Simple random walk in 1D with possibility to stand still

Question

Let $\{\xi_i\}$ be i.i.d. with $\mathbb{P}(\xi_i=1) = \mathbb{P}(\xi_i=-1) = \frac{3}{8}$ and $\mathbb{P}(\xi_i = 0) = \frac{1}{4}$. Let $S_0=0$ and for $n\geq 1$ let $S_n = \xi_1+\ldots+\xi_n$. For $x\in \Bbb{Z}$ let \begin{equation*} \tau_x = \inf \{n\geq 0: S_n = x\}. \end{equation*} For $a<0<b$, compute $\mathbb{P}(\tau_a<\tau_b)$ and $\Bbb{E}[\tau_a\wedge \tau_b]$.

I am thinking about letting $T = \min\{\tau_a,\tau_b\}$ then it is also a stopping time, but then how can I prove that $T<\infty$ a.s.?

Answer 1

As mentioned in the comment the first step has to be to show that $T < \infty$ a.s. For this let $k := 2 (\vert a \vert \lor b)$. Define the independent events $A_n := \{ S_{kn} - S_{k(n-1)} = k\}$. We have $\Bbb P (A_n) = (\frac 3 8)^k$. Thus $$\sum_{n=1}^\infty \Bbb P (A_n) = \infty$$ and Borel-Cantelli yields that $\Bbb P (\limsup_{n\to\infty} A_n)=1$. This shows $T<\infty$ a.s.

Since the dynamic to go down or to go up is symmetric this can be shown by a standard method:

$(S_n)_{n\in\Bbb N}$ is a martingale. By optional stopping theorem $(S_{T\land n})_n$ is a martingale. Further $\vert S_{T\wedge n}\vert \leq \vert a \vert \lor b$. Hence by dominated convergence theorem $$(1-\Bbb P(\tau_a < \tau_b))b -a\Bbb P (\tau_a < \tau_b) = \Bbb E [S_T] = \lim_{n\to\infty} \Bbb E [S_{T\land n}] = 0.$$

For the expectation value of $T$ consider the martingale $S_n^2 - \Bbb E [S_n^2]$ and do similar steps to above with using the monotone convergence theorem additionally.