# On an Internal Dependence
of Simultaneous Measurements

Valentin Vankov Iliev
Institute of Mathematics and Informatics
Bulgarian Academy of Sciences
Sofia, Bulgaria
e-mail<EMAIL_ADDRESS>

###### Abstract

In this paper we show that there exists an internal dependence of the
simultaneous measurements made by the two pairs of linear polarizers operated
in each leg of the apparatus in Aspect’s version of Einstein-Podolsky-Rosen-
Bohm _Gedankenexperiment_. The corresponding Shannon-Kolmogorov’s information
flow linking a polarizer from one leg to a polarizer from the other leg is
proportional to the absolute value of this function of dependence. It turns
out that if Bell’s inequality is violated, then this information flow is
strictly positive, that is, the experiment performed at one leg is
informationally dependent on the experiment at the other leg. By throwing out
the sign of absolute value, we define the signed information flow linking a
polarizer from one leg to a polarizer from the other leg which, in turn,
reproduces the probabilities of the four outcomes of the simultaneous
measurements, predicted by quantum mechanics. We make an attempt to illustrate
the seeming random relation between the total information flow, the total
signed information flow, and the violation of Bell’s inequality in terms of a
kind of uncertainty principle.

##  1 Introduction, Notation

###  1.1 Introduction

In the context of the bipartite quantum system that describes Aspect’s optical
version of Einstein-Podolsky-Rosen-Bohm _Gedankenexperiment_ (see [1] and
[5]), we consider the pairs of linear polarizers operated in each leg of the
apparatus as pairs of self-adjoined linear operators

$A_{\mu_{i}}=\left(\begin{array}[]{ccccc}\cos\mu_{i}&\sin\mu_{i}\\\
\sin\mu_{i}&-\cos\mu_{i}\\\ \end{array}\right),B_{\nu_{j}}=A_{\nu_{j}},$

where $\mu_{i},\nu_{j}\in[0,\pi]$, $i,j=1,2$, are the angles of the
polarizers. Note that each pair has a time switch which interchanges
polarizers, the corresponding time being shorter than the time necessary for a
light signal to travel from one of the pairs of polarizers to the other
(Einstein locality assumption for independence).

Each pair of operators $A_{\mu_{1}},A_{\mu_{2}}$ and $B_{\nu_{1}},B_{\nu_{2}}$
acts on the state space of the corresponding quantum subsystem (a unitary
plane). By tensoring with the unit operator on the other plane, we obtain two
pairs of self-adjoined linear operators $\mathcal{A}_{\mu_{1}}$,
$\mathcal{A}_{\mu_{2}}$ and $\mathcal{B}_{\nu_{1}}$, $\mathcal{B}_{\nu_{2}}$
with spectre $\\{1,-1\\}$ on the state space of the whole quantum system
(tensor product of the two unitary planes). Moreover, for each $i,j=1,2,$ the
operators $\mathcal{A}_{\mu_{i}}$ and $\mathcal{B}_{\nu_{j}}$ commute because
the state space of the whole quantum system has an orthonormal frame
consisting of eigenvectors of both operators. In this case the corresponding
measurements are said to be _simultaneous_.

In accord with the axiom of quantum mechanics about the observables, after
fixing initial state we can consider the members of this frame as outcomes of
a sample space with probability assignment consisting of probabilities
predicted by this axiom. Moreover, with an abuse of the language, we can also
consider the operators as random variables with range $\\{1,-1\\}$ on this
sample space. Under the condition that the singlet state is initial, any one
of these random variables has probability distribution
$(\frac{1}{2},\frac{1}{2})$. Moreover, if $\mu\in\\{\mu_{1},\mu_{2}\\}$ and
$\nu\in\\{\nu_{1},\nu_{2}\\}$, then

$\operatorname{pr}((\mathcal{A}_{\mu}=1)\cap(\mathcal{B}_{\nu}=1))=\operatorname{pr}((\mathcal{A}_{\mu}=-1)\cap(\mathcal{B}_{\nu}=-1))=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right),$
$\operatorname{pr}((\mathcal{A}_{\mu}=1)\cap(\mathcal{B}_{\nu}=-1))=\operatorname{pr}((\mathcal{A}_{\mu}=-1)\cap(\mathcal{B}_{\nu}=1))=\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right).$

Therefore, the product of random variables
$\mathcal{A}_{\mu}\mathcal{B}_{\nu}$ has probability distribution
$\left(\sin^{2}\left(\frac{\mu-\nu}{2}\right),\cos^{2}\left(\frac{\mu-\nu}{2}\right)\right)$
and expected value
$\mathcal{E}(\mathcal{A}_{\mu}\mathcal{B}_{\nu})=-\cos(\mu-\nu)$.

On the other hand, the joint experiment (see [7, Part I, Section 6]) of the
binary trials
$\mathfrak{A}_{\mu}=(\mathcal{A}_{\mu}=1)\cup(\mathcal{A}_{\mu}=-1)$ and
$\mathfrak{B}_{\nu}=(\mathcal{B}_{\nu}=1)\cup(\mathcal{B}_{\nu}=-1)$ produces
the probability distribution

$\left(\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right),\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right)\right)$

with Boltzmann-Shannon entropy $E(\theta_{\mu,\nu})$, where
$E(\theta)=-2\theta\ln\theta-2(\frac{1}{2}-\theta)\ln(\frac{1}{2}-\theta)$ and
$\theta_{\mu,\nu}=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right)$. We
extend the function $E(\theta)$, $\theta\in(0,\frac{1}{2})$, as continuous on
the closed interval $[0,\frac{1}{2}]$.

