Proving that a series of iid random variables diverges

Question

so I am working on a problem and it basically boils down to showing the following:

Let $Y_1, Y_2,...$ be i.i.d. random variables on $(\Omega, \mathcal{F}, \mathbb{P})$ satisfying $\mathbb{P}(Y_1=-1)=\frac{1}{3}$ and $\mathbb{P}(Y_1=1)=\frac{2}{3}$. Show that $M_n :=\sum^{n}_{k=1}Y_k \rightarrow\infty$ as $n\rightarrow\infty$.

This seems very easy but I am a novice and I don't know on how to prove it rigorously. I know that from Borel-Cantelli Lemma for independent events we have that $\mathbb{P}(\limsup_k(\{Y_k=1\})) = 1$ as well as $\mathbb{P}(\limsup_k(\{Y_k=-1\})) = 1$. So thanks to the iid-ness value for every $\omega$ in some set of measure 1 will get incremented infinitely often. This doesn't seem to help since it will also get decremented infinitely often. However we also have that $M_n$ is a submartingale and $\mathbb{E}[M_n]$ is divergent and increasing in $n$.

I have a feeling I need to combine those two pieces of information but I have no idea how. Can anyone give some hint?

Answer 1

Thanks @PhoemueX for the hint. The answer I give here might state some painfully obvious facts but I didn't do math for a long time and I need to keep reassuring myself where certain things come from.

From the SLLN we know that

$\frac{1}{n}M_n\rightarrow\mathbb{E}[Y_n]=\frac{1}{3}$ a.s. as $n\rightarrow\infty$ But this means that $\lim_{n\rightarrow\infty}(\frac{1}{n})\lim_{n\rightarrow\infty}(M_n)=\frac{1}{3}$ and the only way for this to happen if it this expression evaluates to either $0*+\infty$ or $0*-\infty$, i.e. $\lim_{n\rightarrow\infty}(M_n)=+\infty$ or $\lim_{n\rightarrow\infty}(M_n)=-\infty$

To show this must be $+\infty$: we know that $\forall\epsilon>0:\exists N_{\epsilon} \in \mathbb{N}: \forall n>N_{\epsilon}: |\frac{1}{n}M_n-\frac{1}{3}|<\epsilon$. Expanding the last inequality we get for some $\epsilon$ and $n>N_{\epsilon}$: $$|\frac{1}{n}M_n-\frac{1}{3}|<\epsilon \\ M_n<n(\frac{1}{3}+\epsilon) \lor M_n>n(\frac{1}{3}-\epsilon) $$ Here if we pick $\epsilon < \frac{1}{3}$ then $M_n >0$ for all $n > N_{\epsilon}$ hence it must be that $\lim_{n\rightarrow\infty}(M_n)=+\infty$

Answer 2

Here is a direct way to show that $M_n\to \infty$ a.s. Let $M_n':=M_n-n/3$ and consider a positive sequence $\epsilon_n$ which will be determined later. $$ \mathsf{P}(|M_n'|\ge \epsilon_n)\le \frac{\mathsf{E}M_n'^4}{\epsilon_n^4}\le \frac{Cn^2}{\epsilon_n^4}, $$ where $C=\mathsf{E}[Y_1-1/3]^4+\operatorname{Var}(Y_1)^2$. Take $\epsilon_n=n^{0.8}$. Then $$ \sum_{n\ge 1}\mathsf{P}(|M_n'|\ge \epsilon_n)\le C\sum_{n\ge 1}n^{-6/5}<\infty. $$ Therefore, eventually $M_n > n/3-n^{0.8}$ a.s.