Hitting times of a biased continuous time random walk

Question

Let $X_{s \geq 0}$ be a continuous time random walk on $\mathbb{Z}$, i.e. waiting times between jumps are exponentially distributed with mean one. The random walk is biased: $\mathbb{P}(X_s\text{ jumps up}) = 2/3$, $\mathbb{P}(X_s\text{ jumps down}) = 1/3$. Let $a \in \mathbb{Z}_{+}$. I am trying to derive a lower-bound on $\mathbb{P}(\tau_a < s)$ for $a$ and $s$ large, where $\tau_a$ is a hitting time of $a$. I tried embedding this random walk into a Brownian motion and then deal with hitting times of a BM of a line, however the calculations of time rescaling are very technical.

Answer 1

Not a complete answer, but this might give you some ideas about that probability:

If the jumps arrive with exponential waiting time, your process can be expressed as a compound Poisson process: $$ X_t = \sum^{N_t}_{i=1}J_i $$ where $N_t$ is a Poisson process with intensity $\lambda=1$ and $J_i$ are iid random variables which are $1$ with probability $2/3$ and $-1$ with probability $1/3$.

To find the probability, condition on number of jumps in the interval $[0, s]$: $$ \mathbb P(\tau_a <s) = \mathbb E[\mathbb E[1_{\tau_a<s} \mid N_s = n]] = \sum^{\infty}_{n=0}\mathbb E[1_{\tau_a<s} \mid N_s = n] \mathbb P(N_s=n)\\ =\sum^{\infty}_{n=0}\mathbb P(\tau_a<s \mid N_s = n) e^{-\lambda s}\frac{(\lambda s)^n}{n!} $$ where $\lambda=1$ in your case. The probability inside the sum is the same as $$ P(\tau_a<s \mid N_s = n) =\mathbb P \left( \max_{1\leq k\leq n}\sum^k_{i=1} J_i \geq a \right) $$ which can be viewed as the probability that a simple asymmetric random walk reaches the level $a$ within $n$ steps. That is a combinatorial problem which most probably has been solved by someone.