Random Walk Stopping Time

Question

Let $(X_1,X_2,...)$ be i.i.d random variables, with $P(X_t=1)=P(X_t=-1)=1/2$. Then

$S_t= \frac{1}{t}\sum_{i=1}^{t}X_i $ is a zero mean random walk. Let $\tau$ be the stopping time corresponding to the first time that $S_t$ hits one, \begin{equation} \tau=\inf\{t \geq 1 \mid S_t \geq 1 \} \end{equation} I am trying to show that $P(\tau = \infty)>0$.

For starters, clearly $P(\tau < \infty)>0$ since $P(\tau=1)=1/2$. Also, the probability that $S_t$ exceeds $1$ infinitely often is zero, since $S_t$ converges to $0$. But I would like to show that with positive probability, $S_t$ can converge to zero without ever exceeding $1$.

Answer 1

Since $X_t$ takes only the values $1$ and $-1$, we have

$$|S_t| \leq 1$$

for all $t$. In particular, we find

$$\{S_t \geq 1\} = \{S_t = 1\} = \bigcap_{j=1}^t \{X_j = 1\}$$

for all $t \in \mathbb{N}$, i.e. $S_t \geq 1$ can only happen if all $X_j$ equal $1$ for $j=1,\ldots,t$. This implies

$$\mathbb{P}(\tau<\infty) \leq \mathbb{P}(X_1 = 1).$$

On the other hand, if $X_1 = 1$, then $\tau=1$; hence, we actually get

$$\mathbb{P}(\tau<\infty) = \mathbb{P}(\tau=1) =\mathbb{P}(X_1=1) = \frac{1}{2}.$$

Answer 2

You have $\{X_1 = -1\}\subseteq \{S_t \leq \frac{t-1}{t} < 1\ \forall t\in \Bbb{N} \}$ $ \subseteq \{ \tau = +\infty \}$

So

$$ 1/2 = \Bbb{P}(X_1 = -1) \leq \Bbb{P}(S_t \leq \frac{t-1}{t} < 1\ \forall t\in \Bbb{N})\leq \Bbb{P}(\tau = +\infty) $$