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?



Probability density from standard domain - I
$\newcommand{\ii}[1]{\operatorname{\mathbf I}\{#1\}}$


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}


