Extension to higher dimensions \mathbb{S}^n for n>1.

m points on an n-dimensional sphere. What holds here?

Literature

Vaguely related to the Bertrand paradox

Choosing the worst possible set A

Suppose now that your worst enemy can choose the set A under the restriction that \mu (A)=a for some given 1\ge a \ge 0. How should he choose it? How big will a have to be so that we always can be sure of catching m points? We start with the following observation.

Lemma 2. Let A \subseteq \mathbb{T}^1 be a (Lebesgue) measurable set. Suppose there exists points x_1, x_2 \in T^1 so that for all \tau \in T^1 it holds that x_1 and x_2 not simultaneous can be in A + \tau. Then m(A) \le 1/2.

Proof. Let d = x_2 - x_1 \in T^1. The sets A and A - d are disjoint. To see this, assume they are not. Then we can find u \in A \cap (A - d). Let \tau = x1 - u. Note that x_1-\tau=u \in A and x_2-\tau=d+u\in A, hence x_1, x_2 \in A + \tau which contradicts our assumption. Since A, it follows by translation invariance of the Lebesgue measure that

By Lemma 2 we have that if A is a arbitrary subset of \mathbb{T}^1 with |A| > 1/2, then there is a \tau so that \{x_1, x_2\} \subset A + \tau. Obviously, we cannot do better than 1/2, e.g., the set A=[0,1/2-\epsilon] cannot catch all pairs of points. Now one can ask how large the set A needs to be in order for us to always catch m random points. More precisely: What is the smallest length a such that any set A \subset \mathbb{T}^1 satisfying |A| \ge a or |A| > a, there exists a \tau so that \{x_1, \dots, x_m\} \subset A + \tau ?

TO-DO: Extend lemma to |A|=1/2 and to m points |A| = a > 1-1/m.

Choosing the best possible set A

How do one construct a set A of smallest possible measure so that for any given set of m points we can find a \tau so that all m points belong to A+\tau \pmod 1. We start with a well-known fact about the Cantor set that covers the case m=2.

Theorem. Let A be the middle third Cantor set. Then for any two points \{x_1,x_2\} there exists a \tau so that \{x_1,x_2\} \subset A+\tau.

Can we extend this to m>2? The following result tells us that we can, not only catch m random points, but countably many points with a set A with arbitrarily small measure.

Theorem. Lad \epsilon > 0 være givet. Så findes A \subset R / Z med \mu(A) < \epsilon således at givet vilkårlig tællelig mængde X \subset R / Z findes \tau så X \subset A+\tau.

Bevis. Lad A være en åben og tæt delmængde af R / Z med m(A) < \epsilon. Skriv X = \{x_1, x_2,\dots\} og sæt d_i = x_{i+1} - x_1, i = 1, 2, \dots. Så er B = \cap_i (A - d_i) en fællesmængde af åbne tætte mængder, og da R / Z med standard metrik er fuldstændigt metrisk rum følger det af Baire's sætning at B er tæt i R / Z. Derfor findes b \in B.

Sæt nu \tau = x_1 - b. Da b \in A er x_1 = b + \tau et element af A + \tau. For vilkårlig i \in N har vi b + d_i med i A så A + \tau indeholder

Theorem. Der findes A \subset R / Z med m(A) = 0 således at givet en tællelig mængde X \subset R / Z findes \tau så X \subset A + \tau.

Bevis. Lad B_n være åbne tætte mængder med |B_n| < 1/n og sæt A = \cap_{n \in N} B_n. Fortsæt så som i beviset af sidste sætning og bemærk at en tællelig fællesmængde af tællelige fællesmængder er en tællelig fællesmængde. ■

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of measure 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points by either A or B is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [2/6,3/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C. in contrast to B, does not satisfy the 'symmetry' condition C=C+1/2 \pmod 1 from Remark 1.

The following result shows that sets with "similar" symmetry properties have the same covering probabilities.

Lemma 1. Let n \in \mathbb{N} and A_0 \subset T^1 be given. Define A=\cup_{l=0}^{n-1}A_0 + l/n \pmod 1 and B=nA_0 \pmod 1. If |A| =|B| = n |A_0| , then

Proof. Let M be a measurable set such that A_0 \subseteq M , |M|=1/n, and nM=[0,1] (up to sets of measure zero). Let \bar{\mu} be the uniform probability measure on M. Notice that \bar{\mu}=n \mu on M. Therefore, the \bar{\mu}-probability of X_1,\dots,X_m \in A_0 for i.i.d. X_1,\dots,X_m \in U(M), denoted P^{\bar{\mu}}_{A_0}(m), is the same as the \mu-probability of X_1,\dots,X_n \in nA_0=B for i.i.d. X_1,\dots,X_m \in U([0,1]), that is, P_B(m). On the other hand, by Remark 1, the probability P_A(m) is identical to P^{\bar{\mu}}_{A_0}(m). ■

We claim that the sets H_j are essentially disjoint, in the sense that H_i \cap H_j has measure zero on T^n when i \neq j. To see this notice that if (x_k)_{k=1}^m \in H_i \cap H_j and x_i \neq x_j then the distance going counter-clockwise from x_i to x_j is at least |B| \geq \frac{1}{2}, and likewise the distance from x_j to x_i going counter-clockwise is at least \frac{1}{2}. So for (X_1, ..., X_m) to lie in H_i \cap H_j we must have that X_i and X_j either coincide or are antipodal, events with probability zero.

