $q$ and moves up with probability $p.$
It is true that there exists a constant $\rho<1$ such that for all $a, n \in \mathbb{N} \cup \{0\}$,
$$\mathbb{P}(S_{n} > a) \leq \rho^{a}?$$
I believe it should be true, but don't know how to easily prove it. In the case of the biased random walk on $\mathbb{Z}$ (without the barrier at $0$), one can derive this result by noting
$$\mathbb{P}(S_{n} > a) \leq \mathbb{P}(\max_{t}\{S_{t}\} > a)$$
and using the fact that $\max_{t}\{S_{t}\}$ is geometrically distributed with mean $q/p.$ The same doesn't work here, though, since it seems like the maximum of the random walk with barrier is $\infty$ almost surely.
Any help is appreciated!
One approach would be to use generating functions. The general theory of 1D walks normally just allows a space homogeneous step set, i.e., the same set of steps whereever we are. Yet, you could use, e.g., https://arxiv.org/pdf/1605.01687.pdf. Using the terminology therein, you want to count meanders given in Theorem 2.1, and your step sets are $$ P(u) = p u + \frac{q}{u} \qquad \text{ and } \qquad P_0(u) = p u + q.$$ Then, we get an explicit (yet, not too nice, but not too ugly) expression for $$ F(z,u) = \sum_{n,k} \mathbb{P}(S_n=k) z^n u^k.$$ The following simple transformation then gives $$ G(z,u) := \sum_{n,k} \mathbb{P}(S_n>k) z^n u^k = \frac{1}{(1-z)(1-u)} - \frac{F(z,u)}{1-u}.$$ It remains to prove that $[u^k] G(z,u) < \rho^k$ for some $\rho \in (0,1)$. I have done it for $k=0$ and $k=1$, and it seems to be the case that $\rho=p/q$. What is more, the more general bound seems to hold $$ \mathbb{P}(S_n>k) \leq \rho^{k+1},$$ yet it would be some work to prove that (at least I don't see right now, how to get it quickly). What is rather easy to get using Analytic Combinatorics and especially singularity analysis for algebraic functions, is that $$ \lim_{n \to \infty} \mathbb{P}(S_n>k) = \rho^{k+1}.$$ So the above conjecture is true asymptotically (i.e., for large $n$), and here sharp.