Homepage of the HATA group at DTU

## 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]]
Changed lines 88-102 from:
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to:
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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
{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. &#9632;$$}

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 {$|A| = a > 1-1/m$}.
[[#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 {$\epsilon > 0$} v&aelig;re givet. S&aring; findes {$A \subset R / Z$} med {$\mu(A) < \epsilon$} s&aring;ledes at givet vilk&aring;rlig t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A+\tau$}.

''Bevis''. Lad A v&aelig;re en &aring;ben og t&aelig;t delm&aelig;ngde af {$R / Z$} med {$m(A) < \epsilon$}. Skriv {$X = \{x_1, x_2,\dots\}$} og s&aelig;t {$d_i = x_{i+1} - x_1, i = 1, 2, \dots$}. S&aring; er {$B = \cap_i (A - d_i)$} en f&aelig;llesm&aelig;ngde af &aring;bne t&aelig;tte m&aelig;ngder, og da {$R / Z$} med standard metrik er fuldst&aelig;ndigt metrisk rum f&oslash;lger det af Baire's s&aelig;tning at {$B$} er t&aelig;t i {$R / Z$}. Derfor findes {$b \in B$}.

S&aelig;t nu {$\tau = x_1 - b$}. Da {$b \in A$} er {$x_1 = b + \tau$} et element af {$A + \tau$}. For vilk&aring;rlig {$i \in N$} har vi
{$b + d_i$} med i {$A$} s&aring; {$A + \tau$} indeholder {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}
Alts&aring; er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}
&#9632;

'''Theorem'''. Der findes {$A \subset R / Z$} med {$m(A) = 0$} s&aring;ledes at givet en t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A + \tau$}.

''Bevis''. Lad {$B_n$} v&aelig;re &aring;bne t&aelig;tte m&aelig;ngder med {$|B_n| < 1/n$} og s&aelig;t {$A = \cap_{n \in N} B_n$}. Forts&aelig;t s&aring; som i beviset af sidste s&aelig;tning og bem&aelig;rk at en t&aelig;llelig f&aelig;llesm&aelig;ngde af t&aelig;llelige f&aelig;llesm&aelig;ngder er en t&aelig;llelig f&aelig;llesm&aelig;ngde.
&#9632;

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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 {$P^{\bar{\mu}}_{A_0}(m)$}.
&#9632;

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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
{$P((X_1, ..., X_m) \in H) = \sum P(H_k) = m a^{m-1}$}.
&#9632;

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

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 {$U(0,1)$} &mdash; 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 {$U(0,1)$} &mdash; 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
{$P((X_1, ..., X_m) \in H) = \sum P(H_k) = m a^{m-1}$}.
&#9632;

'''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 {$P^{\bar{\mu}}_{A_0}(m)$}.
&#9632;

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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 {$\epsilon > 0$} v&aelig;re givet. S&aring; findes {$A \subset R / Z$} med {$\mu(A) < \epsilon$} s&aring;ledes at givet vilk&aring;rlig t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A+\tau$}.

''Bevis''. Lad A v&aelig;re en &aring;ben og t&aelig;t delm&aelig;ngde af {$R / Z$} med {$m(A) < \epsilon$}. Skriv {$X = \{x_1, x_2,\dots\}$} og s&aelig;t {$d_i = x_{i+1} - x_1, i = 1, 2, \dots$}. S&aring; er {$B = \cap_i (A - d_i)$} en f&aelig;llesm&aelig;ngde af &aring;bne t&aelig;tte m&aelig;ngder, og da {$R / Z$} med standard metrik er fuldst&aelig;ndigt metrisk rum f&oslash;lger det af Baire's s&aelig;tning at {$B$} er t&aelig;t i {$R / Z$}. Derfor findes {$b \in B$}.