We can conclude that P((X_1, ..., X_m) \in H) = \sum P(H_k) = m a^{m-1}. ■

TO-DO: Extend this argument to the case a>1/2.

By Theorem 1 it follows that the probability of covering m random points by a half circle A=[0,1/2) is P_A(m)=m (1/2)^{m-1} which was the question asked in Question 1. One might wonder whether you arrive at the same probability if A consists of two quarter-circles. If the quarter-circles are diametrically opposite, e.g., A=[0,1/4) \cup [1/2,3/4) this is indeed the case. It follows from the following observation:

Remark 1. Let n \in \mathbb{N} be given. If A=A+k/n \pmod 1 for all k=1,\dots,n-1, then

To be more precise we need some notation. We identify \mathbb{S}^1 with \mathbb{T}^1=\mathbb{R}/\mathbb{Z} and do most of our computations in \mathbb{T}^n modulo 1. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on [0,1] . Then P_A(m) = P\{(X_1, X_2, ..., X_m) \in H\}, where we set

Theorem 1. If a \le 1/2, then P_A(m)=m a^{m-1} for all m=1,2,\dots.

Proof. Notice that the complement of A + \tau is B + \tau, where B = [a, 1). Therefore the following three statements are equivalent.

• \exists \tau s.t. x_j \in A + \tau, j = 1, 2, ..., m;
• \exists k s.t. x_j \in A + x_k, j = 1, 2, ..., m;
• \exists k s.t. x_j \notin B + x_k, j = 1, 2, ..., m.

In other words, letting H_k = \{(x_1, x_2, \dots, x_n) \,\vert\, x_j \notin (x_k, x_k + a] \pmod 1, j = 1, ..., n\} , we have H = \cup H_k.

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of measure 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points by either A or B is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [2/6,3/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C. in contrast to B, does not satisfy the 'symmetry' condition C=C+1/2 \pmod 1 from Remark 1.

Theorem. Lad \epsilon > 0 være givet. Så findes A \subset R / Z med \mu(A) < \epsilon således at givet vilkårlig tællelig mængde X \subset R / Z findes \tau så X \subset A+\tau.

Bevis. Lad A være en åben og tæt delmængde af R / Z med m(A) < \epsilon. Skriv X = \{x_1, x_2,\dots\} og sæt d_i = x_{i+1} - x_1, i = 1, 2, \dots. Så er B = \cap_i (A - d_i) en fællesmængde af åbne tætte mængder, og da R / Z med standard metrik er fuldstændigt metrisk rum følger det af Baire's sætning at B er tæt i R / Z. Derfor findes b \in B.

Sæt nu \tau = x_1 - b. Da b \in A er x_1 = b + \tau et element af A + \tau. For vilkårlig i \in N har vi b + d_i med i A så A + \tau indeholder

Theorem. Der findes A \subset R / Z med m(A) = 0 således at givet en tællelig mængde X \subset R / Z findes \tau så X \subset A + \tau.

Bevis. Lad B_n være åbne tætte mængder med |B_n| < 1/n og sæt A = \cap_{n \in N} B_n. Fortsæt så som i beviset af sidste sætning og bemærk at en tællelig fællesmængde af tællelige fællesmængder er en tællelig fællesmængde.

Det er klart at a=1 er en løsning, men kan vi finde et mindre a? Et (vildt) gæt ville være a=1-1/m. Det passer jo både med m=1 og m=2 :). For m=3, vil så |A|=2/3, hvilket også virker rimelig i worst-case-scenario: A=[0,2/3] dækker 3 equidistante punkter.

Det er klart at a=1 er en loesning, men kan vi finde et mindre a? Et (vildt) gæt ville være a=1-1/m. Det passer jo baade med m=1 og m=2 :). For m=3, vil så |A|=2/3, hvilket også virker rimelig i worst-case-scenario: A=[0,2/3] daekker 3 equidistante punkter.

Bevis. Lad B_n være åbne tætte mængder med |B_n| < 1/n og sæt A = \cap_{n \in N} B_n. Fortsæt så som i beviset af sidste sætning og bemærk at en tællelig fællesmængde af tællelige fællesmængder er en tællelig fællesmængde.

b + d_i med i A så A + \tau indeholder

Sæt nu \tau = x_1 - b. Da b \in A er x_1 = b + \tau et element af A + \tau. For vilkårlig i \in N har vi b + d_i med i A så A + \tau indeholder

Theorem. Der findes A \subset R / Z med m(A) = 0 således at givet en tællelig mængde X \subset R / Z findes \tau så X \subset A + \tau.