By modifying the entropy function $E(\theta)$, we obtain the strictly
increasing _degree of dependence function_ $e\colon[0,\frac{1}{2}]\to[-1,1]$,
which mimics the regression coefficient (see [6, 5.2]).

It turns out that the average quantity of information
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})$ (see [3, §1]) of one of the
experiments $\mathfrak{A}_{\mu}$ and $\mathfrak{B}_{\nu}$, relative to the
other can be found by the formula
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})=|e(\theta_{\mu,\nu})|\ln 2$. We can
consider $I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})$ as a measure of the flow
carrying information between these two binary trials (see [6, 5.3]). Since $s$
is an invertible function, the corresponding _signed information flow_
$I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu})(\theta)=e(\theta)\ln
2$ replicates the probability distribution (3.1.2) produced by quantum
mechanics.

In terms of Aspect’s experiment, the sum
$I(\mathfrak{A},\mathfrak{B})=\sum_{i,j}^{2}I(\mathfrak{A}_{\mu_{i}},\mathfrak{B}_{\nu_{j}})$
(called total information flow) can be thought about as a measure of the flow
carrying information between the two pairs of polarizers. In his paper [2]
John Bell deduced under the assumptions of ”locality” and ”realism” that if
measurements are performed independently (Einstein locality assumption for
independence) on the two separated particles (photons in Aspect’s experiment)
of an entangled pair, then the assumption that the outcomes depend upon
”hidden variables” implies constraint condition called Bell’s inequality (see
Subsection 4.1). It comes out that if Bell’s inequality is violated, then the
total information flow is strictly positive. In other words, in this case
there exists an informational dependence between the two legs of apparatus.

In the end of the paper we discuss the relation between the information flow
$I(\mathfrak{A},\mathfrak{B})$ and the violation of Bell’s inequality. Using
Examples 4.1.1 and the Java program from the link that can be found there, we
note that this relation is subject to a kind of uncertainty principle.

###  1.2 Notation

$\mathcal{H}$: $2$-dimensional unitary space with inner product $\langle
x|y\rangle$ which is linear in the second slot and anti-linear in the first
slot;

$\mathbb{I}=\mathbb{I}_{\mathcal{H}}$: the identity linear operator on
$\mathcal{H}$;

$\mathcal{H}^{\otimes 2}=\mathcal{H}\otimes\mathcal{H}$: the unitary tensor
square with inner product $\langle x_{1}\otimes x_{2}|y_{1}\otimes
y_{2}\rangle=\langle x_{1}|y_{1}\rangle\langle x_{2}|y_{2}\rangle$;

$\mathcal{U}^{\left(2\right)}$: the unit sphere in $\mathcal{H}^{\otimes 2}$;

$\operatorname{Spec}(A)$: the real spectre of a self-adjoined linear operator
$A$ on $\mathcal{H}$ with trace zero, having the form
$\operatorname{Spec}(A)=\\{\lambda_{1}^{\left(A\right)},\lambda_{2}^{\left(A\right)}\\}$,
$\lambda_{1}^{\left(A\right)}+\lambda_{2}^{\left(A\right)}=0$;

$u^{\left(A\right)}=\\{u_{1}^{\left(A\right)},u_{2}^{\left(A\right)}\\}$: the
orthonormal frame for $\mathcal{H}$, formed by the corresponding eigenvectors
of $A$;

$\mathcal{H}_{i}^{\left(A\right)}$: the eigenspaces
$\mathbb{C}u_{i}^{\left(A\right)}$ of $A$, $i=1,2$;

##  2 Self-Adjoint Operators on $\mathcal{H}$

###  2.1 Two Special Commuting Operators

We fix an orthonormal frame $h=\\{h_{1},h_{2}\\}$ for $\mathcal{H}$ and
identify the self-adjoined operators with their matrices with respect to $h$.
For any $\mu\in[0,\pi]$ we denote by $A_{\mu}$ the self-adjoined operator

$\left(\begin{array}[]{ccccc}\cos\mu&\sin\mu\\\ \sin\mu&-\cos\mu\\\
\end{array}\right).$

We have
$\lambda_{1}^{\left(A_{\mu}\right)}=1,\lambda_{2}^{\left(A_{\mu}\right)}=-1$,
and

$u_{1}^{\left(A_{\mu}\right)}=(\cos\frac{\mu}{2})h_{1}+(\sin\frac{\mu}{2})h_{2},u_{2}^{\left(A_{\mu}\right)}=(-\sin\frac{\mu}{2})h_{1}+(\cos\frac{\mu}{2})h_{2}.$

For any $\nu\in[0,\pi]$ we set $B_{\nu}=A_{\nu}$.

Note that $\\{h_{1}\otimes h_{1},h_{1}\otimes h_{2},h_{2}\otimes
h_{1},h_{2}\otimes h_{2}\\}$ and $u^{\left(A_{\mu}\right)}\otimes
u^{\left(B_{\nu}\right)}=\\{u_{1}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{1}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}\\}$ are orthonormal frames for
$\mathcal{H}^{\otimes 2}$.

Let us set $\mathcal{A}_{\mu}=A_{\mu}\otimes\mathbb{I}$,
$\mathcal{B}_{\nu}=\mathbb{I}\otimes B_{\nu}$. It is a straightforward check
that the last two linear operators on $\mathcal{H}^{\otimes 2}$ are also self-
adjoined with
$\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)}=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)}=1$,
$\lambda_{2}^{\left(\mathcal{A}_{\mu}\right)}=\lambda_{2}^{\left(\mathcal{B}_{\nu}\right)}=-1$,
the $\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)}$-eigenspace
$\mathcal{H}_{i}^{\left(\mathcal{A}_{\mu}\right)}=\mathcal{H}_{i}^{\left(A_{\mu}\right)}\otimes\mathcal{H}$
has orthonormal frame $\\{u_{i}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{i}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}\\}$, and the
$\lambda_{i}^{\left(\mathcal{B}_{\nu}\right)}$-eigenspace
$\mathcal{H}_{i}^{\left(\mathcal{B}_{\nu}\right)}=\mathcal{H}\otimes\mathcal{H}_{i}^{\left(B_{\nu}\right)}$
has orthonormal frame $\\{u_{1}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}\\}$, $i=1,2$.

