Reichenbachian Common Cause Clusters

Abstract

The principle of the common cause demands that every pair of causally independent but statistically correlated events should be the effect of a common cause. This demand is often supplemented with the requirement that said cause should screen-off the two events from each other. This paper introduces a new probabilistic model for common causes, which generalises this requirement to include sets of distinct but non-disjoint causes. It is demonstrated that the model hereby proposed satisfies the explanatory function generally attributed to common causes and an existential proof is offered. Finally, the model is compared to other known structures intended to perform a similar function, such as Reichenbachian Common Cause Systems and Bayesian Networks.

Appendices

Appendix A: Proof of Proposition 4

Proof

Let $(\varOmega _{0}, {\mathcal {A}}_{0}, p_{0})$ be a classical probability space with $\sigma$-algebra of random events ${\mathcal {A}}_{0}$ and probability measure $p_{0}$.

Let $A,B\in \varOmega _{0}$ be genuinely independent and let $Corr_{0}\left( {A},{B}\right) >0$. It is worth noticing from the outset that these assumptions immediately demand that

$$\begin{aligned} 0< p_{0}\left( {A\cap B}\right)< p_{0}\left( {A}\right) , p_{0}\left( {B}\right) < 1. \end{aligned}$$

(42)

This fact will play a role in numerous passages of the following proof, although for the sake of brevity we will avoid explicitly mentioning it hereafter.

The existence of CCCs for $(A_{0},B_{0})$ in some extension of $(\varOmega _{0}, {\mathcal {A}}_{0}, p_{0})$ has been proved by Hofer-Szabó et al. (1999). Therefore, we shall hereafter consider only RCCCs of finite size greater than one. Let us accordingly choose any integer number $n\ge 2$. Our proof will be articulated in three major steps. First, we shall construct an extension of the chosen classical probability space and four sets of $2^{n}$ disjoint events in said extension that satisfy some specified constraints. Second, we shall combine those sets to obtain a partition $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ of the sample space so constructed and a set $\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}$ so that $\left\{ {X_{i}}\right\} _{{i=1}}^{{2^{n}}} = \rho \left( {\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}}\right)$. Finally, we will define a probability function on the above mentioned extension such that $\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}$ satisfies the necessary and sufficient conditions for an RCCC.

Step 1

The first step of our proof will be largely similar to the second step in of Hofer-Szabó and Rédei’s proof for the existence of RCCSs of arbitrary finite size (Hofer-Szabó and Rédei 2006), later revised in Mazzola and Evans (2017).⁷

To start with, let us take $2^{n}$ identical copies $S_{1}, \ldots , S_{2^{n}}$ of $\varOmega$ and $2^{n}$ identical copies ${\mathcal {A}}_{1}, \ldots , {\mathcal {A}}_{2^{n}}$ of ${\mathcal {A}}$. For each $i = 1, \ldots , 2^{n}$, let ${\mathcal {A}}_{i}$ be the $\sigma$-algebra on $S_{i}$ and let $\phi _{i}: {\mathcal {A}}_{0} \rightarrow {\mathcal {A}}_{i}$ be the $\sigma$-algebra homomorphism from ${\mathcal {A}}_{0}$ to ${\mathcal {A}}_{i}$. We can now define:

$$\begin{aligned}&{\mathcal {A}} = \left\{ {X = \bigcup _{i=1}^{2^{n}} \phi _{i}(X_{0}): X_{0}\in {\mathcal {A}}}\right\}\end{aligned}$$

(43)

$$\begin{aligned}&X\cap Y = \bigcup _{i = 1}^{2^{n}} \phi _{i}(\rho _{i}(X) \cap \rho _{i}(Y)) \qquad \forall X,Y\in {\mathcal {A}} \end{aligned}$$

(44)

$$\begin{aligned}&X\cup Y = \bigcup _{i = 1}^{2^{n}} \phi _{i}(\rho _{i}(X)\cup \rho _{i}(Y)) \qquad \forall X,Y\in {\mathcal {A}} \end{aligned}$$

(45)

$$\begin{aligned}&\overline{X} = \bigcup _{i = 1}^{2^{n}} \phi _{i}(\overline{\rho _{i}(X)}) \qquad \forall X\in {\mathcal {A}} \end{aligned}$$

