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I wish to find the expected area of a triangle with vertices independently and uniformly at random chosen from a unit circle. I solved this problem, but my answer does not agree with the one found here. Where did I make a mistake?

Attempt: Let $O$ be the center of the circle and let $P_1, P_2, P_3$ be the vertices. By linearity of expectation, symmetry, and law of sines, $\mathbb{E}([P_1P_2P_3]) = \mathbb{E}([OP_1P_2] + [OP_2P_3] + [OP_3P_1]) = 3\mathbb{E}([OP_1P_2]) = \\ \frac{3}{2}\mathbb{E}(OP_1 \cdot OP_2 \cdot \sin(\angle P_1 O P_2)).$

Specifying $OP_1, OP_2$ does not affect the distribution of the angle, so $\mathbb{E}(OP_1 \cdot OP_2 \cdot \sin(\angle P_1 O P_2)) = \mathbb{E}(OP_1 \cdot OP_2)\mathbb{E}(\sin(\angle P_1 O P_2)).$

Every angle in $[0,\pi]$ is equally likely, so $\mathbb{E}(\sin(\angle P_1 O P_2)) = \frac{1}{\pi} \int_0^{\pi} \sin(x) \, dx = \frac{2}{\pi}.$ The distances $OP_1, OP_2$ are independent, so $\mathbb{E}(OP_1 \cdot OP_2) = \mathbb{E}(OP_1)^2.$ The ring of distance $r$ from the center has circumference $2\pi r,$ so the pdf of $OP_1$ is proportional to $r.$ Thus, $\mathbb{E}(OP_1) = \frac{\int_0^1 r^2 \, dr}{\int_0^1 r \, dr} = \frac{2}{3}.$ Combining everything gives us an expected value of $\frac{3}{2} \cdot \frac{2}{\pi} \cdot \left(\frac{2}{3}\right)^2 = \frac{4}{3\pi}$ for the area of the triangle.

I suspect that the first 2 paragraphs are solid and the separation of angle and distance is legal. Maybe the issue is that $OP_1, OP_2$ are not independent, but how could that be?

Display name
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  • Did you make sure what you do makes sense when the triangle does not include the center? – Bananach Jul 19 '20 at 10:23
  • @Bananach That's it. So the very 1st line of my solution was where I went wrong. All I need to do is subtract off the term $2Pr(P_1 \text{ obtuse})\mathbb{E}([OP_2P_3] | P_1 \text{ obtuse})+\dots = 3Pr(P_1 \text{ obtuse}) \mathbb{E}[OP_2 OP_3 \cdot \sin(\angle P_2 O P_3) | P_1 \text{ obtuse}).$ I wonder if this will be easy to calculate. – Display name Jul 19 '20 at 10:29
  • I feel like this might all be easier if you think of signed areas? Not sure though, there is probably a reason the answer in the other question doesn't look so easy – Bananach Jul 19 '20 at 10:31
  • @Bananach Signed areas looks like an easier approach to fix this. But how do I assign signs? The sign of $[OP_1 P_2]$ would be positive, negative if $P_3$ is acute, obtuse respectively, so we have to keep track of all $3$ points. Maybe if you let $p$ be the probability $P_i$ is acute, then we could go through all the calculations and obtain $\frac{4}{3\pi} \cdot p - \frac{4}{3\pi} \cdot (1-p) = \frac{4}{3\pi}(2p-1).$ This approach suggests we need to show $p = \frac{297}{384},$ which is quite a wild fraction. – Display name Jul 19 '20 at 10:40
  • I thought you just have everything signed, i.e. compute integrals over $(0, 2\pi )$ instead of $(0,\pi)$ and then take the expectation of the absolute value of the sum of the signed areas. That of course has the major disadvantage that you lose linearity, and I don't know how difficult the computations would become – Bananach Jul 19 '20 at 10:50
  • @Bananach Nevermind. A triangle being obtuse or acute has no connection to it containing or not containing the center. So instead we should let $1-p$ be the probability that $P_1 P_2 P_3$ does not contain the center, and $P_1$ is between $P_2$ and $P_3.$ Showing $p = 99/128$ here still seems daunting. It would entail showing the probability $P_1 P_2 P_3$ does not contain the center is $87/128.$ – Display name Jul 19 '20 at 11:10

2 Answers2

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Not an answer to the question asked

@Bananach has already answered (in the comments) the "where's the problem in my reasoning?" question well. This answer attempts to take that same reasoning and deform it to something that might yield the correct answer with almost equal simplicity.

Here's an alternative approach: Pick your three points; let $s$ be the angle counterclockwise from $P_1$ to $P_2$; let $t$ be the angle counterclockwise from $P_1$ to $P_3$. I claim that $s$ and $t$ are uniformly distributed on $[0, 2\pi]$. I expect you believe this part, so I'll move on.

I'm now going to rotate the picture so that $P_1$ is at the point $U = (1, 0)$, so we have angles $s$ and $t$ measured CCW in the usual way from the positive $x$-axis. Letting $S = (\cos s, \sin s)$ and $T = (\cos t, \sin t)$, we're interested in the area of the triangle $UST$. As in your solution, let $O$ be the origin, and let $H(P, Q, R)$ be the signed area of the triangle $PQR$. Then the signed area of $UST$ is $$ J = H(O,U,S) + H(O, S, T) + H(O, T, U). $$ You'd like to compute the expected value of $|J|$, and this absolute value is a pain in the neck.

Our probability space $X$ consists of uniformly equally likely points $(s, t) \in [0, 2\pi]$. Suppose we split $X$ into two parts, $X_1 = \{(s, t) \in X \mid s \le t \}$, $X_2 = X - X_1$. Then on $X_1$, we know that $J$ is positive, and on $X_2$, we know that $J$ is negative. And in fact the expected values of $J$ on the two parts are opposites (simply swap $s$ and $t$ to see this), so we find that the expected value of $|J|$ on all of $X$ is twice the expected value of $J$ on $X_1$.

I'll bet that you can compute that expected value using linearity of expectation, etc., just as you did before, except that you'll need a double integral where the inner integral's limits depend on the outer integral (because you're integrating over a triangle rather than a square).

John Hughes
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The first place I doubted what you were saying was this:

"Every angle in $[0,\pi]$ is equally likely"

Can you give a reason you think this is true for the uniform distribution of the points $P_1$ and $P_2$?

John Hughes
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    Fix $P_1.$ Given that $P_2$ lies on a ring of radius $r$ around the center, $P_2$ is equally likely to be anywhere on the ring, meaning the angle is uniformly distributed from $[0,\pi].$ This applies for all $r.$ – Display name Jul 19 '20 at 10:26
  • Ah...somehow I'd gotten the idea that you fixed one of the three vertices, say $P_3$, and called this $O$. So the angle I was seeing was between two chords passing through the point $O$ on the circle. I suppose it's possible that angle is uniformly distributed, but I have my doubts. The angle you were describing is clearly uniformly distributed, of course. Thanks for clearing it up. – John Hughes Jul 19 '20 at 11:08