Consider a random path on the surface of a 2-sphere, made of N discrete points (each picked with a uniform distribution across the surface). The path connects two successive points by the shortest line between them, and we connect the last point to the first one to make a closed path.
My question is, what is the statistics of the solid angle subtended by this path? I expect the mean to be zero, but I have no idea for the standard deviation.
Here are important tracks. But it's not a complete solution.
As any solid angle is measured by the corresponding area, we need to use the formula for the area of a triangular triangle. Afterwards, there will be a summation process. Let $A,B,C$ be the angles of the triangle, we have
$$area=A+B+C-\pi$$
(angular excess: see Girard's theorem here).
But we must be able to compute these angles.
This is done by using this formula:
$$\cos c=\cos a\cos b+\sin a\sin b\cos C \ \ \iff \ \ \cos C=\dfrac{\cos c-\cos a\cos b}{\sin a\sin b}$$
and the 2 other formulas obtained by circular permutation:
$$\cos A=\dfrac{\cos a-\cos b\cos c}{\sin b\sin a} \ \ \text{and} \ \ \cos B=\dfrac{\cos b-\cos c\cos a}{\sin c\sin b}$$
When the path $M_1M_2\cdots M_nM_1$ is completed, one computes the sum of areas:
$$\Delta=s_1 \times area(M_1M_2M_3)+ s_2 \times area(M_1M_3M_4)+ \cdots + s_{n-2} \times area(M_1M_{n-1}M_{n})\tag{1}$$
where the sign $s_k$ is $+$ if the triangle is positively oriented, $-$ otherwise.
(1) defines a random variable.
The mathematical expectation of $\Delta$ is the sum of the mathematical expectations of the triangular areas, which are clearly $0$ because there is a balance between the positive and negative areas.
The variance of $\Delta$, assuming mutual independence between the different areas, is the sum of variances of the triangle's areas:
$$E(area(M_1M_kM_{k+1})^{\color{red}{2}})$$
... which have to be computed by taking into account the variances and expectancies of sides lengths $\cos a, \cos b, \cos c$, themselves functions of the number $N$ of points "dropped" on the sphere, which doesn'tt look very easy to do, because we have to take as well into account the shortest path limitation. But I think rather good approximations can be obtained by taking a uniform enough distribution of points on the sphere.
Remark: Given a number $N$ of points on the sphere, set apart exceptional values of $N$, it is not possible to have an even distribution of points on the sphere, for example with $N=5$, but we can get rather close to it.