(46)

$$\begin{aligned}&S = \bigcup _{i=1}^{2^{n}} S_{i}, \end{aligned}$$

(47)

where $\rho _{i}: {\mathcal {A}} \rightarrow {\mathcal {A}}_{i}$ is a projection of ${\mathcal {A}}$ on ${\mathcal {A}}_{i}$ such that $\bigcup _{i=1}^{2^{n}}\phi _{i}(\rho _{i}(X)) = X$, for each $i = 1, \ldots , 2^{n}$ and $X\in {\mathcal {A}}$. It would then be easy to verify that ${\mathcal {A}} = \left( {\mathcal {A}}, \cap , \cup , \overline{\bullet }, S, \emptyset \right)$ is a $\sigma$-algebra and that the map

$$\begin{aligned}&\phi : {\mathcal {A}}_{0}\rightarrow {\mathcal {A}} \nonumber \\&\quad X_{0}\mapsto \phi _{i}(X_{0})\cup \cdots \cup \phi _{2^{n}}(X_{0}) \end{aligned}$$

(48)

is a $\sigma$-algebra embedding of ${\mathcal {A}}_{0}$ in ${\mathcal {A}}$.

Our next task is to define a probability measure p such that $(\varOmega , {\mathcal {A}}, p)$ is an extension of $(\varOmega _{0}, {\mathcal {A}}_{0}, p_{0})$. To this end, it will be sufficient to select $4\times 2^{n}$ non-negative real numbers $r_{i}^{j}\ (i =1, \ldots , 2^{n}$; $j = 1, \ldots , 4$) such that

$$\begin{aligned} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{j}_{i}} = 1 \qquad j = 1, \ldots , 4 \end{aligned}$$

(49)

and to postulate that, for all for all $X\in {\mathcal {A}}$:

$$\begin{aligned} p\left( {X}\right) &= {} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{1}_{i}p_{0}\left( {A\cap B\cap \phi ^{-1}(X)}\right) } + \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{2}_{i}p_{0}\left( {A\cap \overline{B} \cap \phi ^{-1}(X)}\right) } \nonumber \\&+ \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{3}_{i}p_{0}\left( {\overline{A}\cap B\cap \phi ^{-1}(X)}\right) } + \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{4}_{i}p_{0}\left( {\overline{A} \cap \overline{B} \cap \phi ^{-1}(X)}\right) }. \end{aligned}$$

(50)

Because $\left\{ {A\cap B, A\cap \overline{B},\overline{A}\cap {B}, \overline{A}\cap \overline{B}}\right\}$ is a partition of $\varOmega _{0}$, it follows that

$$\begin{aligned} p\left({\phi (X_{0})}\right) = \displaystyle \sum _{{i =1}}^{{2^{n}}}{\displaystyle \sum _{{j = 1}}^{{4}}{r^{j}_{i}} p_{0}\left( {X_{0}}\right) }= p_{0}\left( {X_{0}}\right) \end{aligned}$$

(51)

for all $X_{0}\in {\mathcal {A}}_{0}$. Therefore, by Definition 8, $(\varOmega , {\mathcal {A}}, p)$ is indeed an extension of $(\varOmega _{0}, {\mathcal {A}}_{0}, p_{0})$.

Step 2

The second step of our proof will consist in constructing a partition $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ of $\varOmega$ and a set $\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}$ so that $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}} = \rho \left( {\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}}\right)$.

The first partition will be obtained by simply putting $X_{i} =S_{i}$ for $i=1, \ldots , 2^{n}$. That such set is a partition of $\varOmega$ is evident from the construction carried out in step one. We can therefore immediately proceed to construct $\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}$.

To do this, let us consider all the possible ordered n-tuples whose terms are equal to 0 or 1. There will be exactly $2^{n}$ such n-tuples, so let us label them $x_{1}$, $x_{2}$, ..., $x_{2^{n}}$. In particular, for ease of notation, let us make sure to reserve label $x_{1}$ for the one n-tuple all of whose entries are equal to 0. We can now assign to each element of $\left\{ {X_{i}}\right\} _{{i=1}}^{{2^{n}}}$ exactly one n-tuple in the set $\left\{ {x_{i}}\right\} _{{i=1}}^{{2^{n}}}$, and vice-versa. For notational simplicity, we shall assume that each event $X_{i}$ is assigned the n-tuple $x_{i}$ having the same index.

