How to modify the strength of a reason

Abstract

Kearns and Star have previously recommended that we measure the degree to which a reason supports a conclusion, either about how to act or what to believe, as the conditional probability of the conclusion given the reason. I show how to properly formulate this recommendation to allow for dependencies and conditional dependencies among the considerations being aggregated. This formulation allows us to account for how considerations, which do not themselves favour a specific conclusion, can modify the strength of a reason for that conclusion, and thus to explain the intensifiers and attenuators described by Dancy. The formulation also accounts for the workings of partial undercutters in epistemology. I then show how my account avoids the counterexamples that Brunero levied against probability-based theories of the strengths of reasons. My account supports the theory, suggested by Kearns and Star, that the strengths of reasons are measured by conditional probabilities. If my account is successful, then it will count in favour of the idea that the strengths of reasons are measured on the same scale as are conditional probabilities.

Appendix: How to derive formula (G)

Suppose we have a piece of moral reasoning where two reasons, R and $R^{*}$, each either supports or counts against a conclusion, ${\Phi }$. To calculate the conditional credence of ${\Phi }$ based on the conjunction of R and $R^{*}$, we start with the ratio-formula definition of conditional credence, $Pr(A{|}B{)} = Pr\left( {A\& B} \right)/Pr\left( B \right)$ and substitute ${\Phi }$ for A and $R\& R^{*}$ for B. By commuting the conjunctions in both the numerator and the denominator, we get:

$$\Pr \left( {\Phi |R \& R^{*} } \right) = \frac{{\Pr \left( {\Phi \& \left( {R \& R^{*} } \right)} \right)}}{{\Pr \left( {R \& R^{*} } \right)}} = \frac{{\Pr \left( {R^{*}\& \left( {R \& \Phi } \right)} \right)}}{{\Pr \left( {R^{*} \& R} \right)}}$$

(1)

From the ratio formula for conditional probability, we also get the product rule for the probability of a conjunction, $Pr\left( {A \& B} \right) = Pr(A{|}B{)} \times Pr\left( B \right).$ When we apply the product rule to the numerator of (1) by substituting $R^{*}$ for A and $\left( {R\& {\Phi }} \right)$ for B, we get:

$$\Pr \left( {\Phi |R\& R^{*} } \right) = \frac{{\Pr \left( {R^{*} |R\& \Phi } \right) \times \Pr \left( {R\& \Phi } \right)}}{{\Pr \left( {R^{*} \& R} \right)}}$$

(2)

When we again apply the product rule to ${\text{Pr}}\left( {R\& {\Phi }} \right)$ in the numerator of (2), we get:

$$\Pr \left( {\Phi |R\& R^{*} } \right) = \frac{{\Pr \left( {R^{*} |R\& \Phi } \right) \times \Pr \left( {R|\Phi } \right) \times p\left( \Phi \right)}}{{\Pr \left( {R^{*} \& R} \right)}}$$

(3)

When we apply the product rule to the denominator of (3), we get:

$$\Pr \left( {\Phi |R\& R^{*} } \right) = \frac{{\Pr \left( {R^{*} |R\& \Phi } \right) \times \Pr \left( {R|\Phi } \right) \times \Pr \left( \Phi \right)}}{{\Pr \left( {R^{*} |R} \right) \times \Pr \left( R \right)}}$$

(4)

By rearranging the terms in both the numerator and denominator of (4), we get:

$$\Pr \left( {\Phi |R\& R^{*} } \right) = \frac{{\Pr \left( {R^{*} |R\& \Phi } \right)}}{{\Pr (R^{*} |R)}} \times \frac{{\Pr \left( {R|\Phi } \right)}}{\Pr \left( R \right)} \times \Pr \left( \Phi \right)$$

(5)

Looking back into the main text, we can see that the last two terms on the right-hand side are the right-hand side of Bayes Theorem (BT).

$$\left( {BT} \right)\quad \Pr (\Phi |R) = \frac{{\Pr \left( {R|\Phi } \right)}}{\Pr \left( R \right)} \times \Pr \left( \Phi \right)$$

Substituting this into (5) we get a general mathematical theorem (G) for the case of two reasons.

$$\left( G \right)\quad \Pr \left( {\Phi |R \& R^{*} } \right) = \frac{{\Pr \left( {R^{*} |R \& \Phi } \right)}}{{\Pr (R^{*} |R)}} \times \Pr (\Phi |R)$$

Notice that because all the operations involved in the proof are commutative, the proof will be symmetric in R and R*. We could equally well have derived the inverted formula, (G*).

$$\left( {G^{*} } \right)\quad \Pr \left( {\Phi |R\& R^{*} } \right) = \frac{{\Pr \left( {R|R^{*} \& \Phi } \right)}}{{\Pr (R|R^{*} )}} \times \Pr (\Phi |R^{*} )$$

We can extend the above derivation to produce corollaries of Bayes’ Theorem for inductive arguments that involve as many premises as we may wish, though the corollaries will rapidly become very complicated. (G) is a useful formula for combining the support offered by two moral premises into an overall degree of support for a conclusion. It shows right away that the degrees of support given to a conclusion by each of two considerations combine multiplicatively, not additively (Kagan 1998).