Revision with Conditional Probability Functions: Two Impossibility Results

Abstract

In the general framework which takes conditional probability functions as a primitive notion, we prove two triviality results about the possibility of defining the probability Pr(A > B,Γ) of a counterfactual as the probability of the consequent B after some revision of the probability function that puts the probability of the antecedent A to 1. The paper is divided in three parts. In the first part, we present the context in which this research takes place and provide the formal tools we will use in the two other parts. In the second part, we present our first triviality result: Any probability revision process that satisfies the identity $\textrm{Pr}(A>B, \Gamma) = \textrm{Pr}^{A}(B,\Gamma)$ for all Pr, A, B and Γ (where $\textrm{Pr}^{A})$ is any revision of the probability distribution Pr which sets $\textrm{Pr}^{A}(A, \Gamma) = 1)$ is trivial in a sense that will be specified. Finally, in the third part, we prove another triviality result: Any probability revision process that satisfies the identity $\textrm{Pr}(A>B, \Gamma) = \textrm{Pr}^{A}(B,\Gamma_{A}^{\ast})$ for all Pr, A, B and Γ (where $\Gamma_{A}^{\ast}$ is a revision of the background Γ in the light of A) is also trivial.

Appendix

Lemma 9.1

If Γ is Pr-normal, then $\mathrm{Pr} (A \wedge \neg A,\Gamma ) = 0$.

Proof

Suppose Γ is Pr-normal. From NP.3 and NP.8 we have:

$$1 = \mathrm{Pr} (A \vee \neg A,\Gamma ) = \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (\neg A,\Gamma ) - \mathrm{Pr} (A \wedge \neg A,\Gamma ).$$

Applying NP.5 to the right we have:

$$1 = \mathrm{Pr} (A,\Gamma ) + 1 - \mathrm{Pr} (A,\Gamma ) - \mathrm{Pr} (A \wedge \neg A,\Gamma ).$$

Simple arithmetic gives the desired result.

Lemma 9.2

If Γ is Pr-normal but $\Gamma \cup {\{}A{\}}$ is Pr-abnormal, then $\mathrm{Pr} (A,\Gamma ) = 0$.

Proof

Suppose Γ is Pr-normal but $\Gamma \cup \{A\}$ is Pr-abnormal. From Lemma 9.1 and NP.4, we have:

$$0 = \mathrm{Pr} (A \wedge \neg A,\Gamma ) = \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (\neg A,\Gamma \cup \{A\}).$$

Since $\Gamma \cup \{A\}$ is Pr-abnormal, $\mathrm{Pr} (\neg A,\Gamma \cup \{A\}) = 1$ and thus $\mathrm{Pr} (A,\Gamma ) = 0$.

Lemma 9.3

$\mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (A \wedge B,\Gamma )$.

Proof

If Γ is Pr-abnormal, the result is immediate, so suppose that Γ is Pr-normal. By the definition of ⊃ and applying NP.3, we have:

$$\mathrm{Pr} (A \supset B,\Gamma ) = \mathrm{Pr} (\neg A \vee B,\Gamma ) = \mathrm{Pr} (\neg A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (\neg A \wedge B,\Gamma ).$$

Applying NP.6 and NP.4 yields:

$$\mathrm{Pr} (A \supset B,\Gamma ) = \mathrm{Pr} (\neg A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (B,\Gamma )\times \mathrm{Pr} (\neg A,\Gamma \cup \{B\}).$$

Since Γ is Pr-normal, we can apply NP.5 to obtain:

$$\mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (B,\Gamma )\times \mathrm{Pr} (\neg A,\Gamma \cup \{B\}).$$

((9.1))

On one hand, suppose that $\Gamma \cup \{B\}$ is Pr-abnormal. Then applying Lemma 9.2 to (9.1) yields:

$$\mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ).$$

But Lemma 9.2 and NP.6 and NP.4 ensure that $\mathrm{Pr} (A \wedge B,\Gamma ) = 0$. So we have:

$$\mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (A \wedge B,\Gamma ).$$

On the other hand, suppose that $\Gamma \cup \{B\}$ is Pr-normal. Then applying NP.5 to (9.1) gives:

$$\begin{gathered} \mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (B,\Gamma )\times (1 - \mathrm{Pr} (A,\Gamma \cup \{B\})) \\ = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (A \wedge B,\Gamma ) \end{gathered}$$

So, in either case, the result follows.

Lemma 9.4

$\mathrm{If \ Pr} (A \supset B,\Gamma ) = 1$, then $\mathrm{Pr} (A,\Gamma ) \le \mathrm{Pr} (B,\Gamma )$.

Proof

Suppose $\mathrm{Pr} (A \supset B,\Gamma ) = 1$. Then from Lemma 9.3 we have:

$$1 = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (A \wedge B,\Gamma )$$

So by elementary arithmetic we have:

$$\mathrm{Pr} (A,\Gamma ) = \mathrm{Pr} (A \wedge B,\Gamma )$$

But by NP.1 and NP.4 and NP.6, we know:

$$\mathrm{Pr} (A \wedge B,\Gamma ) \le \mathrm{Pr} (B,\Gamma )$$

Hence the desired result follows.

