Ladder variables, Durrett Exercise 5.4.2

Question

I'm struggle to prove the Exercise 5.4.2 in Durrett's "Probability: Theory and Examples" Fifth edition book, and need some help to enlighten me a bit.

Here's the question:

It's a random walk/renewal process question.

First, let $\Omega^N = \{ \omega_1, \omega_2,...: \forall \omega_k \in \mathbf{R} \}$, i.e. infinite dimensionally sequence space.

Durret defines a shift operator $\theta: \Omega^N \to \Omega^N$. $\theta(\{\omega_1, \omega_2,...\}) = \{ \omega_2,\omega_3,...\}$, and similarly, $\theta^{n}(\{\omega_1, \omega_2,...\}) = \{ \omega_{n+1},\omega_{n+2},...\}$, i.e. shifting $n$ random variables. It also introduce an infinity value $\Delta$, when $\theta^{\infty}(\omega) = \Delta$ for any $\omega \in \Omega^{N}$

Define the Ladder variables: $\alpha(\omega) = \inf\{n: \omega_1 + ... + \omega_n \gt 0\}$, where $\inf\emptyset = \infty$ and set $\alpha(\Delta) = \infty$, and $\omega_1, \omega_2,...$ are iid random variables taking values in $\mathbf{R}$, and let:

$$ \alpha_k(\omega) = \alpha_{k-1}(\omega) + \alpha(\theta^{\alpha_{k-1}}(\omega)) $$

for all $k \ge 1$. At time $\alpha_k$, the random walk is at a record high value.

Prove:

1. If $P(\alpha \lt \infty) \lt 1$ then $P(\sup{S_n} \lt \infty) = 1$
2. If $P(\alpha \lt \infty) = 1$ then $P(\sup{S_n} = \infty) = 1$

Note: Since each renewal step $\alpha$ is independent, we know that $P(\alpha_k \lt \infty) = P(\alpha \lt \infty)^k$, and such will goes to zero if $P(\alpha \lt \infty) \lt 1$, but I'm not able to associate this with the event $\{ \sup{S_n} \lt \infty \}$. Can anyone help me out?

Answer 1

I figured it out eventually. It's actually quite simple:

1. $P(\alpha \lt \infty) \lt 1$ and the fact that $P(\alpha_k \lt \infty) = P(\alpha \lt \infty)^k$ indicates $\sum_{k=1}^{\infty}P(\alpha_k \lt \infty) \lt \infty$, by the Borel-Cantelli lemma, $P(\alpha_k \lt \infty \text{ i.o}) = 0$, or equivalently, $\alpha_k = \infty$ eventually all. which is equivalent to $\sup{S_n} \lt \infty$ almost surely.

2. $P(\alpha \lt \infty) = 1$ indicates $P(\alpha_k \lt \infty) = P(\alpha \lt \infty)^k = 1$, therefore, $\alpha_k \lt \infty$ with probability one for all $k$. $S_{\alpha_k}$ is a increasing for almost all $\omega$. So $\sup_{n \ge 1}{S_{n}} = \sup_{k \ge 1}{S_{\alpha_k}} = \lim_{n\to\infty}{S_n}$. By strong law of large numbers, $\sup_{n \ge 1}{S_{n}} = \infty$ with probability 1.

Looking for opinions on the proofs. Any feedback is appreciated.