# Arc Length of largest arc when $n$ points are chosen at random on the circumference of the unit circle.

I am currently doing a little self-study of A Probability Path by Sidney Resnick, and am having trouble with the following problem:

Points are chosen at random on the circumference of the unit circle. $Y_n$ is the arc length of the largest arc not containing any points when $n$ points are chosen. Show $Y_n \to 0$ almost surely.

So far my strategy is the following:

Since $\{Y_n\}$ is a monotone (decreasing) sequence, convergence almost surely is equivalent to convergence in probability, so it suffices to show that the expected value $E(Y_n) \to 0$.

A naive guess is that $E(Y_n) = \frac{2\pi}{n}$.

We can write $$E(Y_n) = \int_0^{2\pi}P(Y_n > l)\,d l.$$ and $$P(Y_n > l) = \begin{cases} 1, & \text{if } l < \frac{2\pi}{n},\\ \text{??} & \text{otherwise}. \end{cases}$$

How do we determine $P(Y_n > l)$? Or is there another way to proceed?

• $P(Y_n \gt l)$ is relatively easy to calculate if $l \ge \pi$ but harder for smaller $l$. – Henry Jul 11 '15 at 22:00
• After the first point is chosen you can work in the interval, no? – Mariano Suárez-Álvarez Jul 11 '15 at 22:12
• Perhaps you only need some upper bound on this probability. If the upper bound goes to $0$ as $n\to\infty$, that should do it. ${}\qquad{}$ – Michael Hardy Jul 11 '15 at 23:40
• I think what Resnick has in mind here is that you apply Theorem 6.4.1 (quantile estimation). For the uniform distribution it implies that if you fix $n$ and then choose $N\to\infty$ points and put them in order so that $X^{N}_1\le X^{N}_{2}\cdots X^{N}_{N}$, then $X^{N}_{kN/n}$ approach $k/n$. This means the points are roughly uniformly distributed and implies the result you want. – Jeremy Teitelbaum Sep 10 '16 at 20:04

If the length of the longest such arc does not approach $0$ almost surely, then with probability $>0$ there is some arc that remains empty for all $n$. Pick any arc of positive length. The probability that it is empty after $n$ points are chosen is $$\left( \frac{\text{length of the particular arc}}{\text{whole circumference}} \right)^n \to 0 \text{ as }n\to\infty.$$
• I think you may need more than this as I would have thought as you need to show it for every arc of the same length simultaneously or at least for the $\frac{c}{l}$ arcs making up the circle. That would not be difficult if the point/pointless status of each arc was independent as you would have the probability $1 - \left(1 - \left(\frac{l}{c}\right)^n\right) ^ {c/l} \to 0$ as $n$ increases, but they are not quite independent. – Henry Jul 12 '15 at 8:46