Since $u^{\left(A_{\mu}\right)}\otimes u^{\left(B_{\nu}\right)}$ is an
orthonornal frame of $\mathcal{H}^{\otimes 2}$ consisting of eigenvectors of
both $\mathcal{A}_{\mu}$ and $\mathcal{B}_{\nu}$, then the last two operators
commute.

Let $\psi\in\mathcal{U}^{\left(2\right)}$ and let
$S(\psi;\mathcal{A}_{\mu},\mathcal{B}_{\nu})$ be the sample space with set of
outcomes $u^{\left(A_{\mu}\right)}\otimes
u^{\left(B_{\nu}\right)}=\\{u_{1}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{1}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}\\}$ and probability assignment
$\\{p_{11},p_{12},p_{21},p_{22}\\}$ with $p_{ij}=|\langle
u_{i}^{\left(A_{\mu}\right)}\otimes
u_{j}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}$, $i,j=1,2$. With an abuse of
the language, we consider the observable $\mathcal{A}_{\mu}$ as a random
variable $\mathcal{A}_{\mu}\colon u^{\left(A_{\mu}\right)}\otimes
u^{\left(B_{\nu}\right)}\to\mathbb{R}$,
$\mathcal{A}_{\mu}(u_{1}^{\left(A_{\mu}\right)}\otimes
u_{j}^{\left(B_{\nu}\right)})=\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)}$,
$\mathcal{A}_{\mu}(u_{2}^{\left(A_{\mu}\right)}\otimes
u_{j}^{\left(B_{\nu}\right)})=\lambda_{2}^{\left(\mathcal{A}_{\mu}\right)}$,
$j=1,2$, on the sample space $S(\psi;\mathcal{A}_{\mu},\mathcal{B}_{\nu})$
with probability distribution
$p_{\mathcal{A}_{\mu}}(\lambda_{i}^{\left(A\right)})=|\langle
u_{i}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}+|\langle
u_{i}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}$, $i=1,2$, and
$p_{\mathcal{A}_{\mu}}(\lambda)=0$ for
$\lambda\notin\operatorname{Spec}(\mathcal{A}_{\mu})$. Identifying the event
$\\{u_{i}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)},u_{i}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}\\}$ with the ”event”
$\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)}$, we have
$\operatorname{pr}(\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)})=|\langle
u_{i}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}+|\langle
u_{i}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}$, $i=1,2$.

We also consider the observable $\mathcal{B}_{\nu}$ as a random variable
$\mathcal{B}_{\nu}\colon u^{\left(A_{\mu}\right)}\otimes
u^{\left(B_{\nu}\right)}\to\mathbb{R}$,
$\mathcal{B}_{\nu}(u_{j}^{\left(A_{\mu}\right)}\otimes
u_{1}^{\left(B_{\nu}\right)})=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)}$,
$\mathcal{B}_{\nu}(u_{j}^{\left(A_{\mu}\right)}\otimes
u_{2}^{\left(B_{\nu}\right)})=\lambda_{2}^{\left(\mathcal{B}_{\nu}\right)}$,
$j=1,2$, on the sample space $S(\psi;\mathcal{A}_{\mu},\mathcal{B}_{\nu})$
with probability distribution
$p_{\mathcal{B}_{\nu}}(\lambda_{i}^{\left(A\right)})=|\langle
u_{1}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}+|\langle
u_{2}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}$, $i=1,2$, and
$p_{\mathcal{B}_{\nu}}(\lambda)=0$ for
$\lambda\notin\operatorname{Spec}(\mathcal{B}_{\nu})$. Identifying the event
$\\{u_{1}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)},u_{2}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}\\}$ with the ”event”
$\mathcal{B}_{\nu}=\lambda_{i}^{\left(\mathcal{B}_{\nu}\right)}$, we have
$\operatorname{pr}(\mathcal{B}_{\nu}=\lambda_{i}^{\left(\mathcal{B}_{\nu}\right)})=|\langle
u_{1}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}+|\langle
u_{2}^{\left(A_{\mu}\right)}\otimes
u_{i}^{\left(B_{\nu}\right)}|\psi\rangle|^{2}$, $i=1,2$.

In particular, let us set $\psi=\frac{1}{\sqrt{2}}(h_{1}\otimes
h_{2}-h_{2}\otimes h_{1})$. We have

$\operatorname{pr}(\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)})=$
$\frac{1}{2}|\langle u_{i}^{\left(A_{\mu}\right)}|h_{1}\rangle\langle
u_{1}^{\left(B_{\nu}\right)}|h_{2}\rangle-\langle
u_{i}^{\left(A_{\mu}\right)}|h_{2}\rangle\langle
u_{1}^{\left(B_{\nu}\right)}|h_{1}\rangle|^{2}+$ $\frac{1}{2}|\langle
u_{i}^{\left(A_{\mu}\right)}|h_{1}\rangle\langle
u_{2}^{\left(B_{\nu}\right)}|h_{2}\rangle-\langle
u_{i}^{\left(A_{\mu}\right)}|h_{2}\rangle\langle
u_{2}^{\left(B_{\nu}\right)}|h_{1}\rangle|^{2}$

and

$\operatorname{pr}(\mathcal{B}_{\nu}=\lambda_{j}^{\left(\mathcal{B}_{\nu}\right)})=$
$\frac{1}{2}|\langle u_{1}^{\left(A_{\mu}\right)}|h_{1}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{2}\rangle-\langle
u_{1}^{\left(A_{\mu}\right)}|h_{2}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{1}\rangle|^{2}+$ $\frac{1}{2}|\langle
u_{2}^{\left(A_{\mu}\right)}|h_{1}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{2}\rangle-\langle
u_{2}^{\left(A_{\mu}\right)}|h_{2}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{1}\rangle|^{2}.$

