Conjunction of Conditional Events and T-norms

TL;DR

This paper explores the connection between conjunctions of conditional events and Frank t-norms, demonstrating their equivalence in various cases and analyzing coherence conditions, especially for Lukasiewicz t-norm.

Contribution

It establishes a formal link between conjunctions of conditional events and Frank t-norms, including coherence conditions for Lukasiewicz t-norm.

Findings

Conjunctions coincide with Frank t-norms like Product, Minimum, and Lukasiewicz in coherence.

Counterexample shows Lukasiewicz t-norm may lead to incoherent assessments.

Conditions for coherence with Lukasiewicz t-norm are provided.

Abstract

We study the relationship between a notion of conjunction among conditional events, introduced in recent papers, and the notion of Frank t-norm. By examining different cases, in the setting of coherence, we show each time that the conjunction coincides with a suitable Frank t-norm. In particular, the conjunction may coincide with the Product t-norm, the Minimum t-norm, and Lukasiewicz t-norm. We show by a counterexample, that the prevision assessments obtained by Lukasiewicz t-norm may be not coherent. Then, we give some conditions of coherence when using Lukasiewicz t-norm.

Equations

$$(\Sigma)\quad\begin{array}[]{ll}\sum_{h=1}^{m}\lambda_{h}q_{hi}=\mu_{i}\,,\;i\in\{1,\ldots,n\}\,,\sum_{h=1}^{m}\lambda_{h}=1,\;\;\lambda_{h}\geq 0\,,\;\,h\in\{1,\ldots,m\}\,.\end{array}$$

$$\begin{array}[]{ll}I_{0}=\{i:\max_{\Lambda\in S}\sum_{h:C_{h}\subseteq H_{i}}\lambda_{h}=0\},\;\mathcal{F}_{0}=\{X_{i}|H_{i}\,,i\in I_{0}\},\;\;\mathcal{M}_{0}=(\mu_{i},\,i\in I_{0})\,.\end{array}$$

$$(A|H)\wedge(B|K)=\left\{\begin{array}[]{ll}1,&\mbox{if $AHBK$ is true,}\\ 0,&\mbox{if $\widebar{A}H\vee\widebar{B}K$ is true,}\\ x,&\mbox{if $\widebar{H}BK$ is true,}\\ y,&\mbox{if $AH\widebar{K}$ is true,}\\ z,&\mbox{if $\widebar{H}\,\widebar{K}$ is true}.\end{array}\right.$$

$$\max \{x+y-1,0\}=z^{\prime }\le z\le z^{\prime \prime }=\min \{x,y\}.$$

$$\Pi =\{(x,y,z):(x,y)\in [0,1{]}^2,\max \{x+y-1,0\}\le z\le \min \{x,y\}\}.$$

$$\begin{array}[]{lll}\mathcal{C}_{1\cdots n}=\left\{\begin{array}[]{llll}1,&\mbox{ if }\bigwedge_{i=1}^{n}E_{i}H_{i},\mbox{ is true}\\ 0,&\mbox{ if }\bigvee_{i=1}^{n}\widebar{E}_{i}H_{i},\mbox{ is true},\\ x_{S},&\mbox{ if }\bigwedge_{i\in S}\widebar{H}_{i}\bigwedge_{i\notin S}E_{i}{H}_{i}\,\mbox{ is true},\;\emptyset\neq S\subseteq\{1,2\ldots,n\}.\end{array}\right.\end{array}$$

$$\small\begin{array}[]{lll}\mathcal{C}_{123}=\left\{\begin{array}[]{llll}1,&\mbox{ if }E_{1}H_{1}E_{2}H_{2}E_{3}H_{3}\mbox{ is true},\\ 0,&\mbox{ if }\widebar{E}_{1}H_{1}\vee\widebar{E}_{2}H_{2}\vee\widebar{E}_{3}H_{3}\mbox{ is true},\\ x_{1},&\mbox{ if }\widebar{H}_{1}E_{2}H_{2}E_{3}H_{3}\mbox{ is true},\\ x_{2},&\mbox{ if }\widebar{H}_{2}E_{1}H_{1}E_{3}H_{3}\mbox{ is true},\\ x_{3},&\mbox{ if }\widebar{H}_{3}E_{1}H_{1}E_{2}H_{2}\mbox{ is true},\\ x_{12},&\mbox{ if }\widebar{H}_{1}\widebar{H}_{2}E_{3}H_{3}\mbox{ is true},\\ x_{13},&\mbox{ if }\widebar{H}_{1}\widebar{H}_{3}E_{2}H_{2}\mbox{ is true},\\ x_{23},&\mbox{ if }\widebar{H}_{2}\widebar{H}_{3}E_{1}H_{1}\mbox{ is true},\\ x_{123},&\mbox{ if }\widebar{H}_{1}\widebar{H}_{2}\widebar{H}_{3}\mbox{ is true}.\\ \end{array}\right.\end{array}$$

$$\small\left\{\begin{array}[]{l}(x_{1},x_{2},x_{3})\in[0,1]^{3},\\ \max\{x_{1}+x_{2}-1,x_{13}+x_{23}-x_{3},0\}\leq x_{12}\leq\min\{x_{1},x_{2}\},\\ \max\{x_{1}+x_{3}-1,x_{12}+x_{23}-x_{2},0\}\leq x_{13}\leq\min\{x_{1},x_{3}\},\\ \max\{x_{2}+x_{3}-1,x_{12}+x_{13}-x_{1},0\}\leq x_{23}\leq\min\{x_{2},x_{3}\},\\ 1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\geq 0,\\ x_{123}\geq\max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}-x_{3}\},\\ x_{123}\leq\min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}.\end{array}\right.$$

$$\begin{array}[]{ll}\max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}-x_{3}\}\,\;\leq\;x_{123}\;\leq\\ \leq\;\;\min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}.\end{array}$$