Lemma 9.5

If $\mathrm{Pr} (A,\Delta ) \le \mathrm{Pr} (B,\Delta )$ for all Δ, then $\mathrm{Pr} (A \supset B,\Gamma ) = 1$ for all Γ.

Proof

Suppose $\mathrm{Pr} (A,\Delta ) \le \mathrm{Pr} (B,\Delta )$ for all Δ. Let Γ be arbitrary. Then by NP.1 and NP.2 we have:

$$\mathrm{Pr} (B,\Gamma \cup \{A\}) = 1$$

But by Lemma 9.3 and NP.4 we have:

$$\mathrm{Pr} (A \supset B,\Gamma ) = 1 - \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\})$$

The desired result follows immediately from these two equations.

Lemma 9.6

Disjunction is the least upper bound, in the following sense:

• (A) $\mathrm{Pr} (A,\Gamma ) \le \mathrm{Pr} (A \vee B,\Gamma )$ for all Γ.

• (B) $\mathrm{Pr} (B,\Gamma ) \le \mathrm{Pr} (A \vee B,\Gamma )$ for all Γ.

• (C) Let C be any sentence such that:

  • (i) $\mathrm{Pr} (A,\Delta ) \le \mathrm{Pr} (C,\Delta )$ for all Δ, and

  • (ii) $\mathrm{Pr} (B,\Delta ) \le \mathrm{Pr} (C,\Delta )$ for all Δ.

Then for all Γ, $\mathrm{Pr} (A \vee B,\Delta ) \le \mathrm{Pr} (C,\Delta )$.

Proof

For the (B) part, by NP.1 we know the maximum value for $\mathrm{Pr} (B,\Gamma \cup \{A\})$ is 1, so by NP.3 and NP.4 we have:

$$\begin{array}{ll}&\mathrm{Pr} (A \vee B,\Gamma ) = \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\}) \nonumber \\ & \ge \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A,\Gamma )\end{array}$$

((9.2))

and the right hand side just reduces to $\mathrm{Pr} (B,\Gamma )$, as desired. Part (A) follows in a similar fashion by first applying NP.6 to the conjunction. For the (C) part of the Lemma, suppose both of the following hold:

$$\mathrm{Pr} (A,\Delta ) \le \mathrm{Pr} (C,\Delta ) \hbox{for all} \Delta.$$

((9.3))

$$\mathrm{Pr} (B,\Delta ) \le \mathrm{Pr} (C,\Delta ) \hbox{for all} \Delta.$$

((9.4))

Taking Δ to be of the form $\Gamma \cup {\{}A{\}}$ for arbitrary Γ, and using NP.1 and NP.2, we then obtain the following:

$$\mathrm{Pr} (C,\Gamma \cup \{A\}) = 1 \ \hbox{for all} \Gamma.$$

((9.5))

$$\mathrm{Pr} (C,\Gamma \cup \{B\}) = 1 \ \hbox{for all} \Gamma.$$

((9.6))

Next we multiply $\mathrm{Pr} (A \vee B,\Gamma \cup \{C\})$ by $\mathrm{Pr} (C,\Gamma )$ and apply NP.3 and NP.4:

$$\begin{aligned} \mathrm{Pr} (C,\Gamma ) & \times \mathrm{Pr} (A \vee B,\Gamma \cup \{C\}) = \nonumber \\ & \mathrm{Pr} (C,\Gamma )\times \mathrm{Pr} (A,\Gamma \cup \{C\}) + \mathrm{Pr} (C,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{C\}) \nonumber \\ & - \mathrm{Pr} (C,\Gamma )\times \mathrm{Pr} (A,\Gamma \cup \{C\})\times \mathrm{Pr} (B,\Gamma \cup \{A,C\}) \end{aligned}$$

((9.7))

Then applying NP.4 and NP.6 to the right hand side of (9.7), we obtain:

$$\begin{aligned}\mathrm{Pr} (C,\Gamma ) & \times \mathrm{Pr} (A \vee B,\Gamma \cup \{C\}) = \nonumber \\ & \quad \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (C,\Gamma \cup \{A\}) + \mathrm{Pr} (B,\Gamma )\times \mathrm{Pr} (C,\Gamma \cup \{B\}) \nonumber \\ & \quad - \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\})\times \mathrm{Pr} (C,\Gamma \cup \{A,B\})\end{aligned}$$

((9.8))

But then using (9.5) and (9.6) on the right of (9.8) we have:

$$\begin{aligned}\mathrm{Pr} (C,\Gamma ) & \times \mathrm{Pr} (A \vee B,\Gamma \cup \{C\}) = \nonumber \\ & \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\})\end{aligned}$$

((9.9))

Hence from (9.9), NP.1 and NP.3 we have:

$$\mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\}) \le \mathrm{Pr} (C,\Gamma )$$

((9.10))

Finally the desired result follows by applying NP.3 and NP.4 to the left hand side of (9.10)

Lemma 9.7

If $\mathrm{Pr} (A,\Gamma ) = 1$, then $\mathrm{Pr} (A \wedge B,\Gamma ) = \mathrm{Pr} (B,\Gamma )$.

