## MathForum.MathForum History

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[[#nD]]

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

''{$m$} points on an {$n$}-dimensional sphere''. What holds here?

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[[#ref]]

!! Literature

Vaguely related to the [[http://en.wikipedia.org/w/index.php?title=Bertrand_paradox_%28probability%29 | Bertrand paradox]]

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[[#worst]]

!! 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

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. ■ $$}

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 {$|A| = a > 1-1/m$}.

[[#worst]]

!! 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

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. ■ $$}

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 {$|A| = a > 1-1/m$}.

Added lines 62-88:

[[#best]]

!! 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 {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}

Altså er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}

■

'''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.

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!! 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 {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}

Altså er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}

■

'''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.

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Added lines 49-61:

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

{$$ P_A(m) = P_B(m) \text{ for all } m \in \mathbb{N}_0. $$}

''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)$}.

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Added lines 34-48:

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

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

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'''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

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod{\frac1n} $$}

Added lines 19-33:

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

{$$ H = \{(x_1,\dots,x_m)\in \mathbb{T}^m \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

'''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

{$H = \cup H_k$}.

Deleted lines 18-122:

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

{$$ H = \{(x_1,\dots,x_m)\in \mathbb{T}^m \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

'''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

{$H = \cup H_k$}.

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

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod{\frac1n} $$}

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

{$$ P_A(m) = P_B(m) \text{ for all } m \in \mathbb{N}_0. $$}

''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)$}.

■

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[[#best]]

!! 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 {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}

Altså er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}

■

'''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.

■

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[[#worst]]

!! 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

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. ■ $$}

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 {$|A| = a > 1-1/m$}.

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[[#nD]]

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

''{$m$} points on an {$n$}-dimensional sphere''. What holds here?

----

[[#ref]]

!! Literature

Vaguely related to the [[http://en.wikipedia.org/w/index.php?title=Bertrand_paradox_%28probability%29 | Bertrand paradox]]

>>comment<<

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.

>><<

Changed line 50 from:

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.

to:

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.

Changed lines 73-80 from:

'''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 {$i = 1, 2, 3, \dots$}

''Bevis''. Lad A

{$b + d_i$} med i {$A$}

to:

'''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 {$i = 1, 2, 3, \dots$}

''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 {$i = 1, 2, 3, \dots$}

Changed lines 83-85 from:

'''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~~.

''Bevis''. Lad {$B_n$}

to:

'''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.

''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.

Changed line 122 from:

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.

to:

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.

Changed line 122 from:

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.

to:

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.

Changed line 85 from:

''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.

to:

''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.

Changed line 79 from:

{$b + d_i$} med i {$A$} så {$A + \tau$} indeholder {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}$$}~~.~~

to:

{$b + d_i$} med i {$A$} så {$A + \tau$} indeholder {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}

Changed lines 78-80 from:

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,~~...~~

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,

to:

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 {$i = 1, 2, 3, \dots$}

■

{$b + d_i$} 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 {$i = 1, 2, 3, \dots$}

■

Added lines 83-86:

'''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.

■

''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.

■

Added lines 65-66:

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$}.

Changed lines 68-82 from:

'''~~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 {~~$~~m=2$~~}.

to:

'''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, ...

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, ...

Changed line 57 from:

''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)$}.

to:

''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)$}.

Changed lines 57-58 from:

''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)$} ~~

to:

''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)$}.

■

■

Changed lines 48-49 from:

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod~~ ~~\frac1n $$}

to:

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod{\frac1n} $$}

Changed lines 57-60 from:

''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$}.

to:

''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)$}

Changed lines 50-51 from:

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.

to:

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.

Changed line 57 from:

''Proof''. ~~TO-DO~~

to:

''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$}.

Changed line 54 from:

'''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

to:

'''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

Changed lines 54-55 from:

'''Lemma 1'''. Let {$~~A_0~~ \~~subset T^1~~$} ~~be given. Define ~~{$A~~=\cup~~_~~{l=~~0~~}~~^~~{n-~~1} ~~\pmod 1$} and~~ {$~~B~~=~~nA~~_0~~$~~}

to:

'''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

{$$ P_A(m) = P_B(m) \text{ for all } m \in \mathbb{N}_0. $$}

''Proof''. TO-DO

{$$ P_A(m) = P_B(m) \text{ for all } m \in \mathbb{N}_0. $$}

''Proof''. TO-DO

Changed line 50 from:

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.

to:

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.