$$\begin{array}[]{ll}x_{123}^{\prime}=\max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}-x_{3}\},\\ x_{123}^{\prime\prime}=\min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}.\end{array}$$

$$\small T_{\lambda}(u,v)=\left\{\begin{array}[]{ll}T_{M}(u,v)=\min\{u,v\},&\text{ if }\lambda=0,\\ T_{P}(u,v)=uv,&\text{ if }\lambda=1,\\ T_{L}(u,v)=\max\{u+v-1,0\},&\text{ if }\lambda=+\infty,\\ \log_{\lambda}(1+\frac{(\lambda^{u}-1)(\lambda^{v}-1)}{\lambda-1}),&\text{ otherwise}.\end{array}\right.$$

$$\small T_{\lambda}(A|H,B|K)=\left\{\begin{array}[]{ll}1,&\mbox{ if $AHBK$ is true,}\\ 0,&\mbox{ if $\widebar{A}H$ is true or $\widebar{B}K$ is true,}\\ x,&\mbox{ if $\widebar{H}BK$ is true,}\\ y,&\mbox{ if $\widebar{K}AH$ is true,}\\ T_{\lambda}(x,y),&\mbox{ if $\widebar{H}\,\widebar{K}$ is true},\end{array}\right.$$

$$(A|H)\wedge (A|K)=AHK+x\text{widebar}HAK+y\text{widebar}KAH+z\text{widebar}H\text{widebar}K\in \{1,0,x,y,z\}.$$

$$\Pi =\{(x,y,z):(x,y)\in [0,1{]}^2,T_P(x,y)=xy\le z\le \min \{x,y\}=T_M(x,y)\}.$$

$$(\Sigma)\left\{\begin{array}[]{l}\lambda_{1}+x\lambda_{3}+x\lambda_{5}+\lambda_{6}=x,\;\;\lambda_{1}+y\lambda_{4}+\lambda_{5}+y\lambda_{6}=y,\;\;\lambda_{1}+x\lambda_{5}+y\lambda_{6}=z,\\ \sum_{h=1}^{6}\lambda_{h}=1,\;\;\lambda_{h}\geq 0,\;h=1,\ldots,6.\end{array}\right.$$

$$\begin{array}[]{l}\lambda_{1}+\lambda_{6}=0,\;\;\lambda_{1}+\lambda_{5}=0,\;\;\lambda_{1}=0,\;\lambda_{2}+\lambda_{3}+\lambda_{4}=1,\;\;\lambda_{h}\geq 0,\;\;h=1,\ldots,6.\end{array}$$

$$\small(A|H)\wedge(A|K)=\left\{\begin{array}[]{ll}0,&\mbox{ if }\widebar{A}\widebar{H}K\vee\widebar{A}H\widebar{K}\mbox{ is true,}\\ x,&\mbox{ if }\widebar{H}AK\mbox{ is true,}\\ y,&\mbox{ if }AH\widebar{K}\mbox{ is true,}\\ z,&\mbox{ if }\widebar{H}\widebar{K}\mbox{ is true.}\\ \end{array}\right.$$

$$\begin{array}[]{ll}\max\{0,x_{1}(x_{2}+x_{3}-1),x_{2}(x_{1}+x_{3}-1),x_{3}(x_{1}+x_{2}-1)\}\,\;\leq\;x_{1}x_{2}x_{3}\;\leq\\ \leq\;\;\min\{x_{1}x_{2},x_{1}x_{3},x_{2}x_{3},(1-x_{1})(1-x_{2})(1-x_{3})+x_{1}x_{2}x_{3}\}.\end{array}$$

$$\begin{array}[]{ll}\max\{0,x_{1},x_{1}+x_{2}-x_{3}\}=x_{1}\,\;\leq\;x_{1}\;\leq x_{1}=\min\{x_{1},x_{2},1-x_{3}+x_{1}\}.\end{array}$$

Full text

1 Introduction

In this paper we use the coherence-based approach to probability of de Finetti ([1, 2, 7, 9, 10, 13, 14, 16, 17, 18, 22, 25]). We use a notion of conjunction which, differently from other authors, is defined as a suitable conditional random quantity with values in the unit interval (see, e.g. [20, 21, 23, 24, 36]). We study the relationship between our notion of conjunction and the notion of Frank t-norm. For some aspects which relate probability and Frank t-norm see, e.g., [5, 6, 8, 11, 15, 34]. We show that, under the hypothesis of logical independence, if the prevision assessments involved with the conjunction $(A|H)\wedge(B|K)$ of two conditional events are coherent, then the prevision of the conjunction coincides, for a suitable $\lambda\in[0,+\infty]$, with the Frank t-norm $T_{\lambda}(x,y)$, where $x=P(A|H),y=P(B|K)$. Moreover, $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$. Then, we consider the case $A=B$, by determining the set of all coherent assessment $(x,y,z)$ on $\{A|H,A|K,(A|H)\wedge(A|K)\}$. We show that, under coherence, it holds that $(A|H)\wedge(A|K)=T_{\lambda}(A|H,A|K)$, where $\lambda\in[0,1]$. We also study the particular case where $A=B$ and $HK=\emptyset$. Then, we consider conjunctions of three conditional events and we show that to make prevision assignments by means of the Product t-norm, or the Minimum t-norm, is coherent. Finally, we examine the Lukasiewicz t-norm and we show by a counterexample that coherence is in general not assured. We give some conditions for coherence when the prevision assessments are made by using the Lukasiewicz t-norm.

