单位圆随机弦长的期望

设两点对应的圆心夹角（取较小的那个）为 $\Theta\in[0,\pi]$，则可证明 $\Theta$ 在 $[0,\pi]$ 上均匀分布。

弦长为

$$L=2\sin\frac{\Theta}{2}.$$

因此

$$\mathbb{E}[L]=\frac{1}{\pi}\int_0^{\pi}2\sin\frac{\theta}{2}\,d\theta =\frac{4}{\pi}.$$

所以 $x=4$。

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Original Explanation

Let $\theta$ and $\phi$ be IID $\text{Unif}(0, 2\pi)$. Picking the two points uniform on the circumference is equivalent to picking these two angles. The measure of the angle between $\theta$ and $\phi$ is $(\theta - \phi)$. To get the length of the line segment connecting the two points, we can take a shortcut: the length here only depends on the distance between our points we select in terms of the angle.

Regardless of where we end up having our points located, we can just rotate the circle so that one of the two points is located at $(1, 0)$. We can fix $\theta_1 = 0$. Thus, the length between $\theta_1 = 0$ and the point at $\theta_2 = \theta$ is $\sqrt{(1 - \cos(\theta))^2 + \sin^2(\theta)} = \sqrt{2} \sqrt{1 - \cos(\theta)}$. This means that finding the expected length of the chord is equivalent to finding $\sqrt{2}\mathbb{E}[\sqrt{1 - \cos(\theta)}]$, where $\theta \sim \text{Unif}(0,2\pi)$.

Now, $\sqrt{2}\mathbb{E}[\sqrt{1 - \cos\theta}] = \frac{\sqrt{2}}{2\pi}\int_0^{2\pi}\sqrt{1-\cos\theta}\,d\theta$.

A simple approach is to conjugate the interior by multiplying and dividing by $\sqrt{1+\cos\theta}$ so that we get $\sqrt{1-\cos^2\theta} = |\sin\theta|$. Doing this, we get $\frac{\sqrt{2}}{2\pi}\displaystyle \int_0^{2\pi} \frac{|\sin\theta|}{\sqrt{1+\cos\theta}}\,d\theta$.

Note that the region on $(0, \pi)$ and $(\pi, 2\pi)$, our integrand is symmetric, so we can just evaluate over one interval and double it. This means our new integral is $\frac{\sqrt{2}}{\pi} \int_0^{\pi} \frac{\sin\theta}{\sqrt{1+\cos\theta}}\,d\theta$. Let $u = 1 + \cos\theta$. Then $du = -\sin\theta\,d\theta$. Our bounds would respectively become $2$ and $0$, but the negative from the $du$ flips them back. Therefore, our new integral is $\frac{\sqrt{2}}{\pi} \int_0^2 u^{-\frac{1}{2}}\,du$.

Evaluating this, we get $\frac{2\sqrt{2}}{\pi} \cdot \sqrt{u} \Big|_0^2 = \frac{4}{\pi}$.

$x = 4$