For each $i = 1, \ldots , n$, let us now define $C_{i}$ as the union of all elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ that have been assigned to some n-tuple whose i-th term is equal to 1. Let $\left\{ {C_{i}}\right\} _{{i=1}}^{{n}}$ be the set of events so defined. Similarly, for each $i = 1, \ldots , n$, let $\overline{C_{i}}$ be the union of all elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ that have been assigned to some n-tuple whose i-th term is equal to 0, and let $\left\{ {\overline{C_{i}}}\right\} _{{i=1}}^{{n}}$ be the resulting set.

It would be then easy to verify that, by construction, each element of $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ will be the union of $2^{n-1}$ elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$, the intersection of any two $C_{i}\cap C_{j}\in \left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ will be identical to the union of $2^{n-2}$ elements, and so on. By converse, each element of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ will then be identical to the conjunction of n elements of $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}\cup \left\{ {\overline{C_{i}}}\right\} _{{i=1}}^{{n}}$ such that no two conjuncts are the complement of each other. In other words,

$$\begin{aligned} X_{i} = C_{1}^{*}\cap \cdots \cap C_{n}^{*}&\quad C_{i}^{*} \in \left\{ {C_{i},\overline{C_{i}}}\right\} \quad {\text{ for all}} \; i = 1, \ldots , n. \end{aligned}$$

(52)

Therefore, by Definition 5,

$$\begin{aligned} \left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}} = \rho \left( {\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}}\right) . \end{aligned}$$

(53)

Step 3

In step one of our proof, we constructed the probability function p on the $\sigma$-algebra of events ${\mathcal {A}}$ by introducing $4\times 2^{n}$ non-negative real numbers $r^{j}_{i}$ ($i = 1, \ldots , 2^{n}$; $j = 1, \ldots , 4$) whose sum is equal to one. It must be noticed that the result of step one was independent of the exact value of each such number, provided that they satisfied (49). To conclude our proof, we now need to further constrain the above mentioned numbers in such a way as to ensure that $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ meets the necessary and jointly sufficient conditions for an RCCC for $(A,B)= (\phi (A_{0}),\phi (B_{0}))$.

To this end, let us start by defining the following two constants:

$$\begin{aligned} \alpha &= {} \frac{p_{0}\left( {A}\right) }{2^{n}-1}, \end{aligned}$$

(54)

$$\begin{aligned} \beta &= {} \frac{p_{0}\left( {B}\right) }{2^{n}-1}. \end{aligned}$$

(55)

Next, let us consider n positive real numbers $k_{1}, \ldots , k_{n}$ such that:

$$\begin{aligned}&2^{n}-1<k_{1}<k_{2}< \cdots< k_{n-1}<k_{n}<\min {\left( \frac{1}{\alpha }, \frac{1}{\beta }\right) }, \end{aligned}$$

(56)

$$\begin{aligned}&\displaystyle \sum _{{i= 1}}^{{n}}{\frac{n!}{i!(n-i)!}k_{i}} = \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }. \end{aligned}$$

(57)

Some such numbers do exist, but for ease of exposition we shall postpone a proof of their existence to the supplementary section following this appendix.

Let us now select any function $f:\mathbb {R}\rightarrow \mathbb {R}$ such that

$$\begin{aligned} f(i)= \left\{ \begin{array}{ll} 0 &{}\quad i = 0, \\ k_{i} &{}\quad i = 1, \ldots , n. \end{array}\right. \end{aligned}$$

(58)

Finally, Let us consider once again the $2^{n}\ n$-tuples introduced in the previous step. Let us define the rank of each n-tuple $x_{i}$, denoted $r(x_{i})$, as the number of terms in $x_{i}$ that are equal to 1. It can be noted that, by construction, the rank of each n-tuple $x_{i}$ will be equal to the number of elements of $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ that have a non-empty intersection with $X_{i}$. Moreover, the number of n-tuples with rank k will be equal to $\frac{n!}{k! (n-k)!}$. Let us recall, further, that $x_{1}$ will be the sole n-tuple with rank zero.