2 Preliminary Notions and Results

In our approach, given two events $A$ and $H$, with $H\neq\emptyset$, the conditional event $A|H$ is looked at as a three-valued logical entity which is true, or false, or void, according to whether $AH$ is true, or $\widebar{A}H$ is true, or $\widebar{H}$ is true. We observe that the conditional probability and/or conditional prevision values are assessed in the setting of coherence-based probabilistic approach. In numerical terms $A|H$ assumes one of the values $1$, or [math], or $x$, where $x=P(A|H)$ represents the assessed degree of belief on $A|H$. Then, $A|H=AH+x\widebar{H}$. Given a family $\mathcal{F}=\{X_{1}|H_{1},\ldots,X_{n}|H_{n}\}$, for each $i\in\{1,\ldots,n\}$ we denote by $\{x_{i1},\ldots,x_{ir_{i}}\}$ the set of possible values of $X_{i}$ when $H_{i}$ is true; then, for each $i$ and $j=1,\ldots,r_{i}$, we set $A_{ij}=(X_{i}=x_{ij})$. We set $C_{0}=\widebar{H}_{1}\cdots\widebar{H}_{n}$ (it may be $C_{0}=\emptyset$); moreover, we denote by $C_{1},\ldots,C_{m}$ the constituents contained in $H_{1}\vee\cdots\vee H_{n}$. Hence $\bigwedge_{i=1}^{n}(A_{i1}\vee\cdots\vee A_{ir_{i}}\vee\widebar{H}_{i})=\bigvee_{h=0}^{m}C_{h}$. With each $C_{h},\,h\in\{1,\ldots,m\}$, we associate a vector $Q_{h}=(q_{h1},\ldots,q_{hn})$, where $q_{hi}=x_{ij}$ if $C_{h}\subseteq A_{ij},\,j=1,\ldots,r_{i}$, while $q_{hi}=\mu_{i}$ if $C_{h}\subseteq\widebar{H}_{i}$; with $C_{0}$ it is associated $Q_{0}=\mathcal{M}=(\mu_{1},\ldots,\mu_{n})$. Denoting by $\mathcal{I}$ the convex hull of $Q_{1},\ldots,Q_{m}$, the condition $\mathcal{M}\in\mathcal{I}$ amounts to the existence of a vector $(\lambda_{1},\ldots,\lambda_{m})$ such that: $\sum_{h=1}^{m}\lambda_{h}Q_{h}=\mathcal{M}\,,\;\sum_{h=1}^{m}\lambda_{h}=1\,,\;\lambda_{h}\geq 0\,,\;\forall\,h$; in other words, $\mathcal{M}\in\mathcal{I}$ is equivalent to the solvability of the system $(\Sigma)$, associated with $(\mathcal{F},\mathcal{M})$,

[TABLE]

Given the assessment $\mathcal{M}=(\mu_{1},\ldots,\mu_{n})$ on $\mathcal{F}=\{X_{1}|H_{1},\ldots,X_{n}|H_{n}\}$, let $S$ be the set of solutions $\Lambda=(\lambda_{1},\ldots,\lambda_{m})$ of system $(\Sigma)$. We point out that the solvability of system $(\Sigma)$ is a necessary (but not sufficient) condition for coherence of $\mathcal{M}$ on $\mathcal{F}$. When $(\Sigma)$ is solvable, that is $S\neq\emptyset$, we define:

[TABLE]

For what concerns the probabilistic meaning of $I_{0}$, it holds that $i\in I_{0}$ if and only if the (unique) coherent extension of $\mathcal{M}$ to $H_{i}|(\bigvee_{j=1}^{n}H_{j})$ is zero. Then, the following theorem can be proved ([3, Theorem 3])

Theorem 2.1

[Operative characterization of coherence] A conditional prevision assessment ${\mathcal{M}}=(\mu_{1},\ldots,\mu_{n})$ on the family $\mathcal{F}=\{X_{1}|H_{1},\ldots,X_{n}|H_{n}\}$ is coherent if and only if the following conditions are satisfied:

(i) the system $(\Sigma)$ defined in (1) is solvable; (ii) if $I_{0}\neq\emptyset$, then $\mathcal{M}_{0}$ is coherent. **

Coherence can be related to proper scoring rules ([4, 19, 30, 31, 32]).

Definition 1

Given any pair of conditional events $A|H$ and $B|K$, with $P(A|H)=x$ and $P(B|K)=y$, their conjunction is the conditional random quantity $(A|H)\wedge(B|K)$, with $\mathbb{P}[(A|H)\wedge(B|K)]=z$, defined as

[TABLE]

In betting terms, the prevision $z$ represents the amount you agree to pay, with the proviso that you will receive the quantity $(A|H)\wedge(B|K)$. Different approaches to compounded conditionals, not based on coherence, have been developed by other authors (see, e.g., [27, 33]). We recall a result which shows that Fréchet-Hoeffding bounds still hold for the conjunction of conditional events ([23, Theorem 7]).

Theorem 2.2

Given any coherent assessment $(x,y)$ on $\{A|H,B|K\}$, with $A,H,B$, $K$ logically independent, $H\neq\emptyset,K\neq\emptyset$, the extension $z=\mathbb{P}[(A|H)\wedge(B|K)]$ is coherent if and only if the following Fréchet-Hoeffding bounds are satisfied:

[TABLE]

Remark 1

From Theorem 2.2, as the assessment $(x,y)$ on $\{A|H,B|K\}$ is coherent for every $(x,y)\in[0,1]^{2}$, the set $\Pi$ of coherent assessments $(x,y,z)$ on $\{A|H,B|K,(A|H)\wedge(B|K)\}$ is

[TABLE]

The set $\Pi$ is the tetrahedron with vertices the points $(1,1,1)$, $(1,0,0)$, $(0,1,0)$, $(0,0,0)$. For other definition of conjunctions, where the conjunction is a conditional event, some results on lower and upper bounds have been given in [35].

