Consider a biased random walk $(X_s)$ on $\mathbb{Z}$ with the probability of moving to the right equal to $p>1/2$ and the probability of moving to the left equal to $q=1-p$. Here $X_t$ is the position of the walk at time $t$.
I am wanting to know if a concentration inequality of the form: $$\mathbb{P}(|\max_{s\leq t} X_s - X_t|>f(t))<g(t)\,,$$ where $g(t)\to 0$ as $t\to \infty$, such that the above inequality holds? The goal for me is to be able to say $|\max_{s\leq t} X_s - X_t|\leq f(t)$ with high probability (say as $t\to \infty$, this probability goes to one.). If it does, I am wondering what $f(t), g(t)$ could be?
My attempt: I know that the distribution of $\min_{s\leq t} X_s$ is geometric from here but I do not know the distribution of $\max_{s\leq t} X_s$. I could try to apply Markov's inequality and say: \begin{align*} \mathbb{P}(|\max_{s\leq t} X_s - X_t|>f(t)) &\leq \frac{\mathbb{E}|\max_{s\leq t} X_s - X_t|}{f(t)} \\ &\leq \frac{\mathbb{E}|\max_{s\leq t} X_s|+\mathbb{E}|X_t|}{f(t)} \\ &= \frac{h(t) + c_2t}{f(t)} \,, \end{align*} where $h(t)=\mathbb{E}|\max_{s\leq t} X_s|, c_2t=\mathbb{E}|X_t|$. So if I know the distribution of $\max X_s$, then I can choose $f(t)$, such that $$\frac{h(t) + c_2t}{f(t)} \to 0\quad\text{as}\quad t\to \infty\,.$$
But I expect better $f(t),g(t)$ are already known using concentration inequalities more powerful than the Markov's inequality since the biased random walk on $\mathbb{Z}$ is a well-studied problem. Any thoughts? Thanks.