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A point is chosen uniformly at random from the circumference of a circle of diameter 1. Let X be the length of the chord joining the random point to an arbitrary fixed point on the circumference. Find the cdf of X. So X=2rsin(theta/2) r=1/2 X-sin(theta/2) P(sin(theta) < x)=P(theta < 2arcsinx)=F(x). I'm not sure where to go from here. I know that we will have to integrate. Any help would be appreciated. Thanks!

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1 Answer 1

Let the fixed point be $\left(\frac{1}{2}, 0\right)$, and the random point be $\left(\frac{1}{2}\cos(\phi), \frac{1}{2}\sin(\phi)\right)$. The distance $d(\phi) = \frac{1}{2} \sqrt{ (1-\cos(\phi))^2 + \sin^2(\phi)} = \sqrt{\frac{1-\cos\phi}{2}} = \vert \sin \frac{\phi}{2} \vert$.

$$ \begin{eqnarray} F_d(x) &=& \mathbb{P}(d(\phi) \le x) = \mathbb{P}(\vert \sin \frac{\phi}{2} \vert \le x) = \int_0^{2 \pi} \frac{1}{2 \pi} \mathbf{1}( \vert \sin \frac{\phi}{2} \vert \le x) \mathrm{d} \phi \\ &=& \int_0^{\pi} \frac{1}{\pi} \mathbf{1}( \sin \frac{\phi}{2} \le x) \mathrm{d} \phi \\ &=& \left\{ \begin{array}{cl} 0 & x \le 0 \\ \frac{2}{\pi} \arcsin(x) & 0 < x \le 1 \\ 1 & x > 1 \end{array} \right. \end{eqnarray} $$

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