9.12: Bertrand's paradox( "randomness" depends on how you look at) Theorem9.18 (the principle of equal weight) implies that

 $\bullet$ the "randomness" may be related to the invariant probability measure.
However, this is due to the finiteness of the state space. In the case of infinite state space, \begin{align} \mbox{ "randomness" depends on how you look at } \end{align} This is explained in this section.
9.12.1: Bertrand's paradox("randomness" depends on how you look at)

Here, let us review the argument about the Bertrand paradox. Consider the following problem:

Given a circle with the radius 1. Suppose a chord of the circle is chosen { \Large at random}. What is the probability that the chord is shorter than ${\sqrt 3 }$? Define the rotation map $T_{\mbox{ rot}}^\theta : {\mathbb R}^2 \to {\mathbb R}^2$ $(0\le \theta <2\pi)$ and the reverse map $T_{\mbox{ rev}} : {\mathbb R}^2 \to {\mathbb R}^2$ such that

\begin{align} & T_{\mbox{ rot}}^\theta x = \left[\begin{array}{ll} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{array}\right] \cdot \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] , \quad T_{\mbox{ rev}} x = \left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right] \cdot \left[\begin{array}{l} x_1 \\ x_2 \end{array}\right] \end{align}
Problem 9.28 (Bertrand paradox and its answer) Given a circle with the radius 1. Put $\Omega$ $=\{l \;|\; l \mbox{ is a chord} \}$, that is, the set of all chords.
 $(A):$ Can we uniquely define an invariant probability measure on $\Omega$?

Here, "invariant" means "invariant concerning the rotation map $T_{\mbox{ rot}}^\theta$ and reverse map $T_{\mbox{ rev}}$".

In what follows, we show that the above invariant measure exists but it is not determined uniquely. [The first answer (Fig.$1$(in Figure 9.10))].

In Fig.$1$, we see that the chord ℓ is represented by a point $(\alpha, \beta)$ in the rectangle $\Omega_1$ $\equiv$ $\{ (\alpha, \beta) \; | \; 0 < \alpha \le 2 \pi, \; 0 < \beta \le \pi/2 \mbox{(radian)}\}$. That is, we have the following identification:

\begin{align} \Omega ( = \mbox{the set of all chords}) \ni \mbox{ℓ}_{(\alpha, \beta)} \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} (\alpha , \beta) \in \Omega_1 ( \subset {\mathbb R}^2 ). \end{align}

Note that we have the natural probability measure $\nu_1$ on $\Omega_1$ such that $\nu_1 (A) = \frac{ \mbox{Meas}[{}A]}{\mbox{Meas}[{}\Omega_1{}] } = \frac{ \mbox{Meas} [{}A]}{ \pi^2 }$ $(\forall A \in {\cal B}_{\Omega_1})$, where " Meas" = " Lebesgue measure". Transferring the probability measure $\nu_1$ on $\Omega_1$ to $\Omega$, we get $\rho_1$ on $\Omega$. That is,

\begin{align} {\mathcal M}_{+1} (\Omega ) \ni \rho_1 \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} \nu_1 \in {\mathcal M}_{+1} (\Omega_1 ) \end{align}
 $(\sharp):$ It is clear that the measure $\rho_1$ is invariant concerning the rotation map $T_{\mbox{ rot}}^\theta$ and reverse map $T_{\mbox{ rev}}$.

Therefore, we have a natural measurement ${\mathsf M}_{L^\infty ( \Omega, m )}({\mathsf O}_E \equiv (\Omega, {\mathcal B}_{\Omega} , F_E), S_{[\ast]}(\rho_1) )$. Consider the identification:

\begin{align} \Omega \supseteq \Xi_{\sqrt{3}} \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} \{ (\alpha , \beta )\in \Omega_1 \;:\; \mbox{"the length of } \mbox{ℓ}_{(\alpha , \beta)}" < {\sqrt 3} \} \subseteq \Omega_1 \end{align}

