Let $\{\xi_i\}$, $i=1,2,...$, i.i.d., be Bernoulli random variables with $P(\xi_i=1)=p$ and $P(\xi_i=-1)=q=1-p$. Consider the random walk $X_n = \sum_{k=1}^{n} \xi_k$. Let $a < 0 < b$ integers, and $T_a = \inf\{n > 0: X_n = a\}$ and $T_b = \inf\{n > 0: X_n = b\}$. I want to find the distribution of $T = T_a \wedge T_b$.
My idea is to find its generating function $E[z^T]$ first. Let $\lambda$ be a positive number satisfying $\lambda>1$, $\lambda p + \frac{q}{\lambda} > 1$ and $\lambda q + \frac{p}{\lambda} > 1$. I consider this: \begin{align*} E[\lambda^{X_n}] = (E[\lambda^{\xi_1}])^n = (\lambda p + \frac{q}{\lambda})^n \end{align*} so $$ E[\lambda^{X_n}(\lambda p + \frac{q}{\lambda})^{-n}]=1, \forall n $$ Let $Y_n = \lambda^{X_n}(\lambda p + \frac{q}{\lambda})^{-n}$, then $Y_n$ is a martingale and $|Y_{n\wedge T}| \leq \lambda^b$. By Doob Stopping theorem, \begin{align*} E[Y_T] &= E[\lambda^{X_T}(\lambda p + \frac{q}{\lambda})^{-T}] \\&= \lambda^a \int_{T_a < T_b}(\lambda p + \frac{q}{\lambda})^{-T}\ dP + \lambda^b \int_{T_b < T_a}(\lambda p + \frac{q}{\lambda})^{-T}\ dP = 1 \end{align*} Let $Y'_n = \lambda^{-X_n}(\lambda q + \frac{p}{\lambda})^{-n}$, then $Y'_n$ is also a martingale and $|Y'_{n\wedge T}| \leq \lambda^{-a}$. By Doob Stopping theorem, \begin{align*} E[Y'_T] &= E[\lambda^{-X_T}(\lambda q + \frac{p}{\lambda})^{-T}] \\&= \lambda^{-a} \int_{T_a < T_b}(\lambda q + \frac{p}{\lambda})^{-T}\ dP + \lambda^{-b} \int_{T_b < T_a}(\lambda q + \frac{p}{\lambda})^{-T}\ dP = 1 \end{align*} If $p=q=\frac{1}{2}$, let $\frac{1}{z} = \lambda q + \frac{p}{\lambda} = \lambda p + \frac{q}{\lambda} > 1$, then we can get: \begin{align*} \int_{T_a < T_b}(\lambda q + \frac{p}{\lambda})^{-T}\ dP = \frac{\lambda^{-b} - \lambda^b}{\lambda^{a-b} - \lambda^{b-a}}\\ \int_{T_a > T_b}(\lambda q + \frac{p}{\lambda})^{-T}\ dP = \frac{\lambda^{a} - \lambda^{-a}}{\lambda^{a-b} - \lambda^{b-a}} \end{align*} so $$E[z^T] = \frac{\lambda^{a} - \lambda^{-a} - (\lambda^{b} - \lambda^{-b})}{\lambda^{a-b} - \lambda^{b-a}} = \frac{\lambda^{a+b} + 1}{\lambda^a + \lambda^b}$$ Plug in $\lambda = \frac{1+\sqrt{1-z^2}}{2z}$, we get $E[z^T]$, hence we get the distribution of $T$.
But this method cannot be used to solve the case when $p \neq q$. In this case, $\lambda p + \frac{q}{\lambda} \neq \lambda q + \frac{p}{\lambda}$. Is there any way to overcome this difficulty?
Set $\mu p = \frac{q}{\lambda}$, then $\mu = \frac{q}{\lambda p} < 1$, $\frac{1}{z} = \lambda p + \frac{q}{\lambda} = \frac{q}{\mu} + \mu p > 1$. Similarly, $$ E[Y_T] = \mu^a \int_{T_a < T_b}(\lambda p + \frac{q}{\lambda})^{-T}\ dP + \mu^b \int_{T_b < T_a}(\lambda p + \frac{q}{\lambda})^{-T}\ dP = 1 \\ $$ Thus \begin{align*} \int_{T_a < T_b}(\lambda q + \frac{p}{\lambda})^{-T}\ dP &= \frac{{\mu}^b - \lambda^{b}}{\lambda^{a} {\mu}^b - \lambda^{b}{\mu}^a}\\ \int_{T_a > T_b}(\lambda q + \frac{p}{\lambda})^{-T}\ dP &= \frac{\lambda^{a} - {\mu}^a }{\lambda^{a} {\mu}^b - \lambda^{b}{\mu}^a} \end{align*} so $$E[z^T] = \frac{\lambda^{a}- \lambda^{b} - ({\mu}^a - {\mu}^b)}{\lambda^{a} {\mu}^b - \lambda^{b}{\mu}^a}$$ where $\lambda = \frac{1 + \sqrt{1 - 4pqz^2}}{2pz}$ and $\mu = \frac{1 - \sqrt{1 - 4pqz^2}}{2pz}$.