We can now identify the real numbers $r^{j}_{i}$ ($i = 1, \ldots , 2^{n}$; $j = 1, \ldots , 4$) as follows:

$$\begin{aligned}&r_{i}^{1} = \left\{ \begin{array}{ll} 0 &{}\quad i = 1, \\ \frac{\alpha \beta f(r(x_{i})))}{p_{0}\left( {A\cap B}\right) } &{} \quad i= 2, \ldots , 2^{n}. \\ \end{array}\right. \end{aligned}$$

(59)

$$\begin{aligned}&r_{i}^{2} = \left\{ \begin{array}{ll} 0 &{}\quad i = 1, \\ \frac{\alpha \left[ 1- \beta f(r(x_{i}))\right] }{p_{0}\left( {A\cap \overline{B}}\right) } &{} \quad i= 2, \ldots ,2^{n}. \\ \end{array}\right. \end{aligned}$$

(60)

$$\begin{aligned}&r_{i}^{3} = \left\{ \begin{array}{ll} 0 &{}\quad i = 1, \\ \frac{\beta \left[ 1- \alpha f(r(x_{i}))\right] }{p_{0}\left( {\overline{A}\cap B}\right) } &{} \quad i= 2, \ldots ,2^{n}. \\ \end{array}\right. \end{aligned}$$

(61)

$$\begin{aligned}&r_{i}^{4}=\left\{ \begin{array}{ll} 1-\displaystyle \sum _{{i=2}}^{{2^{n}}}{r^{4}_{i}} &{} \quad i = 1 \\ \frac{\frac{1}{f(r(x_{i}))}-\alpha -\beta +\alpha \beta f(r(x_{i}))}{p_{0}\left( {\overline{A} \cap \overline{B}}\right) } &{} \quad i = 2, \ldots , 2^{n}. \\ \end{array}\right. \end{aligned}$$

(62)

Before we can proceed any further, we need to verify that the numbers so defined do meet the criteria set in step one: namely, that each of them is non-negative, and that they obey (49). Let us proceed in order.

That $r^{1}_{i}$ is non-negative for each $i = 1, \ldots , 2^{n}$ is an immediate consequence of (54)–(55), (56), and (58). That their sum is equal to one follows instead from (57) and (58), as follows:

$$\begin{aligned} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{1}_{i}} &= {} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{\frac{\alpha \beta f(r(x_{i})))}{p_{0}\left( {A\cap B}\right) }}=\displaystyle \sum _{{i = 2}}^{{2^{n}}}{\frac{\alpha \beta k_{i}}{p_{0}\left( {A\cap B}\right) }} =\alpha \beta \displaystyle \sum _{{i = 0}}^{{n}}{\frac{n!}{i!(n-i)!}k_{i}\frac{1}{p_{0}\left( {A\cap B}\right) }}\nonumber \\ &= {} \alpha \beta \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta } \frac{1}{p_{0}\left( {A\cap B}\right) } = 1. \end{aligned}$$

(63)

Let us now move to $r^{2}_{i}$ and $r^{3}_{i}$, where $i = 1, \ldots , 2^{n}$. That each of the above mentioned numbers is non-negative (in fact, positive) is a consequence of (56)–(58). That they satisfy (49) can instead be demonstrated by means of (54)–(55), (57) and (58):

$$\begin{aligned} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{2}_{i}} &= {} \frac{ \displaystyle \sum _{{i = 1}}^{{2^{n}}}\alpha \left[ 1- \beta f(r(x_{i}))\right] }{p_{0}\left( {A\cap \overline{B}}\right) } = \frac{(2^{n}-1)\alpha - p_{0}\left( {A\cap B}\right) }{p_{0}\left( {A\cap \overline{B}}\right) }\nonumber \\ &= {} \frac{p_{0}\left( {A}\right) -p_{0}\left( {A\cap B}\right) }{p_{0}\left( {A\cap \overline{B}}\right) }=\frac{p_{0}\left( {A\cap \overline{B}}\right) }{p_{0}\left( {A\cap \overline{B}}\right) } =1, \end{aligned}$$