Definition 2

Let be given $n$ conditional events $E_{1}|H_{1},\ldots,E_{n}|H_{n}$. For each subset $S$, with $\emptyset\neq S\subseteq\{1,\ldots,n\}$, let $x_{S}$ be a prevision assessment on $\bigwedge_{i\in S}(E_{i}|H_{i})$. The conjunction $\mathcal{C}_{1\cdots n}=(E_{1}|H_{1})\wedge\cdots\wedge(E_{n}|H_{n})$ is defined as

[TABLE]

In particular, $\mathcal{C}_{1}=E_{1}|H_{1}$; moreover, for $\mathcal{S}=\{i_{1},\ldots,i_{k}\}\subseteq\{1,\ldots,n\}$, the conjunction $\bigwedge_{i\in S}(E_{i}|H_{i})$ is denoted by $\mathcal{C}_{i_{1}\cdots i_{k}}$ and $x_{\mathcal{S}}$ is also denoted by $x_{i_{1}\cdots i_{k}}$. Moreover, if $\mathcal{S}=\{i_{1},\ldots,i_{k}\}\subseteq\{1,\ldots,n\}$, the conjunction $\bigwedge_{i\in S}(E_{i}|H_{i})$ is denoted by $\mathcal{C}_{i_{1}\cdots i_{k}}$ and $x_{\mathcal{S}}$ is also denoted by $x_{i_{1}\cdots i_{k}}$. In the betting framework, you agree to pay $x_{1\cdots n}=\mathbb{P}(\mathcal{C}_{1\cdots n})$ with the proviso that you will receive: $1$, if all conditional events are true; [math], if at least one of the conditional events is false; the prevision of the conjunction of that conditional events which are void, otherwise. The operation of conjunction is associative and commutative. We observe that, based on Definition 2, when $n=3$ we obtain

[TABLE]

We recall the following result ([24, Theorem 15]).

Theorem 2.3

Assume that the events $E_{1},E_{2},E_{3},H_{1},H_{2},H_{3}$ are logically independent, with $H_{1}\neq\emptyset,H_{2}\neq\emptyset,H_{3}\neq\emptyset$. Then, the set $\Pi$ of all coherent assessments $\mathcal{M}=(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ on $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$ is the set of points $(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ which satisfy the following conditions

[TABLE]

Remark 2

As shown in (8), the coherence of $(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ amounts to the condition

[TABLE]

Then, in particular, the extension $x_{123}$ on $\mathcal{C}_{123}$ is coherent if and only if $x_{123}\in[x_{123}^{\prime},x_{123}^{\prime\prime}]$, where $x_{123}^{\prime}=\max\{0,x_{12}+x_{13}-x_{1},x_{12}+x_{23}-x_{2},x_{13}+x_{23}-x_{3}\}$, $x_{123}^{\prime\prime}=\min\{x_{12},x_{13},x_{23},1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\}.$

Then, by Theorem 2.3 it follows [24, Corollary 1]

Corollary 1

For any coherent assessment $(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23})$ on $\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23}\}$ the extension $x_{123}$ on $\mathcal{C}_{123}$ is coherent if and only if $x_{123}\in[x_{123}^{\prime},x_{123}^{\prime\prime}]$, where

[TABLE]

We recall that in case of logical dependencies, the set of all coherent assessments may be smaller than that one associated with the case of logical independence. However (see [24, Theorem 16]) the set of coherent assessments is the same when $H_{1}=H_{2}=H_{3}=H$ (where possibly $H=\Omega$; see also [26, p. 232]) and a corollary similar to Corollary 1 also holds in this case. For a similar result based on copulas see [12].

3 Representation by Frank t-norms for $(A|H)\wedge(B|K)$

We recall that for every $\lambda\in[0,+\infty]$ the Frank t-norm $T_{\lambda}:[0,1]^{2}\rightarrow[0,1]$ with parameter $\lambda$ is defined as

[TABLE]

We recall that $T_{\lambda}$ is continuous with respect to $\lambda$; moreover, for every $\lambda\in[0,+\infty]$, it holds that $T_{L}(u,v)\leq T_{\lambda}(u,v)\leq T_{M}(u,v)$, for every $(u,v)\in[0,1]^{2}$ (see, e.g., [28],[29]). In the next result we study the relation between our notion of conjunction and t-norms.

Theorem 3.1

Let us consider the conjunction $(A|H)\wedge(B|K)$, with $A,B,H,K$ logically independent and with $P(A|H)=x$, $P(B|K)=y$. Moreover, given any $\lambda\in[0,+\infty]$, let $T_{\lambda}$ be the Frank t-norm with parameter $\lambda$. Then, the assessment $z=T_{\lambda}(x,y)$ on $(A|H)\wedge(B|K)$ is a coherent extension of $(x,y)$ on $\{A|H,B|K\}$; moreover $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$. Conversely, given any coherent extension $z=\mathbb{P}[(A|H)\wedge(B|K)]$ of $(x,y)$, there exists $\lambda\in[0,+\infty]$ such that $z=T_{\lambda}(x,y)$.

Proof

We observe that from Theorem 2.2, for any given $\lambda$, the assessment $z=T_{\lambda}(x,y)$ is a coherent extension of $(x,y)$ on $\{A|H,B|K\}$. Moreover, from (11) it holds that $T_{\lambda}(1,1)=1$, $T_{\lambda}(u,0)=T_{\lambda}(0,v)=0$, $T_{\lambda}(u,1)=u$, $T_{\lambda}(1,v)=v$. Hence,

[TABLE]

and, if we choose $z=T_{\lambda}(x,y)$, from (3) and (12) it follows that $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$.

