Accuracy-dominance and conditionalization

Abstract

Epistemic decision theory produces arguments with both normative and mathematical premises. I begin by arguing that philosophers should care about whether the mathematical premises (1) are true, (2) are strong, and (3) admit simple proofs. I then discuss a theorem that Briggs and Pettigrew (Noûs 54(1):162–181, 2020) use as a premise in a novel accuracy-dominance argument for conditionalization. I argue that the theorem and its proof can be improved in a number of ways. First, I present a counterexample that shows that one of the theorem’s claims is false. As a result of this, Briggs and Pettigrew’s argument for conditionalization is unsound. I go on to explore how a sound accuracy-dominance argument for conditionalization might be recovered. In the course of doing this, I prove two new theorems that correct and strengthen the result reported by Briggs and Pettigrew. I show how my results can be combined with various normative premises to produce sound arguments for conditionalization. I also show that my results can be used to support normative conclusions that are stronger than the one that Briggs and Pettigrew’s argument supports. Finally, I show that Briggs and Pettigrew’s proofs can be simplified considerably.

Appendices

Appendix 1

Theorem 3

Assume that $\mathcal J$ satisfies Separability, Continuity, Extensionality, Strict Propriety, and Temporal Separability.

• (a) Let $\langle c,a \rangle$ be a credal strategy. (1) If c is not probabilistic, then there is a probabilistic and conditionalizing credal strategy that strongly accuracy-dominates $\langle c,a \rangle$. (2) If c is probabilistic and $a_E$ is not probabilistic for some $E \in \mathcal E$ with $c(E)>0$, then there is a probabilistic and conditionalizing credal strategy that strongly accuracy-dominates $\langle c,a \rangle$. (3) If c is probabilistic and $a_E$ is not probabilistic for some $E \in \mathcal E$ with $c(E)=0$, then there is a probabilistic and conditionalizing credal strategy that weakly accuracy-dominates $\langle c,a \rangle$.

• (b) For each credal strategy that is not conditionalizing, there is an alternative credal strategy that is probabilistic and conditionalizing that strongly accuracy-dominates it.

• (c) For each credal strategy that is probabilistic and conditionalizing, (1) there is no alternative credal strategy whatsoever that strongly accuracy-dominates it, and (2) there is no non-conditionalizing credal strategy that even weakly accuracy-dominates it.

1.1 Proof of Theorem 3

Assume throughout that $\mathcal J$ satisfies Separability, Continuity, Extensionality, Strict Propriety, and Temporal Separability.

The proofs of parts (a) and (b) rely on the following lemma, which builds on the observations about arbitrary behavior on zero credence propositions that are expressed in Lemma 1.

Lemma 2

Let c be a probabilistic credence function, and let $\langle c,a \rangle$ be a conditionalizing strategy. If $\langle c,a \rangle$ is not probabilistic, then it is weakly accuracy-dominated by a credal strategy $\langle c, a' \rangle$ that is both probabilistic and conditionalizing.

Proof

If $\langle c,a \rangle$ is not probabilistic, then $a_{E}$ is not probabilistic for at least one $E \in \mathcal E$. For all such E, $c(E)=0$, by Lemma 1(ii); and, by Theorem 1, there is a probabilistic credence function $p_E$ that strongly accuracy-dominates $a_E$. For all $E \in \mathcal E$, define the credal act $a'$ by

$$\begin{aligned} a'_E = {\left\{ \begin{array}{ll} a_E & \text {if }a_E\text { is probabilistic};\\ p_E & \text {otherwise}. \end{array}\right. }\end{aligned}$$

Then, $a'$ is probabilistic and weakly accuracy-dominates a. By Lemma 1(i), $a'$ is a conditionalization of c because it differs from a only on propositions with zero c-credence. By Temporal Separability, $\langle c,a' \rangle$ weakly accuracy-dominates $\langle c,a \rangle$. $\square$

We now give the proof of Theorem 3. It will be convenient to prove (b) first.

Proof of Theorem 3 (b)

The argument follows the one for (II) given by Briggs and Pettigrew (2020), which, in turn, relies on some technical propositions in Predd et al. (2009). My aim here is simply to emphasize that the last step of Briggs and Pettigrew’s argument (7.4, pp. 178–179) involving a transfinite induction can be eliminated. I will state some lemmas that summarize the key steps of Briggs and Pettigrew’s argument. But first we need to introduce some notation.