(64)

$$\begin{aligned} \displaystyle \sum _{{i = 1}}^{{2^{n}}}{r^{3}_{i}} &= {} \frac{ \displaystyle \sum _{{i = 1}}^{{2^{n}}}\beta \left[ 1- \alpha f(r(x_{i}))\right] }{p_{0}\left( {\overline{A}\cap {B}}\right) } =\frac{(2^{n}-1)\beta - p_{0}\left( {A\cap B}\right) }{p_{0}\left( {\overline{A}\cap {B}}\right) }\nonumber \\ &= {} \frac{p_{0}\left( {B}\right) -p_{0}\left( {A\cap B}\right) }{p_{0}\left( {\overline{A}\cap {B}}\right) }=\frac{p_{0}\left( {\overline{A}\cap {B}}\right) }{p_{0}\left( {\overline{A}\cap {B}}\right) } =1. \end{aligned}$$

(65)

Lastly, let us consider numbers $r^{4}_{i}$ for $i = 1, \ldots , 2^{n}$. That their sum is equal to one can be immediately seen from the very definition of $r^{4}_{1}$. To show that $r^{4}_{1}$ is non-negative (indeed, positive) let us observe that, owing to (56)–(58) and to the construction offered in the Supplement:

$$\begin{aligned} \displaystyle \sum _{{i=2}}^{{2^{n}}}{\frac{1}{f(r(x_{i}))}}\le \displaystyle \sum _{{i=2}}^{{2^{n}}}{\frac{1}{f(1)}} =\displaystyle \sum _{{i=2}}^{{2^{n}}}{\frac{1}{k_{1}}} =\frac{2^{n}-1}{k_{1}} =\frac{2^{n}-1}{k_{0}-\varepsilon _{1}} < \frac{2^{n}-1}{2^{n}-1} = 1, \end{aligned}$$

(66)

Hence:

$$\begin{aligned} r_{1}^{4} &= {} \frac{p_{0}\left( {\overline{A}\cap \overline{B}}\right) - \displaystyle \sum _{{i =2}}^{{2^{n}}}{\left( \frac{1}{f(r(x_{i}))}-\alpha -\beta +\alpha \beta f(r(x_{i}))\right) }}{p_{0}\left( {\overline{A}\cap \overline{B}}\right) }\nonumber \\ &> {} \frac{p_{0}\left( {\overline{A}\cap \overline{B}}\right) - \left[ 1 + \displaystyle \sum _{{i = 2}}^{{2^{n}}}{\left( -\alpha -\beta +\alpha \beta f(r(x_{i}))\right) }\right] }{p_{0}\left( {\overline{A} \cap \overline{B}}\right) } \end{aligned}$$

(67)

$$\begin{aligned} &= {} \frac{p_{0}\left( {\overline{A}\cap \overline{B}}\right) - 1 + p_{0}\left( {A}\right) + p_{0}\left( {B}\right) -p_{0}\left( {A\cap B}\right) }{p_{0}\left( {\overline{A}\cap \overline{B}}\right) } = 0. \end{aligned}$$

(68)

Having so established that the numbers $r^{j}_{i}$ ($i = 1, \ldots , 2^{n}; j = 1, \ldots , 4$) are well-defined, we can finally proceed to show that, according to the probability function p defined by (50) and (59)–(62), $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ is in fact an RCCC for (A, B).

Condition (24) is satisfied by construction, as (50), (56), and (59)–(62) imply that, for every $X_{i}\in \left\{ {X_{i}}\right\} _{{i =1}}^{{2^{n}}}$,

$$\begin{aligned} p\left( {X_{i}}\right) =\left\{ \begin{array}{ll} \left( 1-\displaystyle \sum _{{i=2}}^{{2^{n}}}{r^{4}_{i}}\right) p_{0}\left( {\overline{A} \cap \overline{B}}\right)> 0 &{} \quad i = 1 \\ \frac{1}{f(r(x_{i}))}> 0 &{} \quad i = 2, \ldots , 2^{n}\\ \end{array}\right. \end{aligned}$$

(69)