S&aelig;t nu {$\tau = x_1 - b$}. Da {$b \in A$} er {$x_1 = b + \tau$} et element af {$A + \tau$}. For vilk&aring;rlig {$i \in N$} har vi
{$b + d_i$} med i {$A$} s&aring; {$A + \tau$} indeholder {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}
Alts&aring; er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}
&#9632;

'''Theorem'''. Der findes {$A \subset R / Z$} med {$m(A) = 0$} s&aring;ledes at givet en t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A + \tau$}.

''Bevis''. Lad {$B_n$} v&aelig;re &aring;bne t&aelig;tte m&aelig;ngder med {$|B_n| < 1/n$} og s&aelig;t {$A = \cap_{n \in N} B_n$}. Forts&aelig;t s&aring; som i beviset af sidste s&aelig;tning og bem&aelig;rk at en t&aelig;llelig f&aelig;llesm&aelig;ngde af t&aelig;llelige f&aelig;llesm&aelig;ngder er en t&aelig;llelig f&aelig;llesm&aelig;ngde.
&#9632;

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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
{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. &#9632;$$}

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?

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[[#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 l&oslash;sning, men kan vi finde et mindre a? Et (vildt) g&aelig;t ville v&aelig;re a=1-1/m. Det passer jo b&aring;de med m=1 og m=2 :). For m=3, vil s&aring; |A|=2/3, hvilket også virker rimelig i worst-case-scenario: A=[0,2/3] d&aelig;kker 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. findes {$A \subset R / Z$} med {$\mu(A) < \epsilon$} således at givet vilkårlig tællelig mængde {$X \subset R / Z$} findes {$\tau$} {$X \subset A+\tau$}.

''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 {$d_i = x_{i+1} - x_1, i = 1, 2, \dots$}. 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 {$A$} {$A + \tau$} 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 {$i = 1, 2, 3, \dots$}
to:
'''Theorem'''. Lad {$\epsilon > 0$} v&aelig;re givet. S&aring; findes {$A \subset R / Z$} med {$\mu(A) < \epsilon$} s&aring;ledes at givet vilk&aring;rlig t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A+\tau$}.

''Bevis''. Lad A v&aelig;re en &aring;ben og t&aelig;t delm&aelig;ngde af {$R / Z$} med {$m(A) < \epsilon$}. Skriv {$X = \{x_1, x_2,\dots\}$} og s&aelig;t {$d_i = x_{i+1} - x_1, i = 1, 2, \dots$}. S&aring; er {$B = \cap_i (A - d_i)$} en f&aelig;llesm&aelig;ngde af &aring;bne t&aelig;tte m&aelig;ngder, og da {$R / Z$} med standard metrik er fuldst&aelig;ndigt metrisk rum f&oslash;lger det af Baire's s&aelig;tning at {$B$} er t&aelig;t i {$R / Z$}. Derfor findes {$b \in B$}.

S&aelig;t nu {$\tau = x_1 - b$}. Da {$b \in A$} er {$x_1 = b + \tau$} et element af {$A + \tau$}. For vilk&aring;rlig {$i \in N$} har vi
{$b + d_i$} med i {$A$} s&aring; {$A + \tau$} indeholder {$$b + d_i + \tau = b + (x_{i+1} - x_1) + x_1 - b = x_{i+1}.$$}
Alts&aring; er {$x_i \in A + \tau$} for alle {$i = 1, 2, 3, \dots$}
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'''Theorem'''. Der findes {$A \subset R / Z$} med {$m(A) = 0$} således at givet en tællelig mængde {$X \subset R / Z$} findes {$\tau$} {$X \subset A + \tau$}.

''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.
to:
'''Theorem'''. Der findes {$A \subset R / Z$} med {$m(A) = 0$} s&aring;ledes at givet en t&aelig;llelig m&aelig;ngde {$X \subset R / Z$} findes {$\tau$} s&aring; {$X \subset A + \tau$}.