Let $\mathcal A = \{A _1,..., A _m\}$. Note that every credence function can be identified with a vector in $\mathbb R^m$ in the following way:

$$\begin{aligned}c = \big \langle c\{ A _1\},...,c\{ A _m\}\big \rangle .\end{aligned}$$

Similarly, with $\mathcal E = \{E_1,...,E_n\}$, every credal strategy $\langle c,a \rangle$ can be identified with a vector in $\mathbb R^{m + mn}$:

$$\begin{aligned}\langle c,a \rangle = \Big \langle \underbrace{c(A _1),...,c(A _m)}_c, \underbrace{a_{E_1}(A _1),...,a_{E_1}( A _m)}_{a_{E_1}},...,\underbrace{a_{E_n}(A _1),...,a_{E_n}(A _m)}_{a_{E_n}}\Big \rangle .\end{aligned}$$

Now, given a credal strategy $\langle c,a \rangle$ and $\omega \in \Omega$, define $\langle c,a \rangle _\omega$ by replacing both c and $a_{\mathcal E_\omega }$ with the valuation function $v_\omega$ in the vector representation of $\langle c,a \rangle$. More formally, let $i_\omega$ be that $i \in \{1,...,n\}$ for which $\mathcal E_\omega = E_i$, i.e. that i for which $\omega \in E_i$, and define

$$\begin{aligned}{\langle c,a \rangle} _\omega = \Big \langle&\underbrace{v_\omega ({A}_1),...,v_\omega ({A}_m)}_{v_\omega \,\text { replacing } \, c},...,a_{E_{i_\omega -1}}({A}_1),...,a_{E_{i_\omega - 1}}({A}_m), \underbrace{v_\omega ({A}_1),...,v_\omega ({A}_m)}_{v_\omega \,\text { replacing } \, a_{\mathcal E_\omega }},\\&a_{E_{i_\omega +1}}({A}_1),...,a_{E_{i_\omega + 1}}({A}_m),...,a_{E_n}({A}_1),...,a_{E_n}({A}_m) \Big \rangle. \end{aligned}$$

$\square$  

Briggs and Pettigrew show the following.

Lemma 3

(Briggs and Pettigrew (2020), Lemma 2) Let $\langle c,a \rangle$ be a credal strategy. Then, c is probabilistic and $\langle c,a \rangle$ is a conditionalizing strategy if and only if $\langle c,a \rangle$ is in the convex hull of $\{\langle c,a \rangle _\omega : \omega \in \Omega \}$.

Lemma 4

(Briggs and Pettigrew (2020), p. 178) If the credal strategy $\langle c,a \rangle$ is not in the convex hull of $\{\langle c,a \rangle _\omega : \omega \in \Omega \}$, then there is some credal strategy $\langle c',a' \rangle$ in the convex hull of $\{\langle c,a \rangle _\omega : \omega \in \Omega \}$ such that

$$\begin{aligned} \mathcal J_\omega (\langle c',a' \rangle ) < \mathcal J_\omega (\langle c,a \rangle ) \end{aligned}$$

(5)

for all $\omega \in \Omega$.

So, if $\langle c,a \rangle$ is a credal strategy that is not conditionalizing, then, by Lemma 3, $\langle c,a \rangle$ is not in the convex hull of $\{\langle c,a \rangle _\omega : \omega \in \Omega \}$. By Lemma 4, $\langle c,a \rangle$ is strongly accuracy-dominated by a credal strategy $\langle c',a' \rangle$ that is in the convex hull of $\{\langle c,a \rangle _\omega : \omega \in \Omega \}$. Calling on Lemma 3 once more, we find that $c'$ is probabilistic and $\langle c',a' \rangle$ is a conditionalizing strategy. If $\langle c',a' \rangle$ is probabilistic, then we are done. If not, apply Lemma 2 to find a credal strategy $\langle c'',a'' \rangle$ that is probabilistic and conditionalizing and weakly accuracy-dominates $\langle c',a' \rangle$. Then, $\langle c'',a'' \rangle$ strongly accuracy-dominates $\langle c,a \rangle$.