Changed lines 45-50 from:

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:

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

{$$ \{x_i\} \subset~~B~~ + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset ~~B~~ + \tau \pmod \frac1n $$}

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~~. ~~ ~~

'''Remark 1'''

{$$ \{x_i\} \subset

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

to:

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

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod \frac1n $$}

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 {$B=nA_0$}

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

{$$ \{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod \frac1n $$}

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 {$B=nA_0$}

Changed line 50 from:

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.

to:

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.

Changed lines 50-51 from:

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,~~

to:

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.

Added lines 50-51:

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,

The "symmetry" condition {$B=B+k/n \pmod 1$} is necessary, in the sense,

Added lines 46-49:

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

{$$ \{x_i\} \subset B + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset B + \tau \pmod \frac1n $$}

Changed line 45 from:

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:

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:

Changed line 45 from:

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., {$A=[0,1/4) \cup [1/2,3/4)$}

to:

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., {$A=[0,1/4) \cup [1/2,3/4)$}

Changed line 20 from:

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

to:

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

Changed line 45 from:

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''.

to:

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., {$A=[0,1/4) \cup [1/2,3/4)$}

Changed lines 24-25 from:

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

to:

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

Added lines 44-45:

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''.

Changed lines 63-66 from:

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

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:''' Extend lemma to {$~~|A|=1/2$~~} ~~and~~ to {$~~m~~$} ~~points ~~{$~~P=1-1/m~~$}.

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 {$|A| \ge a$}

'''TO-DO:''' Extend lemma to {

to:

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 {$|A| = a > 1-1/m$}.

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 {$|A| = a > 1-1/m$}.

Changed line 20 from:

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

to:

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

Changed lines 22-23 from:

{$$ H = \{(x_1,\dots,x_m)~~ ~~\~~,~~\~~vert~~\,~~ ~~\~~exists ~~\~~tau~~ \~~text{ s.t. } x_j \in~~ A+\tau~~=[\tau,\tau+a)~~ \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

to:

{$$ H = \{(x_1,\dots,x_m)\in \mathbb{T}^m \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

Changed lines 36-38 from:

{$(x)~~\in H~~_~~i~~ \~~cap~~ H_~~j$}~~

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.

and {$x_i

to:

{$(x_k)_{k=1}^m \in H_i \cap H_j$}

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.

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.

Changed line 40 from:

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

to:

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

Changed line 58 from:

'''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$}.

to:

'''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$}.

Changed line 63 from:

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

to:

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

Changed line 58 from:

'''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$}.

to:

'''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$}.

Changed line 18 from:

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:

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 {$\mathbb{S}^1$}.

Deleted lines 80-84:

!! Misc./Notes

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

{$a\not=b \neq b$}

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{$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a) \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

to:

{$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau=[\tau,\tau+a) \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

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{$P_m~~(a~~) = P\{(X_1, X_2, ..., X_m) \in H\}$}, where we set

to:

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

Changed lines 24-25 from:

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

to:

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

Changed line 32 from:

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

to:

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

Changed line 60 from:

''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

to:

''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

27.03.2012, at 08:45 UTC
by - typos

Changed lines 37-38 from:

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.

to:

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.

Changed line 40 from:

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

to:

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

27.03.2012, at 08:41 UTC
by - Rettede lidt hist og her mest i bevis for theorem. Kan ikke få \neq til at virke!

Changed lines 18-34 from:

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 {$\mathbb{S}^1$}.

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

{~~$P((X_1,\dots,X_m) \in H)=m a^{m-1~~}~~$}, where {~~$$ ~~H = \~~{~~(x_~~1,~~\dots,x_m) \,\vert\, \exists \tau \text~~{ ~~s.t. ~~} ~~x_j \in [\tau~~,\~~tau+a] \pmod 1 \text{ for all } j=1,~~\~~dots, m\}.~~$~~$}~~

''Proof''.

~~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$} (w.r.t. the positive direction on {$~~ \~~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~~$}~~. ~~

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 {$~~P_a(m) = m a^~~{~~m-~~1}$}. ~~■~~

'''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. ~~

'''Theorem

''Proof''.

{$ H_k = \

The sets

{$$P((X

'''TO-DO:''' Extend this argument

'''Example 1'''. If {

to:

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

{$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a) \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

'''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 {$a>1/2$}.