To show that (25) holds, it is sufficient to notice that, owing to (50) and (59)–(62):

$$\begin{aligned}&p\left( {A\cap B \cap X_{i}}\right) =\left\{ \begin{array}{ll} 0 &{} \quad : i = 1\\ \alpha \beta f(r(x_{i})) &{} \quad : i= 2, \ldots , 2^{n}\\ \end{array}\right. \end{aligned}$$

(70)

$$\begin{aligned}&p\left( {A\cap X_{i}}\right) = \left\{ \begin{array}{ll} 0 &{} \quad : i = 1\\ \alpha &{} \quad : i= 2, \ldots , 2^{n}\\ \end{array}\right. \end{aligned}$$

(71)

$$\begin{aligned}&p\left( {B\cap X_{i}}\right) =\left\{ \begin{array}{ll} 0 &{} \quad : i = 1\\ \beta &{} \quad : i= 2, \ldots , 2^{n}\\ \end{array}\right. \end{aligned}$$

(72)

Along with (69), this implies that

$$\begin{aligned}&p\left( {A\cap B\cap X_{i}}\right) p\left( {X_{i}}\right) = 0 = p\left( {A\cap X_{i}}\right) p\left( {B\cap X_{i}}\right) \quad i = 1 \end{aligned}$$

(73)

$$\begin{aligned}&p\left( {A\cap B\cap X_{i}}\right) p\left( {X_{i}}\right) = \alpha \beta f(r(x_{i}))\frac{1}{f(r(x_{i}))} = \alpha \beta = p\left( {A\cap X_{i}}\right) p\left( {B\cap X_{i}}\right) \quad i = 2, \ldots , 2^{n}, \end{aligned}$$

(74)

from which (25) immediately ensues.

Let us finally move to condition (26). Let $C_{j}\in \left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ be arbitrarily chosen, and so let $X\in \gamma \left( \left\{ {C_{i}}\right\} _{\begin{array}{c} i = 1\\ i\ne j \end{array}}^{n}\right)$. By Definition 6, X will be identical to the conjunction of k elements of $\left\{ {C_{i}}\right\} _{{\begin{array}{c} i =1\\ i\ne j \end{array}}}^{{n}}\cup \left\{ {\overline{C_{i}}}\right\} _{{\begin{array}{c} i = 1\\ i\ne j \end{array}}}^{{n}}$, with $k = 0, \ldots , n-1$, such that no two conjuncts are the complement of each other. Hence, by construction, X will be also identical to the union of $2^{n -k}$ elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$, with $k = 0, \ldots , n-1$. Similarly, $C_{j}\cap X$ will be identical to the union of $2^{n -k-1}$ elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ of the form

$$\begin{aligned} C_{1}^{*}\cap \cdots \cap C_{j} \cap \cdots C_{2^{n}}^{*} \quad C^{*}_{i}\in \left\{ {C_{i}, \overline{C_{i}}}\right\} _{{\begin{array}{c} i = 1\\ i\ne j \end{array}}}^{{2^{n}}}, \end{aligned}$$

(75)

while $\overline{C_{j}}\cap X$ will be identical to the union of $2^{n -k-1}$ elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ of the form

$$\begin{aligned} C_{1}^{*}\cap \cdots \cap \overline{C_{j}} \cap \cdots C_{2^{n}}^{*} \quad C^{*}_{i}\in \left\{ {C_{i}, \overline{C_{i}}}\right\} _{{\begin{array}{c} i = 1\\ i\ne j \end{array}}}^{{2^{n}}}. \end{aligned}$$

(76)

For notational convenience, let us relabel all elements of $\left\{ {X_{i}}\right\} _{{i = 1}}^{{2^{n}}}$ so that:

$$\begin{aligned}&C_{j}\cap X = \bigcup _{i = 1}{2^{n-k-1}}X^{1}_{i}, \end{aligned}$$

(77)

$$\begin{aligned}&\overline{C_{j}}\cap X = \bigcup _{i = 1}{2^{n-k-1}}X^{0}_{i}. \end{aligned}$$

(78)

Moreover, let us similarly relabel all the n-tuples of zeroes and ones associated with the above events.