Bevis. Lad B_n være åbne tætte mængder med |B_n| < 1/n og sæt A = \cap B_n. Fortsæt så som i beviset af sidste sætning og bemærk at en tællelig fællesmængde af tællelige fællesmængder er en tællelig fællesmængde. ■

How do one construct a set A of smallest possible measure so that for any given set of m points we can find a \tau so that all m points belong to A+\tau \pmod 1. We start with a well-known fact about the Cantor set that covers the case m=2.

Theorem. Let A be the middle third Cantor set. Then for any two points \{x_1,x_2\} there exists a \tau so that \{x_1,x_2\} \subset A+\tau.

Can we extend this to m>2? The following result tells us that we can, not only catch m random points, but countably many points with a set A with arbitrarily small measure.

Theorem. Lad \epsilon > 0 være givet. Så findes A \subset R / Z med \mu(A) < \epsilon således at givet vilkårlig tællelig mængde X \subset R / Z findes \tau så X \subset A+\tau.

Bevis. Lad A være en åben og tæt delmængde af R / Z med m(A) < \epsilon. Skriv X = \{x_1, x_2,\dots\} og sæt d_i = x_{i+1} - x_1, i = 1, 2, \dots. Så er B = \cap_i (A - d_i) en fællesmængde af åbne tætte mængder, og da R / Z med standard metrik er fuldstændigt metrisk rum følger det af Baire's sætning at B er tæt i R / Z. Derfor findes b \in B.

Sæt nu \tau = x_1 - b. Da b \in A er x_1 = b + \tau et element af A + tau. For vilkårlig i \in N har vi b + d_i med i A så A + tau indeholder b + d_i + tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}. Altså er x_i in A + tau for alle i = 1, 2, 3, ...

Proof. Let M be a measurable set such that A_0 \subseteq M , |M|=1/n, and nM=[0,1] (up to sets of measure zero). Let \bar{\mu} be the uniform probability measure on M. Notice that \bar{\mu}=n \mu on M. Therefore, the \bar{\mu}-probability of X_1,\dots,X_m \in A_0 for i.i.d. X_1,\dots,X_m \in U(M), denoted P^{\bar{\mu}}_{A_0}(m), is the same as the \mu-probability of X_1,\dots,X_n \in nA_0=B for i.i.d. X_1,\dots,X_m \in U([0,1]), that is, P_B(m). On the other hand, by Remark 1, the probability P_A(m) is identical to P^{\bar{\mu}}_{A_0}(m).

Proof. Let M be a measurable set such that A_0 \subseteq M , |M|=1/n, and nM=[0,1] (up to sets of measure zero). Let \bar{\mu} be a probability measure on M. Notice that \bar{\mu}=n \mu on M. Therefore, the \bar{\mu}-probability of X_1,\dots,X_m \in A_0 for i.i.d. X_1,\dots,X_m \in U(M), denoted P^{\bar{\mu}}_{A_0}(m), is the same as the \mu-probability of X_1,\dots,X_n \in nA_0=B for i.i.d. X_1,\dots,X_m \in U([0,1]), that is, P_B(m). On the other hand, by Remark 1, the probability P_A(m) is identical to P^{\bar{\mu}}_{A_0}(m). ■

Proof. Let M be a measurable set such that A_0 \subseteq M , |M|=1/n, and nM=[0,1] (up to sets of measure zero). Let \bar{\mu} be a probability measure on M. Notice that \bar{\mu}=n \mu on M. Therefore, the \bar{\mu}-probability of X_1,\dots,X_m \in A_0 for i.i.d. X_1,\dots,X_m \in U(M) is identical to the \mu-probability of X_1,\dots,X_n \in nA_0 for i.i.d. X_1,\dots,X_m \in U([0,1]), that is, P_{nA_0}(m)

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of measure 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points by either A or B is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [2/6,3/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C. in contrast to B, does not satisfy the "symmetry" condition C=C+1/2 \pmod 1 from Remark 1.

Proof. Let M be a measurable set such that A_0 \subseteq M , |M|=1/n, and nM=[0,1] (up to sets of measure zero). Let \bar{\mu} be a probability measure on M.

Lemma 1. Let n \in \mathbb{N} and A_0 \subset T^1 be given. Define A=\cup_{l=0}^{n-1}A_0 + l/n \pmod 1 and B=nA_0 \pmod 1. If |A| =|B| = n |A_0| , then

Lemma 1. Let n \in \mathbb{N} and A_0 \subset T^1 be given. Define A=\cup_{l=0}^{n-1}A_0 + l/n \pmod 1 and B=nA_0. If |A| =|B| = n |A_0| , then

Proof. TO-DO

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of measure 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points by either A or B is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [2/6,3/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C does not satisfy the "symmetry" condition C=C+k/n \pmod 1 from Remark 1.

By Theorem 1 it follows that the probability of covering m random points by a half circle A=[0,1/2) is P_A(m)=m (1/2)^{m-1} which was the question asked in Question 1. One might wonder whether you arrive at the same probability if A consists of two quarter-circles. If the quarter-circles are diametrically opposite, e.g., A=[0,1/4) \cup [1/2,3/4) this is indeed the case. It follows from the following observation:

Remark 1. Let n \in \mathbb{N} be given. If A=A+k/n \pmod 1 for all k=1,\dots,n-1, then

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of measure 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points by either A or B is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [0,1/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C does not satisfy the "symmetry" condition C=C+k/n \pmod 1 from Remark 1.

