Let $X_1, X_2,…$ be iid r.v.s with $E(X_1)=0$, and $0<var(X_1)<+\infty$. I would like to proof:
\begin{equation}
\limsup\frac{S_n}{\sqrt{n}}=+\infty
\end{equation}
Since for each $c$, $\left\{\limsup \frac{S_n}{\sqrt{n}}\geq c\right\}$ is a tail event. By kolmogorov's $0$–$1$ law, we have $P\left(\left\{\limsup \frac{S_n}{\sqrt{n}}\geq c\right\}\right)=0$ or $1$.
Also, by Fatou's lemma and (Central Limit Theorem) CLT, we have
\begin{equation}
P\left(\limsup\frac{S_n}{\sqrt{n}} \geq c\right)\geq \limsup P\left(\frac{S_n}{\sqrt{n}} \geq c\right)=P(N(0,var(X_1)\geq c)>0.
\end{equation}
Then we have $P\left(\limsup\frac{S_n}{\sqrt{n}} \geq c\right)=1$ for each $c$. Then we have $P\left(\limsup\frac{S_n}{\sqrt{n}}=+\infty \right)=1$. Which means the conclusion holds.
However, I have a question.
By Fatou's lemma and (Central Limit Theorem) CLT, we can also have
\begin{equation}
P\left(\limsup\frac{S_n}{\sqrt{n}} \leq c\right)\geq \limsup P\left(\frac{S_n}{\sqrt{n}} \leq c\right)=P(N(0,var(X_1)\leq c)>0.
\end{equation}
This means $P\left(\limsup\frac{S_n}{\sqrt{n}} \leq c\right)=1$ for each $c$. Then we have $P\left(\limsup\frac{S_n}{\sqrt{n}}=-\infty \right)=1$. Which makes a contradiction. I think there are some mistakes in the proof of part II, but I cannot figure it out.
Thanks in advance for any help!
Best Answer
Consider a sequence of r.v.s. $\{X_n\}$. Since $$ \{\limsup X_n \ge c\}\supseteq \{X_n\ge c\text{ i.o.}\}, $$ $$ \mathsf{P}(\limsup X_n\ge c)\ge \mathsf{E}[\limsup 1\{X_n\ge c\}]\ge \limsup \mathsf{P}(X_n\ge c) $$ by (reverse) Fatou's lemma. However, $$ \{\limsup X_n \le c\}\not\supseteq \{X_n\le c\text{ i.o.}\}, $$ and you cannot proceed as in the previous case.