Taking into account the form of the eigenvectors of the matrices $A_{\mu}$ and
$B_{\nu}$, we obtain

$\operatorname{pr}(\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)})=\operatorname{pr}(\mathcal{B}_{\nu}=\lambda_{j}^{\left(\mathcal{B}_{\nu}\right)})=\frac{1}{2},i,j=1,2.$

We identify the intersection
$(\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{j}^{\left(\mathcal{B}_{\nu}\right)})$
with the event $\\{u_{i}^{\left(A_{\mu}\right)}\otimes
u_{j}^{\left(B_{\nu}\right)}\\}$, $i,j=1,2$, in the sample space
$S(\psi;\mathcal{A}_{\mu},\mathcal{B}_{\nu})$ and obtain

$\operatorname{pr}((\mathcal{A}_{\mu}=\lambda_{i}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{j}^{\left(\mathcal{B}_{\nu}\right)}))=$
$\frac{1}{2}|\langle u_{i}^{\left(A_{\mu}\right)}|h_{1}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{2}\rangle-\langle
u_{i}^{\left(A_{\mu}\right)}|h_{2}\rangle\langle
u_{j}^{\left(B_{\nu}\right)}|h_{1}\rangle|^{2}.$

In particular, we have

$\operatorname{pr}((\mathcal{A}_{\mu}=\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)}))=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right),$
$\operatorname{pr}((\mathcal{A}_{\mu}=\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{2}^{\left(\mathcal{B}_{\nu}\right)}))=\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),$
$\operatorname{pr}((\mathcal{A}_{\mu}=\lambda_{2}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)}))=\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),$
$\operatorname{pr}((\mathcal{A}_{\mu}=\lambda_{2}^{\left(\mathcal{A}_{\mu}\right)})\cap(\mathcal{B}_{\nu}=\lambda_{2}^{\left(\mathcal{B}_{\nu}\right)}))=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right).$

The random variable $\mathcal{A}_{\mu}\mathcal{B}_{\nu}$ has probability
distribution

$p_{\mathcal{A}_{\mu}\mathcal{B}_{\nu}}(1)=\sin^{2}\left(\frac{\mu-\nu}{2}\right),p_{\mathcal{A}_{\mu}\mathcal{B}_{\nu}}(-1)=\cos^{2}\left(\frac{\mu-\nu}{2}\right),$

and $p_{\mathcal{A}_{\mu}\mathcal{B}_{\nu}}(\lambda)=0$ for $\lambda\neq\pm
1$. The expected value of this random variable is
$\mathcal{E}(\mathcal{A}_{\mu}\mathcal{B}_{\nu})=-\cos(\mu-\nu)$.

##  3 Entropy and Degree of Dependence

###  3.1 Entropy

Now, we combine the terminology and notation of this paper with those of [6].
Let us set
$A=(\mathcal{A}_{\mu}=\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)})$,
$B=(\mathcal{B}_{\nu}=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)})$,
$A^{c}=(\mathcal{A}_{\mu}=\lambda_{2}^{\left(\mathcal{A}_{\mu}\right)})$,
$B^{c}=(\mathcal{B}_{\nu}=\lambda_{2}^{\left(\mathcal{B}_{\nu}\right)})$.
$\alpha=\operatorname{pr}(A)=\frac{1}{2}$,
$\beta=\operatorname{pr}(B)=\frac{1}{2}$.

The pair $(A,B)$ of events in the sample space
$S(\psi;\mathcal{A},\mathcal{B})$ produces an experiment

$\mathfrak{J}=(A\cap B)\cup(A\cap B^{c})\cup(A^{c}\cap B)\cup(A^{c}\cap
B^{c})$ (3.1.1)

(cf. [4, I,§3]) and the probabilities of its results:

$\xi_{1}=\operatorname{pr}(A\cap
B)=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right),\
\xi_{2}=\operatorname{pr}(A\cap
B^{c})=\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),$
$\xi_{3}=\operatorname{pr}(A^{c}\cap
B)=\frac{1}{2}\cos^{2}\left(\frac{\mu-\nu}{2}\right),\
\xi_{4}=\operatorname{pr}(A^{c}\cap
B^{c})=\frac{1}{2}\sin^{2}\left(\frac{\mu-\nu}{2}\right).$ (3.1.2)