The following result shows that sets with "similar" symmetry properties have the same covering probabilities.

Lemma 1. Let A_0 \subset T^1 be given. Define A=\cup_{l=0}^{n-1} \pmod 1 and B=nA_0

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6), and C=[0,1/6) \cup [2/6,3/6), all of length 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [0,1/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C does not satisfy the "symmetry" condition C=C+k/n \pmod 1 from Remark 1.

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6)$, and {$C=[0,1/6) \cup [2/6,3/6), all of length 1/3. By Remark 1 we see that P_A(m)=P_B(m)=m (1/3)^{m-1} hence, in paricular, the probability of catching two random points is P_A(2)=P_B(2)=2/3. For the set C, however, we set that P_C(2)=1 since if the distance from x_1 to x_2 is less that 1/6, we can find \tau so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. On the other hand, if the distance between x_1 and x_2 is greater than 1/6, but less that 1/2, we can find \tau so that x_1 \in [0,1/6)+\tau and x_2 \in [0,1/6)+\tau, hence \{x_1,x_2\}\subset C+\tau. The event that the distance between x_1 and x_2 is exactly either 1/6 or 1/2 has probability zero. We remark that the set C does not satisfy the "symmetry" condition C=C+k/n \pmod 1 from Remark 1.

Consider the three sets A=[0,1/3), B=[0,1/6) \cup [3/6,4/6)$, and {$C=[0,1/6) \cup [2/6,3/6), all of length 1/3. By Remark 1, P_A(m)=P_B(m)=m (1/3)^{m-1}, hence in paricular, the probability of catching two random points is P_A(2)=P_B(2)=2/3. For the set C however we set that P_C(2)=1 since if the distance from $x_1$ to $x_2$ is less that 1/6, we can find $\tau$ so that x_1,x_2 \in [0,1/6)+\tau \subset C+\tau. The "symmetry" condition B=B+k/n \pmod 1 is necessary, in the sense,

Remark 1: Let n \in \mathbb{N} be given. If B=B+k/n \pmod 1 for all k=1,\dots,n-1, then

By Theorem 1 it follows that the probability of covering m random points by a half circle A=[0,1/2) is P_A(m)=m (1/2)^{m-1}, which was the question asked in Question 1. One might wonder whether you arrive at the same probability if A consists of two quarter-circles. If the quarter-circles diametrically opposite, e.g., A=[0,1/4) \cup [1/2,3/4) this is indeed the case. It follows from the following observation:

By Theorem 1 it follows that the probability of covering m random points by a half circle A=[0,1/2) is P_A(m)=m (1/2)^{m-1}, which was the question asked in Question 1. We might wonder whether one arrives at the same probability if A consists of two quarter-circles. If the quarter-circles diametrically opposite, e.g., A=[0,1/4) \cup [1/2,3/4)

To be more precise we need some notation. We identify \mathbb{S}^1 with \mathbb{T}^1=\mathbb{R}/\mathbb{Z} and do most of our computations in \mathbb{T}^n modulo 1. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on [0,1] . Then

By Theorem 1 we have that the probability of covering m random points by a half circle A=[0,1/2) is P_A(m)=m (1/2)^{m-1}, which was the question asked in Question 1. We might wonder whether one arrives at the same probability if A consists of two quarter-circles. If the quarter-circles diametrically opposite, e.g., A=[0,1/4) \cup [1/2,3/4)

Theorem 1. If a \le 1/2, then P_A(m)=m a^{m-1} for all m=1,2,\dots.

By Theorem 1 we have that the probability of covering m random points by a half circle A=[0,1/2] is P_A(m)=m (1/2)^{m-1}, which was the question asked in Question 1.

By Lemma 2 we have that if A is a arbitrary subset of \mathbb{T}^1 with |A| > 1/2, then there is a \tau so that \{x_1, x_2\} \subset A + \tau. Obviously, we cannot do better than 1/2, e.g., the set A=[0,1/2-\epsilon] cannot catch all pairs of points. Now one can ask how large the set A needs to be in order for us to always catch m random points. More precisely: What is the smallest length a such that any set A \subset \mathbb{T}^1 satisfying |A| \ge a or |A| > a, there exists a \tau so that \{x_1, \dots, x_m\} \subset A + \tau ?

TO-DO: Extend lemma to |A|=1/2 and to m points |A| = a > 1-1/m.

To be more precise we need some notation. We identify \mathbb{S}^1 with \mathbb{T}^1=\mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on [0,1] . Then

(x_k)_{k=1}^m \in H_i \cap H_j and x_i \neq x_j then the distance going counter-clockwise from x_i to x_j is at least |B| \geq \frac{1}{2}, and likewise the distance from x_j to x_i going counter-clockwise is at least \frac{1}{2}. So for (X_1, ..., X_m) to lie in H_i \cap H_j we must have that X_i and X_j either coincide or are antipodal, events with probability zero.