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

{$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a) \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

'''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 {$a>1/2$}.

Changed lines 18-19 from:

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:

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 {$\mathbb{S}^1$}.

Changed line 22 from:

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

to:

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

Changed lines 32-34 from:

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

to:

'''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.

'''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.

Changed line 55 from:

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

to:

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

Changed lines 47-48 from:

'''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$}.

to:

'''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$}.

Changed line 52 from:

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

to:

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

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! On the probability of ~~catching~~ {$m$} random points with a given subset of the unit circle

to:

! On the probability of covering {$m$} random points with a given subset of the unit circle

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[[#best]]

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[[#best]]

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[[#nD]]

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[[#ref]]

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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.

to:

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.

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''{$m$} on an {$n$}-dimensional sphere''. What holds here?

to:

''{$m$} points on an {$n$}-dimensional sphere''. What holds here?

Changed lines 48-49 from:

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. ~~$$} ~~■

to:

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. ■ $$}

Added lines 60-65:

----

!! Literature

Vaguely related to the [[http://en.wikipedia.org/w/index.php?title=Bertrand_paradox_%28probability%29 | Bertrand paradox]]

Changed lines 18-19 from:

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 {$\mathbb{S}^1$}.

to:

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 {$\mathbb{S}^1$}.

Changed lines 55-59 from:

-~~ will be updated soon ~~-

to:

----

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

''{$m$} on an {$n$}-dimensional sphere''. What holds here?

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

''{$m$} on an {$n$}-dimensional sphere''. What holds here?

Changed lines 51-56 from:

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

to:

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 -

>>comment<<

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

- will be updated soon -

>>comment<<

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to:

>><<

Changed lines 30-31 from:

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:

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 {$P_a(m) = m a^{m-1}$}. ■

Changed lines 48-50 from:

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. $$}

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:

Så nu ved vi at hvis A er en vilkårlig delmængde af T^1 med |A|

to:

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. $$} ■

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:

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:

Changed lines 38-39 from:

'''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 {$m=2$}.

to:

'''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$}.

Changed lines 43-44 from:

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.

to:

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.

Changed lines 47-48 from:

''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~~

{$$2 m(A) = m(A~~) + m(A - d) <= m(T^1) = 1~~ $$}

{$$2 m(A) = m(

to:

''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

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. $$}

{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. $$}

Changed line 27 from:

{$ 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$}. ~~\\~~

to:

{$ 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$}.

Added lines 34-35:

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Changed lines 43-47 from:

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?

'''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

'''Lemma'''. Lad {$A \subseteq T^1 := R / Z

Bevis. Sæt

to:

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

'''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

Changed line 29 from:

{$$P((X_1, \dots, X_m) \in H) = \sum_k P((X_1, ..., X_m) \in H_k).$}

to:

{$$P((X_1, \dots, X_m) \in H) = \sum_k P((X_1, ..., X_m) \in H_k).$$}

Changed lines 27-28 from:

{$ 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$}.

to:

{$ 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$}. \\

Changed lines 30-33 from:

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 {$P_a(m) = m a^{m-1}$}.~~ {$\box$}~~

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

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

to:

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:''' Extend this argument to the case {$a>1/2$}.

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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

!!

to:

'''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?

!! 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?

Changed line 31 from:

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 {$P_a(m) = m a^{m-1}$}. {$\~~Box~~$}

to:

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 {$P_a(m) = m a^{m-1}$}. {$\box$}

Changed line 31 from:

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:

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 {$P_a(m) = m a^{m-1}$}. {$\Box$}

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to:

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Changed lines 31-38 from:

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 {$P_a(m) = m a^{m-1}$}.

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...

Hvis a > 1/2 er H_k'erne ikke disjunkte og

P((X_

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

I dette tilfælde ser problemet sværere ud...

to:

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

!!

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

!!

Changed line 27 from:

{$ 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$}.

to:

{$ 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$}.

Changed line 20 from:

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 {$[0,1]$}.

to:

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] $}.

Changed lines 18-19 from:

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 {$\mathbb{S}^1$}.

to:

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 {$\mathbb{S}^1$}.

Changed lines 29-34 from:

P((X_1,

P(

Og da vi summerer over n ens termer får vi

P_

to:

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 {$P_a(m) = m a^{m-1}$}.

