The expected length of the arc containing the point (1,0) is; π
How to solve Unit Circle Problems?
First of all, it is pertinent to note that the lengths of the four arcs in which the circle is divided by the three points and (1, 0) have identical distributions.
The problem is same as dividing a circle of length 2π at four points chosen at random, and labelling one of them (1, 0). Now, let us make L₁, L₂, L₃, L₄ to be the lengths of the four arcs determined by those four points.
Thus, we have;
L₁ + L₂ + L₃ + L₄ = 2π
Whereas, the expected value of several random variables is additive as;
E[ L₁ + L₂ + L₃ + L₄] = [EL₁] + [EL₂] + [EL₃] + [EL₄]
By symmetry the lengths have the same probability distribution and as such; [EL₁] = [EL₂] = [EL₃] = [EL₄] and the sum must be 2π. Thus, we can say that each expected value is 2π/4 = π/4.
Finally, the expected value of the arc containing the point (1, 0) is the sum of the expected values of the two arcs that are adjacent to (1, 0), i.e.,
π/2 + π/2 = π
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