There have been several recent problems about expected values and unit circles. I came up with another problem like this, but I noticed something interesting. Here's the problem.

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius \(1,\) find the expected area of such a triangle.

Seems pretty simple, right? Actually, it has two different answers, depending on how you interpret the problem. Consider the following.

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius \(1,\) a diameter and a random point not on the diameter are randomly picked and connected to form a triangle. Find the expected area of this triangle.

And

It is a well known fact that the triangle with vertices on opposite ends of the diameter of a circle and a third point on the circle is a right triangle. In a circle of radius \(1,\) an \(x\text{-value}\) is randomly selected from the diameter and its corresponding \(y\text{-value}\) is plotted on the circle. Find the expected area of this triangle.

Assume WLOG that the diameters of these circles are on the \(x\text{-axis}\)

These seem like the exact same problem. But there's two different ways to solve them, given the wording of the problem.

\(\textbf{Solving the first problem}\)

The points on the diameter are \((-1,0)\) and \((1,0).\) These are the base of the triangle. We just have to find the average height of the circle. Any point on the unit circle can be represented by \((\cos\theta,\sin\theta),\) where \(\theta\) is the angle formed by drawing the ray starting at the origin and passing through \((1,0)\) and the ray starting at the origin and passing through the point. The height of the triangle is going to be \(\sin\theta.\) The angle will be anywhere in the range \([0,2\pi).\) We can just find the average of \(|\sin\theta|\) over the range \([0,2\pi)\) to find the expected height. This will be the same as the average value of \(\sin\theta\) over \([0,\pi).\) \[ \begin{align} \dfrac{1}{\pi}\int_0^\pi\sin\theta\text{ }d\theta&=\dfrac{-1}{\pi}\left.\cos\theta\right|^\pi_0\\ &=\dfrac{-1}{\pi}(\cos\pi-\cos0)\\ &=\dfrac{-1}{\pi}(-1-1)\\ &=\dfrac{2}{\pi} \end{align} \] The height of this triangle has an average value of \(\frac{2}{\pi}\) and the base is a constant \(2,\) so the expected area of the triangle is \(\frac{1}{2}\times2\times\frac{2}{\pi}=\boxed{\frac{2}{\pi}}\)

\(\textbf{Solving the second problem}\)

In this problem, you are randomly picking an \(x\text{-value}\) in the range \((-1,1)\) and plotting it on the circle, which we can again remove the parts of the circle in the third and fourth quadrants because of symmetry. We can model the semicircle by \(y=\sqrt{1-x^2}.\) The average value of this function is the area of the semicircle divided by \(2,\) which is \(\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}.\) The base of the triangle is \(2\) and the average height is \(\frac{\pi}{4},\) so the expected area of the triangle is \(\frac{1}{2}\times2\times\frac{\pi}{4}=\boxed{\frac{\pi}{4}}.\)

Wait, we've gotten two different answers to the same problem. Why is that? Consider the length of the arc we are picking a point from. In the first problem, we are picking any point from an arc of length \(\pi.\) In the second problem, the arc has length \(2.\) But these points are still going to result in the same triangles. The variable arc in the first problem \((A_1)\) and the variable arc in the second problem \((A_2)\) are different; \(\frac{A_1}{A_2}=\frac{\pi}{2}.\) So we can then expect to see a difference in area.

See if you can try to explain the difference mathematically!

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TopNewestThe difference in the two cases is that the probability distributions are different. Hence, the expected value is different in each case.

In the first case, you are picking the third point with uniform distribution on the circular arc, whereas in second case, the uniform distribution in on the x-axis (diameter). In the first case, the distribution over x-axis would be a bowl shaped distribution, i.e. it would produce higher probability for smaller-area triangles, whereas in second case, as stated earlier, the distribution is uniform over x-axis. Hence, you get lower expected area in first case.

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In the first case \( \theta \) is uniformly distributed over [0,\( \pi\)] so by transform of variable distribution function of \(x= \cos \theta \) will be= \( \frac {1}{\pi \sqrt{1-x^2}} \) over [-1,1]

In the second case x is distributed uniformly over [-1,1] .

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Can i ask you one thing....how much time did it take to write such a long note with latex

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This took maybe about \(15\) minutes. LaTeX isn't that hard if you've used it a lot.

Do you have anything to say about the paradox?

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