The probability distribution $(\xi_{1},\xi_{2},\xi_{3},\xi_{4})$ satisfies the
linear system [6, 4.1. (3)] whose solutions form a straight line with
parametric representation $\xi_{1}=\theta$, $\xi_{2}=\frac{1}{2}-\theta$,
$\xi_{3}=\frac{1}{2}-\theta$, $\xi_{4}=\theta$ in the hyperplane
$\xi_{1}+\xi_{2}+\xi_{3}+\xi_{4}=1$. Note that the parameter $\theta=\xi_{1}$
runs within the closed interval $[0,\frac{1}{2}]$. The entropy of
$(\xi_{1},\xi_{2},\xi_{3},\xi_{4})$ is
$E(\theta)=-\sum_{k=1}^{4}\xi_{k}(\theta)\ln(\xi_{k}(\theta))=-2\theta\ln\theta-2(\frac{1}{2}-\theta)\ln(\frac{1}{2}-\theta)$
and the function $E(\theta)$ can be extended as continuous on the interval
$[0,\frac{1}{2}]$. It strictly increases on the interval $[0,\frac{1}{4}]$,
strictly decreases on the interval $[\frac{1}{4},\frac{1}{2}]$ and has a
global maximum at $\theta=\frac{1}{4}$. In particular,
$\max_{\theta\in\left[0,\frac{1}{2}\right]}E(\theta)=E(\frac{1}{4})=2\ln 2$.
Since $\min_{\theta\in\left[0,\frac{1}{4}\right]}E(\theta)=E(0)=\ln
2=E(\frac{1}{2})=\min_{\theta\in\left[\frac{1}{4},\frac{1}{2}\right]}E(\theta)$,
we obtain $\min_{\theta\in\left[0,\frac{1}{2}\right]}E(\theta)=\ln 2$.

###  3.2 Degree of Dependence

It is more useful to modify the entropy function, thus obtaining the strictly
increasing _degree of dependence function_ $e\colon[0,\frac{1}{2}]\to[-1,1]$,

$e(\theta)=\left\\{\begin{array}[]{ccccccccccccccccccccccccccccccc}-\frac{E(\frac{1}{4})-E(\theta)}{E(\frac{1}{4})-E(0)}&\hbox{\rm\
if\ }0\leq\theta\leq\frac{1}{4}\\\
\frac{E(\frac{1}{4})-E(\theta)}{E(\frac{1}{4})-E(\frac{1}{2})}&\hbox{\rm\ if\
}\frac{1}{4}\leq\theta\leq\frac{1}{2}.\\\ \end{array}\right.$

Taking into account the values of extrema of entropy function, we obtain

$e(\theta)=\left\\{\begin{array}[]{ccccccccccccccccccccccccccccccc}-2+\frac{E(\theta)}{\ln
2}&\hbox{\rm\ if\ }0\leq\theta\leq\frac{1}{4}\\\ 2-\frac{E(\theta)}{\ln
2}&\hbox{\rm\ if\ }\frac{1}{4}\leq\theta\leq\frac{1}{2}.\\\
\end{array}\right.$

The events $A$ and $B$ are _independent_ exactly when the entropy is maximal
(equal to $2\ln 2$), that is, when $e(\theta)=0$ and this, in turn, is
equivalent to the equality $|\mu-\nu|=\frac{\pi}{2}$. We have $e(\theta)=-1$
or $e(\theta)=1$ if and only if $|\mu-\nu|=0$ or $|\mu-\nu|=\pi$,
respectively, and in these two cases the entropy is minimal and equal to $\ln
2$. Now, let, in addition, assume that $A$ and $B$ are events in a sample
space with equally likely outcomes. If $e(\theta)=-1$, then one of $A$ and $B$
is a subset of the complement of the other (maximal _negative dependence_),
and if $e(\theta)=1$ one of them is a subset of the other (maximal _positive
dependence_).

###  3.3 The Information Flow

The experiment $\mathfrak{J}$ from (3.1.1) is the joint experiment (see [7,
Part I, Section 6]) of two simple binary trials: $\mathfrak{A}_{\mu}=A\cup
A^{c}$ and $\mathfrak{B}_{\nu}=B\cup B^{c}$ with
$\operatorname{pr}(A)=\operatorname{pr}(B)=\frac{1}{2}$. The _average quantity
of information of one of the experiments $\mathfrak{A}_{\mu}$ and
$\mathfrak{B}_{\nu}$, relative to the other_, (see [3, §1]), is defined in
this particular case by the formula
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\theta)=\xi_{1}(\theta)\ln
4\xi_{1}(\theta)+\xi_{2}(\theta)\ln 4\xi_{2}(\theta)+\xi_{3}(\theta)\ln
4\xi_{3}(\theta)+\xi_{4}(\theta)\ln 4\xi_{4}(\theta)$. The above notation is
correct since the interchanges of $A$ and $A^{c}$ or $B$ and $B^{c}$ causes
permutations of $\xi_{i}$’s. Thus, we obtain
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\theta)=\max_{\theta\in\left[0,\frac{1}{2}\right]}E(\theta)-E(\theta)$.
Now, the definition of the degree function $e(\theta)$ yields immediately
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\theta)=|e(\theta)|\ln 2$ for
$\theta\in[0,\frac{1}{2}]$.

Translating into the language of information theory, we have $e(\theta)=-1$ or
$e(\theta)=1$ if and only if
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\theta)=\max_{0\leq\tau\leq\frac{1}{2}}I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\tau)=\ln
2$. Finally, we have $e(\theta)=0$ if and only if
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})(\theta)=0$, and under this condition
the experiments $\mathfrak{A}_{\mu}$ and $\mathfrak{B}_{\nu}$ are said to be
_informationally independent_.

###  3.4 The Signed Information Flow

Let us set
$I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu})(\theta)=e(\theta)\ln
2$ for $\theta\in[0,\frac{1}{2}]$ and call this quantity _average quantity of
signed information of one of the events
$\mathcal{A}_{\mu}=\lambda_{1}^{\left(\mathcal{A}_{\mu}\right)}$ and
$\mathcal{B}_{\nu}=\lambda_{1}^{\left(\mathcal{B}_{\nu}\right)}$, relative to
the other_. Then
$I(\mathfrak{A}_{\mu},\mathfrak{B}_{\nu})=|I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu})|$
and since the function $e$ is invertible, we obtain
$\theta=e^{-1}(\frac{1}{\ln
2}I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu}))$. In particular,
the value of the signed information flow
$I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu})$ reproduces the
probability distribution (3.1.2) predicted by quantum theory.