P((X_1, ..., X_m) \in H) = \sum P(H_k) = m a^{m-1}.

Lemma 2. Let A \subseteq \mathbb{T}^1 be a (Lebesgue) measurable set. Suppose there exists points x_1, x_2 \in T^1 so that for all \tau \in T^1 it holds that x_1 and x_2 not simultaneous can be in A + \tau. Then m(A) \le 1/2.

By the lemma we have that if A is a arbitrary subset of \mathbb{T}^1 with |A| > 1/2, then there is a \tau so that \{x_1, x_2\} \subset A + \tau. Obviously, we cannot do better than 1/2, e.g., the set A=[0,1/2-\epsilon] cannot catch all pairs of points. Now one can ask how large the set A needs to be

Lemma. Let A \subseteq \mathbb{S}^1 \cong \mathbb{T}^1 be a (Lebesgue) measurable set. Suppose there exists points x_1, x_2 \in T^1 so that for all \tau \in T^1 it holds that x_1 and x_2 not simultaneous can be in A + \tau. Then m(A) \le 1/2.

We consider a generalized version of Question 1. We want to find the probability P_A(m) that m points falls within an arc A of \mathbb{S}^1 of length a, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu (also denoted |\cdot|) being the (probability Lebesgue) surface measure on \mathbb{S}^1.

a\not=b \neq b

P_A(m) = P\{(X_1, X_2, ..., X_m) \in H\}, where we set

Theorem. If a \le 1/2, then P_A(m)=m a^{m-1} for all m=1,2,\dots.

In other words, letting H_k = \{(x_1, x_2, \dots, x_n) \,\vert\, x_j \notin (x_k, x_k + a] \pmod 1, j = 1, ..., n\} , we have

Proof. Let d = x_2 - x_1 \in T^1. The sets A and A - d are disjoint. To see this, assume they are not. Then we can find u \in A \cap (A - d). Let \tau = x1 - u. Note that x_1-\tau=u \in A and x_2-\tau=d+u\in A, hence x_1, x_2 \in A + \tau which contradicts our assumption. Since A, it follows by translation invariance of the Lebesgue measure that

Christian: Ikke-lig-med tegn kan laves således:

Misc./Notes

a\not{=}b

and x_i \neq x_j then the distance going counter-clockwise from x_i to x_j is at least |B| \geq \frac{1}{2}, and likewise the distance from x_j to x_i going counter-clockwise is at least \frac{1}{2}. So for X_1, ..., X_m to lie in H_i \cap H_j we must have that X_i and X_j either coincide or are antipodal, events with probability zero.

P(X_1, ..., X_m \in H) = \sum P(H_k) = m a^{m-1}.

We consider a generalized version of Question 1. We want to find the probability P_a(m) that m points falls within an arc of \mathbb{S}^1 of length a, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu (also denoted |\cdot|) being the (probability Lebesgue) surface measure on \mathbb{S}^1.

To be more precise we need some notation. We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on [0,1] . Then P_m(a) = P\{(X_1, X_2, ..., X_m) \in H\}, where we set

Theorem. If a \le 1/2, then P_a(m)=m a^{m-1} for all m=1,2,\dots.

Proof. Notice that the complement of A + \tau is B + \tau, where B = [a, 1). Therefore the following three statements are equivalent.

• \exists \tau s.t. x_j \in A + \tau, j = 1, 2, ..., m;
• \exists k s.t. x_j \in A + x_k, j = 1, 2, ..., m;
• \exists k s.t. x_j \notin B + x_k, j = 1, 2, ..., m.

In other words, letting H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \notin (x_k, x_k + a] \pmod 1, j = 1, ..., n\} , we have H = \cup H_k.

We claim that the sets H_j are essentially disjoint, in the sense that H_i \cap H_j has measure zero on T^n when i \neq j. To see this notice that if (x)\in H_i \cap H_j and x_i \neq x_j then the distance going counter-clockwise from x_i to x_j is at least |B| \geq \frac{1}{2}, and likewise the distance from x_j to x_i going counter-clockwise is at least \frac{1}{2}. So for X_1, ..., X_m to lie in $H_i \cap H_j$ we must have that $X_i$ and $X_j$ either coincide or are antipodal, events with probability zero.

We can conclude that $P(X_1, ..., X_m \in H) = \sum P(H_k) = m a^{m-1}$. ■

TO-DO: Extend this argument to the case a>1/2.

We consider a generalized version of Question 1. We want to find the probability P(m)=P_a(m) that m points falls within an arc of \mathbb{S}^1 of length a, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu (also denoted |\cdot|) being the (probability Lebesgue) surface measure on \mathbb{S}^1.

Theorem 1. If a \le 1/2, then P_a(m)=m a^{m-1} for all m=0,1,\dots, that is,

TO-DO: Extend this argument to the case a>1/2 and to (symmetric) unions of intervals.

Example 1. If A=[0,1/6) \cup [2/6,3/6), then the (extension) of Theorem 1 tells us that P(2)=2 (1/3)^1 = 2/3. However, if we change A to a non-symmetric version, e.g., A=[0,1/6) \cup [2/6,3/6), then P(2)=1. Hence, Theorem 1 cannot be extended to non-symmetric sets.

