Proving $P\left(\mu \not \to 0\right)=P\left(\bigcup_{N=1}^\infty\bigcap_{m=1}^\infty\bigcup_{n>m}\left|\mu\right| \geq \frac1N\right)$

Question

I have the following problem regarding random walks where I need to show that:

$$P\left(\frac{\sum_{k=1}^n Y_k}{n} \not \to 0\right)=P\left(\bigcup_{N=1}^\infty\bigcap_{m=1}^\infty\bigcup_{n>m}\left|\frac{\sum_{k=1}^n Y_k}{n}\right| \geq \frac1N\right)$$

where the $Y_i$s are i.i.d. random variables. I'm given a hint that I should use the definition of convergence here, but I don't see at least yet how that would help me.

My question is: How do I prove this equality?

This is what I have tried so far:

I denote $X_n=\sum_{k=1}^n Y_k$. I began by thinking that $P\left(\frac{X_n}{n} \not \to 0\right)$ consists from those events where the average $\frac{X_n}{n}$ either converges to some $L\neq0$ or does not converge at all. That is, the set of events

$$\left\{\frac{X_n}{n} \not \to 0\right\}=\left\{\frac{X_n}{n}\to L\neq0\right\}\bigcup\left\{\frac{X_n}{n} \to \pm\infty\right\}\triangleq A.$$

My idea was that if I can show that $A=\left\{\bigcup_{N=1}^\infty\bigcap_{m=1}^\infty\bigcup_{n>m}\left|\frac{X_n}{n}\right| \geq \frac1N\right\}$ then I would have solved this problem.

Next I took a closer look at the set $\left\{\bigcap_{m=1}^\infty\bigcup_{n>m}\left|\frac{X_n}{n}\right| \geq \frac1N\right\}$. It seemed to me that:

$$\left\{\bigcup_{n>1}\left|\frac{X_n}{n}\right| \geq \frac1N\right\}\supseteq \left\{\bigcup_{n>2}\left|\frac{X_n}{n}\right| \geq \frac1N\right\}\supseteq \left\{\bigcup_{n>3}\left|\frac{X_n}{n}\right| \geq \frac1N\right\}\supseteq \cdots,$$

but now I got stucked and don't know yet how to proceed.

Answer 1

Let $\displaystyle Z_n = \frac{\sum_{k=1}^n Y_k}{n}$.

By the very definition of convergence of a sequence, $$\begin{align}\{w\in \Omega,\; \lim_{n\to \infty}Z_n(w) = 0\} &= \{w\in \Omega,\; \forall \epsilon >0, \exists m, \forall n>m, |Z_n(w)|< \epsilon \} \\ &= \{w\in \Omega,\; \forall N\geq 1, \exists m, \forall n>N, |Z_n(w)|< \frac{1}{N} \} \end{align} $$

Then $$\begin{align}\{w\in \Omega,\; \lim_{n\to \infty}Z_n(w) = 0\}^c &=\{w\in \Omega,\; \exists N\geq 1, \forall m, \exists n>N, |Z_n(w)|\geq \frac{1}{N} \} \\ &= \bigcup_{N=1}^\infty\bigcap_{m=1}^\infty\bigcup_{n>m}\left|Z_n\right| \geq \frac1N \end{align}$$

Answer 2

Recall that, for a sequence of real numbers $a_n$, $$\limsup_{n\to\infty} \,a_n>\alpha\iff \exists\beta>\alpha,\forall n,\exists m\ge n, a_m>\beta$$ Therefore $$P(U_n\not\to 0)=P\left(\limsup_{n\to\infty}\,\lvert U_n\rvert>0\right)=P\{\omega\,:\, \exists m\in\Bbb N,\forall n\in\Bbb N,\exists h\ge n, \lvert U_h(\omega)\rvert>1/m\}=\\=P\left(\bigcup_{m\in\Bbb N}\bigcap_{n\in\Bbb N}\bigcup_{h\ge n}\{\lvert U_h\rvert>1/m\}\right)$$

In the first passage I mean, of course, the random variable $\limsup\limits_{n\to\infty}\, f_n:\Omega\to\Bbb R\ (\text{or }\Bbb R\cup\{\pm\infty\})$, with $\left[\limsup\limits_{n\to\infty}\, f_n\right](\omega):=\limsup\limits_{n\to\infty}\,(f_n(\omega))$.