Show that $\limsup|Y_{1}+...+Y_{n}|/n = \infty$ almost surely

Question

Can someone help me with part c) of question 2.8 located here (a 2005 probability course from Warwick University): https://homepages.warwick.ac.uk/~masgav/teaching/pm05_sheet2.pdf

The question is:

Let $Y_{1},Y_{2},..$ be sequence of i.i.d random variables with $\mathbb{E}(|Y_{i}|) = \infty$. Then show that $\limsup\limits_{n\rightarrow\infty}\dfrac{|Y_{n}|}{n} = \infty$ a.s and indeed, $\limsup\limits_{n\rightarrow\infty}\dfrac{|\sum_{i=1}^{n}Y_{i}|}{n} = \infty$ a.s.

Would it make sense to show that $\mathbb{P}\{\dfrac{|\sum_{i=1}^{n}Y_{i}|}{n} > c \;\; \text{infinitely often}\} = 1$ a.s for all $c > 0$?

Answer 1

First part of letter c follows directly from b.

Second part of letter c follows from triangle inequality.

I think? Wait...does it follow that if X and Y are identically distributed that |X| and |Y| are identically distributed?

Answer 2

Proving that statement seems daunting, but thankfully we have a work-around solution. Notice the following deterministic result:

Lemma. If $\limsup |a_1 + \cdots + a_n|/n < \infty$, then $\limsup |a_n| / n < \infty$.

Proof. Notice that we also have $M := \sup |a_1 + \cdots + a_n|/n < \infty$. Let $s_n = a_1 + \cdots + a_n$. Then $$ \frac{|a_n|}{n} = \frac{|s_n - s_{n-1}|}{n} \leq \frac{|s_n|}{n} + \frac{n-1}{n} \frac{|s_{n-1}|}{n-1} \leq 2M. $$ This proves the claim. ////

You can utilize the contrapositive of this.