Given that $f_\Phi(\phi)=\dfrac{\sin(\phi)}{2}$ for $\Phi\in[0,\pi]$, and $f_\Theta(\theta)=\dfrac{1}{2\pi}$ for $\Theta\in[0,2\pi]$, where $\Phi$ and $\Theta$ are independent. What is the PDF of $X=\sin(\Phi)\cos(\Theta)$?

So far, I have managed to obtain the PDFs of $U=\sin(\Phi)$ and $W=\cos(\Theta)$ as $$f_U(u)=\frac{u}{\sqrt{1-u^2}}, \quad\text{and}\quad f_W(w)=\frac{1}{\pi\sqrt{1-w^2}}.$$

I have attempted to follow the product distribution as described here. I know the answer should be $f_X(x)=\frac{1}{2}$, but I have been unable to progress further. How should I proceed?

  • $\begingroup$ Is there an independence assumption? $\endgroup$ – Kavi Rama Murthy Dec 13 '18 at 7:15
  • $\begingroup$ Yes, sorry, phi and theta are independent. $\endgroup$ – Dracolich56 Dec 13 '18 at 7:19

The easiest solution returns to the sphere's geometry. The part of the surface ranging from $x$ to $x+dx$ is a ribbon of radius $\sqrt{1-x^2}$ and thickness $d\arcsin x=\frac{dx}{\sqrt{1-x^2}}$, so its area is $2\pi dx$. This can be used to prove the classical fact that a sphere's area matches that of a cylinder just long and wide enough to hold it, because each ribbon on the sphere's surface has the same area as its shadow on the cylinder. 3Blue1Brown has a great video on it here.

  • $\begingroup$ That makes sense to me. However, when I first approached this problem, I did not realize the physical significance of the PDFs (that is, how they pertain to a distribution on a sphere), so I was wondering how I might go about it otherwise. Thanks for the help! $\endgroup$ – Dracolich56 Dec 13 '18 at 9:32

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