''Bevis''. Lad {$B_n$} v&aelig;re &aring;bne t&aelig;tte m&aelig;ngder med {$|B_n| < 1/n$} og s&aelig;t {$A = \cap_{n \in N} B_n$}. Forts&aelig;t s&aring; som i beviset af sidste s&aelig;tning og bem&aelig;rk at en t&aelig;llelig f&aelig;llesm&aelig;ngde af t&aelig;llelige f&aelig;llesm&aelig;ngder er en t&aelig;llelig f&aelig;llesm&aelig;ngde.
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 |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 l&oslash;sning, men kan vi finde et mindre a? Et (vildt) g&aelig;t ville v&aelig;re a=1-1/m. Det passer jo b&aring;de med m=1 og m=2 :). For m=3, vil s&aring; |A|=2/3, hvilket også virker rimelig i worst-case-scenario: A=[0,2/3] d&aelig;kker 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, ...
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 {$A$}{$A + \tau$} 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 {$i = 1, 2, 3, \dots$}
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'''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.
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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 {$m=2$}.
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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 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, ...
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''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 {$P^{\bar{\mu}}_{A_0}(m)$}.
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{$$\{x_i\} \subset A + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset A + \tau \pmod \frac1n$$}
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, {$P_{nA_0}(m)$}

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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.
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''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 {$M$}.
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'''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
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'''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
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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.
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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.
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 {$B=nA_0$}
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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.
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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.
Consider the three sets {$A=[0,1/3)$}, {$B=[0,1/6) \cup [3/6,4/6)$, and {$C=[0,1/6) \cup [2/6,3/6)$}, all of length {$1/3$}. By Remark 1, {$P_A(m)=P_B(m)=m (1/3)^{m-1}$}, hence in paricular, the probability of catching two random points is {$P_A(2)=P_B(2)=2/3$}. For the set {$C$} however we set that {$P_C(2)=1$} since if the distance from $x_1$ to $x_2$ is less that 1/6, we can find $\tau$ so that {$x_1,x_2 \in [0,1/6)+\tau \subset C+\tau$}.
The "symmetry" condition {$B=B+k/n \pmod 1$} is necessary, in the sense,

'''Remark 1''': Let {$n \in \mathbb{N}$} be given. If {$B=B+k/n \pmod 1$} for all {$k=1,\dots,n-1$}, then
{$$\{x_i\} \subset B + \tau \pmod 1 \Leftrightarrow \{x_i\} \subset B + \tau \pmod \frac1n$$}
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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:
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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., {$A=[0,1/4) \cup [1/2,3/4)$}
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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 {$U(0,1)$} &mdash; 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 {$U(0,1)$} &mdash; the uniform distribution on {$[0,1]$}. Then
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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., {$A=[0,1/4) \cup [1/2,3/4)$}
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'''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 {$m=1,2,\dots$}.

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''.
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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 {$P=1-1/m$}.
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 {$|A| = a > 1-1/m$}.
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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 {$U(0,1)$} &mdash; 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 {$U(0,1)$} &mdash; the uniform distribution on {$[0,1]$}. Then
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{$$H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau=[\tau,\tau+a) \pmod 1 \text{ for all } j=1,\dots, m\}.$$}
to:
{$$H = \{(x_1,\dots,x_m)\in \mathbb{T}^m \,\vert\, \exists \tau \text{ s.t. } x_j \in A+\tau \pmod 1 \text{ for all } j=1,\dots, m\}.$$}
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{$(x)\in H_i \cap H_j$}
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.
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.
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{$P(X_1, ..., X_m \in H) = \sum P(H_k) = m a^{m-1}$}.
to:
{$P((X_1, ..., X_m) \in H) = \sum P(H_k) = m a^{m-1}$}.
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'''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 {$m(A) \le 1/2$}.
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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
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'''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 {$m(A) \le 1/2$}.
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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 {$\mathbb{S}^1$}.
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!! Misc./Notes

