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TLDR;

What's the probability law of the area of a random cut on a sphere, with the random cut being perpendicular to a fixed segment connecting two antipodal points on the sphere, and sampled uniformly?

More precise description and my results

Suppose to have a sphere of radius 1, to choose a segment connecting two antipodal points and to randomly sample (with uniform distribution) a point on this segment. Now draw the 2d plane orthogonal to the segment and passing by the point.

This plan will cut the sphere in two, determining a circle. I would like to find the distribution of the areas of such circles.

Using the transfer operator, unless I did some trivial algebra errors, I have showed that called $P(A)$ the random variable of the area of such circle, we have that (the support is $[0, \pi]$), $P(A)$ has a density, and it is

$$ \frac{1}{2\pi\sqrt{(1 - A/\pi)}}. $$

Since I need this result for a computational problem I would be fine with this, but I wonder it this density is coming from a know probability law. Any help is appreciated :)

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    $\begingroup$ It is a stretched (by a factor of $\pi$) version of a $\text{Beta}\left(1,\frac12\right)$ distribution $\endgroup$
    – Henry
    Jun 24, 2021 at 14:40
  • $\begingroup$ Brillian! Thank you. If you make it as an answer I'll give you a deserved upvote :) $\endgroup$
    – Nisba
    Jun 24, 2021 at 14:47

1 Answer 1

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As requested in comments:

It is a stretched (by a factor of $\pi$) version of a $\text{Beta}\left(1,\frac12\right)$ distribution

See Wikipedia on the Beta distribution for some of the statistics, remembering to scale where necessary

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  • $\begingroup$ Am I misunderstanding something? ... It seems to me that we have the area of the base of a spherical cap (like in this picture), so $A=\pi\sin^2\theta$, $\theta\sim\text{Uniform}(0,\pi)$. If so, I think @Nisba's density for $A$ is missing a factor that turns it into a (stretched) $\text{Beta}(1/2,1/2)$ distribution. $\endgroup$
    – r.e.s.
    Jun 24, 2021 at 15:43
  • $\begingroup$ @r.e.s. Traces of the Bertrand paradox. What seems to be uniformly distributed is the point on a diagonal of the sphere (I think segment is used here as line segment rather than circular or spherical segment). So the point halfway from the centre to the surface corresponds to a median area of $\pi\left(1-\frac1{2^2}\right)$. This is consistent with $\text{Beta}(1,1/2)$ but not with $\text{Beta}(1/2,1/2)$. $\endgroup$
    – Henry
    Jun 24, 2021 at 16:05

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