Then, Axiom$^{\mbox{ (m)}}$ 1 ( in $\S$9.1) says that the probability that a measured value belongs to $\Xi_{\sqrt{3}}$ is given by

\begin{align} & \int_{\Omega} [F_E (\Xi_{\sqrt{3}})](\omega) \; \rho_1( d \omega )= \int_{\Xi_{\sqrt{3}}}\;\; 1 \;\; \rho_1( d \omega ) \\ = & m_1 (\{ \mbox{ℓ}_{(\alpha , \beta)} \approx (\alpha, \beta) \in \Omega_1 \; | \; \mbox{ "the length of } \mbox{ℓ}_{(\alpha , \beta)}" \le {\sqrt 3} \}) \\ = & \frac{\mbox{Meas}[{}\{ (\alpha, \beta) \; | \; 0 \le \alpha \le 2 \pi, \;\pi/6 \le \beta \le \pi/2 \}]} {\mbox{Meas}[{}\{ (\alpha, \beta) \; | \; 0 \le \alpha \le 2 \pi, \;0 \le \beta \le \pi/2\}]} \\ = & \frac{2 \pi \times ( \pi/3)} {\pi^2} = \frac{2}{3}. \end{align}

[The second answer (Fig.$2$ (in Figure 9.10))].

In Fig.$2$, we see that the chord ℓ is represented by a point $(x,y)$ in the circle $\Omega_2$ $\equiv$ $\{ (x,y) \; | \; x^2 + y^2 < 1 \}$.

That is, we have the following identification:

\begin{align} \Omega ( = \mbox{the set of all chords}) \ni \mbox{ℓ}_{(x,y)} \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} (x,y) \in \Omega_2 ( \subset {\mathbb R}^2 ). \end{align}

We have the natural probability measure $\nu_2$ on $\Omega_2$ such that $\nu_2 (A) = \frac{ \mbox{Meas}[{}A]}{\mbox{Meas}[{}\Omega_2{}] } = \frac{ \mbox{Meas} [{}A]}{ \pi }$ $(\forall A \in {\cal B}_{\Omega_2})$. Transferring the probability measure $\nu_2$ on $\Omega_2$ to $\Omega$, we get $\rho_2$ on $\Omega$. That is,

\begin{align} {\mathcal M}_{+1} (\Omega ) \ni \rho_2 \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} \nu_2 \in {\mathcal M}_{+1} (\Omega_2 ) \end{align}
 $(\sharp):$ It is clear that the measure $\rho_2$ is invariant concerning the rotation map $T_{\mbox{ rot}}^\theta$ and reverse map $T_{\mbox{ rev}}$.}

Therefore, we have a natural measurement ${\mathsf M}_{L^\infty ( \Omega, m )}({\mathsf O}_E \equiv (\Omega, {\mathcal B}_{\Omega} , F_E), S_{[\ast]}(\rho_2) )$.

Consider the identification:

\begin{align} \Omega \supseteq \Xi_{\sqrt{3}} \underset{\scriptsize{\mbox{identification}}}{\longleftrightarrow} \{ (x , y )\in \Omega_2 \;:\; \mbox{"the length of } \mbox{ℓ}_{(\alpha , \beta)}" < {\sqrt 3} \} \subseteq \Omega_1 \end{align}

Then, Axiom$^{\mbox{ (m)}}$ 1 ( in $\S$9.1) says that the probability that a measured value belongs to $\Xi_{\sqrt{3}}$ is given by

\begin{align} & \int_{\Omega} [F_E (\Xi_{\sqrt{3}})](\omega) \; \rho_2( d \omega )= \int_{\Xi_{\sqrt{3}}}\;\; 1 \;\; \rho_2( d \omega ) \\ = & \nu_2 (\{ \mbox{ℓ}_{(x,y)} \approx (x,y) \in \Omega_2 \; | \; \mbox{ "the length of } \mbox{ℓ}_{(x,y)}" \le {\sqrt 3} \}) \\ = & \frac{\mbox{Meas} [{} \{ (x,y) \; | \; 1/4 \le x^2+y^2 \le 1 \} ]} {\pi} = \frac{3}{4}. \end{align}

Conclusion 9.29

Thus, even if there is a custom to regard a natural probability measure (i.e., an invariant measure concerning natural maps) as "random", the first answer and the second answer say that

 $(\sharp):$ $\quad$ the uniqueness in (A) of Problem 9.28 is denied.