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Three points are uniformly thrown on a circumference of circle of radius 1, find the mathematical expectation of the area of the triangle formed by them.

I've tried to use that formula: $S = \frac{1}{2}(\sin(a)+\sin(b)+\sin(c))$, where a,b,c - arc lengths

Then $$E[S] = \frac{1}{2}(E[\sin(a)]+E[\sin(b)] + E[\sin(2\pi-a-b)]$$

$$E[\sin(a)] = \frac{1}{4\pi^2}\int_{0}^{2\pi}\sin(x)dx\int_{0}^{2\pi-x}dy = \frac{1}{2\pi}$$

$$E[\sin(b)] = \frac{1}{2\pi}$$

$$E[\sin(2pi-a-b)] = \frac{1}{4\pi^2}\int_{0}^{2\pi}dx\int_{0}^{2\pi-x}\sin(2\pi-x-y)dy = \frac{1}{2\pi}$$

$$E[S] = \frac{3}{4\pi}$$

But the correct answer is $\frac{3}{2\pi}$

Where did I make a mistake?

joriki
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Strike
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    How are you getting the normalization factor of $\frac{1}{4\pi}$? Also, your double integral doesn't seem entirely well-formed. What is $x$ in the upper limit $2\pi-x$ in the second integral? ¶ Being a bit more careful with your normalization might point to the issue here. – Brian Tung Feb 15 '24 at 19:48
  • @BrianTung It's not a normalization factor, it is a probability density of arcs a and b, since they are uniformly distributed in $[0, 2\pi]$. Integral is taken over $x+y < 2\pi$ – Strike Feb 15 '24 at 19:55
  • See here – Jean Marie Feb 15 '24 at 21:03
  • @Strike: You're right—although that is what I meant, I just spoke wrongly. – Brian Tung Feb 15 '24 at 21:29
  • Maybe probability density can be obtained like this: Denote points on circumference as 1,2,3: $$P(a<x) = 1 - P(2,3 \notin x) = 1 - (1-\frac{x}{2\pi})^2$$ $$p_a (x) = \frac{1}{\pi} (1-\frac{x}{2\pi}) $$ $$E[sin(a)] = \int_0^{2\pi} sin(x)p_a (x)dx = \frac{1}{\pi}$$ Now I'm not sure, but if I assume that $E[sin(a)]=E[sin(b)]=E[sin(c)]$, I can obtain correct answer $E[S] = \frac{3}{2\pi}$ – Strike Feb 16 '24 at 00:19

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