Probability density from standard domain - I

Question

Pick $x+iy$ at random with respect to hyperbolic measure from $\{z:|z|\geq1,|\mathcal R(z)|\leq\frac12\}$. What does the probability distribution function of $\frac1{\sqrt y}$ look like?

Answer 1

$\newcommand{\ii}[1]{\operatorname{\mathbf I}\{#1\}}$

The density (with respect to the Lebesgue measure) of the hyperbolic measure in your region is \begin{equation} f(x,y)=\frac3{\pi y^2}\,\ii{y>0,\ x^2+y^2>1,\ |x|<1/2} \end{equation} for $(x,y)\in\mathbb R^2$, where $\ii\cdot$ denotes the indicator, and has been normalized for the region.

Introducing the new variables $(u,v):=(x,1/\sqrt y)$, so that $(x,y)=(u,v^{-2})$, we have the joint density of $(u,v)$: \begin{equation} g(u,v)=f(u,v^{-2})\left|\frac{\partial(u,v^{-2})}{\partial(u,v)}\right| =\frac{6v}{\pi} \ii{v>0,\ u^2+v^{-4}>1,\ |u|<1/2}. \end{equation} Hence, the density of $v=1/\sqrt y$ is \begin{equation} h(v):= \int_{-\infty}^\infty g(u,v)\,du= \begin{cases} 6v/\pi &\text{ if } 0<v<1, \\ 6v\left(1-2 \sqrt{1-v^{-4}}\right)/\pi &\text{ if } 1\leq v<\sqrt[4]{4/3}, \\ 0&\text{ if }v\ge \sqrt[4]{4/3}. \end{cases} \end{equation}