Conversely, given any coherent extension $z$ of $(x,y)$, there exists $\lambda$ such that $z=T_{\lambda}(x,y)$. Indeed, if $z=\min\{x,y\}$, then $\lambda=0$; if $z=\max\{x+y-1,0\}$, then $\lambda=+\infty$; if $\max\{x+y-1,0\}<z<\min\{x,y\}$, then by continuity of $T_{\lambda}$ with respect to $\lambda$ it holds that $z=T_{\lambda}(x,y)$ for some $\lambda\in\,]0,\infty[$ (for instance, if $z=xy$, then $z=T_{1}(x,y)$) and hence $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$. ∎

Remark 3

As we can see from (3) and Theorem 3.1, in case of logically independent events, if the assessed values $x,y,z$ are such that $z=T_{\lambda}(x,y)$ for a given $\lambda$, then the conjunction $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$. For instance, if $z=T_{1}(x,y)=xy$, then $(A|H)\wedge(B|K)=T_{1}(A|H,B|K)=(A|H)\cdot(B|K)$. Conversely, if $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$ for a given $\lambda$, then $z=T_{\lambda}(x,y)$. Then, the set $\Pi$ given in (5) can be written as $\Pi=\{(x,y,z):(x,y)\in[0,1]^{2},z=T_{\lambda}(x,y),\lambda\in[0,+\infty]\}.$

4 Conjunction of $(A|H)$ and $(A|K)$

In this section we examine the conjunction of two conditional events in the particular case when $A=B$, that is $(A|H)\wedge(A|K)$. By setting $P(A|H)=x$, $P(A|K)=y$ and $\mathbb{P}[(A|H)\wedge(A|K)]=z$, it holds that

[TABLE]

Theorem 4.1

Let $A,H,K$ be three logically independent events, with $H\neq\emptyset$, $K\neq\emptyset$. The set $\Pi$ of all coherent assessments $(x,y,z)$ on the family $\mathcal{F}=\{A|H,A|K,(A|H)\wedge(A|K)\}$ is given by

[TABLE]

Proof

Let $\mathcal{M}=(x,y,z)$ be a prevision assessment on $\mathcal{F}$. The constituents associated with the pair $(\mathcal{F},\mathcal{M})$ and contained in $H\vee K$ are: $C_{1}=AHK$, $C_{2}=\widebar{A}HK$, $C_{3}=\widebar{A}\widebar{H}K$, $C_{4}=\widebar{A}H\widebar{K}$, $C_{5}=A\widebar{H}K$, $C_{6}=AH\widebar{K}$. The associated points $Q_{h}$’s are $Q_{1}=(1,1,1),Q_{2}=(0,0,0),Q_{3}=(x,0,0),Q_{4}=(0,y,0),Q_{5}=(x,1,x),Q_{6}=(1,y,y)$. With the further constituent $C_{0}=\widebar{H}\widebar{K}$ it is associated the point $Q_{0}=\mathcal{M}=(x,y,z)$. Considering the convex hull $\mathcal{I}$ (see Figure 1) of $Q_{1},\ldots,Q_{6}$, a necessary condition for the coherence of the prevision assessment $\mathcal{M}=(x,y,z)$ on $\mathcal{F}$ is that $\mathcal{M}\in\mathcal{I}$, that is the following system must be solvable

[TABLE]

First of all, we observe that solvability of $(\Sigma)$ requires that $z\leq x$ and $z\leq y$, that is $z\leq\min\{x,y\}$. We now verify that $(x,y,z)$, with $(x,y)\in[0,1]^{2}$ and $z=\min\{x,y\}$, is coherent. We distinguish two cases: $(i)$ $x\leq y$ and $(ii)$ $x>y$.

Case $(i)$. In this case $z=\min\{x,y\}=x$. If $y=0$ the system $(\Sigma)$ becomes

[TABLE]

which is clearly solvable. In particular there exist solutions with $\lambda_{2}>0,\lambda_{3}>0,\lambda_{4}>0$, by Theorem 2.1, as the set $I_{0}$ is empty the solvability of $(\Sigma)$ is sufficient for coherence of the assessment $(0,0,0)$. If $y>0$ the system $(\Sigma)$ is solvable and a solution is $\Lambda=(\lambda_{1},\ldots,\lambda_{6})=(x,\frac{x(1-y)}{y},0,\frac{y-x}{y},0,0)$. We observe that, if $x>0$, then $\lambda_{1}>0$ and $I_{0}=\emptyset$ because $\mathcal{C}_{1}=HK\subseteq H\vee K$, so that $\mathcal{M}=(x,y,x)$ is coherent. If $x=0$ (and hence $z=0$), then $\lambda_{4}=1$ and $I_{0}\subseteq\{2\}$. Then, as the sub-assessment $P(A|K)=y$ is coherent, it follows that the assessment $\mathcal{M}=(0,y,0)$ is coherent too.

Case $(ii)$. The system is solvable and a solution is $\Lambda=(\lambda_{1},\ldots,\lambda_{6})=(y,\frac{y(1-x)}{x},\frac{x-y}{x},0,0,0).$ We observe that, if $y>0$, then $\lambda_{1}>0$ and $I_{0}=\emptyset$ because $\mathcal{C}_{1}=HK\subseteq H\vee K$, so that $\mathcal{M}=(x,y,y)$ is coherent. If $y=0$ (and hence $z=0$), then $\lambda_{3}=1$ and $I_{0}\subseteq\{1\}$. Then, as the sub-assessment $P(A|H)=x$ is coherent, it follows that the assessment $\mathcal{M}=(x,0,0)$ is coherent too. Thus, for every $(x,y)\in[0,1]^{2}$, the assessment $(x,y,\min\{x,y\})$ is coherent and, as $z\leq\min\{x,y\}$, the upper bound on $z$ is $\min\{x,y\}=T_{M}(x,y)$.