Proof of Theorem 3 (a)

• (1) If c is not probabilistic, then the argument proceeds as in the proof of (b) above: by Lemmas 3 and 4, $\langle c,a \rangle$ is strongly accuracy-dominated by a credal strategy $\langle c',a' \rangle$ such that $c'$ is probabilistic and $\langle c',a' \rangle$ is a conditionalizing strategy; if $\langle c',a' \rangle$ is not already probabilistic, then, by Lemma 2, it is weakly accuracy-dominated by a probabilistic and conditionalizing strategy $\langle c'',a'' \rangle$, and $\langle c'',a'' \rangle$ strongly accuracy-dominates $\langle c,a \rangle$.

• (2) If c is probabilistic and $a_E$ is not probabilistic for some $E \in \mathcal E$ with $c(E)>0$, then $\langle c,a \rangle$ is not a conditionalizing strategy, by Lemma 1(ii). By part (b), $\langle c,a \rangle$ is strongly accuracy-dominated by a credal strategy that is probabilistic and conditionalizing.

• (3) Suppose that c is probabilistic and $a_E$ is not probabilistic for some $E \in \mathcal E$ with $c(E)=0$. If $\langle c,a \rangle$ is not a conditionalizing strategy, then it is strongly accuracy-dominated by (b), so assume that $\langle c,a \rangle$ is a conditionalizing strategy. Then, by Lemma 2, $\langle c,a \rangle$ is weakly accuracy-dominated.

$\square$

Proof of Theorem 3(c)

Assume that $\langle c,a \rangle$ is probabilistic and conditionalizing. Using Temporal Separability, we can compute the c-expected inaccuracy of $\langle c,a \rangle$:

$$\begin{aligned}exp_c(\mathcal J(\langle c,a \rangle ))=_{df} \sum _{\omega \in \Omega }c\{\omega \}\mathcal J_\omega (\langle c,a \rangle ) = \sum _{\omega \in \Omega }c\{\omega \}\mathcal J_\omega (c) + exp_c(\mathcal J(a)).\end{aligned}$$

For (1), use Strict Propriety and Theorem 2 to observe that

$$\begin{aligned} \sum _{\omega \in \Omega }c\{\omega \}\mathcal J_\omega (c) + exp_c(\mathcal J(a)) \le \sum _{\omega \in \Omega }c\{\omega \}\mathcal J_\omega (c') + exp_c(\mathcal J(a')) \end{aligned}$$

(6)

for all credal strategies $\langle c',a' \rangle$. But if $\langle c',a' \rangle$ strongly accuracy-dominates $\langle c,a \rangle$, so that $\mathcal J_\omega (\langle c',a' \rangle ) < \mathcal J_\omega (\langle c,a \rangle )$ for all $\omega \in \Omega$, then

$$\begin{aligned}exp_c(\mathcal J(\langle c',a' \rangle )) < exp_c(\mathcal J(\langle c,a \rangle )).\end{aligned}$$

So, no probabilistic and conditionalizing credal strategy is strongly accuracy-dominated.⁹

For (2), suppose there were a non-conditionalizing credal strategy $\langle c',a' \rangle$ that weakly accuracy-dominated $\langle c, a \rangle$. By part (b) of the theorem, $\langle c', a' \rangle$ would be strongly accuracy-dominated, and therefore so would $\langle c,a \rangle$, contradicting what we have just shown. $\square$

Appendix 2

Theorem 4

Assume that $\mathcal J$ satisfies Separability, Continuity, Extensionality, Strict Propriety, and Temporal Separability. Let c be a probabilistic credence function, and let $\langle c,a \rangle$ be a conditionalizing strategy.

• (a) If $\langle c,a \rangle$ is not Blackwell, then it is weakly accuracy-dominated by a credal strategy that is probabilistic, conditionalizing, and Blackwell.

• (b) Conversely, if $\langle c,a \rangle$ is Blackwell and probabilistic, then it is not weakly accuracy-dominated by any credal strategy whatsoever.