##  4 Four Operators and Bell’s Map

For any $\mu_{1},\mu_{2},\nu_{1},\nu_{2}\in[0,\pi]$ we consider the self-
adjoined operators $A_{\mu_{i}}$, $B_{\nu_{j}}$, $i,j=1,2$, see Subsection
2.1. We extend notation introduced in Sections 2 and 3 in a natural way:
$\theta_{ij}=\frac{1}{2}\sin^{2}\left(\frac{\mu_{i}-\nu_{j}}{2}\right)$,
$\theta_{ij}\in[0,\frac{1}{2}]$, $\mathfrak{A}_{\mu_{i}}$,
$\mathfrak{B}_{\nu_{j}}$,
$I(\mathfrak{A}_{\mu_{i}},\mathfrak{B}_{\nu_{j}})=|e(\theta_{ij})|\ln 2$,
$i,j=1,2$. The sum
$I(\mathfrak{A},\mathfrak{B})=\sum_{i,j=1}^{2}I(\mathfrak{A}_{\mu_{i}},\mathfrak{B}_{\nu_{j}})$
is said to be the _average quantity of information of one of the pairs of
experiments $\mathfrak{A}=\\{\mathfrak{A}_{\mu_{1}},\mathfrak{A}_{\mu_{2}}\\}$
and $\mathfrak{B}=\\{\mathfrak{B}_{\nu_{1}},\mathfrak{B}_{\nu_{2}}\\}$
relative to the other_, or, _total information flow_. The sum
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=\sum_{i,j=1}^{2}I^{\left(s\right)}(\mathcal{A}_{\mu},\mathcal{B}_{\nu})$
is said to be the _average quantity of signed information of one of the pairs
of experiments
$\mathfrak{A}=\\{\mathfrak{A}_{\mu_{1}},\mathfrak{A}_{\mu_{2}}\\}$ and
$\mathfrak{B}=\\{\mathfrak{B}_{\nu_{1}},\mathfrak{B}_{\nu_{2}}\\}$ relative to
the other_, or, _total signed information flow_.

Thus, we obtain the functions

$I(\mathfrak{A},\mathfrak{B})\colon[0,\pi]^{4}\to\mathbb{R},(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\mapsto(\ln
2)\sum_{i,j=1}^{2}|e(\theta_{ij})|,$

and

$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})\colon[0,\pi]^{4}\to\mathbb{R},(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\mapsto(\ln
2)\sum_{i,j=1}^{2}e(\theta_{ij}),$

which represents the intensity of information flow (respectively, signed
information flow) between the pairs of experiments $\mathfrak{A}$ and
$\mathfrak{B}$. We note that $0\leq
I(\mathfrak{A},\mathfrak{B})(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\leq 4\ln 2$ and
$-4\ln 2\leq
I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\leq
4\ln 2$.

In case $\mu_{1}=\mu_{2}=\nu_{1}=\nu_{2}$ we have
$\theta_{11}=\theta_{12}=\theta_{21}=\theta_{22}=0$,
$e(\theta_{11})=e(\theta_{12})=e(\theta_{21})=e(\theta_{22})=-1$, hence
$I(\mathfrak{A},\mathfrak{B})=4\ln 2$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=-4\ln 2$. In case
$\mu_{1}=\mu_{2}=\frac{\pi}{2}$, $\nu_{1}=\nu_{2}=0$ we have
$\theta_{11}=\theta_{12}=\theta_{21}=\theta_{22}=\frac{1}{4}$,
$e(\theta_{11})=e(\theta_{12})=e(\theta_{21})=e(\theta_{22})=0$, and
$I(\mathfrak{A},\mathfrak{B})=0$. Finally, in case $\mu_{1}=\mu_{2}=\pi$,
$\nu_{1}=\nu_{2}=0$ we have
$\theta_{11}=\theta_{12}=\theta_{21}=\theta_{22}=\frac{1}{2}$,
$e(\theta_{11})=e(\theta_{12})=e(\theta_{21})=e(\theta_{22})=1$, and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=4\ln 2$.

Since the image of a compact and connected set via continuous function
$I(\mathfrak{A},\mathfrak{B})$ (respectively, the continuous function
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})$) is a compact and connected
subset of $\mathbb{R}$, we obtain that the range of
$I(\mathfrak{A},\mathfrak{B})$ (respectively,
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})$) coincides with the interval
$[0,4\ln 2]$ (respectively, with the interval $[-4\ln 2,4\ln 2]$).

###  4.1 Bell’s Inequality

The equality
$|\mathcal{A}_{\mu_{1}}\mathcal{B}_{\nu_{1}}+\mathcal{A}_{\mu_{1}}\mathcal{B}_{\nu_{2}}+\mathcal{A}_{\mu_{2}}\mathcal{B}_{\nu_{1}}-\mathcal{A}_{\mu_{2}}\mathcal{B}_{\nu_{2}}|=2$
yields (with an abuse of the probability theory) Bell’s inequality

$|\mathcal{E}(\mathcal{A}_{\mu_{1}}\mathcal{B}_{\nu_{1}})+\mathcal{E}(\mathcal{A}_{\mu_{1}}\mathcal{B}_{\nu_{2}})+\mathcal{E}(\mathcal{A}_{\mu_{2}}\mathcal{B}_{\nu_{1}})-\mathcal{E}(\mathcal{A}_{\mu_{2}}\mathcal{B}_{\nu_{2}}|\leq
2,$

that is, $|b(\mu_{1},\mu_{2},\nu_{1},\nu_{2})|\leq 2$, where
$b(\mu_{1},\mu_{2},\nu_{1},\nu_{2})=\cos(\mu_{1}-\nu_{1})+\cos(\mu_{1}-\nu_{2})+\cos(\mu_{2}-\nu_{1})-\cos(\mu_{2}-\nu_{2})$.

