Conditionals, Conditional Probabilities, and Conditionalization

Abstract

Philosophers investigating the interpretation and use of conditional sentences have long been intrigued by the intuitive correspondence between the probability of a conditional ‘if A, then C’ and the conditional probability of C, given A. Attempts to account for this intuition within a general probabilistic theory of belief, meaning and use have been plagued by a danger of trivialization, which has proven to be remarkably recalcitrant and absorbed much of the creative effort in the area. But there is a strategy for avoiding triviality that has been known for almost as long as the triviality results themselves. What is lacking is a straightforward integration of this approach in a larger framework of belief representation and dynamics. This paper discusses some of the issues involved and proposes an account of belief update by conditionalization.

Thanks to Hans-Christian Schmitz and Henk Zeevat for organizing the ESSLLI 2014 workshop on Bayesian Natural-Language Semantics and Pragmatics, where I presented an early version of this paper. Other venues at which I presented related work include the research group “What if” at the University of Konstanz and the Logic Group at the University of Connecticut. I am grateful to the audiences at all these events for stimulating discussion and feedback. Thanks also to Hans-Christian Schmitz and Henk Zeevat for their patience during the preparation of this manuscript. All errors and misrepresentations are my own.

Appendix: Proofs

Proposition

1 For $X\in \mathcal {F}$, if $\mathrm {Pr} (X)>0$, then ${\mathrm {Pr}}^*\left( \bigcup _{n\in \mathbb {N}} \left( \overline{X} ^n \times X \times \varOmega ^* \right) \right) = 1$.

Proof

Notice that $\bigcup _{n\in \mathbb {N}} \left( \overline{X} ^n \times X \times \varOmega ^* \right)$ is the set of all sequences containing at least one X-world, thus its complement is $\overline{X} ^*$. Now

${\mathrm {Pr}}^*\left( \bigcup _{n\in \mathbb {N}} \left( \overline{X} ^n \times X \times \varOmega ^* \right) \right) = 1 - {\mathrm {Pr}}^*\left( \overline{X} ^*\right)$

$= 1 - \lim _{n\rightarrow \infty }{\mathrm {Pr}}^*\left( \overline{X} ^n \times \varOmega ^* \right) = 1 - \lim _{n\rightarrow \infty }\mathrm {Pr} \left( \overline{X} \right) ^n = 1 \text { since } \mathrm {Pr} (\overline{X}) < 1$.$\square$

Lemma

1 If $\mathrm {Pr} (X)>0$, then $\sum _{n\in \mathbb {N}} \mathrm {Pr} \left( \overline{X} \right) ^n = 1/\mathrm {Pr} (X)$.

Proof

$\sum _{n\in \mathbb {N}} \mathrm {Pr} \left( \overline{X} \right) ^n \times \mathrm {Pr} \left( X\right) = \sum _{n\in \mathbb {N}} \left( \mathrm {Pr} \left( \overline{X} \right) ^n \times \mathrm {Pr} \left( X\right) \right)$

$= \sum _{n\in \mathbb {N}} {\mathrm {Pr}}^*\left( \overline{X} ^n \times X \times \varOmega ^* \right) = {\mathrm {Pr}}^*\left( \bigcup _{n\in \mathbb {N}} \left( \overline{X} ^n \times X \times \varOmega ^* \right) \right) = 1$ by Proposition 1.$\square$

Theorem

1 For $A,C \in \mathcal {L}_{A}^{0}$, if $\mathrm {P} (A)>0$, then $\mathrm {P} ^* \left( A \rightarrow C \right) = \mathrm {P} (C|A)$.

Proof

By Definition 6, the set of sequences $\omega ^*$ such that $V ^* (A\rightarrow C)(\omega ^*)=1$ is the union Since the sets for different values of n are mutually disjoint, the probability of the union is the sum of the probabilities for all n. Now, for all $X,Y\in \mathcal {F}$,

${\mathrm {Pr}}^*\left( \bigcup _{n\in \mathbb {N}} \left( \overline{Y} ^n \times (X \cap Y) \times \varOmega ^* \right) \right) = \sum _{n\in \mathbb {N}} {\mathrm {Pr}}^*\left( \overline{Y} ^n \times (X \cap Y) \times \varOmega ^* \right)$

$= \sum _{n\in \mathbb {N}} \left( \mathrm {Pr} \left( \overline{Y} \right) ^n \times \mathrm {Pr} (X \cap Y) \right) = \sum _{n\in \mathbb {N}} \mathrm {Pr} \left( \overline{Y} \right) ^n \times \mathrm {Pr} (X \cap Y)$

$= \mathrm {Pr} (X \cap Y) / \mathrm {Pr} (Y) \text { by Lemma}$ 1.

