1
$\begingroup$

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

1 Answers 1

4

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] 
  • 0
    Super. Thank you very much, Sasha.2011-08-04