Why didn't $\Pr \left( A \rightarrow B \right)$ catch on?

Question

Students are conditioned to thinking in terms of IF-THEN statements even before high school, and courses offered at the university level often lead to the formalization of material implication. Material implication is a function that two Boolean values and returns/outputs a single Boolean value. One can use 'True' and 'False', but here I will use '1' and '0' respectively. Material implication can be written as a piecewise function (especially since I am not aware of a nice way to format truth tables here).

$$p \rightarrow q \triangleq \begin{cases} 1 & p = 1 \land q = 1 \\ 0 & p = 1 \land q = 0 \\ 1 & p = 0 \land q = 1 \\ 1 & p = 0 \land q = 0 \end{cases}$$

An indicator function of $ A \subseteq \Omega$ is a map $\mathbb{I}_A: \Omega \mapsto \{ 0, 1 \}$, and can be written explicitly in a piecewise fashion.

$$\mathbb{I}_A(\omega) \triangleq \begin{cases} 1 & \omega \in A \\ 0 & \omega \not\in A \end{cases}$$

We will interpret $\Pr(A)$ to be $\int_{\Omega} \mathbb{I}_A(\omega)dP$, and $\Pr(A \rightarrow B)$ to be $\int_{\Omega} \mathbb{I}_A(\omega) \rightarrow \mathbb{I}_B(\omega) dP$. Seeing an implication operator in the integrand is a novelty, but this implication on indicators can be re-written in terms of a new indicator. Thus it is an expression compatible with our existing understanding.

$$\mathbb{I}_A(\omega) \rightarrow \mathbb{I}_B(\omega) = \mathbb{I}_{C}(\omega) \triangleq \begin{cases} 1 & \omega \in A \land \omega \in B \\ 0 & \omega \in A \land \omega \not\in B \\ 1 & \omega \not\in A \land \omega \in B \\ 1 & \omega \not\in A \land \omega \not\in B \end{cases}$$

$$\Pr(C) = \int_{\Omega} \mathbb{I}_C(\omega) dP$$

At some point students encounter the mathematical fact that $\Pr \left( A \rightarrow B \right) \neq \Pr \left( B | A \right)$. Some mathematicians were motivated to develop conditional event algebras that attempt to define these as equal, and I've asked for applications of such algebras in the past. The Goodman–Nguyen–Van Fraasen algebra is an example of a conditional event algebra.

But even without getting into conditional event algebras, $\Pr \left( A \rightarrow B \right)$ is a defined quantity in the standard Kolmogorov treatment of probability that I have yet to see used. Why are we not using it?

Answer 1

Assuming that $A$ and $B$ are events, this seems in effect to be saying $\Pr(A \rightarrow B)$ would mean $\Pr(A^c \cup B)$,

which is equal to $\Pr(B\mid A)+\Pr(B^c\mid A) \Pr(A^c)$

and greater than $\Pr(B\mid A)$ unless one of the terms is $0$.

But the question is whether this notation would be useful in any sense.

Answer 2

Based on your notational definitions, your probability statement is equivalent to:

$$\begin{align} \mathbb{P}(A \rightarrow B) \equiv \mathbb{P}(\bar{A} \cup B) = 1- \mathbb{P}(A \cap \bar{B}), \end{align}$$

which can already be framed perfectly adequately in terms of existing probability notation.

Answer 3

"Why didn't Pr(A→B) catch on?" begs the question. Who ever asserted such notation was necessary or useful in the first place? The only thing that makes sense inside of a probability operator is an event. So if A and B are events, you can consider $Pr(A \cup B)$ or $P(B|A)$ or any type of set theoretic operation, and then you have Bayesian type operations as you mentioned which is just a measure theoretic result. However, $\rightarrow$ is no set theory operator. If $A \rightarrow B$ is an event, then you can WLOG call it $Z$ in some separate event in a separate event space $C$, where you can have events like 1. A causes B 2. Not A causes not B 3. A doesn't cause not B (etc, for 2^3 permutations).

But defining causation as an event is difficult to wrap one's mind around. Nonetheless, I've seen examples of this in clinical trials, where based on physician review, they can look at proportions of patients on drugs who have adverse outcomes, and whether those outcomes are (in a blinded fashion) determined to have been related to the drug or not. Typically this is based entirely on subjective review, and is most strongly argued by a lack of any other explanation, which does not suffice to prove causality.

EDIT: as a final comment, since the OP's edit about "material implication" significantly simplified the question: it is quite confusing to use overlapping notation from logic and statistics. For instance, we already use the $\bar{X}$ to represent the sample average, rather than the complement of event $X$. $\rightarrow$ is already convergence/limit operator. As others have pointed out, it's just one additional typeset to create the event of material implication as $A^C \cup B$.

Answer 4

I think the answer's already in your question. It's fairly natural to use the notation of the sentential calculus within probability statements when what we'd be happy to take as the probability of the sentence's being true corresponds to what you get from totting up the probabilities for all the possible states of affairs for which the sentence is true. And so you sometimes see $\Pr(\lnot p)$, $\Pr(p \lor q)$, & the like, where $p$ & $q$ are sentences; more often $\Pr(\lnot A)$, $\Pr(A \lor B)$, & the like, where $A$ & $B$ are sets^†. Material implication is exceptional, however: $\Pr(p \rightarrow q)$ ought to mean $\Pr(q|p)$, we feel—the probability of "if there are dark clouds now, then it will rain soon" depends only on how often it tends to rain soon given that there are dark clouds now—but that conditional probability isn't what we end up with by treating $\Pr(p \rightarrow q)$ in the same way as the others. If it hasn't caught on in standard Probability Theory, it's some combination of its being unintuitive as explained above, its already being "taken" by conditional events algebras, & lack of demand for an especially concise notation for what it represents in standard P.T.

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† The latter usage is a generally harmless abuse of notation, but I wonder if it doesn't contribute to puzzlement about what $\Pr(A \rightarrow B)$ means.