In particular, let X, Y be the set of worlds in $\varOmega$ at which $V (C)$ and $V (A)$ are true, respectively.$\square$

Proposition

2 If $\mathrm {Pr} (Z) > 0$, then

Proof

$\square$

Proposition

3 If is defined, then

Proof

Since is defined, $\mathrm {Pr} (Y_n) > 0$ for all n. Thus

$= \lim _{n\rightarrow \infty } \dfrac{{\mathrm {Pr}}^*\left( X_1 \cap Y_1 \times \cdots \times X_n \cap Y_n \times \varOmega ^* \right) }{{\mathrm {Pr}}^*\left( Y_1 \times \cdots \times Y_n \times \varOmega ^* \right) } \times \lim _{n\rightarrow \infty } {\mathrm {Pr}}^*\left( Y_1 \times \cdots \times Y_n \times \varOmega ^* \right)$

$= \lim _{n\rightarrow \infty } \left( \dfrac{{\mathrm {Pr}}^*\left( X_1 \cap Y_1 \times \cdots \times X_n \cap Y_n \times \varOmega ^* \right) }{{\mathrm {Pr}}^*\left( Y_1 \times \cdots \times Y_n \times \varOmega ^* \right) } \times {\mathrm {Pr}}^*\left( Y_1 \times \cdots \times Y_n \times \varOmega ^* \right) \right)$

$= \lim _{n\rightarrow \infty } {\mathrm {Pr}}^*\left( X_1 \cap Y_1 \times \cdots \times X_n \cap Y_n \times \varOmega ^* \right) = {\mathrm {Pr}}^*\left( \mathbf {X} \cap \mathbf {Y} \right)$ $\square$

Proposition

4 If is defined and $\mathbf {Y} \subseteq \mathbf {X}$, then

Proof

For all $i\ge 1$, $X_i \cap Y_i = Y_i$ since $\mathbf {Y} \subseteq \mathbf {X}$, and $\mathrm {Pr} (Y_i) > 0$ since is defined. Thus $\square$

The following auxiliary result will be useful in the subsequent proofs.

Proposition 9

If $\mathrm {Pr} (Z) > 0$, then for $X_i \in \mathcal {F}$ and $n \in \mathbb {N}$,

Proof

Immediate because $Z \subseteq \varOmega$.$\square$

The significance of Proposition 9 derives from the fact that the sets of sequences at which a given sentence in $\mathcal {L}_{A}^{}$ is true can be constructed (using set operations under which $\mathcal {F} ^*$ is closed) out of sequence sets ending in $\varOmega ^*$. This is obvious for the non-conditional sentences in $\mathcal {L}_{A}^{0}$. For conditionals $\varphi \rightarrow \psi$ the relevant set is the union of sets of sequences consisting of n $\overline{\varphi }$-worlds followed by a $\varphi \psi$-world. For each n, the corresponding set ends in $\varOmega ^*$ and therefore can be conditioned upon $Z^*$ as shown in Proposition 9. Since these sets for different numbers n are mutually disjoint, the probability of their union is just the sum of their individual probabilities.

Proposition

5 If $\mathrm {Pr} (X \cap Z) > 0$, then

Proof

Since $\mathrm {Pr} (X \cap Z)>0$, $\mathrm {Pr} (\overline{X} \cap Z) < \mathrm {Pr} (Z)$. Thus

$\square$

Lemma

2 If $\mathrm {Pr} (X \cap Z) > 0$, then $\sum _{n\in \mathbb {N}} \mathrm {Pr} \left( \overline{X} |Z\right) ^n = 1 / \mathrm {Pr} (X|Z)$.

Proof

Since $\mathrm {Pr} (X \cap Z) = \mathrm {Pr} (Z) \times \mathrm {Pr} (X|Z)$, both $\mathrm {Pr} (Z) > 0$ and $\mathrm {Pr} (X|Z) > 0$.

$\sum _{n\in \mathbb {N}} \mathrm {Pr} \left( \overline{X} |Z\right) ^n \times \mathrm {Pr} \left( X|Z\right) = \sum _{n\in \mathbb {N}} \left( \mathrm {Pr} \left( \overline{X} |Z\right) ^n \times \mathrm {Pr} \left( X|Z\right) \right)$

$= \sum _{n\in \mathbb {N}} \dfrac{\mathrm {Pr} \left( \overline{X} \cap Z\right) ^n \times \mathrm {Pr} \left( X \cap Z\right) }{\mathrm {Pr} \left( Z\right) ^{n+1}} = \sum _{n\in \mathbb {N}} \dfrac{{\mathrm {Pr}}^*\left( (\overline{X} \cap Z)^n \times (X \cap Z) \times \varOmega ^* \right) }{{\mathrm {Pr}}^*\left( Z^{n+1} \times \varOmega ^* \right) }$

by Proposition 9

by Proposition 5.$\square$

Theorem

2 If $\mathrm {P} \left( BC\right) > 0$, then $\mathrm {P} ^* \left( C \rightarrow D|B^* \right) = \mathrm {P} \left( D|BC \right)$.

Proof

$\mathrm {P} ^* \left( C \rightarrow D|B^* \right) = \lim _{n\rightarrow \infty } \dfrac{\mathrm {P} ^* (((C\rightarrow D) \wedge B)^n \times \varOmega ^*)}{\mathrm {P} ^* \left( B^n \times \varOmega ^* \right) }$

$= \lim _{n\rightarrow \infty }\dfrac{\sum _{i=0}^{n-1}\mathrm {P} (\overline{C} B)^i \times \mathrm {P} (CDB)}{\mathrm {P} (B)^n}$

$= \lim _{n\rightarrow \infty }\sum _{i=0}^{n-1}\mathrm {P} (\overline{C} |B)^i \times \mathrm {P} (CD|B)$

$= \mathrm {P} (CD|B) / \mathrm {P} (C|B)$ by Lemma 2

$= \mathrm {P} (D|BC)$ $\square$