In the theory of Brownian motion, Spitzer's Law states that: $$\lim_{t\rightarrow\infty}\mathbb{P}\bigg\{\frac{2\theta(t)}{\log(t)}\leq x\bigg\} = \int_{-\infty}^x\frac{1}{\pi (1+y^2)}dy$$ where $\theta(t)$ denotes the angular component of the skew-product representation of a planar Brownian motion: $$B(t) = \exp\{W_1(H(t)) + iW_2(H(t))\},\hspace{3pt}H(t) = \int_0^t\frac{1}{|B(s)|^2}ds$$ and where $W_i(t)$ is a linear Brownian motion.
My question is simply how to interpret this result. If we consider $\theta(t)$ to be the "angle" of $B(t)$ with respect to the line $y=0$ up to time $t$, then what this result seems to say is that a rescaled version of it - namely $2\theta(t)/\log(t)$ - asymptotically follows a Cauchy distribution. Sure. So what does this actually say about $\theta(t)$ itself? We know that the Cauchy distribution has infinite variance, so we know that this would mean that $2\theta(t)/\log(t)$ would have infinite variance and infinite mean, meaning that $B(t)$ is going to "spin around itself a lot after a very long time." This is because $\theta(t) > 2\theta(t)/\log(t)$ for $t\rightarrow\infty$. This is intuitive, but it's troubling to me that this is only an asymptotic result for $t\rightarrow\infty$. Would we not expect Brownian motion to spin around itself infinitely often even in finite time, especially given Blumenthal's 0-1 law?
I suppose then that I actually have two questions: is my coarse interpretation of Spitzer's law the "correct intuition" for it, and is there an analogous result to it for the nonasymptotic scenario? Or is it obvious that there cannot be such an analogous result? Any insight as to the meaning of Spitzer's law and its implications are welcome.