{$$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 {$P_a(m) = m a^{m-1}$}.

Changed lines 27-30 from:

{$ 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 ~~.

for if r hvis der findes

to:

{$ 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$}.

Changed lines 26-29 from:

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 .

for

to:

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 .

{$ 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 .

Changed lines 18-19 from:

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 {$\mathbb{S}^1$}.

to:

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 {$\mathbb{S}^1$}.

Added lines 26-41:

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...

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...

Changed line 10 from:

'''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?

to:

'''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?

Deleted line 24:

Added lines 26-37:

'''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

{$$2 m(A) = m(A) + m(A - d) <= m(T^1) = 1 $$}

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.

Changed line 23 from:

{$P(~~\{~~X_1,\dots,X_m~~\}~~ \in H)=m a^{m-1}$}, where {$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a] \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

to:

{$P((X_1,\dots,X_m) \in H)=m a^{m-1}$}, where {$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a] \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

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to:

{$P(\{X_1,\dots,X_m\} \in H)=m a^{m-1}$}, where {$$ H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a] \pmod 1 \text{ for all } j=1,\dots, m\}.$$}

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(:latex:)

\begin{equation}

\prod_{n=2}^N\frac{n}{n-1} = N

\end{equation}

(:latexend:)

(:latex displaycode+=true verbose=true forcerender=true paperwidth='5' ::eqncounter++:)

\begin{equation}

\prod_{n=2}^N\frac{n}{n-1} = N

\end{equation}

(:latexend:)

(: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)

to:

- will be updated soon -

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(:latex displaycode+=true verbose=true forcerender=true paperwidth='5' ::eqncounter++:)

\begin{equation}

\prod_{n=2}^N\frac{n}{n-1} = N

\end{equation}

(:latexend:)

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(:latex:)

\begin{equation}

\prod_{n=2}^N\frac{n}{n-1} = N

\end{equation}

(:latexend:)

\begin{equation}

\prod_{n=2}^N\frac{n}{n-1} = N

\end{equation}

(:latexend:)

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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 {$[0,1]$}.

to:

\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 {$[0,1]$}.

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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 {$m=0,1,\dots$}~~.~~

- will be updated soon-

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

- will be updated soon

to:

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)

'''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)

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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.~~

to:

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 {$[0,1]$}.

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'''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?

to:

'''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?

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! On the probability of catching {$m$} random points with a subset of the unit circle

to:

! On the probability of catching {$m$} random points with a given subset of the unit circle

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! On the probability of catching {$m$} random points ~~on~~ the unit circle

to:

! On the probability of catching {$m$} random points with a subset of the unit circle

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'''''Afternoon tea discussions at DTU Mathematics'''''

to:

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

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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 {$A+\tau \pmod 1$}.

to:

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 {$m=0,1,\dots$}.

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

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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 {$A+\tau ~~(~~\~~mod~~ 1~~)~~$}.

to:

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 {$A+\tau \pmod 1$}.

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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 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 $A+\tau (\mod 1)$.

to:

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 {$A+\tau (\mod 1)$}.

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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

to:

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 $A+\tau (\mod 1)$.

Changed lines 18-22 from:

to:

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

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

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'''''Afternoon tea discussions at DTU ~~Math~~'''''

to:

'''''Afternoon tea discussions at DTU Mathematics'''''

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! On the probability of ~~cacthing~~ {$m$} random points on the unit circle

to:

! On the probability of catching {$m$} random points on the unit circle

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! On the probability of ~~covering~~ {$m$} random points on the unit circle

to:

! On the probability of cacthing {$m$} random points on the unit circle

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'''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?

to:

'''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?

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'''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?

to:

'''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 {$1 \ge a \ge 0$} 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?

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'''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?

to:

'''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?

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

to:

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

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'''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?

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

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'''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 {$A \subset \~~mattbb~~{S}^1$}?

to:

'''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 {$A \subset \mathbb{S}^1$}?

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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?

to:

'''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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- will be updated soon -

to:

- will be updated soon -

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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?

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to:

'''''Afternoon tea discussions at DTU Math'''''

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to:

! On the probability of covering {$m$} random points on the unit circle

-> by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.

- will be updated soon -

-> by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.

- will be updated soon -

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-~~<~~ by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.

to:

-> by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.

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! ~~Math Discussions~~

to:

! Afternoon tea discussions at DTU Math

-< by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.

-< by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.