We now verify that $(x,y,xy)$, with $(x,y)\in[0,1]^{2}$ is coherent; moreover we will show that $(x,y,z)$, with $z<xy$, is not coherent, in other words the lower bound for $z$ is $xy$. First of all, we observe that $\mathcal{M}=(1-x)Q_{4}+xQ_{6}$, so that a solution of $(\Sigma)$ is $\Lambda_{1}=(0,0,0,1-x,0,x)$. Moreover, $\mathcal{M}=(1-y)Q_{3}+yQ_{5}$, so that another solution is $\Lambda_{2}=(0,0,1-y,0,y,0)$. Then $\Lambda=\frac{\Lambda_{1}+\Lambda_{2}}{2}=(0,0,\frac{1-y}{2},\frac{1-x}{2},\frac{y}{2},\frac{x}{2})$ is a solution of $(\Sigma)$ such that $I_{0}=\emptyset$. Thus the assessment $(x,y,xy)$ is coherent for every $(x,y)\in[0,1]^{2}$. In order to verify that $xy$ is the lower bound on $z$ we observe that the points $Q_{3},Q_{4},Q_{5},Q_{6}$ belong to a plane $\pi$ of equation: $yX+xY-Z=xy$, where $X,Y,Z$ are the axis’ coordinates. Now, by considering the function $f(X,Y,Z)=yX+xY-Z$, we observe that for each constant $k$ the equation $f(X,Y,Z)=k$ represents a plane which is parallel to $\pi$ and coincides with $\pi$ when $k=xy$. We also observe that $f(Q_{1})=f(1,1,1)=x+y-1=T_{L}(x,y)\leq xy=T_{P}(x,y)$, $f(Q_{2})=f(0,0,0)=0\leq xy=T_{P}(x,y)$, and $f(Q_{3})=f(Q_{4})=f(Q_{5})=f(Q_{6})=xy=T_{P}(x,y)$. Then, for every $\mathcal{P}=\sum_{h=1}^{6}\lambda_{h}Q_{h}$, with $\lambda_{h}\geq 0$ and $\sum_{h=1}^{6}\lambda_{h}=1$, that is $\mathcal{P}\in\mathcal{I}$, it holds that f(\mathcal{P})=f\big{(}\sum_{h=1}^{6}\lambda_{h}Q_{h}\big{)}=\sum_{h=1}^{6}\lambda_{h}f(Q_{h})\leq xy. On the other hand, given any $a>0$, by considering $\mathcal{P}=(x,y,xy-a)$ it holds that $f(\mathcal{P})=f(x,y,xy-a)=xy+xy-xy+a=xy+a>xy.$ Therefore, for any given $a>0$ the assessment $(x,y,xy-a)$ is not coherent because $(x,y,xy-a)\notin\mathcal{I}$. Then, the lower bound on $z$ is $xy=T_{P}(x,y)$. Finally, the set of all coherent assessments $(x,y,z)$ on $\mathcal{F}$ is the set $\Pi$ in (13). ∎

Based on Theorem 4.1, we can give an analogous version for the Theorem 3.1 (when $A=B$).

Theorem 4.2

Let us consider the conjunction $(A|H)\wedge(A|K)$, with $A,H,K$ logically independent and with $P(A|H)=x$, $P(A|K)=y$. Moreover, given any $\lambda\in[0,1]$, let $T_{\lambda}$ be the Frank t-norm with parameter $\lambda$. Then, the assessment $z=T_{\lambda}(x,y)$ on $(A|H)\wedge(A|K)$ is a coherent extension of $(x,y)$ on $\{A|H,A|K\}$; moreover $(A|H)\wedge(A|K)=T_{\lambda}(A|H,A|K)$. Conversely, given any coherent extension $z=\mathbb{P}[(A|H)\wedge(A|K)]$ of $(x,y)$, there exists $\lambda\in[0,1]$ such that $z=T_{\lambda}(x,y)$.

The next result follows from Theorem 4.1 when $H$, $K$ are incompatible.

Theorem 4.3

Let $A,H,K$ be three events, with $A$ logically independent from both $H$ and $K$, with $H\neq\emptyset$, $K\neq\emptyset$, $HK=\emptyset$. The set $\Pi$ of all coherent assessments $(x,y,z)$ on the family $\mathcal{F}=\{A|H,A|K,(A|H)\wedge(A|K)\}$ is given by $\Pi=\{(x,y,z):(x,y)\in[0,1]^{2},z=xy=T_{P}(x,y)\}.$

Proof

We observe that

[TABLE]

Moreover, as $HK=\emptyset$, the points $Q_{h}$’s are $(x,0,0),(0,y,0),(x,1,x),(1,y,y)$, which coincide with the points $Q_{3},\ldots,Q_{6}$ of the case $HK\neq\emptyset$. Then, as shown in the proof of Theorem 4.1, the condition $\mathcal{M}=(x,y,z)$ belongs to the convex hull of $(x,0,0),(0,y,0),(x,1,x),(1,y,y)$ amounts to the condition $z=xy$. ∎

Remark 4

From Theorem 4.3, when $HK=\emptyset$ it holds that $(A|H)\wedge(A|K)=(A|H)\cdot(A|K)=T_{P}(A|H,A|K),$ where $x=P(A|H)$ and $y=P(A|K)$.

5 Further Results on Frank t-norms

In this section we give some results which concern Frank t-norms and the family $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$. We recall that, given any t-norm $T(x_{1},x_{2})$ it holds that $T(x_{1},x_{2},x_{3})=T(T(x_{1},x_{2}),x_{3})$.

5.1 On the Product t-norm

Theorem 5.1

Assume that the events $E_{1},E_{2},E_{3},H_{1},H_{2},H_{3}$ are logically independent, with $H_{1}\neq\emptyset,H_{2}\neq\emptyset,H_{3}\neq\emptyset$. If the assessment $\mathcal{M}=(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ on $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$ is such that $(x_{1},x_{2},x_{3})\in[0,1]^{3}$, $x_{ij}=T_{1}(x_{i},x_{j})=x_{i}x_{j}$, $i\neq j$, and $x_{123}=T_{1}(x_{1},x_{2},x_{3})=x_{1}x_{2}x_{3}$, then $\mathcal{M}$ is coherent. Moreover, $\mathcal{C}_{ij}=T_{1}(\mathcal{C}_{i},\mathcal{C}_{j})=\mathcal{C}_{i}\mathcal{C}_{j}$, $i\neq j$, and $\mathcal{C}_{123}=T_{1}(\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3})=\mathcal{C}_{1}\mathcal{C}_{2}\mathcal{C}_{3}$.