TO-DO: Extend lemma to |A|=1/2 and to m points P=1-1/m.

Lemma. Let A \subseteq \mathbb{S}^1 \cong \mathbb{T}^1 be a (Lebesgue) measurable set. Suppose there exists points x1, x2 \in T^1 so that for all \tau \in T^1 it holds that x1 and x2 not simultaneous can be in A + \tau. Then m(A) \le 1/2.

By the lemma we have that if A is a arbitrary subset of \mathbb{T}^1 with |A| \ge 1/2, then there is a \tau so that \{x_1, x_2\} \subset A + \tau. Obviously, we cannot do better than 1/2, e.g., the set A=[0,1/2-\epsilon] cannot catch all pairs of points. Now one can ask how large the set A needs to be

On the probability of covering m random points with a given subset of the unit circle

Suppose now that your worst enemy can choose the set A under the restriction that \mu (A)=a for some given 1\ge a \ge 0. How should he choose it? How big will a have to be so that we always can be sure of catching m points? We start with the following observation.

m points on an n-dimensional sphere. What holds here?

Literature

Vaguely related to the Bertrand paradox

We consider a generalized version of Question 1. We want to find the probability P_a(m) that m points falls within an arc of \mathbb{S}^1 of length a, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu (also denoted |\cdot|) being the (probability Lebesgue) surface measure on \mathbb{S}^1.

Extension to higher dimensions \mathbb{S}^n for n>1.

m on an n-dimensional sphere. What holds here?

in order for us to always catch m random points. More precisely: What is the smallest length a such that any set A \subset \mathbb{T}^1 satisfying |A| \ge a, there exists a \tau so that \{x_1, \dots, x_m\} \subset A + \tau ?

TO-DO: Prove P=1-1/m.

- will be updated soon -

Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}. ■

By the lemma we have that if A is a arbitrary subset of \mathbb{T}^1 with |A| \ge 1/2, then there is a \tau so that x1, x2 \in A + \tau. Obviously, we cannot do better than 1/2, e.g., the set A=[0,1/2-\epsilon] cannot catch all pairs of points. Now one can ask how large the set A needs to be

Nu kan man spørge om hvor stor mængden A skal være før at vi altid kan fange m punkter. Mere præcist:

TO-DO: Construct a set A of smallest possible measure (zero?) so that for any given set of m points we can find a \tau so that all m points belong to A+\tau \pmod 1. Start with m=2.

Suppose now that your worst enemy can choose the set A under the restriction that \mu(A)=a for some given 1\ge a \ge 0. How should he choose it? How big will a have to be so that we always can be sure of catching m points? We start with the following observation.

Proof. Let d = x_2 - x_1 \in T^1. The sets A and A - d are disjoint. To see this, assume they are not. Then we can find u \in A \cap (A - d). Let tau = x1 - u. Note that x_1-\tau=u \in A and x_2-\tau=d+u\in A, hence x_1, x_2 \in $A + \tau which contradicts our assumption. Since A, it follows by translation invariance of the Lebesgue measure that

H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \in [x_k, x_k + a[ \pmod 1, j = 1, ..., n\} . We claim that H = \cup_k H_k . To see this claim note that if x = (x_j) \in H , then pick the first x_k after K (w.r.t. the positive direction on \mathbb{S}^1 . Now, clearly x=(x_j) \in H_k. On the other hand, if there exists a k so that (x_j) \in H_k, then pick K=x_k and hence x=(x_j) \in H_k.

Suppose now that your worst enemy can choose the set A under the restriction that \mu(A)=a for some given 1\le a \le 0. How should he choose it? How big will a have to be so that we always can be sure of catching m points? We start with the following observation.

Lemma. Let A \subseteq \mathbb{S}^1 \cong \mathbb{T}^1 be a (Lebesgue) measurable set. Suppose there exists points x1, x2 \in T^1 so that for all tau \in T^1 it holds that x1 and x2 not simultaneous can be in A + \tau. Then m(A) \le 1/2.

Proof. Let d = x_2 - x_1 \in T^1. Der må gælde at A og A - d er disjunkte, for ellers findes u \in A \cap (A - d), vi kan sætte tau = x1 - u og bemærker at både x1 og x2 ligger i A + \tau. Da de to mængder er disjunkte og Lebesgue målet er translationsinvariant har vi

H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \in [x_k, x_k + a[ \pmod 1, j = 1, ..., n\} . We claim that H = \cup_k H_k . To see this claim note that if x = (x_j) \in H , then pick the first x_k after K (w.r.t. the positive direction on \mathbb{S}^1 . Now, clearly x=(x_j) \in H_k. On the other hand, if there exists a k so that (x_j) \in H_k, then pick K=x_k and hence x=(x_j) \in H_k. \\

Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}.

TO-DO: Extend this argument to the case a>1/2.

TO-DO: Construct a set A of smallest possible measure (zero?) so that for any given set of m points we can find a $\tau$ so that all m points belong to A+\tau \pmod 1. Start with m=2.