Christian: Ikke-lig-med tegn kan laves således:
{$a\not=b \neq b$}
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{$a\not{=}b$}
to:
{$a\not=b \neq b$}
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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
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'''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 {$m=1,2,\dots$}.
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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
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''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$}, it follows by translation invariance of the Lebesgue measure that Added line 84: Christian: Ikke-lig-med tegn kan laves således: Added lines 81-86: !! Misc./Notes {$a\not{=}b$} Changed lines 37-38 from: and {$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}$}. 27.03.2012, at 08:41 UTC by 130.225.84.218 - Rettede lidt hist og her mest i bevis for theorem. Kan ikke få \neq til at virke! Changed lines 18-34 from: We consider a generalized version of Question 1. We want to find the probability {$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) $} &mdash; the uniform distribution on {$ [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 {$P_a(m) = m a^{m-1}$}. &#9632; '''TO-DO:''' Extend this argument to the case {$a>1/2$} and to (symmetric) unions of intervals. '''Example 1'''. If {$A=[0,1/6) \cup [2/6,3/6)$}, then the (extension) of Theorem 1 tells us that {$P(2)=2 (1/3)^1 = 2/3$}. However, if we change {$A$} to a non-symmetric version, e.g., {$A=[0,1/6) \cup [2/6,3/6)$}, then {$P(2)=1$}. Hence, Theorem 1 cannot be extended to non-symmetric sets. to: 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) $} &mdash; the uniform distribution on {$ [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}$. &#9632; '''TO-DO:''' Extend this argument to the case {$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$}. Hence, Theorem 1 cannot be extended to non-symmetric sets. Changed line 55 from: '''TO-DO:''' Prove P=1-1/m. to: '''TO-DO:''' Extend lemma to {$|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$} needs to be Changed line 5 from: ! On the probability of catching {$m$} random points with a given subset of the unit circle to: ! On the probability of covering {$m$} random points with a given subset of the unit circle Changed line 43 from: [[#best]] to: [[#worst]] Added line 36: [[#best]] Added line 43: [[#best]] Added line 59: [[#nD]] Added line 66: [[#ref]] Changed lines 43-44 from: Suppose now that your worst enemy can choose the set {$ A $} under the restriction that {$ \mu(A)=a $} for some given {$1\ge a \ge 0$}. 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. Changed line 59 from: ''{$m$} on an {$n$}-dimensional sphere''. What holds here? to: ''{$m$} points on an {$n$}-dimensional sphere''. What holds here? Changed lines 48-49 from: {$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1.$$} &#9632; to: {$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1. &#9632;$$} 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? 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 \subset T^1 med |A| \ge a, så eksisterer der et \tau så {x1, ..., xm} \subset A + \tau? 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 $}? '''TO-DO:''' Prove P=1-1/m. - will be updated soon - >>comment<< Changed lines 59-60 from: - will be updated soon - to: >><< Changed lines 30-31 from: Since {$(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}$}. &#9632; 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: to: {$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1.$$} &#9632; 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 {$$2 m(A) = m( A) + m(A - d) <= m(T^1) = 1$$} 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
{$$2 m(A) = m(A) + m(A - d) \le m(\mathbb{T}^1) = 1.$$}
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 {$x=(x_j) \in H_k$}.
----
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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
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
