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I'm trying to prove that $$\lim_{t\to\infty}\frac{R_t^{(\nu)}}{t}=0 \quad \text{ a.s}$$ where $R_t^{(\nu)}$ is a Bessel process of order $\nu$. According to my source we obtain this directly from knowing that $$R^{(\nu)}\sim\{Z_t:t\geq0\}$$ where $$Z(t):=\begin{cases} 0, &t=0\\ tR^{(\nu)}\left(\frac{1}{t}\right), &t>0. \end{cases}$$

I just don't see how I'm suposed to do that. I'm using this definition for Bessel processes:$$R^{(n)}_t = \sqrt{(W^{(1)}_t)^2+(W^{(2)}_t)^2+\ldots+(W^{(n)}_t)^2},\quad t\geq 0$$ where $\nu:=n/2-1$ and $W^{(i)}_t$ are Wiener processes/Brownian motions.

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The event $$A:=\left\{\omega \mid \lim_{t\to\infty}\frac{R_t^{(\nu)}(\omega)}{t}=0\right\}$$ can be written as $A=\left\{\omega \mid \left(\frac{R_t^{(\nu)}(\omega)}{t}\right)_t\in C\right\}$, where $C$ denotes the set families $(A_t)_{t \geqslant 0}$ such that $\lim_{t\to +\infty}A_t=0$. Since $$\left(\frac{R_t^{(\nu)}(\omega)}{t}\right)_{t\gt 0}\overset{D}{=} \left(R_{1/t}^{(\nu)}(\omega)\right)_{t\gt 0}$$ it suffices to prove that $\lim_{s\to 0}R_s^{(\nu)}=0$, which in turn reduces to prove that $\lim_{s\to 0}W_s=0$ for any Brownian motion $W$.

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