Strong Law of Large Numbers - Converse

Question

Probability with Martingales:

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I want to try to show the last one

$$\left[\limsup \frac{|S_n|}{n}\right] = \infty \ \text{a.s.}$$

which is equivalent to

$$\forall k \in \mathbb N$$

$$\left[\limsup \frac{|S_n|}{n}\right] \ge k \ \text{a.s.}$$

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What I tried:

I guess I can't use BCL2 like I did for

$$\left[\limsup \frac{|X_n|}{n}\right] = \infty \ \text{a.s.} \tag{*}$$

since even though $\{X_n\}_{n \in \mathbb N}$ is independent, $\{S_n\}_{n \in \mathbb N}$ isn't. However, I guess I don't need to. I tried using $(*)$:

$$(*) \iff \forall k \in \mathbb N$$

$$\left[\limsup \frac{|X_n|}{n}\right] \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n - S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n| + |S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n|}{n} + \limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

Now:

$$\limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_{n} - X_n|}{n} \ge k \ \text{a.s.}$$

$$\to \limsup \frac{|S_n|}{n} + \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

$$\iff \limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

I'm stuck. It seems that the most I can conclude is

$$\limsup \frac{|S_n|}{n} \ge k \ \text{a.s.} \ \text{or} \ \limsup \frac{|X_n|}{n} \ge k \ \text{a.s.}$$

How can I approach this please?

Can I somehow conclude

$$\limsup \frac{|S_n|}{n} \ge k \ \text{a.s.}$$

from

$$\limsup \frac{|S_{n-1}|}{n} \ge k \ \text{a.s.}$$

?

Answer 1

As I pointed out in my comment, $\limsup_n|X_n|/n\le 2\limsup_n|S_n|/n$. Consequently, $\{\limsup_n|X_n|/n\}\subset\{\limsup_n|S_n|/n\}$, so if the former has probability 1 then so does the latter.

Answer 2

Answer to last question: Yes.

I was able to show a.s. that

$$\limsup \frac{|S_{n-1}|}{n} \ge k$$

Now observe that

$$LHS = \limsup \frac{|S_{n-1}|}{n-1} \frac{n-1}{n} \le \limsup \frac{|S_{n-1}|}{n-1} = \limsup \frac{|S_{n}|}{n}$$

$$\to \limsup \frac{|S_{n}|}{n} \ge k$$

QED