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).$$} 26.03.2012, at 22:18 UTC by 90.184.159.203 - 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: 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}. Changed lines 36-38 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 !! 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? 26.03.2012, at 22:12 UTC by 90.184.159.203 - 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} 26.03.2012, at 22:11 UTC by 90.184.159.203 - 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} 26.03.2012, at 22:10 UTC by 90.184.159.203 - Changed lines 14-15 from: to: ---- 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... 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 !! 26.03.2012, at 22:04 UTC by 90.184.159.203 - 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}. 26.03.2012, at 22:03 UTC by 90.184.159.203 - 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) &mdash; 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) } &mdash; the uniform distribution on { [0,1] }. 26.03.2012, at 22:03 UTC by 90.184.159.203 - 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: 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} . 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}$}. 26.03.2012, at 21:58 UTC by 90.184.159.203 - 26.03.2012, at 21:57 UTC by 90.184.159.203 - Changed lines 27-30 from: {$ 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$} (w.r.t. the positive for if r hvis der findes k så (x_j) \in H_k kan vi vælge K = x_k og ser (x_j) \in H . Omvendt hvis (x_j) \in H findes et (og muligvis flere) første x_k der ligger efter K eller er lig med K, med hensyn den ordning der svarer til positiv omløbsretning på cirklen. Nu er (x_j) \in H_k . to: {$ 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$}. 26.03.2012, at 21:53 UTC by 90.184.159.203 - Changed lines 26-29 from: For k = 1, ..., n, sæt H_k = {x_1, x_2, ..., x_n | x_j \in [x_k; x_k + a[ modulo 1, j = 1, ..., n} . Det er klart at der gælder H = \cup_k H_k for hvis der findes k så (x_j) \in H_k kan vi vælge K = x_k og ser (x_j) \in H . Omvendt hvis (x_j) \in H findes et (og muligvis flere) første x_k der ligger efter K eller er lig med K, med hensyn den ordning der svarer til positiv omløbsretning på cirklen. Nu er (x_j) \in H_k . to: For {$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$} (w.r.t. the positive for if r hvis der findes k så (x_j) \in H_k kan vi vælge K = x_k og ser (x_j) \in H . Omvendt hvis (x_j) \in H findes et (og muligvis flere) første x_k der ligger efter K eller er lig med K, med hensyn den ordning der svarer til positiv omløbsretning på cirklen. Nu er (x_j) \in H_k . 26.03.2012, at 21:46 UTC by 90.184.159.203 - Changed lines 18-19 from: We consider a generalized version of Question 1. We want to find the probability {$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... 26.03.2012, at 21:43 UTC by 90.184.159.203 - 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$} random points? 26.03.2012, at 21:42 UTC by 90.184.159.203 - Deleted line 24: Added lines 26-37: '''Lemma'''. Lad {$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. 26.03.2012, at 17:25 UTC by 130.225.84.242 - 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 {$$H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a] \pmod 1 \text{ for all } j=1,\dots, m\}.$$} 26.03.2012, at 17:24 UTC by 130.225.84.242 - Changed line 23 from: to: {$P(\{X_1,\dots,X_m\} \in H)=m a^{m-1}$}, where {$$H = \{(x_1,\dots,x_m) \,\vert\, \exists \tau \text{ s.t. } x_j \in [\tau,\tau+a] \pmod 1 \text{ for all } j=1,\dots, m\}.$$} 26.03.2012, at 17:20 UTC by 130.225.84.242 - Changed lines 28-50 from: - will be updated soon - (:latex:) \prod_{n=2}^N\frac{n}{n-1} = N (:latexend:) (:latex displaycode+=true verbose=true forcerender=true paperwidth='5' ::eqncounter++:) \prod_{n=2}^N\frac{n}{n-1} = N (: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) &mdash; the uniform distribution on {$[0,1]$}. (:latexend) to: - will be updated soon - 26.03.2012, at 17:16 UTC by 130.225.84.242 - Added lines 36-42: (:latex displaycode+=true verbose=true forcerender=true paperwidth='5' ::eqncounter++:) \prod_{n=2}^N\frac{n}{n-1} = N (:latexend:) 26.03.2012, at 17:15 UTC by 130.225.84.242 - Added lines 31-36: (:latex:) \prod_{n=2}^N\frac{n}{n-1} = N (:latexend:) Changed lines 42-43 from: 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) &mdash; 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) &mdash; the uniform distribution on {$[0,1]$}. 