I am reading a paper that uses a fact about Brownian excursion which I don't understand.
Let $(E_t)$ be a standard Brownian excursion, i.e. $E_t = X_t + i R_t$, where $X$ is a standard real Brownian motion and $R_t= |W_t|$, where $W$ is a standard Brownian motion in $\mathbb{R}^3$.
Then, the following fact is claimed:
$\, \mathbb{P} \bigg( \cap_{n \in \mathbb{N}} \big\{ \arg (E_s) \in (0, \frac{\pi}{3}) \text{ for some } s \in (0, \frac{1}{n}) \big\} \bigg) = 1.$
Can I use scaling to deduce that $\mathbb{P} \bigg( \cap_{n \in \mathbb{N}} \big\{ \arg (E_s) \in (0, \frac{\pi}{3}) \text{ for some } s \in (0, \frac{1}{n}) \big\} \bigg) \geq \mathbb{P} ( \text{arg } E_1 \in (0, \frac{\pi}{3})) >0$, and then use Blumenthal's zero one law to arrive at the claim?
Yup, this works. By Brownian scaling, $arg(E_{s})\overset{d}=arg(E_{1})$ for all $s>0$. Then exactly as you've said: \begin{align} \mathbb{P}\big(\cap_{n=1}^{\infty}\{arg(E_{s})\in{(0,\frac{\pi}{3})} \hspace{5pt}\text{for some }s\in{[0,\frac{1}{n}]} \}\big)=\lim_{n\rightarrow{\infty}}\mathbb{P}\big(arg(E_{s})\in{(0,\frac{\pi}{3})} \hspace{5pt}\text{for some }s\in{[0,\frac{1}{n}]}\big)\geq{\limsup_{n\rightarrow{\infty}}\hspace{2pt}\mathbb{P}(arg(E_{\frac{1}{n}})\in{(0,\frac{\pi}{3})})}=\mathbb{P}(arg(E_{1})\in{(0,\frac{\pi}{3})})>0\end{align} So by Blumenthal's 0-1 law, the relevant event has probability 1.