J. S. Bell in [2] proves that if there exist ”…additional variables which
restore to the (quantum) theory causality and locality”, then the above
inequality is satisfied. Since $I(\mathfrak{A},\mathfrak{B})=0$ is equivalent
to the equalities $|\mu_{i}-\nu_{j}|=\frac{\pi}{2}$, $i,j=1,2$, this yields
$b=0$. Thus, we obtain that if Bell’s inequality is violated, then the total
information flow $I(\mathfrak{A},\mathfrak{B})$ is strictly positive, that is,
the experiments $\mathfrak{A}$ and $\mathfrak{B}$ are informationally
dependent.

###### Examples 4.1.1

Note that the results of all calculations below are rounded up to the 7-th
digit.

1) (Aspect’s experiment) $\mu_{1}=\frac{\pi}{8},\mu_{2}=\frac{3\pi}{8}$,
$\nu_{1}=\frac{\pi}{4}$, $\nu_{2}=0$. Then we obtain
$\cos(\frac{\pi}{8})=0.9238795$, $\cos(\frac{3\pi}{8})=0.3826834$, and
therefore $b(\frac{\pi}{8},\frac{3\pi}{8},\frac{\pi}{4},0)=2.3889551$. On the
other hand,
$\theta_{11}=\theta_{12}=\theta_{21}=\frac{1}{2}\sin^{2}(\frac{\pi}{16})=0.0190301$,
$e(\theta_{11})=e(\theta_{12})=e(\theta_{21})=-0.0415353$,
$\theta_{22}=\frac{1}{2}\sin^{2}(\frac{3\pi}{16})=0.154329$,
$e(\theta_{22})=-0.1084492$. Hence we have
$I(\mathfrak{A},\mathfrak{B})=0.1615415$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=-0.2330551$.

2) $\mu_{1}=\pi,\mu_{2}=\frac{2\pi}{3}$, $\nu_{1}=0$, $\nu_{2}=\frac{\pi}{3}$.
Then we have $b(\pi,\frac{2\pi}{3},0,\frac{\pi}{3})=-2.5$. On the other hand,
$\theta_{11}=\frac{1}{2}\sin^{2}(\frac{\pi}{2})=0.5$, $e(\theta_{11})=1$,
$\theta_{12}=\theta_{21}=\frac{1}{2}\sin^{2}(\frac{\pi}{3})=\frac{3}{8}=0.375$,
$e(\theta_{12})=e(\theta_{21})=0.1887219$,
$\theta_{22}=\frac{1}{2}\sin^{2}(\frac{\pi}{6})=\frac{1}{8}=0.125$,
$e(\theta_{22})=-0.1887219$. Hence we obtain
$I(\mathfrak{A},\mathfrak{B})=1.0855833$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=1.1887219$

3) $\mu_{1}=\frac{\pi}{2},\mu_{2}=0$, $\nu_{1}=\frac{\pi}{4}$,
$\nu_{2}=\frac{3\pi}{4}$. Then
$b(\frac{\pi}{2},0,\frac{\pi}{4},\frac{3\pi}{4})=2\sqrt{2}$. On the other
hand,
$\theta_{11}=\theta_{12}=\theta_{21}=\frac{1}{2}\sin^{2}(\frac{\pi}{8})=0.0732233$,
$e(\theta_{11})=e(\theta_{12})=e(\theta_{21})=-0.3994425$,
$\theta_{22}=\frac{1}{2}\sin^{2}(\frac{3\pi}{16})=0.154329$,
$e(\theta_{22})=-0.10844492$. Hence we have
$I(\mathfrak{A},\mathfrak{B})=0.9053727$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=-0.22827767$.

4) $\mu_{1}=\pi,\mu_{2}=0$, $\nu_{1}=0$, $\nu_{2}=\pi$. Then we have
$b(\pi,0,0,-\pi)=2$. On the other hand,
$\theta_{11}=\theta_{22}=\frac{1}{2}\sin^{2}(\frac{\pi}{2})=0.5$,
$e(\theta_{11})=e(\theta_{22})=1$,
$\theta_{12}=\theta_{21}=\frac{1}{2}\sin^{2}(0)=0$,
$e(\theta_{12})=e(\theta_{21})=-1$. Therefore we obtain
$I(\mathfrak{A},\mathfrak{B})=4\ln 2=2.7725887=\max
I(\mathfrak{A},\mathfrak{B})$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=0$.

5) $\mu_{1}=\frac{5\pi}{6},\mu_{2}=\frac{2\pi}{3}$, $\nu_{1}=\frac{\pi}{3}$,
$\nu_{2}=\frac{\pi}{2}$. In this case we have
$b(\frac{5\pi}{6},\frac{2\pi}{3},\frac{\pi}{3},\frac{\pi}{2})=1-\frac{\sqrt{3}}{2}$.
On the other hand,
$\theta_{11}=\frac{1}{2}\sin^{2}(\frac{\pi}{4})=\frac{1}{4}$,
$e(\theta_{11})=0$,
$\theta_{12}=\theta_{21}=\frac{1}{2}\sin^{2}(\frac{\pi}{6})=\frac{1}{8}$,
$e(\theta_{12})=e(\theta_{21})=-0.1887219$,
$\theta_{22}=\frac{1}{2}\sin^{2}(\frac{\pi}{12})=0.0334936$,
$e(\theta_{22})=-0.6453728$. Hence we have
$I(\mathfrak{A},\mathfrak{B})=0.7089624$ and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})=-0.267929$.