Choosing the worst possible set A

Suppose now that your worst enemy can choose the set A under the restriction that \mu(A)=a for some given 1\le a \le 0. How should he choose it? How big will a have to be so that we always can be sure of catching m points?

Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}. \box

Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}. \Box

Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}. \qed

TO-DO: Extend this argument to the case a>1/2.

Choosing the best possible set A

TO-DO: Construct a set A of smallest possible measure (zero?) so that for any given set of m points we can find a $\tau$ so that

H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \in [x_k, x_k + a[ \pmod 1, j = 1, ..., n\} . We claim that H = \cup_k H_k . To see this claim note that if x = (x_j) \in H , then pick the first x_k after K (w.r.t. the positive direction on \mathbb{S}^1 . Now, clearly x=(x_j) \in H_k. On the other hand, if there exists a k so that (x_j) \in H_k, then pick K=x_k and hence x=(x_j) \in H_k.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on [0,1] .

We consider a generalized version of Question 1. We want to find the probability P_a(m) that m points falls within an arc of \mathbb{S}^1 of length a, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu being the (probability Lebesgue) surface measure on \mathbb{S}^1.

The sets H_k are disjoint if (and only if) a \le 1/2. Hence, $P((X_1, \dots, X_m) \in H) = \sum_k P((X_1, ..., X_m) \in H_k). Since (X_1, \dots, X_m) are i.i.d. uniformly, we have that P(X_1, ..., X_m) = a^{m-1}. By the above equation, we arrive at P_a(m) = m a^{m-1}.

H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \in [x_k, x_k + a[ \pmod 1, j = 1, ..., n\} . We claim that H = \cup_k H_k . To see this claim note that if x = (x_j) in H, then pick the first x_k after K (w.r.t. the positive direction on \mathbb{S}^1. Now, clearly x=(x_j) \in H_k. On the other hand, if there exists a k so that (x_j) \in H_k, then pick K=x_k and hence x=(x_j) \in H_k.

For k = 1, \dots, n define H_k = \{x_1, x_2, \dots, x_n \,\vert\, x_j \in [x_k, x_k + a[ \pmod 1, j = 1, ..., n\} . We claim that H = \cup_k H_k . To see this claim note that if x = (x_j) in H, then pick the first x_k after K (w.r.t. the positive

 for if r hvis der findes k så (x_j) \in H_k kan vi vælge K = x_k og ser (x_j) \in H . Omvendt hvis (x_j) \in H findes et (og muligvis flere) første x_k der ligger efter K eller er lig med K, med hensyn den ordning der svarer til positiv omløbsretning på cirklen. Nu er (x_j) \in H_k .

We consider a generalized version of Question 1. We want to find the probability P_a(m) that m points falls within an arc of length a of \mathbb{S}^1, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu being the (probability Lebesgue) surface measure on \mathbb{S}^1.

For k = 1, ..., n, sæt

 H_k = {x_1, x_2, ..., x_n | x_j \in [x_k; x_k + a[ modulo 1, j = 1, ..., n} . Det er klart at der gælder
 H = \cup_k H_k

for hvis der findes k så (x_j) \in H_k kan vi vælge K = x_k og ser (x_j) \in H . Omvendt hvis (x_j) \in H findes et (og muligvis flere) første x_k der ligger efter K eller er lig med K, med hensyn den ordning der svarer til positiv omløbsretning på cirklen. Nu er (x_j) \in H_k .

Hvis (og kun hvis) a <= 1/2 gælder at mængderne H_k er parvis disjunkte. Derfor har vi P((X_1, ..., X_n) \in H) = \sum_k P((X_1, ..., X_n) \in H_k). Pga uniform fordeling og uafhængighed gælder

 P(X_1, ..., X_n) = a^{n-1}

Og da vi summerer over n ens termer får vi

 P_a(n) = n a^{n-1} .

Hvis a > 1/2 er H_k'erne ikke disjunkte og P((X_1, ..., X_n) \in H) < \sum_k P((X_1, ..., X_n) \in H_K), så

 P_a(n) < n a^{n-1} .

I dette tilfælde ser problemet sværere ud...

Question 2: Let 1 \ge a \ge 0. Suppose A \subset \mathbb{S}^1 is a Borel measurable set of measure \mu(A)=a , where \mu(\mathbb{S}^1)=a . What is probability that A can cover m random points?

Lemma. Lad A \subseteq T^1 := R / Z være en (Lebesgue) målelig mængde. Hvis der findes punkter x1, x2 \in T^1 så til alle tau \in T^1 gælder at ikke både x1 og x2 er indeholdt i A + \tau, da er m(A) <= 1/2.

