# Simple random walk in 1D with possibility to stand still

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 $$a<0, compute $$\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.?

• Hint: Compute $\mathbb{P}_x(\tau_a<\tau_b)$ and $\mathbb{E}_x[\tau_a\wedge \tau_b]$ for every $a\leqslant x\leqslant b$.
– Did
Commented Dec 12, 2018 at 21:24
• I have no idea what do you mean by the subscript $x$?
– Sean
Commented Dec 12, 2018 at 21:27
• @Did In his question the initial state of the random walk is $S_0 = 0$ Commented Dec 12, 2018 at 21:37
• @Falrach I know, and?
– Did
Commented Dec 12, 2018 at 22:24
• Then I do not understand your hint or why it is helpful. Commented Dec 12, 2018 at 22:33

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.

• To get $\lim_{n \to \infty} \mathbb{E}(S_{T \wedge n}) = \mathbb{E}(S_T)$ you need to know that $\mathbb{P}(T<\infty)=1$ (... which the OP doesn't know, as his question shows).
– saz
Commented Dec 13, 2018 at 8:23