Take the uniform distribution on a sphere, and project it to a plane in the Riemann sphere's way, what's the resulting distribution? Title explains it 90%.
"Uniform distribution on a sphere" means the continuous distribution, not Fibonacci lattice.
There might be two possible interpretations of Riemann sphere (plane crosses the sphere's origin, or plane crosses P(0)), but the result should only differ by a pair of scaling factors.
Is it 2D Gaussian distribution?
I'm aware that something has to be said about the complex infinity point... We should be able to ignore it since it has zero measure, please help define it correctly.
Edit:
Stereographic Projection of Uniform Distribution on Sphere
It's like with the term “Stereographic Projection” it's just a matter of one click....
 A: I am not hopeful that the resulting distribution will be two$-$dimensional Gaussian. To understand my doubts, let's examine a simpler one$-$dimensional case where points from $S=\{(x,y)\neq (0,1):x^2+(y-1/2)^2=1/4\}$ are stereographically projected onto the $x$ axis.
$\text{FYI}$: The "light source" for this stereographic projection is located at $(0,1)$.
Fix any $t\in \mathbb{R}$.
You can show the line connecting $(0,1)$ and $(t,0)$ intersects $S$ precisely at the point $\left(\frac{t}{t^2+1},\frac{t^2}{t^2+1}\right)$.
If $\tilde{X}=\frac{X}{1-Y}$ is the $x-$coordinate of the stereographic projection of $(X,Y)\sim \mathcal{U}(S)$ onto the $x-$axis, the event that $\tilde{X}\leq t$ is precisely the event that $(X,Y)$ belongs to the arc of $S$ connecting $(0,1)$ and $\left(\frac{t}{t^2+1},\frac{t^2}{t^2+1}\right)$ in a counterclockwise manner. This event occurs with probability $1-\frac{1}{2\pi}\arccos\left(\frac{t^2-1}{t^2+1}\right)$ if $t\geq 0$ and with probability if $\frac{1}{2\pi}\arccos\left(\frac{t^2-1}{t^2+1}\right)$ if $t<0$.
The cdf of $\tilde{X}$ explicitly stated below.
$$\begin{equation*}
F_{\tilde{X}}(t)= \left\{
        \begin{array}{ll}
            \frac{1}{2\pi}\arccos\left(\frac{t^2-1}{t^2+1}\right) & \quad t <0 \\
            1-\frac{1}{2\pi}\arccos\left(\frac{t^2-1}{t^2+1}\right) & \quad t \geq 0
        \end{array}
    \right.
\end{equation*}
$$
The pdf $f_{\tilde{X}}=F'_{\tilde{X}}$ of $\tilde{X}$ exhibits an attractive bell shape but is certainly not Gaussian.
