What is the probability density function (PDF) of $f(x)=1/\sin(x)$ when $x$ is uniformly distributed in $(0,90)$?
$f(x)=\sin(x)$ has a known PDF, which has the form $2(\pi\sqrt{1-\sin(x)^2})^{-1}$, but I cannot find the PDF for $1/\sin(x)$. The latter would have interesting applications in astronomy, specially for so-called "luminosity functions".
Thank you very much
Sebastian
Let $z=1/\sin(x)$, then $x=\sin^{-1}(1/z)$ on $(0,\frac{\pi}{2})$ interval. The pdf you seek, is the differential of uniform c.d.f., which is $\frac{2}{\pi} dx$
$$ \frac{2}{\pi} d( \sin^{-1}( z^{-1} ) ) = \frac{2}{\pi z} \frac{1}{\sqrt{z^2-1}},$$ where variable $z>1$.
By the way, you could have used Mathematica to directly find this:
In[100]:= FullSimplify[PDF[TransformedDistribution[1/Sin[x],
x \[Distributed] UniformDistribution[{0, Pi/2}]], z]]
Out[100]= Piecewise[{{2/(Pi*z*Sqrt[-1 + z^2]), z > 1}}, 0]