26.03.2012, at 17:12 UTC by 130.225.84.242 - Changed lines 20-25 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) &mdash; 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 - 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) &mdash; 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) &mdash; the uniform distribution on {$[0,1]$}. (:latexend) 26.03.2012, at 17:03 UTC by 130.225.84.242 - 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 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) &mdash; the uniform distribution on {$[0,1]$}. 26.03.2012, at 17:01 UTC by 130.225.84.242 - 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 {$\mu(A)=a$}. 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? 26.03.2012, at 16:59 UTC by 130.225.84.242 - Changed line 5 from: ! 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 26.03.2012, at 16:59 UTC by 130.225.84.242 - Changed line 5 from: ! 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 26.03.2012, at 16:58 UTC by 130.225.84.242 - Changed lines 1-2 from: '''''Afternoon tea discussions at DTU Mathematics''''' to: '''''Afternoon tea discussions at DTU Mathematics''''' &mdash; discussions on non-applied mathematics. Changed lines 20-23 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 {$A+\tau \pmod 1$}. 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 {$m=0,1,\dots$}. 26.03.2012, at 16:51 UTC by 130.225.84.242 - 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 {$A+\tau (\mod 1)$}. 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 {$A+\tau \pmod 1$}. 26.03.2012, at 16:47 UTC by 130.225.84.242 - Changed lines 18-20 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 (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$A+\tau (\mod 1)$. 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 {$A+\tau (\mod 1)$}. 26.03.2012, at 16:45 UTC by 130.225.84.242 - Changed lines 20-22 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 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$A+\tau (\mod 1)$. 26.03.2012, at 16:43 UTC by 130.225.84.242 - Changed lines 18-22 from: Let {$a \in \mathbb{R} $} be given such that {$0 \le a \le 1$}, and let {$A=[0,a)$}. 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 26.03.2012, at 15:38 UTC by 130.225.84.242 - Changed line 1 from: '''''Afternoon tea discussions at DTU Math''''' to: '''''Afternoon tea discussions at DTU Mathematics''''' 26.03.2012, at 15:38 UTC by 130.225.84.242 - Changed line 5 from: ! 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 26.03.2012, at 14:55 UTC by 130.225.84.242 - Changed line 5 from: ! 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 Changed line 12 from: '''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? Changed lines 10-13 from: '''Question 2''': Let {$1 \ge a \ge 0$}. Suppose {$A \subset \mathbb{S}^1$} is a Borel measurable set of measure {$\mu(A)=a$}. 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? Added lines 2-3: ---- Changed lines 8-10 from: '''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? Changed line 13 from: 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 {$A=[0,a)$}. 26.03.2012, at 09:46 UTC by 127.0.0.1 - Added lines 8-10: '''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? Added line 13: Let {$a \in \mathbb{R} $} be given such that {$0 \ge a \ge 1$}, and let {$A=[0,a)$}. 26.03.2012, at 09:41 UTC by 127.0.0.1 - Changed line 6 from: '''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 {$A \subset \mathbb{S}^1$}? 26.03.2012, at 09:41 UTC by 127.0.0.1 - Changed lines 6-7 from: '''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 {$A \subset \mattbb{S}^1$}? Changed line 11 from: - will be updated soon - to: - will be updated soon - 26.03.2012, at 09:40 UTC by 127.0.0.1 - Added lines 5-6: '''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? 26.03.2012, at 09:38 UTC by 127.0.0.1 - Added lines 5-7: [[#interval]] !! Covering points with an interval 26.03.2012, at 09:37 UTC by 127.0.0.1 - Changed line 1 from: ! Afternoon tea discussions at DTU Math to: '''''Afternoon tea discussions at DTU Math''''' 26.03.2012, at 09:36 UTC by 127.0.0.1 - Changed lines 2-6 from: -> by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen. 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 -
26.03.2012, at 09:34 UTC by 127.0.0.1 -
Changed line 2 from:
-< by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.
to:
-> by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.
26.03.2012, at 09:33 UTC by 127.0.0.1 -
Changed lines 1-2 from:
! Math Discussions
to:
! Afternoon tea discussions at DTU Math
-< by Christian Henriksen, Jakob Lemvig, and Johan Sebastian Rosenkilde Nielsen.
26.03.2012, at 08:09 UTC by 127.0.0.1 -