From this and (59)–(62), along with the theorem of total probability, it follows that:

$$\begin{aligned}&p\left( {A\cap C_{j}\cap X}\right) = \displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{\alpha \beta f(r(x^{1}_{i}))+ \alpha [1-\beta f(r(x^{1}_{i}))]} =\displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{\alpha } = 2^{n-k-1}\alpha , \end{aligned}$$

(79)

$$\begin{aligned}&p\left( {A\cap \overline{C_{j}}\cap X}\right) = \displaystyle \sum _{{i = 1}}^{{2^{n-k-2}}}{\alpha \beta f(r(x^{0}_{i}))+ \alpha [1-\beta f(r(x^{0}_{i}))]} = \displaystyle \sum _{{i = 1}}^{{2^{n-k-2}}}{\alpha } =2^{n-k-2}\alpha , \end{aligned}$$

(80)

and

$$\begin{aligned}&p\left( {B\cap C_{j}\cap X}\right) = \displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{\alpha \beta f(r(x^{1}_{i}))+ \beta [1-\alpha f(r(x^{1}_{i}))]} = \displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{\beta } = 2^{n-k-1}\beta , \end{aligned}$$

(81)

$$\begin{aligned}&p\left( {B\cap \overline{C_{j}}\cap X}\right) = \displaystyle \sum _{{i = 1}}^{{2^{n-k-2}}}{\alpha \beta f(r(x^{0}_{i}))+ \beta [1-\alpha f(r(x^{0}_{i}))]} = \displaystyle \sum _{{i = 1}}^{{2^{n-k-2}}}{\beta } = 2^{n-k-2}\beta . \end{aligned}$$

(82)

It is then a matter of brute calculation to prove that:

$$\begin{aligned}&p\left( {A}\left| {C_{j}}\left| {X}\right. \right. \right) - p\left( {A}\left| {\overline{C_{j}}}\left| {X}\right. \right. \right) =2^{n-k-2}\alpha \left[ \frac{2}{p\left( {C_{j}\cap X}\right) }-\frac{1}{p\left( {\overline{C_{j}}\cap X}\right) }\right] , \end{aligned}$$

(83)

$$\begin{aligned}&p\left( {B}\left| {C_{j}}\left| {X}\right. \right. \right) - p\left( {B}\left| {\overline{C_{j}}}\left| {X}\right. \right. \right) = 2^{n-k-2}\beta \left[ \frac{2}{p\left( {C_{j}\cap X}\right) } -\frac{1}{p\left( {\overline{C_{j}}\cap X}\right) }\right] . \end{aligned}$$

(84)

Due to (54)–(55), each of the above differences is greater than zero in case

$$\begin{aligned} p\left( {\overline{C_{j}}\cap X}\right) - p\left( {C_{j}\cap X}\right) = \displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{p\left( {X^{0}_{i}}\right) } - \displaystyle \sum _{{i = 1}}^{{2^{n-k-1}}}{p\left( {X^{1}_{i}}\right) } >0. \end{aligned}$$

(85)

On the other hand, it would be easy to check that

$$\begin{aligned} r(x^{1}_{i})- r(x^{0}_{i}) =1 \quad i = 1, \ldots , 2^{n-k-1}. \end{aligned}$$

(86)

Given (56), (58) and (69), inequality (85) obtains as a consequence. Hence,

$$\begin{aligned}&p\left( {A}\left| {C_{j}}\left| {X}\right. \right. \right) - p\left( {A}\left| {\overline{C_{j}}}\left| {X}\right. \right. \right) > 0 \end{aligned}$$

(87)

$$\begin{aligned}&p\left( {B}\left| {C_{j}}\left| {X}\right. \right. \right) - p\left( {B}\left| {\overline{C_{j}}}\left| {X}\right. \right. \right) > 0 \end{aligned}$$

(88)

for any $X\in\gamma\left(\left\{C_{i}\right\}^{n}_{\substack{i = 1\\ i\neq j}}\right)$. Because $C_{j}$ and X were chosen arbitrarily, this suffices to establish (26).

