# Probability that randomly chosen n points on a circle are covered by semicircle with order statistics

I am aware of many forms of the solutions posted on SE. I am only interested in an approach using order statistics. The correct probability is $\frac{n}{2^{n-1}}$.

So n points are chosen randomly on a circle and I am interested in the probability that there exists a semi circle that covers all the points. We can think of it as unit circumference circle, and the sample of $n$ points can be modeled as $n$ iid standard uniform random variables, $U_1, ..., U_n$. Also let the two order statistics of this sample to be $U_\textrm{max}$ and $U_\textrm{min}$.

Here is my logic. In order for $n$ points to fall in a semicircle, $U_\textrm{max} - U_\textrm{min} \le 0.5$.

According to the First course in probability by Ross (sec. 6.6), $$\mathbb{P} \left(U_\textrm{max} - U_\textrm{min} \le x \right) = n (1 - x) x^{n-1} + x^n.$$

But using $x = 0.5$, I get probability of $\frac{n+1}{2^{n}}$, which is not the correct answer. In fact using the same distribution for order statistics of $U_1, ..., U_{n-1}$ leads to a correct answer according to this link, which says that I need to condition on $U_\textrm{max}$. But I don't get this idea. Why do we need to condition on the maximum value?

• I do not agree with your logic. If e.g. $U_{min}=0.1$, $U_{max}=0.9$ and all other points take a value in $[0.1,0.2]$ then also a semicircle exists that contains all points. This in spite of $U_{max}-U_{min}=0.8>0.5$. – drhab Oct 12 '17 at 14:17
• Oh, I see. But I am still not clear how conditioning on $U_\textrm{max}$ provides a correct logic. I think the link I provided says we can condition on $U_\textrm{max} = 1$. In this case, if the remaining points take values in [0.4, 0.7]. Among these $n-1$ points, let's say one takes 0.4 and another takes 0.7. In this case, $U_\textrm{max} - U_\textrm{min}\le 0.5$ but we can't find a semicircle. – zcadqe Oct 12 '17 at 14:51
• Are the U(i) you're generating the distance of these points from the center of the circle? If so, that's not a uniform-area distribution. However, I think I'm not understanding what you're doing. – barrycarter Oct 12 '17 at 15:35
• $U_i$'s are the positions in the circumference. Since I am assuming the circumference is 1, the position is distributed Uniform(0, 1). – zcadqe Oct 12 '17 at 15:38

If we let one of the n points be 0.5, then the semi circle would never cross the zero point, because any semi circle crossing the zero point would not contains 0.5 . Furthermore, any semi circle not crossing the zero point would surely contains 0.5 .

The n points in a semi circle.

= The other n-1 points in a semi circle which contains 0.5

= The other n-1 points in a semi circle not cross zero.

= The other n-1 points satisfy Umax−Umin≤0.5

• So I sample $n$ random points. Define the coordinate of the circle such that any one of the $n$ points to be at 0.5. Then remaining $n-1$ points have to satisfy $U_\textrm{max} - U_\textrm{min} \le 0.5$. Is this what you mean? But this depends on which of the $n$ points is chosen to be placed at 0.5. – zcadqe Oct 12 '17 at 21:20
• It seems that I should condition on $U_\textrm{max} = 1$ such that the joint distribution of order statistics of remaining $n-1$ order statistics are distributed same as the order statistics of a sample of $n-1$ iid standard uniform variables. Then I can just plug in $n-1$ in place of $n$ in $\mathbb{P} \left(U_\textrm{max} - U_\textrm{min} \le x \right) = n (1 - x) x^{n-1} + x^n$ – zcadqe Oct 12 '17 at 21:22
• We can randomly choose one point to be 0.5, it is not a condition. For example, we choose the first point. After the first point generated, we define the coordinate such that the first point is 0.5. Now the semi-circle requirement is equivalent to the following n-1 points satisfying Umax−Umin≤0.5. Making U1 = 0.5 will not affect the distribution of U2 to Un, they are still IID uniform between 0 and 1. Making Umax = 1 will change the distribution of the other points since it implies they are not the largest one. – Jeffrey Chen Oct 13 '17 at 1:11