Bevis. Sæt d = x2 - x1 \in T^1. Der må gælde at A og A - d er disjunkte, for ellers findes u \in A \cap (A - d), vi kan sætte tau = x1 - u og bemærker at både x1 og x2 ligger i A + \tau. Da de to mængder er disjunkte og Lebesgue målet er translationsinvariant har vi

Så nu ved vi at hvis A er en vilkårlig delmængde af T^1 med |A| \ge 1/2, så eksisterer der et \tau så {x1, x2} \subset A + \tau. Det er åbenlyst at vi ikke kan gøre det bedre end 1/2, fx, vil A=[0,1/2-\epsilon] ikke kunne fange alle par af punkter. Nu kan man spørge om hvor stor mængden A skal være før at vi altid kan fange m punkter. Mere præcist:

Hvor lille kan man vælge a således at for en vilkårlig mængde A \subset T^1 med |A| \ge a, så eksisterer der et \tau så {x1, ..., xm} \subset A + \tau?

Det er klart at a=1 er en løsning, men kan vi finde et mindre a? Et (vildt) gæt ville være a=1-1/m. Det passer jo både med m=1 og m=2 :). For m=3, vil så |A|=2/3, hvilket også virker rimelig i worst-case-scenario: A=[0,2/3] dækker 3 equidistante punkter.

P((X_1,\dots,X_m) \in H)=m a^{m-1}, where

P(\{X_1,\dots,X_m\} \in H)=m a^{m-1}, where

- will be updated soon -

\section{One} We identify $\mathbb{S}^1$ with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on {0,1$}.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on {0,1$}.

Theorem. If a \le 1/2, then P_a(m)=m a^{m-1} for all m=0,1,\dots, that is,

Proof.

- will be updated soon -

     (:latex [DisplayCode[+] = true|false]
              [Verbose = true|false]
               [ForceRender = true|false]
                [ClearCache = true|false] 
                 [... template parameters ...] :)

section{One} We identify $\mathbb{S}^1$ with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on {0,1$}.

     (:latexend)

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let \{X_1, \dots, X_m \in [0,1] be an independent and identically distributed (i.i.d.) sample from U(0,1) — the uniform distribution on {0,1$}.

Question 2: Let 1 \ge a \ge 0. Suppose A \subset \mathbb{S}^1 is a Borel measurable set of measure \mu(A)=a, where \mu(\mathbb{S}^1)=a. What is probability that A can cover m random points?

On the probability of catching m random points with a given subset of the unit circle

On the probability of catching m random points with a subset of the unit circle

Afternoon tea discussions at DTU Mathematics — discussions on non-applied mathematics.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1. Let \{X_1, \dots, X_m \in [0,1) be independent and identically distributed (i.i.d.) uniformly distributed points.

Theorem. If a \le 1/2, then P_a(m)=m a^{m-1} for all m=0,1,\dots.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau \pmod 1.

We consider a generalized version of Question 1. We want to find the probability P(m)=P_a(m) that m points falls within an arc of length a of \mathbb{S}^1, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu being the (probability Lebesgue) surface measure on \mathbb{S}^1.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists \tau \in [0,1) such that the m random points belong to A+\tau (\mod 1).

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a). We are then asking whether there exists $\tau \in [0,1)$ such that the m random points belong to $A+\tau (\mod 1)$.

We consider a generalized version of Question 1. We want to find the probability P(m)=P_a(m)$ that {$m points falls within an arc of length a of \mathbb{S}^1, where the total length of \mathbb{S}^1 is one, i.e., \mu(\mathbb{S}^1)=1 with \mu being the (Lebesgue) surface measure on \mathbb{S}^1.

We identify \mathbb{S}^1 with \mathbb{R}/\mathbb{Z}. Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).

 We are then asking 

Afternoon tea discussions at DTU Mathematics

On the probability of catching m random points on the unit circle

On the probability of cacthing m random points on the unit circle

Question 3: Let 1 \ge a \ge 0 be given. What is the best subset A \subset \mathbb{S}^1 to choose with \mu(A)=a if we want to cover m random points?

Question 2: Let 1 \ge a \ge 0. Suppose A \subset \mathbb{S}^1 is a Borel measurable set of measure \mu(A)=a. What is probability that A can cover m random points?

Question 2: Let 1 \ge a \ge 0 be given. What is the best subset A \subset \mathbb{S}^1 to choose with \mu(A)=a if we want to cover m random points?

Question 2: Let 1 \ge a \ge 0. Suppose A \subset \mathbb{S}^1 is a Borel measurable set of measure \mu(A)=a. What is probability that A can cover m random points?

Let a \in \mathbb{R} be given such that 0 \le a \le 1, and let A=[0,a).

Question 2: Let 0 \ge a \ge 1. Suppose A \subset \mathbb{S}^1 is a Borel measurable set of measure \mu(A)=a. What is probability that A can cover m random points?

Let a \in \mathbb{R} be given such that 0 \ge a \ge 1, and let A=[0,a).

Question 1: If one picks m points at random on the unit circle \mathbb{S}^1, what is the probability that they can be covered by a half circle A \subset \mathbb{S}^1?

Question 1: If one picks m points at random on the unit circle \mattbb{S}^1, what is the probability that they can be covered by a half circle A \subset \mattbb{S}^1?

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Question 1: If one picks m points at random on the unit circle, what is the probability that they can be covered by a half circle?

Covering points with an interval

Afternoon tea discussions at DTU Math

On the probability of covering m random points on the unit circle

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Afternoon tea discussions at DTU Math

Math Discussions