Proof

From Remark 2, the coherence of $\mathcal{M}$ amounts to the inequalities in (9). As $x_{ij}=T_{1}(x_{i},x_{j})=x_{i}x_{j}$, $i\neq j$, and $x_{123}=T_{1}(x_{1},x_{2},x_{3})=x_{1}x_{2}x_{3}$, the inequalities (9) become

[TABLE]

Thus, by recalling that $x_{i}+x_{j}-1\leq x_{i}x_{j}$, the inequalities are satisfied and hence $\mathcal{M}$ is coherent. Moreover, from (3) and (7) it follows that $\mathcal{C}_{ij}=T_{1}(\mathcal{C}_{i},\mathcal{C}_{j})=\mathcal{C}_{i}\mathcal{C}_{j}$, $i\neq j$, and $\mathcal{C}_{123}=T_{1}(\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3})=\mathcal{C}_{1}\mathcal{C}_{2}\mathcal{C}_{3}$.∎

5.2 On the Minimum t-norm

Theorem 5.2

Assume that the events $E_{1},E_{2},E_{3},H_{1},H_{2},H_{3}$ are logically independent, with $H_{1}\neq\emptyset,H_{2}\neq\emptyset,H_{3}\neq\emptyset$. If the assessment $\mathcal{M}=(x_{1},x_{2},x_{3},x_{12},x_{13},x_{23},x_{123})$ on $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$ is such that $(x_{1},x_{2},x_{3})\in[0,1]^{3}$, $x_{ij}=T_{M}(x_{i},x_{j})=\min\{x_{i},x_{j}\}$, $i\neq j$, and $x_{123}=T_{M}(x_{1},x_{2},x_{3})=\min\{x_{1},x_{2},x_{3}\}$, then $\mathcal{M}$ is coherent. Moreover, $\mathcal{C}_{ij}=T_{M}(\mathcal{C}_{i},\mathcal{C}_{j})=\min\{\mathcal{C}_{i},\mathcal{C}_{j}\}$, $i\neq j$, and $\mathcal{C}_{123}=T_{M}(\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3})=\min\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}\}$.

Proof

From Remark 2, the coherence of $\mathcal{M}$ amounts to the inequalities in (9). Without loss of generality, we assume that $x_{1}\leq x_{2}\leq x_{3}$. Then $x_{12}=T_{M}(x_{1},x_{2})=x_{1}$, $x_{13}=T_{M}(x_{1},x_{3})=x_{1}$, $x_{23}=T_{M}(x_{2},x_{3})=x_{2}$, and $x_{123}=T_{M}(x_{1},x_{2},x_{3})=x_{1}$. The inequalities (9) become

[TABLE]

Thus, the inequalities are satisfied and hence $\mathcal{M}$ is coherent. Moreover, from (3) and (7) it follows that $\mathcal{C}_{ij}=T_{M}(\mathcal{C}_{i},\mathcal{C}_{j})=\min\{\mathcal{C}_{i},\mathcal{C}_{j}\}$, $i\neq j$, and $\mathcal{C}_{123}=T_{M}(\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3})=\min\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3}\}$. ∎

Remark 5

As we can see from $(\ref{EQ:MIN})$ and Corollary 1, the assessment $x_{123}=\min\{x_{1},x_{2},x_{3}\}$ is the unique coherent extension on $\mathcal{C}_{123}$ of the assessment $(x_{1},x_{2},x_{3},\min\{x_{1},x_{2}\},\min\{x_{1},x_{3}\},\min\{x_{2},x_{3}\})$ on $\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23}\}$.

We also notice that, if $\mathcal{C}_{1}\leq\mathcal{C}_{2}\leq\mathcal{C}_{3}$, then $\mathcal{C}_{12}=\mathcal{C}_{1}$, $\mathcal{C}_{13}=\mathcal{C}_{1}$, $\mathcal{C}_{23}=\mathcal{C}_{2}$, and $\mathcal{C}_{123}=\mathcal{C}_{1}$. Moreover, $x_{12}=x_{1}$, $x_{13}=x_{1}$, $x_{23}=x_{2}$, and $x_{123}=x_{1}$.

5.3 On Lukasiewicz t-norm

We observe that in general the results of Theorems 5.1 and 5.2 do not hold for the Lukasiewicz t-norm (and hence for any given Frank t-norm), as shown in the example below. We recall that $T_{L}(x_{1},x_{2},x_{3})=\max\{x_{1}+x_{2}+x_{3}-2,0\}$.

Example 1

The assessment $(x_{1},x_{2},x_{3},T_{L}(x_{1},x_{2}),T_{L}(x_{1},x_{3}),T_{L}(x_{2},x_{3})$, $T_{L}(x_{1},x_{2},x_{3}))$ on the family $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$, with $(x_{1},x_{2},x_{3})=(0.5,0.6,0.7)$ is not coherent. Indeed, by observing that $T_{L}(x_{1},x_{2})=0.1$ $T_{L}(x_{1},x_{3})=0.2$, $T_{L}(x_{2},x_{3})=0.3$, and $T_{L}(x_{1},x_{2},x_{3})=0$, formula (9) becomes $\max\{0,0.1+0.2-0.5,0.1+0.3-0.6,0.2+0.3-0.7\}\,\;\leq\;0\;\leq\;\;\min\{0.1,0.2,0.3,1-0.5-0.6-0.7+0.1+0.2+0.3\}$, that is: $\max\{0,-0.2\}\,\;\leq\;0\;\leq\min\{0.1,0.2,0.3,-0.2\};$ thus the inequalities are not satisfied and the assessment is not coherent.