1.1 Proof of Theorem 4

Proof of Theorem 4(a)

By Lemma 2, we can assume without loss of generality that $\langle c,a \rangle$ is probabilistic. If $\langle c,a \rangle$ is not Blackwell, then there is some subset $\{E_1,...,E_k\}$ of $\mathcal E$ such that $c(E_i)=0$ and $a_{E_i}(E_i) < 1$ for all $i \in \{1,...,k \}$. For each $i \in \{1,...,k\}$, it follows that $a_{E_i}$ is not in the convex hull of $\{v_{w}: w \in E_i\}$, which in turn implies that there is some $p_i$ in the convex hull of $\{v_{w}: w \in E_i\}$ such that

$$\begin{aligned} \mathcal J_\omega (p_i) < \mathcal J_\omega (a_{E_i}) \end{aligned}$$

(7)

for all $\omega \in E_i$ (the proof of this latter fact is exactly like Briggs and Pettigrew’s proof of Lemma 4 and is omitted). Since $p_i$ is in the convex hull of $\{v_{w}: w \in E_i\}$, it is probabilistic and $p_i(E_i)=1$. Now define a new credal act $a'$ by

$$\begin{aligned}a'_E = {\left\{ \begin{array}{ll} p_i & \text {if }E = E_i\text { for some }i \in \{1,...k \};\\ a_E & \text {otherwise}. \end{array}\right. }\end{aligned}$$

By Lemma 1(i), $\langle c,a'\rangle$ is conditionalizing, and, by construction, it is probabilistic and Blackwell. By Temporal Separability and (7), if $\omega \in \bigcup _{i=1}^k E_i$, then, for some $i \in \{1,...,k\}$,

$$\begin{aligned}\mathcal J_\omega (\langle c,a'\rangle ) = \mathcal J_\omega (c) + \mathcal J_\omega (p_i) < \mathcal J_\omega (c) + \mathcal J_\omega (a_{E_i}) = \mathcal J_\omega (\langle c,a \rangle );\end{aligned}$$

otherwise $\mathcal J_\omega (\langle c,a'\rangle ) = \mathcal J_\omega (\langle c,a \rangle )$. So $\langle c,a' \rangle$ weakly accuracy-dominates $\langle c,a \rangle$. $\square$

Proof of Theorem 4(b)

Suppose that $\langle c,a \rangle$ is probabilistic and weakly accuracy-dominated by $\langle c',a'\rangle$. We aim to show that $\langle c,a \rangle$ is not Blackwell, that is, that $a_E(E)<1$ for some $E \in \mathcal E$.

We begin by claiming that $c = c'$. Indeed, due to the weak accuracy-dominance, we have

$$\begin{aligned}exp_c(\mathcal J(\langle c',a' \rangle )) \le exp_c (\mathcal J(\langle c,a \rangle )).\end{aligned}$$

But Theorem 2 implies that $exp_c(\mathcal J(a')) \ge exp_c(\mathcal J(a))$, so

$$\begin{aligned}\sum _{\omega \in \Omega } c\{\omega \}\mathcal J_\omega (c') \le \sum _{\omega \in \Omega } c\{\omega \} \mathcal J_\omega (c),\end{aligned}$$

which implies that $c = c'$, by Strict Propriety.

Weak accuracy-dominance now implies that $a \ne a'$, so $a_E \ne a'_E$ for some $E \in \mathcal E$. It also implies, together with Temporal Separability, that $\mathcal J_\omega (a') \le \mathcal J_\omega (a)$ for all $\omega \in \Omega$. In particular, $\mathcal J_\omega (a'_E) \le \mathcal J_\omega (a_E)$ for all $\omega \in E$, and therefore

$$\begin{aligned} \sum _{\omega \in E} a_E\{\omega \} \mathcal J_\omega (a'_E) \le \sum _{\omega \in E} a_E\{\omega \} \mathcal J_\omega (a_E). \end{aligned}$$

(8)

If $a_E(E)=1$, then, because $a_E$ is probabilistic, (8) implies

$$\begin{aligned}\sum _{\omega \in \Omega } a_E\{\omega \} \mathcal J_\omega (a'_E) \le \sum _{\omega \in \Omega } a_E\{\omega \} \mathcal J_\omega (a_E).\end{aligned}$$

And it follows from Strict Propriety that $a'_E = a_E$, which is false. Thus, $a_E(E)<1$, and $\langle c,a \rangle$ is not Blackwell. $\square$