Before concluding our proof let us finally observe that, owing to the above construction, all elements of $\left\{ {C_{i}}\right\} _{{i = 1}}^{{n}}$ have a non-zero probability intersection with the complements of A and B and, as a consequence, differ from either event by more than a zero probability event. $\square$

Appendix B: Supplement

In the course of the preceding proof, we assumed the existence of real numbers $k_{1}, \ldots , k_{n}$ satisfying conditions (56)–(57). This section will outline one possible construction to obtain some such numbers.

Let all of the assumptions made in the course of our proof hold. Few calculations would then show that:

$$\begin{aligned} 2^{n}-1<\frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }\frac{1}{2^{n}-1} <\min {\left( \frac{1}{\alpha },\frac{1}{\beta }, \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta } -(2^{n}-1)(2^{n}-2)\right) }. \end{aligned}$$

(89)

So, first, let us choose any number $k_{n}$ such that

$$\begin{aligned} \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }\frac{1}{2^{n}-1}< k_{n} <\min {\left( \frac{1}{\alpha },\frac{1}{\beta }, \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }-(2^{n}-1)(2^{n}-2)\right) }. \end{aligned}$$

(90)

Next, let us take the number

$$\begin{aligned} k_{0}=\frac{\frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }-k_{n}}{2^{n}-2}. \end{aligned}$$

(91)

Because any summation of the form $\displaystyle \sum _{{i=1}}^{{n-1}}{\frac{n!}{i!(n-i)!}k_{i}}$ comprises exactly $2^{n}-2$ terms, the following equality must then obtain:

$$\begin{aligned} \displaystyle \sum _{{i= 1}}^{{n-1}}{\frac{n!}{i!(n-i)!}k_{0}}+k_{n} =(2^{n}-2) k_{0} +k_{n} = \frac{p_{0}\left( {A\cap B}\right) }{\alpha \beta }. \end{aligned}$$

(92)

Moreover, owing to our choice of $k_{n}$, it will also be the case that:

$$\begin{aligned} 2^{n}-1<k_{0}<k_{n}. \end{aligned}$$

(93)

Let now posit that $j = \frac{n-1}{2}$ in case n is odd, whereas $j = \frac{n-2}{2}$ if n is even. It is then possible to choose some positive real numbers $\varepsilon _{1}, ..., \varepsilon_{j}$ such that

$$\begin{aligned} 2^{n}-1<k_{0}-\varepsilon _{i}<k_{0}-\varepsilon _{i+1}<k_{0}< k_{0}+\varepsilon _{i+1}<k_{0}+\varepsilon _{i}< k_{n} \quad i = 1, \ldots , j-1. \end{aligned}$$

(94)

On this basis, we can define:

$$\begin{aligned}&k_{i} = k_{0}-\varepsilon _{i} \quad i = 1, \ldots , j \end{aligned}$$

(95)

$$\begin{aligned}&k_{n-i} = k_{0}+\varepsilon _{i} \quad i = 1, \ldots , j \end{aligned}$$

(96)

$$\begin{aligned}&k* = \left\{ \begin{array}{lll} 0 &{} &{} \qquad n\; {\text{ odd }}.\\ k_{j+1} &{} = k_{0} &{} \qquad n \; {\text{ even }}.\\ \end{array}\right. \end{aligned}$$

(97)

Owing to the properties of combinations, the following equality will then obtain:

$$\begin{aligned} \displaystyle \sum _{{i= 1}}^{{n-1}}{\frac{n!}{i!(n-i)!}k_{i}}= k*+ \displaystyle \sum _{{i= 1}}^{{j}}{\frac{n!}{i!(n-i)!}(k_{i}+k_{n-i}})=k*+ \displaystyle \sum _{{i=1}}^{{j}}{\frac{n!}{i!(n-i)!}2k_{0}}. \end{aligned}$$

(98)

This in turn implies:

$$\begin{aligned} \displaystyle \sum _{{i= 1}}^{{n}}{\frac{n!}{i!(n-i)!}k_{i}} = k_{n}+k*+\displaystyle \sum _{{i=1}}^{{j}}{\frac{n!}{i!(n-i)!}2k_{0}} = (2^{n}-2)k_{0}+k_{n}. \end{aligned}$$

(99)

Given (91), this proves that the numbers $k_{i}$ ($i = 1, \ldots , n$) satisfy (57). Because of how they were selected, moreover, they satisfy condition (56).