I came about the following statement:
Let $X$ and $Y$ be independent $N(0,1)$-distributed random variables, $T > 0$ and $t_0 \in [0, T]$. Then $$ \mathbb{P}(\sqrt{t_0} \lvert X \rvert \geq \sqrt{T-t_0} \lvert Y \rvert) = \frac{2}{\pi} \arcsin \sqrt{\frac{t_0}{T}}. $$
How do I prove this? I tried to do it by calculating the density of $\sqrt{t_0}\lvert X \rvert - \sqrt{T-t_0} \lvert Y \rvert$. However, some nasty integrals appeared there.
Using the common density of two independent random normals we can write the probability as $$\begin{align*} &\frac{1}{2\pi} \int_{\mathbb{R}^2} e^{-(x^2+y^2)/2} I\{\sqrt{t_0}|x| \ge \sqrt{T - t_0} |y|\}\, dx\, dy \\ ={}&\frac{1}{2\pi} \int_{\mathbb{R}^2} e^{-(x^2+y^2)/2} I\{t_0(x^2+y^2) \ge Ty^2\}\, dx\, dy\end{align*}.$$
Converting to polar coordinates $x = r\sin(\phi)$, $y = r\cos(\phi)$ then yields $$\begin{align*}&\frac{1}{2\pi} \int_0^{2\pi} \int_0^\infty e^{-r^2/2} r I\{t_0r^2 \ge T r^2 \sin^2(\phi)\} \, dr\, d\phi \\ ={}& \frac{1}{2\pi} \int_0^{2\pi} I\{t_0 \ge T \sin^2(\phi)\}\int_0^\infty e^{-r^2/2} r \, dr\, d\phi \\ ={}& \frac{1}{2\pi} \int_0^{2\pi} I\Big\{|\sin(\phi)| \le \sqrt{t_0/T}\Big\}\, d\phi \\ ={}& \frac{2}{\pi} \arcsin\sqrt{\frac{t_0}{T}}\end{align*}.$$
Where the last equation follows from observing that the subset of $[0, 2\pi]$ on which $|\sin(x)| \le K$ (with $0 < K < 1$) is given by $[0, l] \cup [\pi - l, \pi + l] \cup [2\pi - l, 2\pi]$ with $l = \arcsin(K)$.