More in general we have

Theorem 5.3

The assessment $(x_{1},x_{2},x_{3},T_{L}(x_{1},x_{2}),T_{L}(x_{1},x_{3}),T_{L}(x_{2},x_{3}))$ on the family $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23}\}$, with $T_{L}(x_{1},x_{2})>0$, ${T_{L}(x_{1},x_{3})>0}$, ${T_{L}(x_{2},x_{3})>0}$ is coherent if and only if $x_{1}+x_{2}+x_{3}-2\geq 0$. Moreover, when $x_{1}+x_{2}+x_{3}-2\geq 0$ the unique coherent extension $x_{123}$ on $\mathcal{C}_{123}$ is $x_{123}=T_{L}(x_{1},x_{2},x_{3})$.

Proof

We distinguish two cases: $(i)$ $x_{1}+x_{2}+x_{3}-2<0$; $(ii)$ $x_{1}+x_{2}+x_{3}-2\geq 0$.

Case $(i)$. From (8) the inequality $1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}\geq 0$ is not satisfied because $1-x_{1}-x_{2}-x_{3}+x_{12}+x_{13}+x_{23}=x_{1}+x_{2}+x_{3}-2<0.$ Therefore the assessment is not coherent.

Case $(ii)$. We set $x_{123}=T_{L}(x_{1},x_{2},x_{3})=x_{1}+x_{2}+x_{3}-2$. Then, by observing that $0<x_{i}+x_{j}-1\leq x_{1}+x_{2}+x_{3}-2$, $i\neq j$, formula (9) becomes $\max\{0,x_{1}+x_{2}+x_{3}-2\}\,\;\leq\;x_{1}+x_{2}+x_{3}-2\;\leq\;\;\min\{x_{1}+x_{2}-1,x_{1}+x_{3}-1,x_{2}+x_{3}-1,x_{1}+x_{2}+x_{3}-2\}$, that is: $\;x_{1}+x_{2}+x_{3}-2\;\leq\;x_{1}+x_{2}+x_{3}-2\;\leq\;x_{1}+x_{2}+x_{3}-2.$ Thus, the inequalities are satisfied and the assessment $(x_{1},x_{2},x_{3},T_{L}(x_{1},x_{2}),T_{L}(x_{1},x_{3}),T_{L}(x_{2},x_{3}),T_{L}(x_{1},x_{2},x_{3}))$ on $\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$ is coherent and the sub-assessment $(x_{1},x_{2},x_{3},T_{L}(x_{1},x_{2}),T_{L}(x_{1},x_{3}),T_{L}(x_{2},x_{3}))$ on $\mathcal{F}$ is coherent too. ∎

A result related with Theorem 5.3 is given below.

Theorem 5.4

If the assessment $(x_{1},x_{2},x_{3},T_{L}(x_{1},x_{2}),T_{L}(x_{1},x_{3}),T_{L}(x_{2},x_{3})$, $T_{L}(x_{1},x_{2},x_{3}))$ on the family $\mathcal{F}=\{\mathcal{C}_{1},\mathcal{C}_{2},\mathcal{C}_{3},\mathcal{C}_{12},\mathcal{C}_{13},\mathcal{C}_{23},\mathcal{C}_{123}\}$, is such that $T_{L}(x_{1},x_{2},x_{3})>0$, then the assessment is coherent.

Proof

We observe that $T_{L}(x_{1},x_{2},x_{3})=x_{1}+x_{2}+x_{3}-2>0$; then $x_{i}>0$, $i=1,2,3$, and $0<x_{i}+x_{j}-1\leq x_{1}+x_{2}+x_{3}-2$, $i\neq j$. Then formula (9)) becomes: $\;\;\max\{0,x_{1}+x_{2}+x_{3}-2\}\,\;\leq\;x_{1}+x_{2}+x_{3}-2\;\leq\\ \leq\;\;\min\{x_{1}+x_{2}-1,x_{1}+x_{3}-1,x_{2}+x_{3}-1,x_{1}+x_{2}+x_{3}-2\},$ that is:

$x_{1}+x_{2}+x_{3}-2\;\leq\;x_{1}+x_{2}+x_{3}-2\;\leq\;x_{1}+x_{2}+x_{3}-2.$

Thus, the inequalities are satisfied and the assessment is coherent. ∎

6 Conclusions

We have studied the relationship between the notions of conjunction and of Frank t-norms. We have shown that, under logical independence of events and coherence of prevision assessments, for a suitable $\lambda\in[0,+\infty]$ it holds that $\mathbb{P}((A|H)\wedge(B|K))=T_{\lambda}(x,y)$ and $(A|H)\wedge(B|K)=T_{\lambda}(A|H,B|K)$. Then, we have considered the case $A=B$, by determining the set of all coherent assessment $(x,y,z)$ on $(A|H,B|K,(A|H)\wedge(A|K))$. We have shown that, under coherence, for a suitable $\lambda\in[0,1]$ it holds that $(A|H)\wedge(A|K)=T_{\lambda}(A|H,A|K)$. We have also studied the particular case where $A=B$ and $HK=\emptyset$. Then, we have considered the conjunction of three conditional events and we have shown that the prevision assessments produced by the Product t-norm, or the Minimum t-norm, are coherent. Finally, we have examined the Lukasiewicz t-norm and we have shown, by a counterexample, that coherence in general is not assured. We have given some conditions for coherence when the prevision assessments are based on the Lukasiewicz t-norm. Future work should concern the deepening and generalization of the results of this paper.

 

Acknowledgments. We thank three anonymous referees for their useful comments.