= Solution
For every $a>0$, define
$$
\widehat X_t=a^{-1}X_{a^2t},
\qquad
\widehat Y_t=a^{-1}Y_{a^2t},
\qquad
\widehat B_t=a^{-1}B_{a^2t}.
$$
The <Brownian scaling> theorem makes $\widehat B$ a standard Brownian motion, and substitution shows that $(\widehat X,\widehat Y)$ satisfies the same coupled <Bessel process> equations from initial values $(x/a,y/a)$. Both hitting times are divided by $a^2$, so their order is unchanged. Taking $a=x$ or comparing any two pairs with the same ratio proves that $\mathbb P(\tau_x>\tau_y)$ depends only on $x/y$.
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