6) The link

http://www.math.bas.bg/algebra/valentiniliev/

contains a Java experimental implementation ”dependencemeasurements2”
depending on five parameters: an non-negative integer $n$ and four real
numbers $\mu_{1},\mu_{2},\nu_{1},\nu_{2}$ from the closed interval $[0,\pi]$.
One can also find the description of this program at the above link.

Examples 4.1.1 and, especially, example 6), yield that the relations between
$I(\mathfrak{A},\mathfrak{B})$ and $b$, and
$I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})$ and $b$ seem to be random.
Below we present an attempt to explain the uncertainty of this relation by
refereing to [6, 5.4, 5.5]. We define the events

$U=\\{(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\in[0,\pi]^{4}||b(\mu_{1},\mu_{2},\nu_{1},\nu_{2})|\leq
2\\},$
$V=\\{(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\in[0,\pi]^{4}|I(\mathfrak{A},\mathfrak{B})(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\in[0,2\ln
2]\\},$
$VS=\\{(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\in[0,\pi]^{4}|I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})(\mu_{1},\mu_{2},\nu_{1},\nu_{2})\in[0,4\ln
2]\\},$

with complements $U^{c}$, $V^{c}$, and $VS^{c}$ in $[0,\pi]^{4}$. We suppose
that the probabilities $\alpha=\operatorname{pr}(U)$,
$\beta=\operatorname{pr}(V)$, and
$\beta^{\left(s\right)}=\operatorname{pr}(VS)$ in the sample space
$[0,\pi]^{4}$ furnished with normalized Borel measure are known. The
probabilities $\tau=\operatorname{pr}(U\cap V)$ and
$\tau^{\left(s\right)}=\operatorname{pr}(U\cap VS)$ run through the closed
intervals $I(\alpha,\beta)=[\max(0,\alpha+\beta-1),\min(\alpha,\beta)]$ and
$I(\alpha,\beta^{\left(s\right)})$, respectively. In case
$\tau=\min(\alpha,\beta)]$ (respectively,
$\tau^{\left(s\right)}=\min(\alpha,\beta^{\left(s\right)})]$) there exists a
relation of inclusion (up to a set of probability $0$) between $U$ and $V$
(respectively, between $U$ and $VS$). Otherwise, both conditional
probabilities $\operatorname{pr}(V^{c}|U)$ and $\operatorname{pr}(V|U^{c})$
(respectively, $\operatorname{pr}(VS^{c}|U)$ and
$\operatorname{pr}(VS|U^{c})$) can not be simultaneously as small as one wants
(a kind of uncertainty principle).

###### Remark 4.1.2

The probabilities $\alpha=\operatorname{pr}(U)$, $\beta=\operatorname{pr}(V)$,
$\tau=\operatorname{pr}(U\cap V)$, and
$\tau^{\left(s\right)}=\operatorname{pr}(U\cap VS)$ can be approximated by
using Examples 4.1.2, 6). We draw a random sample $X$ of size $n$ from the
sample space $[0,\pi]^{4}$ and consider $X$ as a sample space with $n$ equally
likely outcomes. Then the probabilities $\hat{\alpha}(n)$, $\hat{\beta}(n)$,
$\hat{\tau}(n)$, and $\widehat{\tau^{\left(s\right)}}(n)$ of the traces of
$U$,$V$, $U\cap V$, and $U\cap VS$ on $X$ (the sample proportions) are
unbiased estimators for $\alpha$, $\beta$, $\tau$, and $\tau^{\left(s\right)}$
when $n$ is large.

Note that as an output of $n$ iterations of the random process from example 6)
we can also find: a) the approximation $[L(n),J(n)]$ of the range of
$I(\mathfrak{A},\mathfrak{B})$ and the approximation $[LS(n),JS(n)]$ of the
range of $I^{\left(s\right)}(\mathfrak{A},\mathfrak{B})$ under the condition
$|b|\leq 2$, b) the above sample proportions, and c) the approximations
$\operatorname{pr}(V^{c}|U)(n)=\frac{\hat{\alpha}(n)-\hat{\tau}(n)}{\hat{\alpha}(n)}$,
$\operatorname{pr}(V|U^{c})(n)=\frac{\hat{\beta}(n)-\hat{\tau}(n)}{1-\hat{\alpha}(n)}$,
and
$\operatorname{pr}(VS^{c}|U)(n)=\frac{\hat{\alpha}(n)-\widehat{\tau^{\left(s\right)}}(n)}{\hat{\alpha}(n)}$,
$\operatorname{pr}(VS|U^{c})(n)=\frac{\widehat{\beta^{\left(s\right)}}(n)-\widehat{\tau^{\left(s\right)}}(n)}{1-\hat{\alpha}(n)}$.

Below are the results obtained by drawing a random sample of size $n=1000$:

$\hat{\alpha}(1000)=0.838$, $\hat{\beta}(1000)=0.704$,
$\hat{\tau}(1000)=0.614$, $\widehat{\tau^{\left(s\right)}}(1000)=0.087$,
$\operatorname{pr}(V^{c}|U)(1000)=0.2673031$,
$\operatorname{pr}(V|U^{c})(1000)=0.5555555$,
$\operatorname{pr}(VS^{c}|U)(1000)=0.8961814$,
$\operatorname{pr}(VS|U^{c})(1000)=0.0185185$.

## Acknowledgements

I thank Dimitar Guelev for making an experimental Java implementation of the
evaluation of dependence of simultaneous measurements and the approximation of
various parameters via a random process. The numerical examples thus produced
were invaluable for my work. I would like to express my gratitude to
administration of the Institute of Mathematics and Informatics at the
Bulgarian Academy of Sciences for creating perfect and safe conditions of
work.

## References

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