Lemma 9.8

Suppose $\mathrm{Pr} (A,\Gamma ) = 1$. Then by Thm 1.8, $\mathrm{Pr} (A \vee B,\Gamma ) = 1$. So, by NP.3, we have:

$$1 = \mathrm{Pr} (A \vee B,\Gamma ) = \mathrm{Pr} (A,\Gamma ) + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A \wedge B,\Gamma )$$

Hence it follows that:

$$1 = 1 + \mathrm{Pr} (B,\Gamma ) - \mathrm{Pr} (A \wedge B,\Gamma )$$

The desired result then follows by elementary arithmetic.

Lemma 9.8

If $\mathrm{Pr} (A,\Gamma ) = 1$, then $\mathrm{Pr} (B,\Gamma ) = \mathrm{Pr} (B,\Gamma \cup \{A\})$

Proof

Suppose $\mathrm{Pr} (A,\Gamma ) = 1$. From NP.4, we have:

$$\begin{aligned} \mathrm{Pr} (A \wedge B,\Gamma ) & = \mathrm{Pr} (A,\Gamma )\times \mathrm{Pr} (B,\Gamma \cup \{A\}) \\ & = \mathrm{Pr} (B,\Gamma \cup \{A\}) \end{aligned}$$

The desired result then follows immediately by Lemma 9.7.

Lemma 9.9

$\mathrm{Pr} (A > B,\Gamma ) \le \mathrm{Pr} (A \supset B,\Gamma )$, for all Γ.

Proof

By (3). and soundness, we have:

$$\mathrm{Pr} ((A > B) \supset (A \supset B),\Gamma ) = 1$$

The desired result then follows by Lemma 9.4.

Lemma 9.10

$\mathrm{Pr} (B,\Gamma ) \le \mathrm{Pr} (A > B,\Gamma )$, for all Γ.

Proof

This result follows immediately from Lemma 9.4, Lemma 9.9, and Proposition 9.1.

Lemma 9.11

$\Gamma \cup \{\neg A\} \ \mathrm{is \ Pr} ^A$-abnormal.

Proof

By elementary arithmetic, we have:

$$\mathrm{Pr} ^A(A,\Gamma \cup \{\neg A\}) = 1 - (1 - \mathrm{Pr} ^A(A,\Gamma \cup \{\neg A\}))$$

For proof by contradiction, suppose $\Gamma \cup \{\neg A\}$ is $\mathrm{Pr} _A$-normal. Then applying NP.5 to the right hand side, we obtain:

$$\mathrm{Pr} ^A(A,\Gamma \cup \{\neg A\}) = 1 - \mathrm{Pr} ^A(\neg A,\Gamma \cup \{\neg A\})$$

Applying NP.2 to the right hand side yields:

$$\mathrm{Pr} ^A(A,\Gamma \cup \{\neg A\}) = 1 - 1 = 0$$

But we know by (1)

$$\mathrm{Pr} ^A(A,\Gamma \cup \{\neg A\}) = 1$$

Thus the supposition is in error, and it must be the case that $\Gamma \cup \{\neg A\}$ is Pr^A-abnormal.

Lemma 9.12

$\mathrm{Pr} (\neg A > (A > B),\Gamma ) = 1$, for all Γ.

Proof

Applying (2), we have:

$$\mathrm{Pr} (\neg A > (A > B),\Gamma ) = \mathrm{Pr} ^{_{\neg A}}(A > B,\Gamma ), \hbox{for all}\ \Gamma .$$

But then applying Lemma 9.8 and (1) on the right hand side gives:

$$\mathrm{Pr} (\neg A > (A > B),\Gamma ) = \mathrm{Pr} ^{_{\neg A}}(A > B,\Gamma \cup \{\neg A\}), \ \hbox{for all}\ \Gamma .$$

Another application of (2) to the right hand side gives:

$$\mathrm{Pr} (\neg A > (A > B),\Gamma ) = \mathrm{Pr} ^{_{\neg AA}}(B,\Gamma \cup \{\neg A\}), \ \hbox{for all}\ \Gamma.$$

But then by Lemma 9.11, the right hand side must be 1.

Lemma 9.13

$\mathrm{Pr} (\neg A,\Gamma ) \le \mathrm{Pr} (A > B,\Gamma )$, for all Γ.

Proof

This result follows immediately from Lemma 9.4, Lemma 9.9, and Lemma 9.12

Lemma 9.14

$\mathrm{Pr} (\neg A \vee B,\Gamma ) \le \mathrm{Pr} (A > B,\Gamma )$, for all Γ.

Proof

The desired result is an immediate result of the part (C) of Lemma 9.6, along with Lemma 9.10 and Lemma 9.13.